Hailstone sequence

Hailstone sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Hailstone sequence of numbers can be generated from a starting positive integer,   n   by:

•   If   n   is     1     then the sequence ends.
•   If   n   is   even then the next   n   of the sequence   = n/2
•   If   n   is   odd   then the next   n   of the sequence   = (3 * n) + 1

The (unproven) Collatz conjecture is that the hailstone sequence for any starting number always terminates.

This sequence was named by Lothar Collatz in 1937   (or possibly in 1939),   and is also known as (the):

•   hailstone sequence,   hailstone numbers
•   3x + 2 mapping,   3n + 1 problem
•   Collatz sequence
•   Hasse's algorithm
•   Kakutani's problem
•   Syracuse algorithm,   Syracuse problem
•   Thwaites conjecture
•   Ulam's problem

The hailstone sequence is also known as   hailstone numbers   (because the values are usually subject to multiple descents and ascents like hailstones in a cloud).

Task

1. Create a routine to generate the hailstone sequence for a number.
2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.
     (But don't show the actual sequence!)

See also

•   xkcd (humourous).
•   The Notorious Collatz conjecture Terence Tao, UCLA (Presentation, pdf).
•   The Simplest Math Problem No One Can Solve Veritasium (video, sponsored).

11l

Translation of: Python

F hailstone(=n)
   V seq = [n]
   L n > 1
      n = I n % 2 != 0 {3 * n + 1} E n I/ 2
      seq.append(n)
   R seq

V h = hailstone(27)
assert(h.len == 112 & h[0.<4] == [27, 82, 41, 124] & h[(len)-4 ..] == [8, 4, 2, 1])

V m = max((1..99999).map(i -> (hailstone(i).len, i)))
print(‘Maximum length #. was found for hailstone(#.) for numbers <100,000’.format(m[0], m[1]))

Output:

Maximum length 351 was found for hailstone(77031) for numbers <100,000

360 Assembly

*        Hailstone sequence        16/08/2015
HAILSTON CSECT
         USING  HAILSTON,R12
         LR     R12,R15
         ST     R14,SAVER14
BEGIN    L      R11,=F'100000'     nmax
         LA     R8,27              n=27
         LR     R1,R8
         MVI    FTAB,X'01'         ftab=true
         BAL    R14,COLLATZ
         LR     R10,R1             p
         XDECO  R8,XDEC            n
         MVC    BUF1+10(6),XDEC+6
         XDECO  R10,XDEC           p
         MVC    BUF1+18(5),XDEC+7
         LA     R5,6
         LA     R3,0               i
         LA     R4,BUF1+25
LOOPED   L      R2,TAB(R3)         tab(i)
         XDECO  R2,XDEC
         MVC    0(7,R4),XDEC+5
         LA     R3,4(R3)           i=i+1
         LA     R4,7(R4)
         C      R5,=F'4'
         BNE    BCT
         LA     R4,7(R4) 
BCT      BCT    R5,LOOPED
         XPRNT  BUF1,80            print hailstone(n)=p,tab(*)
         MVC    LONGEST,=F'0'      longest=0
         MVI    FTAB,X'00'         ftab=true
         LA     R8,1               i
LOOPI    CR     R8,R11             do i=1 to nmax
         BH     ELOOPI
         LR     R1,R8              n
         BAL    R14,COLLATZ
         LR     R10,R1             p
         L      R4,LONGEST
         CR     R4,R10             if longest<p
         BNL    NOTSUP
         ST     R8,IVAL            ival=i
         ST     R10,LONGEST        longest=p
NOTSUP   LA     R8,1(R8)           i=i+1
         B      LOOPI
ELOOPI   EQU    *                  end i
         XDECO  R11,XDEC           maxn
         MVC    BUF2+9(6),XDEC+6
         L      R1,IVAL            ival
         XDECO  R1,XDEC
         MVC    BUF2+28(6),XDEC+6
         L      R1,LONGEST         longest
         XDECO  R1,XDEC
         MVC    BUF2+36(5),XDEC+7
         XPRNT  BUF2,80            print maxn,hailstone(ival)=longest
         B      RETURN
*        *      *                  r1=collatz(r1)
COLLATZ  LR     R7,R1              m=n  (R7)
         LA     R6,1               p=1  (R6)
LOOPP    C      R7,=F'1'           do p=1 by 1 while(m>1)
         BNH    ELOOPP
         CLI    FTAB,X'01'         if ftab
         BNE    NONOK
         C      R6,=F'1'           if p>=1
         BL     NONOK
         C      R6,=F'3'           & p<=3
         BH     NONOK
         LR     R1,R6              then
         BCTR   R1,0
         SLA    R1,2
         ST     R7,TAB(R1)         tab(p)=m
NONOK    LR     R4,R7              m
         N      R4,=F'1'           m&1
         LTR    R4,R4              if m//2=0  (if not(m&1))
         BNZ    ODD
EVEN     SRA    R7,1               m=m/2
         B      EIFM
ODD      LA     R3,3
         MR     R2,R7              *m
         LA     R7,1(R3)           m=m*3+1
EIFM     CLI    FTAB,X'01'         if ftab
         BNE    NEXTP
         MVC    TAB+12,TAB+16      tab(4)=tab(5)
         MVC    TAB+16,TAB+20      tab(5)=tab(6)
         ST     R7,TAB+20          tab(6)=m
NEXTP    LA     R6,1(R6)           p=p+1
         B      LOOPP
ELOOPP   LR     R1,R6              end p; return(p)
         BR     R14                end collatz
*                
RETURN   L      R14,SAVER14        restore caller address
         XR     R15,R15            set return code
         BR     R14                return to caller
SAVER14  DS     F
IVAL     DS     F
LONGEST  DS     F
N        DS     F
TAB      DS     6F
FTAB     DS     X
BUF1     DC     CL80'hailstone(nnnnnn)=nnnnn : nnnnnn nnnnnn nnnnnn ...*
               ... nnnnnn nnnnnn nnnnnn'
BUF2     DC     CL80'longest <nnnnnn : hailstone(nnnnnn)=nnnnn'
XDEC     DS     CL12
         YREGS
         END    HAILSTON

Output:

hailstone(    27)=  112 :     27     82     41 ......      4      2      1
longest <100000 : hailstone( 77031)=  351

ABAP

CLASS lcl_hailstone DEFINITION.
  PUBLIC SECTION.
    TYPES: tty_sequence TYPE STANDARD TABLE OF i
                             WITH NON-UNIQUE EMPTY KEY,
           BEGIN OF ty_seq_len,
             start TYPE i,
             len   TYPE i,
           END OF ty_seq_len,
           tty_seq_len TYPE HASHED TABLE OF ty_seq_len
                            WITH UNIQUE KEY start.

    CLASS-METHODS:
      get_next
        IMPORTING
          n                           TYPE i
        RETURNING
          VALUE(r_next_hailstone_num) TYPE i,

      get_sequence
        IMPORTING
          start             TYPE i
        RETURNING
          VALUE(r_sequence) TYPE tty_sequence,

      get_longest_sequence_upto
        IMPORTING
          limit                     TYPE i
        RETURNING
          VALUE(r_longest_sequence) TYPE ty_seq_len.

  PRIVATE SECTION.
    TYPES: BEGIN OF ty_seq,
             start TYPE i,
             seq   TYPE tty_sequence,
           END OF ty_seq.
    CLASS-DATA: sequence_buffer TYPE HASHED TABLE OF ty_seq
                                     WITH UNIQUE KEY start.
ENDCLASS.

CLASS lcl_hailstone IMPLEMENTATION.
  METHOD get_next.
    r_next_hailstone_num = COND #( WHEN n MOD 2 = 0 THEN n / 2
                                   ELSE ( 3 * n ) + 1 ).
  ENDMETHOD.

  METHOD get_sequence.
    INSERT start INTO TABLE r_sequence.
    IF start = 1.
      RETURN.
    ENDIF.

    READ TABLE sequence_buffer ASSIGNING FIELD-SYMBOL(<buff>)
                               WITH TABLE KEY start = start.
    IF sy-subrc = 0.
      INSERT LINES OF <buff>-seq INTO TABLE r_sequence.
    ELSE.
      DATA(seq) = get_sequence( get_next( start ) ).
      INSERT LINES OF seq INTO TABLE r_sequence.
      INSERT VALUE ty_seq( start = start
                           seq   = seq ) INTO TABLE sequence_buffer.
    ENDIF.
  ENDMETHOD.

  METHOD get_longest_sequence_upto.
    DATA: max_seq TYPE ty_seq_len,
          act_seq TYPE ty_seq_len.

    DO limit TIMES.
      act_seq-len = lines( get_sequence( sy-index ) ).

      IF act_seq-len > max_seq-len.
        max_seq-len   = act_seq-len.
        max_seq-start = sy-index.
      ENDIF.
    ENDDO.

    r_longest_sequence = max_seq.
  ENDMETHOD.
ENDCLASS.

START-OF-SELECTION.
  cl_demo_output=>begin_section( |Hailstone sequence of 27 is: | ).
  cl_demo_output=>write( REDUCE string( INIT result = ``
                                        FOR item IN lcl_hailstone=>get_sequence( 27 )
                                        NEXT result = |{ result } { item }| ) ).
  cl_demo_output=>write( |With length: { lines( lcl_hailstone=>get_sequence( 27 ) ) }| ).
  cl_demo_output=>begin_section( |Longest hailstone sequence upto 100k| ).
  cl_demo_output=>write( lcl_hailstone=>get_longest_sequence_upto( 100000 ) ).
  cl_demo_output=>display( ).

Output:

Hailstone sequence of 27 is: 

27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

With length: 112

Longest hailstone sequence upto 100k

Structure 
START LEN 
77031 351 

ABC

HOW TO RETURN hailstone n:
    PUT {} IN seq
    WHILE 1=1:
        PUT n IN seq[#seq]
        SELECT:
            n=1: RETURN seq
            n mod 2=0: PUT floor(n/2) IN n
            n mod 2=1: PUT 3*n+1 IN n
    RETURN seq

PUT hailstone 27 IN h27
WRITE "Length of Hailstone sequence for 27:", #h27/
WRITE "First 4 elements:", h27[0], h27[1], h27[2], h27[3]/
WRITE "Last 4 elements:", h27[#h27-4], h27[#h27-3], h27[#h27-2], h27[#h27-1]/

PUT 0, 0 IN longest, length
FOR n IN {1..100000}:
    PUT hailstone n IN hn
    IF #hn > length:
        PUT n, #hn IN longest, length

WRITE longest, "has the longest hailstone sequence < 100,000, of length:", length/

Output:

Length of Hailstone sequence for 27: 112
First 4 elements: 27 82 41 124
Last 4 elements: 8 4 2 1
77031 has the longest hailstone sequence < 100,000, of length: 351

ACL2

(defun hailstone (len)
    (loop for x = len 
             then (if (evenp x) 
                         (/ x 2) 
                         (+ 1 (* 3 x))) 
        collect x until (= x 1)))

;; Must be tail recursive
(defun max-hailstone-start (limit mx curr)
   (declare (xargs :mode :program))
   (if (zp limit)
       (mv mx curr)
       (let ((new-mx (len (hailstone limit))))
          (if (> new-mx mx)
              (max-hailstone-start (1- limit) new-mx limit)
              (max-hailstone-start (1- limit) mx curr)))))

Output:

> (take 4 (hailstone 27))
(27 82 41 124)
> (nthcdr 108 (hailstone 27))
(8 4 2 1)
> (len (hailstone 27))
112
> (max-hailstone-start 100000 0 0)
(351 77031)

Ada

Similar to C method:

with Ada.Text_IO; use Ada.Text_IO;
procedure hailstone is
	type int_arr is array(Positive range <>) of Integer;
	type int_arr_pt is access all int_arr;

	function hailstones(num:Integer; pt:int_arr_pt) return Integer is
		stones : Integer := 1;
		n : Integer := num;
		begin
		if pt /= null then pt(1) := num; end if;
		while (n/=1) loop
			stones := stones + 1;
			if n mod 2 = 0 then n := n/2;
			else n := (3*n)+1;
			end if;
			if pt /= null then pt(stones) := n; end if;
		end loop;
		return stones;
	end hailstones;
	
	nmax,stonemax,stones : Integer := 0;
	list : int_arr_pt;
begin
	stones := hailstones(27,null);
	list := new int_arr(1..stones);
	stones := hailstones(27,list);
	put(" 27: "&Integer'Image(stones)); new_line;
	for n in 1..4 loop put(Integer'Image(list(n)));	end loop;
	put(" .... ");
	for n in stones-3..stones loop put(Integer'Image(list(n))); end loop;
	new_line;
	for n in 1..100000 loop
		stones := hailstones(n,null);
		if stones>stonemax then
			nmax := n; stonemax := stones;
		end if;
	end loop;
	put_line(Integer'Image(nmax)&" max @ n= "&Integer'Image(stonemax));
end hailstone;

Output:

 27:  112
 27 82 41 124 ....  8 4 2 1
 77031 max @ n=  351

Alternative method

A method without pointers or dynamic memory allocation, but slower for simply counting. This is also used for the "executable library" task Executable library#Ada.

hailstones.ads:

package Hailstones is
   type Integer_Sequence is array(Positive range <>) of Integer;
   function Create_Sequence (N : Positive) return Integer_Sequence;
end Hailstones;

hailstones.adb:

package body Hailstones is
   function Create_Sequence (N : Positive) return Integer_Sequence is
   begin
      if N = 1 then
         -- terminate
         return (1 => N);
      elsif N mod 2 = 0 then
         -- even
         return (1 => N) & Create_Sequence (N / 2);
      else
         -- odd
         return (1 => N) & Create_Sequence (3 * N + 1);
      end if;
   end Create_Sequence;
end Hailstones;

example main.adb:

with Ada.Text_IO;
with Hailstones;

procedure Main is
   package Integer_IO is new Ada.Text_IO.Integer_IO (Integer);

   procedure Print_Sequence (X : Hailstones.Integer_Sequence) is
   begin
      for I in X'Range loop
         Integer_IO.Put (Item => X (I), Width => 0);
         if I < X'Last then
            Ada.Text_IO.Put (", ");
         end if;
      end loop;
      Ada.Text_IO.New_Line;
   end Print_Sequence;

   Hailstone_27 : constant Hailstones.Integer_Sequence :=
     Hailstones.Create_Sequence (N => 27);

begin
   Ada.Text_IO.Put_Line ("Length of 27:" & Integer'Image (Hailstone_27'Length));
   Ada.Text_IO.Put ("First four: ");
   Print_Sequence (Hailstone_27 (Hailstone_27'First .. Hailstone_27'First + 3));
   Ada.Text_IO.Put ("Last four: ");
   Print_Sequence (Hailstone_27 (Hailstone_27'Last - 3 .. Hailstone_27'Last));

   declare
      Longest_Length : Natural := 0;
      Longest_N      : Positive;
      Length         : Natural;
   begin
      for I in 1 .. 99_999 loop
         Length := Hailstones.Create_Sequence (N => I)'Length;
         if Length > Longest_Length then
            Longest_Length := Length;
            Longest_N := I;
         end if;
      end loop;
      Ada.Text_IO.Put_Line ("Longest length is" & Integer'Image (Longest_Length));
      Ada.Text_IO.Put_Line ("with N =" & Integer'Image (Longest_N));
   end;
end Main;

Output:

Length of 27: 112
First four: 27, 82, 41, 124
Last four: 8, 4, 2, 1
Longest length is 351
with N = 77031

Aime

void
print_hailstone(integer h)
{
    list l;

    while (h ^ 1) {
        lb_p_integer(l, h);
        h = h & 1 ? 3 * h + 1 : h / 2;
    }

    o_form("hailstone sequence for ~ is ~1 ~ ~ ~ .. ~ ~ ~ ~, it is ~ long\n",
           l[0], l[1], l[2], l[3], l[-3], l[-2], l[-1], 1, ~l + 1);
}

void
max_hailstone(integer x)
{
    integer e, i, m;
    index r;

    m = 0;
    i = 1;
    while (i < x) {
        integer h, k, l;

        h = i;
        l = 1;
        while (h ^ 1) {
            if (i_j_integer(k, r, h)) {
                l += k;
                break;
            } else {
                l += 1;
                h = h & 1 ? 3 * h + 1 : h / 2;
            }
        }

        r[i] = l - 1;

        if (m < l) {
            m = l;
            e = i;
        }

        i += 1;
    }

    o_form("hailstone sequence length for ~ is ~\n", e, m);
}

integer
main(void)
{
    print_hailstone(27);
    max_hailstone(100000);

    return 0;
}

Output:

hailstone sequence for 27 is 27 82 41 124 .. 8 4 2 1, it is 112 long
hailstone sequence length for 77031 is 351

ALGOL 60

Works with: A60

begin 
    comment Hailstone sequence - Algol 60;
    integer array collatz[1:400]; integer icollatz;
    
    integer procedure mod(i,j); value i,j; integer i,j;
    mod:=i-(i div j)*j;
    
    integer procedure hailstone(num);
    value num; integer num;
    begin
        integer i,n;
        icollatz:=1; n:=num; i:=0;
        collatz[icollatz]:=n;
        for i:=i+1 while n notequal 1 do begin
            if mod(n,2)=0 then n:=n div 2
                          else n:=(3*n)+1;
            icollatz:=icollatz+1;
            collatz[icollatz]:=n
        end;
        hailstone:=icollatz
    end hailstone;
 
    integer i,nn,ncollatz,count,nlongest,nel,nelcur,nnn;
    nn:=27;
    ncollatz:=hailstone(nn);
    outstring(1,"sequence for"); outinteger(1,nn); outstring(1," :\n");
    for i:=1 step 1 until ncollatz do outinteger(1,collatz[i]);
    outstring(1,"\n");
    outstring(1,"number of elements:"); outinteger(1,ncollatz);
    outstring(1,"\n\n");
    nlongest:=0; nel:=0; nnn:=100000;
    for count:=1, count+1 while count<nnn do begin
        nelcur:=hailstone(count);
        if nelcur>nel then begin
            nel:=nelcur;
            nlongest:=count
        end
    end;
    outstring(1,"number <"); outinteger(1,nnn); 
    outstring(1,"with the longest sequence:"); outinteger(1,nlongest);
    outstring(1,", with"); outinteger(1,nel); outstring(1,"elements.");
    outstring(1,"\n")
end

Output:

sequence for 27  :
 27  82  41  124  62  31  94  47  142  71  214  107  322  161  484  242  121  364  182  91  274  137  412  206  103  310  155  466  233  700  350  175  526  263  790  395  1186  593  1780  890  445  1336  668  334  167  502  251  754  377  1132  566  283  850  425  1276  638  319  958  479  1438  719  2158  1079  3238  1619  4858  2429  7288  3644  1822  911  2734  1367  4102  2051  6154  3077  9232  4616  2308  1154  577  1732  866  433  1300  650  325  976  488  244  122  61  184  92  46  23  70  35  106  53  160  80  40  20  10  5  16  8  4  2  1
number of elements: 112

number < 100000 with the longest sequence: 77031 , with 351 elements.

ALGOL 68

Translation of: C

- note: This specimen retains the original C coding style.

Works with: ALGOL 68 version Standard - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - using the print routine rather than printf

MODE LINT = # LONG ... # INT;

PROC hailstone = (INT in n, REF[]LINT array)INT:
(
    INT hs := 1;
    INT index := 0;
    LINT n := in n;
 
    WHILE n /= 1 DO
        hs +:= 1;
        IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
        n := IF ODD n THEN 3*n+1 ELSE n OVER 2 FI
    OD;
    IF array ISNT REF[]LINT(NIL) THEN array[index +:= 1] := n FI;
    hs
);
 
main:
(
    INT j, hmax := 0;
    INT jatmax, n;
    INT border = 4;
 
    FOR j TO 100000-1 DO 
       n := hailstone(j, NIL);
       IF hmax < n THEN
           hmax := n;
           jatmax := j
       FI
    OD;
 
    [2]INT test := (27, jatmax);
    FOR key TO UPB test DO
        INT val = test[key];
        n := hailstone(val, NIL);
        [n]LINT array;
        n := hailstone(val, array);
 
        printf(($"[ "n(border)(g(0)", ")" ..."n(border)(", "g(0))"] len="g(0)l$,
            array[:border], array[n-border+1:], n))
        #;free(array) #
    OD;
    printf(($"Max "g(0)" at j="g(0)l$, hmax, jatmax))
# ELLA Algol68RS:
    print(("Max",hmax," at j=",jatmax, new line))
#
)

Output:

[ 27, 82, 41, 124,  ..., 8, 4, 2, 1] len=112
[ 77031, 231094, 115547, 346642,  ..., 8, 4, 2, 1] len=351
Max 351 at j=77031

ALGOL-M

The limitations of ALGOL-M's 15-bit integer data type will not allow the required search up to 100000 for the longest sequence, so we stick with what is possible.

BEGIN

INTEGER N, LEN, YES, NO, LIMIT, LONGEST, NLONG;

% RETURN P MOD Q %
INTEGER FUNCTION MOD(P, Q);
INTEGER P, Q;
BEGIN
  MOD := P - Q * (P / Q);
END;

% COMPUTE AND OPTIONALLY DISPLAY HAILSTONE SEQUENCE FOR N. %
% RETURN LENGTH OF SEQUENCE OR ZERO ON OVERFLOW. %
INTEGER FUNCTION HAILSTONE(N, DISPLAY);
INTEGER N, DISPLAY;
BEGIN
  INTEGER LEN;
  LEN := 1;
  IF DISPLAY = 1 THEN WRITE("");
  WHILE (N <> 1) AND (N > 0) DO
    BEGIN
      IF DISPLAY = 1 THEN WRITEON(N,"  ");
      IF MOD(N,2) = 0 THEN
         N := N / 2
      ELSE
         N := (N * 3) + 1;
      LEN := LEN + 1;
    END;
  IF DISPLAY = 1 THEN WRITEON(N);
  HAILSTONE := (IF N < 0 THEN 0 ELSE LEN);
END;

% EXERCISE THE FUNCTION %
YES := 1; NO := 0;
WRITE("DISPLAY HAILSTONE SEQUENCE FOR WHAT NUMBER?");
READ(N);
LEN := HAILSTONE(N, YES);
WRITE("SEQUENCE LENGTH =", LEN);

% FIND LONGEST SEQUENCE BEFORE OVERFLOW %
N := 2;
LONGEST := 1;
LEN := 2;
NLONG := 2;
LIMIT := 1000;
WRITE("SEARCHING FOR LONGEST SEQUENCE UP TO N =",LIMIT," ...");
WHILE (N < LIMIT) AND (LEN <> 0) DO
  BEGIN
    LEN := HAILSTONE(N, NO);
    IF LEN > LONGEST THEN
       BEGIN
          LONGEST := LEN;
          NLONG := N;
       END;
    N := N + 1;
  END;
IF LEN = 0 THEN WRITE("SEARCH TERMINATED WITH OVERFLOW AT N =",N-1);
WRITE("MAXIMUM SEQUENCE LENGTH =", LONGEST, " FOR N =", NLONG);   

END

Output:

DISPLAY HAILSTONE SEQUENCE FOR WHAT NUMBER?
-> 27
    27      82      41     124      62      31      94      47     142      71
   214     107     322     161     484     242     121     364     182      91
   274     137     412     206     103     310     155     466     233     700
   350     175     526     263     790     395    1186     593    1780     890
   445    1336     668     334     167     502     251     754     377    1132
   566     283     850     425    1276     638     319     958     479    1438
   719    2158    1079    3238    1619    4858    2429    7288    3644    1822
   911    2734    1367    4102    2051    6154    3077    9232    4616    2308
  1154     577    1732     866     433    1300     650     325     976     488
   244     122      61     184      92      46      23      70      35     106
    53     160      80      40      20      10       5      16       8       4
     2       1
SEQUENCE LENGTH =   112
SEARCHING FOR LONGEST SEQUENCE UP TO N = 10000 ...
SEARCH TERMINATED WITH OVERFLOW AT N =   447
MAXIMUM SEQUENCE LENGTH =   144 FOR N = 327

ALGOL W

begin
    % show some Hailstone Sequence related information                       %
    % calculates the length of the sequence generated by n,                  %
    % if showFirstAndLast is true, the first and last 4 elements of the      %
    % sequence are stored in first and last                                  %
    % hs holds a cache of the upbHs previously calculated sequence lengths   %
    % if showFirstAndLast is false, the cache will be used                   %
    procedure hailstone ( integer value  n
                        ; integer array  first, last ( * )
                        ; integer result length
                        ; integer array  hs          ( * )
                        ; integer value  upbHs
                        ; logical value  showFirstAndLast
                        ) ;
    if not showFirstAndLast and n <= upbHs and hs( n ) not = 0 then begin
        % no need to store the start and end of the sequence and we already  %
        % know the length of the sequence for n                              %
        length := hs( n )
        end
    else begin
        % must calculate the sequence length                                 %
        integer sv;
        for i := 1 until 4 do first( i ) := last( i ) := 0;
        length := 0;
        sv     := n;
        if sv > 0 then begin
            while begin
                length := length + 1;
                if showFirstAndLast then begin
                    if length <= 4 then first( length ) := sv;
                    for lPos := 1 until 3 do last( lPos ) := last( lPos + 1 );
                    last( 4 ) := sv
                    end
                else if sv <= upbHs and hs( sv ) not = 0 then begin
                    % have a known value                                 %
                    length := ( length + hs( sv ) ) - 1;
                    sv     := 1
                end ;
                sv not = 1
            end do begin
                sv := if odd( sv ) then ( 3 * sv ) + 1 else sv div 2
            end while_sv_ne_1 ;
            if n < upbHs then hs( n ) := length
        end if_sv_gt_0
    end hailstone ;
    begin
        % test the hailstone procedure                                       %
        integer HS_CACHE_SIZE;
        HS_CACHE_SIZE := 100000;
        begin
            integer array first, last ( 1 :: 4 );
            integer       length, maxLength, maxNumber;
            integer array hs          ( 1 :: HS_CACHE_SIZE );
            for i := 1 until HS_CACHE_SIZE do hs( i ) := 0;
            hailstone( 27, first, last, length, hs, HS_CACHE_SIZE, true );
            write( i_w := 1, s_w := 0
                 , "27: length ", length, ", first: ["
                 , first( 1 ), " ", first( 2 ), " ", first( 3 ), " ", first( 4 )
                 , "] last: ["
                 , last( 1 ), " ", last( 2 ), " ", last( 3 ), " ", last( 4 )
                 , "]"
                 );
            maxNumber := 0;
            maxLength := 0;
            for n := 1 until 100000 do begin
                hailstone( n, first, last, length, hs, HS_CACHE_SIZE, false );
                if length > maxLength then begin
                    maxNumber := n;
                    maxLength := length
                end if_length_gt_maxLength
            end for_n ;
            write( i_w := 1, s_w := 1, "Maximum sequence length: ", maxLength, " for: ", maxNumber )
        end
    end
end.

Output:

27: length 112, first: [27 82 41 124] last: [8 4 2 1]
Maximum sequence length: 351  for: 77031

Amazing Hopper

#include <basico.h>

#proto Hailstone(_X_,_SW_)

algoritmo

    valor=27, máxima secuencia=0, vtemp=0
    imprimir ( _Hailstone(27,1) ---copiar en 'máxima secuencia'--- , NL )
    
    i=28
    iterar
        _Hailstone(i,0), copiar en 'vtemp'
        cuando( sea mayor que 'máxima secuencia' ) {
             máxima secuencia = vtemp
             valor=i
        }
        ++i
    hasta que ' #(i==100000) '
    imprimir ( #(utf8("Máxima longitud ")),máxima secuencia,\
               " fue encontrada para Hailstone(",valor,\
               #(utf8(") para números <100,000")), NL )

terminar

subrutinas

Hailstone(n, sw)
    largo_de_secuencia = 0
    v={}, n, mete(v)
    iterar
        tomar si ( es par(n), #(n/2), \
                   tomar si ( #(n<>1), #(3*n+1), 1) )
        ---copiar en 'n'--- mete(v)
    hasta que ' #(n==1) '
    #(length(v)), mover a 'largo_de_secuencia'
    cuando (sw){
         decimales '0'
         #( v[1:4] ), ",...,", 
         #( v[largo_de_secuencia-4 : largo_de_secuencia] )
         NL, #(utf8("Tamaño de la secuencia: ")) 
         imprime esto; decimales normales
    }
retornar ' largo_de_secuencia '

Output:

27,82,41,124,...,16,8,4,2,1
Tamaño de la secuencia: 112
Máxima longitud 351 fue encontrada para Hailstone(77031) para números <100,000

APL

Works with: Dyalog APL

⍝ recursive dfn:
dfnHailstone←{
    c←⊃⌽⍵ ⍝ last element
    1=c:1 ⍝ if it is 1, stop.
    ⍵,∇(1+2|c)⊃(c÷2)(1+3×c) ⍝ otherwise pick the next step, and append the result of the recursive call
}

⍝ tradfn version:
∇seq←hailstone n;next
⍝ Returns the hailstone sequence for a given number

seq←n                   ⍝ Init the sequence
:While n≠1
    next←(n÷2) (1+3×n)  ⍝ Compute both possibilities
    n←next[1+2|n]       ⍝ Pick the appropriate next step
    seq,←n              ⍝ Append that to the sequence
:EndWhile
∇

Output:

 dfnHailstone 5
5 16 8 4 2 1
 5↑hailstone 27
27 82 41 124 62
 ¯5↑hailstone 27
16 8 4 2 1
 ⍴hailstone 27
112
 1↑{⍵[⍒↑(⍴∘hailstone)¨⍵]}⍳100000
77031

AppleScript

on hailstoneSequence(n)
    script o
        property sequence : {n}
    end script
    
    repeat until (n = 1)
        if (n mod 2 is 0) then
            set n to n div 2
        else
            set n to 3 * n + 1
        end if
        set end of o's sequence to n
    end repeat
    
    return o's sequence
end hailstoneSequence

set n to 27
tell hailstoneSequence(n)
    return {n:n, |length of sequence|:(its length), |first 4 numbers|:items 1 thru 4, |last 4 numbers|:items -4 thru -1}
end tell

Output:

{|length of sequence|:112, |first 4 numbers|:{27, 82, 41, 124}, |last 4 numbers|:{8, 4, 2, 1}}

-- Number(s) below 100,000 giving the longest sequence length, using the hailstoneSequence(n) handler above.
set nums to {}
set longestLength to 1
repeat with n from 2 to 99999
    set thisLength to (count hailstoneSequence(n))
    if (thisLength < longestLength) then
    else if (thisLength > longestLength) then
        set nums to {n}
        set longestLength to thisLength
    else
        set end of nums to n
    end if
end repeat
return {|number(s) giving longest sequence length|:nums, |length of sequence|:longestLength}

Output:

{|number(s) giving longest sequence length|:{77031}, |length of sequence|:351}

ARM Assembly

Output is in hexadecimal but is otherwise correct. Because of the Game Boy Advance's limited screen size, only the first 4 and last 4 entries are printed to the screen. The emulator's memory dump can show the rest. In addition, the task was split into two separate programs.

Hailstone Sequence of N equal to 27

        .org 0x08000000

        b ProgramStart

        ;cartridge header goes here


;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;	
; Program Start

	.equ ramarea, 0x02000000
	.equ CursorX,ramarea		
	.equ CursorY,ramarea+1
	.equ hailstoneram,0x02000004
	
	
ProgramStart:
	mov sp,#0x03000000			;Init Stack Pointer
	
	mov r4,#0x04000000  		        ;DISPCNT - LCD Video Controller
	mov r2,#0x403    			;4= Layer 2 on / 3= ScreenMode 3
	str	r2,[r4]         	        ;now the user can see the screen

	bl ResetTextCursors			;set text cursors to top left of screen. This routine, as well as the other I/O
	                                        ;     routines, were omitted to keep this entry short.
	mov r0,#27
	adr r1,HailStoneMessage_Init
	bl PrintString
	bl NewLine
	bl ShowHex32
	bl NewLine
	bl NewLine
	
	
	bl Hailstone
	        
                ;function is complete, return the output
	        adr r1,HailStoneMessage_0
		bl PrintString
		bl NewLine
		ldr r1,HailStoneRam_Mirror		;mov r2,0x02000004
		
		
		ldr r0,[r1],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r1],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r1],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r1],#4
		bl ShowHex32
		bl NewLine
		bl NewLine
		
		adr r1,HailStoneMessage_1
		bl PrintString
		bl NewLine
		
		ldr r0,[r2],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r2],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r2],#4
		bl ShowHex32
		bl NewLine
		
		ldr r0,[r2],#4
		bl ShowHex32
		bl NewLine
		bl NewLine
		
		adr r1,HailStoneMessage_2
		bl PrintString
		bl NewLine
		mov r0,r3
		bl ShowHex32	

forever:
	b forever              ;we're done, so trap the program counter. 
	
Hailstone:
	;input: R0 = n.
	;out: 	r2 = pointer to last 4 entries
	;	r3 = length of sequence
	
	
	;reg usage:
	;R1 = scratchpad
	;R3 = counter for entries in the sequence.
	;R5 = pointer to output ram
	stmfd sp!,{r4-r12,lr}

		mov r5,#0x02000000
		add r5,r5,#4
		str r0,[r5],#4				;store in hailstone ram and post-inc by 4
		mov r3,#0
loop_hailstone:
		add r3,r3,#1				;represents number of entries in the sequence
		cmp r0,#1
		beq hailstone_end
		tst r0,#1
		;;;; executes only if r0 was even
		moveq r0,r0,lsr #1			;divide 		
		
		;;;; executes only if r0 was odd
		movne r1,r0
		movne r0,r0,lsl #1
		addne r0,r1,r0
		addne r0,r0,#1
		
		str r0,[r5],#4			;store in hailstone ram, post inc by 4
		
		b loop_hailstone
		
		
hailstone_end:
		sub r5,r5,#16			;subtract 16 to get pointer to last 4 entries.
		mov r2,r5				;output ptr to last 4 entries to r2.
		;pointer to first 4 entries is 0x02000004
		ldmfd sp!,{r4-r12,pc}
	
HailStoneRam_Mirror:
	.long 0x02000004
HailstoneMessage_Init:
	.byte "Your input was:",255
	.align 4
HailstoneMessage_0:
	.byte "First 4 numbers are:",255
	.align 4
HailstoneMessage_1:
	.byte "Last 4 numbers are:",255
	.align 4	
HailstoneMessage_2:
	.byte "Sequence length is:",255
	.align 4

;;;;;;;;;;; EVERYTHING PAST THIS POINT IS JUST I/O ROUTINES FOR PRINTING NUMBERS AND WORDS TO THE GAME BOY ADVANCE'S SCREEN
;;;;;;;;;;; I ASSURE YOU THAT ALL OF IT WORKS BUT CHANCES ARE YOU DIDN'T COME HERE TO SEE THAT.
;;;;;;;;;;; THANKS TO KEITH OF CHIBIAKUMAS.COM FOR WRITING THEM!

Output:

Your input was:
0000001B

First 4 numbers are:
0000001B
00000052
00000029
0000007C

Last 4 numbers are:
00000008
00000004
00000002
00000001

Sequence length is:
00000070

Picture of output on VisualBoyAdvance screen

Biggest Sequence Between 2 and 100,000

To keep this short, I'm only including the part that changed, and the output. This goes after the call to ResetTextCursors but before the infinite loop:

	mov r0,#1
	bl Hailstone
	mov r6,r3
	
	
	mov r0,#2
	mov r8,#100000
	
	
loop_getBiggestHailstone:
	mov r10,r0
		bl Hailstone
	mov r0,r10
	
	cmp r3,r6
		movgt r6,r3				;if greater than, store in r6
		movgt r7,r0				;if greater than, store in r7
	add r0,r0,#1
	cmp r0,r8
	blt loop_getBiggestHailstone
	
	adr r1,HailstoneMessage_0
	bl PrintString
	bl NewLine
	adr r1,HailStoneMessage_1
	bl PrintString
	bl NewLine
	
	mov r0,r7
	bl ShowHex32
	bl NewLine
	
	adr r1,HailStoneMessage_2
	bl PrintString
	bl NewLine
	
	mov r0,r6
	bl ShowHex32
	bl NewLine

Output:

The number that makes the 
biggest sequence is:
00012CE7
And that sequence has a length
 of:
0000015F

Output of second program on VisualBoyAdvance's screen

Arturo

hailstone: function [n][
	ret: @[n]
	while [n>1][
		if? 1 = and n 1 -> n: 1+3*n
		else -> n: n/2

		'ret ++ n
	]
	ret
]

print "Hailstone sequence for 27:"
print hailstone 27

maxHailstoneLength: 0
maxHailstone: 0

loop 2..1000 'x [
	l: size hailstone x
	if l>maxHailstoneLength [
		maxHailstoneLength: l
		maxHailstone: x
	]
]

print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]

Output:

Hailstone sequence for 27:
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
max hailstone sequence found (<100000): of length 351 for 77031

AutoHotkey

; Submitted by MasterFocus --- http://tiny.cc/iTunis

; [1] Generate the Hailstone Seq. for a number

List := varNum := 7 ; starting number is 7, not counting elements
While ( varNum > 1 )
  List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List

; [2] Seq. for starting number 27 has 112 elements

Count := 1, List := varNum := 27 ; starting number is 27, counting elements
While ( varNum > 1 )
  Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count

; [3] Find number<100000 with longest seq. and show both values

MaxNum := Max := 0 ; reset the Maximum variables
TimesToLoop := 100000 ; limit number here is 100000
Offset := 70000 ; offset - use 0 to process from 0 to 100000
Loop, %TimesToLoop%
{
  If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
    Break
  text := "Processing...`n-------------------`n"
  text .= "Current starting number: " Index "`n"
  text .= "Current sequence count: " Count
  text .= "`n-------------------`n"
  text .= "Maximum starting number: " MaxNum "`n"
  text .= "Maximum sequence count: " Max " <<" ; text split to avoid long code lines
  ToolTip, %text%
  Count := 1 ; going to count the elements, but no "List" required
  While ( varNum > 1 )
    Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
  If ( Count > Max )
    Max := Count , MaxNum := Index ; set the new maximum values, if necessary
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max

AutoIt

$Hail = Hailstone(27)
ConsoleWrite("Sequence-Lenght: "&$Hail&@CRLF)
$Big = -1
$Sequenzlenght = -1
For $I = 1 To 100000
	$Hail = Hailstone($i, False)
	If Number($Hail) > $Sequenzlenght Then
	$Sequenzlenght = Number($Hail)
	$Big = $i
	EndIf
Next
ConsoleWrite("Longest Sequence : "&$Sequenzlenght&" from number "&$Big&@CRLF)
Func Hailstone($int, $sequence = True)
	$Counter = 0
	While True
		$Counter += 1
		If $sequence = True Then ConsoleWrite($int & ",")
		If $int = 1 Then ExitLoop
		If Not Mod($int, 2) Then
			$int = $int / 2
		Else
			$int = 3 * $int + 1
		EndIf
		If Not Mod($Counter, 25) AND $sequence = True Then ConsoleWrite(@CRLF)
	WEnd
	If $sequence = True Then ConsoleWrite(@CRLF)
	Return $Counter
EndFunc   ;==>Hailstone

Output:

27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,
310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,
566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,
6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,
53,160,80,40,20,10,5,16,8,4,2,1,
Sequence-Lenght: 112
Longest Sequence : 351 from number 77031

AWK

#!/usr/bin/awk -f
function hailstone(v, verbose) {
	n = 1;
	u = v; 
	while (1) { 
		if (verbose) printf " "u;
		if (u==1) return(n); 
		n++;
		if (u%2 > 0 )
			u = 3*u+1;
		else
			u = u/2;
	} 
}

BEGIN {
	i = 27;
	printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
	ix=0; 
	m=0; 
	for (i=1; i<100000; i++) {
		n = hailstone(i,0);
		if (m<n) {
			m=n; 
			ix=i;
		}
	}
	printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}

Output:

27 82 41 124 ....... 8 4 2 1
hailstone(27) has 112 elements
longest hailstone sequence is 77031 and has 351 elements

BASIC

Applesoft BASIC

10 HOME

100 N = 27
110 GOSUB 400"HAILSTONE
120 DEF FN L(I) = E(I + 4 * (I < 0))
130IFL=112AND(S(0)=27ANDS(1)=82ANDS(2)=41ANDS(3)=124)AND(FNL(M-3)=8ANDFNL(M-2)=4ANDFNL(M-1)=2ANDFNL(M)=1)THENPRINT"THE HAILSTONE SEQUENCE FOR THE NUMBER 27 HAS 112 ELEMENTS STARTING WITH 27, 82, 41, 124 AND ENDING WITH 8, 4, 2, 1"
140 PRINT
150 V = PEEK(37) + 1

200 N = 1
210 GOSUB 400"HAILSTONE
220 MN = 1
230 ML = L
240 FOR I = 2 TO 99999
250     N = I
260     GOSUB 400"HAILSTONE
270     IFL>MLTHENMN=I:ML=L:VTABV:HTAB1:PRINT "THE NUMBER " MN " HAS A HAILSTONE SEQUENCE LENGTH OF "L" WHICH IS THE LONGEST HAILSTONE SEQUENCE OF NUMBERS LESS THAN ";:Y=PEEK(37)+1:X=PEEK(36)+1
280     IF Y THEN VTAB Y : HTAB X : PRINTI+1;
290 NEXT I

300 END

400 M = 0
410 FOR L = 1 TO 1E38
420     IF L < 5 THEN S(L-1) = N
430     M = (M + 1) * (M < 3)
440     E(M) = N
450     IF N = 1 THEN RETURN
460     EVEN = INT(N/2)=N/2
470     IF EVEN THEN N=N/2
480     IF NOT EVEN THEN N = (3 * N) + 1
490 NEXT L : STOP

BBC BASIC

      seqlen% = FNhailstone(27, TRUE)
      PRINT '"Sequence length = "; seqlen%
      maxlen% = 0
      FOR number% = 2 TO 100000
        seqlen% = FNhailstone(number%, FALSE)
        IF seqlen% > maxlen% THEN
          maxlen% = seqlen%
          maxnum% = number%
        ENDIF
      NEXT
      PRINT "The number with the longest hailstone sequence is " ; maxnum%
      PRINT "Its sequence length is " ; maxlen%
      END
      
      DEF FNhailstone(N%, S%)
      LOCAL L%
      IF S% THEN PRINT N%;
      WHILE N% <> 1
        IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
        IF S% THEN PRINT N%;
        L% += 1
      ENDWHILE
      = L% + 1

Output:

        27        82        41       124        62        31        94        47
       142        71       214       107       322       161       484       242
       121       364       182        91       274       137       412       206
       103       310       155       466       233       700       350       175
       526       263       790       395      1186       593      1780       890
       445      1336       668       334       167       502       251       754
       377      1132       566       283       850       425      1276       638
       319       958       479      1438       719      2158      1079      3238
      1619      4858      2429      7288      3644      1822       911      2734
      1367      4102      2051      6154      3077      9232      4616      2308
      1154       577      1732       866       433      1300       650       325
       976       488       244       122        61       184        92        46
        23        70        35       106        53       160        80        40
        20        10         5        16         8         4         2         1

Sequence length = 112
The number with the longest hailstone sequence is 77031
Its sequence length is 351

Commodore BASIC

100 PRINT : PRINT "HAILSTONE SEQUENCE FOR N = 27:"
110 N=27 : SHOW=1
120 GOSUB 1000
130 PRINT X"ELEMENTS"
140 PRINT : PRINT "FINDING N WITH THE LONGEST HAILSTONE SEQUENCE"
150 SHOW=0
160 T0 = TI
170 FOR N=2 TO 100000
180 : GOSUB 1000
190 : IF X>MAX THEN MAX=X : NMAX = N
200 : REM' PRINT N,X,MAX
210 NEXT
230 PRINT "LONGEST HAILSTONE SEQUENCE STARTS WITH "NMAX"."
240 PRINT "IT HAS"MAX"ELEMENTS"
260 END
1000 REM '*** HAILSTONE SEQUENCE SUBROUTINE ***
1010 X = 0 : S = N
1020 IF SHOW THEN PRINT S,
1030 X = X+1
1040 IF S=1 THEN RETURN
1050 IF INT(S/2)=S/2 THEN S = S/2 : GOTO 1020
1060 S = 3*S+1
1070 GOTO 1020

FreeBASIC

' version 17-06-2015
' compile with: fbc -s console

Function hailstone_fast(number As ULongInt) As ULongInt
    ' faster version
    ' only counts the sequence

    Dim As ULongInt count = 1

    While number <> 1
        If (number And 1) = 1 Then
            number += number Shr 1 + 1  ' 3*n+1 and n/2 in one
            count += 2
        Else
            number Shr= 1 ' divide number by 2
            count += 1
        End If
    Wend

    Return count

End Function

Sub hailstone_print(number As ULongInt)
    ' print the number and sequence

    Dim As ULongInt count = 1

    Print "sequence for number "; number
    Print Using "########"; number;   'starting number

    While number <> 1
        If (number And 1) = 1 Then
            number = number * 3 + 1   ' n * 3 + 1
            count += 1
        Else
            number = number \ 2       ' n \ 2
            count += 1
        End If
        Print Using "########"; number;
    Wend

    Print : Print
    Print "sequence length = "; count
    Print
    Print String(79,"-")

End Sub

Function hailstone(number As ULongInt) As ULongInt
    ' normal version
    ' only counts the sequence

    Dim As ULongInt count = 1

    While number <> 1
        If (number And 1) = 1 Then
            number = number * 3 + 1 ' n * 3 + 1
            count += 1
        End If
        number = number \ 2 ' divide number by 2
        count += 1
    Wend

    Return count

End Function

' ------=< MAIN >=------

Dim As ULongInt number
Dim As UInteger x, max_x, max_seq

hailstone_print(27)
Print

For x As UInteger = 1 To 100000
    number = hailstone(x)
    If number > max_seq Then
        max_x = x
        max_seq = number
    End If
Next

Print  "The longest sequence is for "; max_x; ", it has a sequence length of "; max_seq

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print : Print "hit any key to end program"
Sleep
End

Output:

sequence for number 27
     27      82      41     124      62      31      94      47     142      71
    214     107     322     161     484     242     121     364     182      91
    274     137     412     206     103     310     155     466     233     700
    350     175     526     263     790     395    1186     593    1780     890
    445    1336     668     334     167     502     251     754     377    1132
    566     283     850     425    1276     638     319     958     479    1438
    719    2158    1079    3238    1619    4858    2429    7288    3644    1822
    911    2734    1367    4102    2051    6154    3077    9232    4616    2308
   1154     577    1732     866     433    1300     650     325     976     488
    244     122      61     184      92      46      23      70      35     106
     53     160      80      40      20      10       5      16       8       4
      2       1

sequence length = 112
-------------------------------------------------------------------------------
The longest sequence is for 77031, it has a sequence length of 351

GW-BASIC

10 N# = 27
20 P = 1
30 GOSUB 130
40 PRINT "That took";C;"steps."
50 P = 0 : A = 0 : B = 0
60 FOR M = 1 TO 99999!
70 N# = M
80 GOSUB 130
90 IF C > B THEN B = C: A = M
100 NEXT M
110 PRINT "The longest sequence is for n=";A;" and is ";B;" steps long."
120 END
130 C = 1
140 IF P = 1 THEN PRINT N#
150 IF N# < 2 THEN RETURN
160 IF N#/2 = INT(N#/2) THEN N# = N#/2 ELSE N# = 3*N# + 1
170 C = C + 1
180 GOTO 140

Liberty BASIC

print "Part 1: Create a routine to generate the hailstone sequence for a number."
print ""
while hailstone < 1 or hailstone <> int(hailstone)
    input "Please enter a positive integer: "; hailstone
wend
print ""
print "The following is the 'Hailstone Sequence' for your number..."
print ""
print hailstone
while hailstone <> 1
    if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
    print hailstone
wend
print ""
input "Hit 'Enter' to continue to part 2...";dummy$
cls
print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1."
print ""
print "No. in Seq.","Hailstone Sequence Number for 27"
print ""
c = 1: hailstone = 27
print c, hailstone
while hailstone <> 1
    c = c + 1
    if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
    print c, hailstone
wend
print ""
input "Hit 'Enter' to continue to part 3...";dummy$
cls
print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!"
print ""
print "Calculating result... Please wait... This could take a little while..."
print ""
print "Percent Done", "Start Number", "Seq. Length", "Maximum Sequence So Far"
print ""
for cc = 1 to 99999
    hailstone = cc: c = 1
    while hailstone <> 1
        c = c + 1
        if hailstone / 2 = int(hailstone / 2) then hailstone = hailstone / 2 else hailstone = (3 * hailstone) + 1
    wend
    if c > max then max = c: largesthailstone = cc
    locate 1, 7
    print "                                                                    "
    locate 1, 7
    print using("###.###", cc / 99999 * 100);"%", cc, c, max
    scan
next cc
print ""
print "The number less than 100,000 with the longest 'Hailstone Sequence' is "; largesthailstone;". It's sequence length is "; max;"."
end

OxygenBasic

function Hailstone(sys *n)
'=========================
if n and 1
  n=n*3+1
else
  n=n>>1
end if
end function

function HailstoneSequence(sys n) as sys
'=======================================
count=1
do
  Hailstone n
  Count++
  if n=1 then exit do
end do
return count
end function

'MAIN
'====

maxc=0
maxn=0
e=100000
for n=1 to e
 c=HailstoneSequence n
  if c>maxc
    maxc=c
    maxn=n
  end if
next

print e ", " maxn ", " maxc

'result 100000, 77031, 351

PureBasic

NewList Hailstones.i() ; Make a linked list to use as we do not know the numbers of elements needed for an Array

Procedure.i FillHailstones(n) ; Fills the list & returns the amount of elements in the list
  Shared Hailstones()         ; Get access to the Hailstones-List
  ClearList(Hailstones())     ; Remove old data
  Repeat
    AddElement(Hailstones())  ; Add an element to the list
    Hailstones()=n            ; Fill current value in the new list element
    If n=1
      ProcedureReturn ListSize(Hailstones())
    ElseIf n%2=0
      n/2
    Else
      n=(3*n)+1
    EndIf
  ForEver
EndProcedure

If OpenConsole()
  Define i, l, maxl, maxi
  l=FillHailstones(27)
  Print("#27 has "+Str(l)+" elements and the sequence is: "+#CRLF$)
  ForEach Hailstones()
    If i=6
      Print(#CRLF$)
      i=0
    EndIf    
    i+1
    Print(RSet(Str(Hailstones()),5))
    If Hailstones()<>1
      Print(", ")
    EndIf
  Next
  
  i=1
  Repeat
    l=FillHailstones(i)
    If l>maxl
      maxl=l
      maxi=i
    EndIf
    i+1
  Until i>=100000
  Print(#CRLF$+#CRLF$+"The longest sequence below 100000 is #"+Str(maxi)+", and it has "+Str(maxl)+" elements.")
 
  Print(#CRLF$+#CRLF$+"Press ENTER to exit."): Input()
  CloseConsole()
EndIf

Output:

 #27 has 112 elements and the sequence is:
    27,    82,    41,   124,    62,    31,
    94,    47,   142,    71,   214,   107,
   322,   161,   484,   242,   121,   364,
   182,    91,   274,   137,   412,   206,
   103,   310,   155,   466,   233,   700,
   350,   175,   526,   263,   790,   395,
  1186,   593,  1780,   890,   445,  1336,
   668,   334,   167,   502,   251,   754,
   377,  1132,   566,   283,   850,   425,
  1276,   638,   319,   958,   479,  1438,
   719,  2158,  1079,  3238,  1619,  4858,
  2429,  7288,  3644,  1822,   911,  2734,
  1367,  4102,  2051,  6154,  3077,  9232,
  4616,  2308,  1154,   577,  1732,   866,
   433,  1300,   650,   325,   976,   488,
   244,   122,    61,   184,    92,    46,
    23,    70,    35,   106,    53,   160,
    80,    40,    20,    10,     5,    16,
     8,     4,     2,     1
 
 The longest sequence found up to 100000 is #77031 which has 351 elements.
 
 Press ENTER to exit.

Run BASIC

print "Part 1: Create a routine to generate the hailstone sequence for a number."
print ""
 
while hailstone < 1 or hailstone <> int(hailstone)
    input "Please enter a positive integer: "; hailstone
wend
count = doHailstone(hailstone,"Y")
 
print: print "Part 2: Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1."
count = doHailstone(27,"Y")
 
print: print "Part 3: Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.(But don't show the actual sequence)!"
print "Calculating result... Please wait... This could take a little while..."
print "Stone Percent Count"
for i = 1 to 99999
   count = doHailstone(i,"N")
	if count > maxCount then
	   theBigStone = i
	   maxCount	= count
     print using("#####",i);" ";using("###.#", i / 99999 * 100);"% ";using("####",count)
     end if
next i
end
 
'---------------------------------------------
' pass number and print (Y/N)
FUNCTION doHailstone(hailstone,prnt$)
if prnt$ = "Y" then
 print
 print "The following is the 'Hailstone Sequence' for number:";hailstone
end if
while hailstone <> 1
   if (hailstone and 1) then hailstone = (hailstone * 3) + 1 else hailstone = hailstone / 2
   doHailstone = doHailstone + 1
   if prnt$ = "Y" then 
    print hailstone;chr$(9);
    if (doHailstone mod 10) = 0 then print
   end if
wend
END FUNCTION

Tiny BASIC

Tiny BASIC is limited to signed integers from -32768 to 32767. This code combines two integers into one: number = 32766A + B, to emulate integers up to 1.07 billion. Dealing with integer truncation, carries, and avoiding overflows requires some finesse. Even so one number, namely 77671, causes an overflow because one of its steps exceeds 1.07 billion.

    PRINT "Enter a positive integer"
    INPUT N        REM unit column
    LET M = 0      REM 32766 column
    LET C = 1      REM count
    LET P = 1      REM print the sequence?
    LET L = 1      REM finite state label
    GOSUB 10
    LET F = 1      REM current champion
    LET E = 0      REM 32766 part of current champ
    LET Y = 1      REM length of current longest sequence
    LET P = 0      REM no more printing
    LET W = 0      REM currently testing this number
    LET V = 0      REM 32766 column of the number
    PRINT "Testing for longest chain for n<100000..."
 5  LET W = W + 1
    REM PRINT V, " ", W
    LET N = W
    LET M = V
    LET C = 1      REM reset count
    IF W = 32766 THEN GOSUB 50
    GOSUB 10
    IF C > Y THEN GOSUB 60
    IF V = 3 THEN IF W = 1702 THEN GOTO 8
    GOTO 5
 8  PRINT "The longest sequence starts at 32766x",E," + ",F
    PRINT "And goes for ",Y," steps."
    END
    
10  IF P = 1 THEN IF M > 0 THEN PRINT C,"   32766x",M," + ",N
    IF P = 1 THEN IF M = 0 THEN PRINT C,"             ",N
    IF M = 0 THEN IF N = 1 THEN RETURN
    LET C = C + 1
    IF 2*(N/2)=N THEN GOTO 20
    IF M > 10922 THEN GOTO 100
    IF N > 21844 THEN GOTO 30
    IF N > 10922 THEN GOTO 40

    LET N = 3*N + 1
    LET M = 3*M
    GOTO 10
    
20  LET N = N/2
    IF (M/2)*2<>M THEN LET N = N + 16383   REM account for integer truncation
    LET M=M/2
    GOTO 10

30  LET N = N - 21844     REM two ways of accounting for overflow
    LET N = 3*N + 1
    LET M = 3*M + 2
    GOTO 10
    
40  LET N = N - 10922
    LET N = 3*N + 1
    LET M = 3*M + 1
    GOTO 10
    
50  LET W = 0         REM addition with carry
    LET V = V + 1
    RETURN
    
60  LET Y = C         REM tracking current champion
    LET F = W
    LET E = V
    RETURN

100 PRINT "Uh oh, getting an overflow for 32766x",V," + ",W
    PRINT "at step number ",C
    PRINT "trying to triple 32766x",M," + ",N
    RETURN

Output:

Enter a positive integer
27
1             27
2             82
3             41
....
110             4
111             2
112             1
Testing for longest chain for n<100000...
Uh oh, getting an overflow for 32766x2 + 12139
at step number 72
trying to triple 32766x15980 + 7565
The longest sequence starts at 32766x2 + 11499
And goes for 351 steps.

Batch File

1. Create a routine to generate the hailstone sequence for a number.
2. Show that the hailstone sequence for the number 27 has 112 elements...

@echo off
setlocal enabledelayedexpansion
echo.
::Task #1
call :hailstone 111
echo Task #1: (Start:!sav!)
echo !seq!
echo.
echo Sequence has !cnt! elements.
echo.
::Task #2
call :hailstone 27
echo Task #2: (Start:!sav!)
echo !seq!
echo.
echo Sequence has !cnt! elements.
echo.
pause>nul
exit /b 0

::The Function
:hailstone
set num=%1
set seq=%1
set sav=%1
set cnt=0

:loop
set /a cnt+=1
if !num! equ 1 goto :eof
set /a isodd=%num%%%2
if !isodd! equ 0 goto divideby2

set /a num=(3*%num%)+1
set seq=!seq! %num%
goto loop

:divideby2
set /a num/=2
set seq=!seq! %num%
goto loop

Output:

Task #1: (Start:111)
111 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

Sequence has 70 elements.

Task #2: (Start:27)
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

Sequence has 112 elements.

The script above could only be used in smaller inputs. Thus, for the third task, a slightly different script will be used. However, this script is still slow. I tried this on a fast computer and it took about 40-45 minutes to complete.

@echo off
setlocal enableDelayedExpansion
if "%~1"=="test" (
  for /l %%. in () do (
    set /a "test1=num %% 2, cnt=cnt+1"
    if !test1! equ 0 (set /a num/=2 & if !num! equ 1 exit !cnt!) else (set /a num=3*num+1)
  )
)

set max=0
set record=0

for /l %%X in (2,1,100000) do (
	set num=%%X & cmd /c "%~f0" test
	if !errorlevel! gtr !max! (set /a "max=!errorlevel!,record=%%X")
)
set /a max+=1

echo.Number less than 100000 with longest sequence: %record%
echo.With length %max%.
pause>nul

exit /b 0

Output:

Number less than 100000 with longest sequence: 77031
With length 351.

beeswax

This approach reuses the main hailstone sequence function for all three tasks.

The pure hailstone sequence function, returning the sequence for any number entered in the console:

   >@:N  q
>%"d3~@.PNp
d~2~pL~1F{<T_

Returning the sequence for the starting value 27

   >@:N  q
>%"d3~@.PNq
d~2~qL~1Ff{<BF3_
{NNgA<

Output of the sequence, followed by the length of the sequence:

27
82
41
124
62
31
94
47

...

2158
1079
3238
1619
4858
2429
7288
3644
1822

...

16
8
4
2
1

112

Number below 100,000 with the longest hailstone sequence, and the length of that sequence:

   >@:  q pf1_#
>%"d3~@.Pqf#{g?` `{gpK@~BP9~5@P@q'M<
d~2~pL~1Ff<         <            >?d
    >zAg?MM@1~y@~gLpz2~yg@~3~hAg?M d
                   >?~fz1~y?yg@hhAg?Mb

Output:

77031 351

Befunge

93*:.    v   
> :2%v  >
v+1*3_2/
>" ",:.v   v<
<v v-1:< <
+1\_$1+v^ \  
v .,+94<>^>::v
>" "03pv  :* p
v67:" "<  0: 1
>p78p25  *^*p0
  v!-1:  <<*^<
9$_:0\  ^-^< v
v01g00:< 1   4
>g"@"*+`v^  <+
v01/"@":_ $ ^,
>p"@"%00p\$:^.
vg01g00  ,+49<
>"@"*+.@

Output:

27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
112
77031
351

BQN

Works in: CBQN

Collatz ← ⥊⊸{
  𝕨𝕊1: 𝕨;
  (𝕨⊸∾ 𝕊 ⊢) (2|𝕩)⊑⟨𝕩÷2⋄1+3×𝕩⟩
}

Collatz1 ← ⌽∘{
  1: ⟨1⟩;
  𝕩∾˜𝕊(2|𝕩)⊑⟨𝕩÷2⋄1+3×𝕩⟩
}

•Show Collatz1 5

•Show (⊑∾≠){𝕩⊑˜⊑⍒≠¨𝕩}Collatz1¨1+↕99999

⟨ 5 16 8 4 2 1 ⟩
⟨ 77031 351 ⟩

Bracmat

(
  ( hailstone
  =   L len
    .   !arg:?L
      &   whl
        ' ( !arg:~1
          & (!arg*1/2:~/|3*!arg+1):?arg
          & !arg !L:?L
          )
      & (!L:? [?len&!len.!L)
  )
& ( reverse
  =   L e
    .   :?L
      & whl'(!arg:%?e ?arg&!e !L:?L)
      & !L
  )
& hailstone$27:(?len.?list)
& reverse$!list:?first4 [4 ? [-5 ?last4
& put$"Hailstone sequence starting with "
& put$!first4
& put$(str$(" has " !len " elements and ends with "))
& put$(!last4 \n)
& 1:?N
& 0:?max:?Nmax
&   whl
  ' ( !N+1:<100000:?N
    &   hailstone$!N
      : (   >!max:?max&!N:?Nmax
          | ?
        . ?
        )
    )
&   out
  $ ( str
    $ ( "The number <100000 with the longest hailstone sequence is "
        !Nmax
        " with "
        !max
        " elements."
      )
    )
);

Brainf***

This example is incomplete. Please ensure that it meets all task requirements and remove this message.

>>>>>>,>,>,<<

[
 .[-<+>]
]
>
[
 .[-<+>]
]
>
[
 .[-<+>]
]
<<<<


>------------------------------------------------[<<+>>-]>
[
    <<<
    [<+>-]<
    [>++++++++++<-]>
    >>>
    ------------------------------------------------
    [<<<+>>>-]>
    [
        <<<<
        [<+>-]<
        [>++++++++++<-]>
        >>>>
        ------------------------------------------------
        [<<<<+>>>>-]
    ]
    <

<<<[>+<<<+>>-]>[-<+>]>>>>>>>>>++++[>+++++++++++<-]++++[>>++++++++<<-]<<<<<<<<<<

[
    >>>>>>>>>>+>.>.<<<<<<<<<<<<
    >>+>+<<<        
    [-[->]<]+       
    >>>[>]          
    <[-<]<[-]<      

    [>+>+<<-]>[<+>-]+
    >[
    <<<[->>>>+>+>+<<<<<<]>>>>>>
    [-<<<<<<+>>>>>>]<[-<<<<<+>>>>>]<[-<<<<+>>>>]
    <<<<+>>
    -
    >[-]]
    <<[-]>[
    <<[-<+>[-<->>>>>+>]<<<<<]>>>>[-<<<<+>>>>]<<
    -]

    <<[->+>+<<]>[-<+>]>
    
    [>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]
    ++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[.[-]<]<
    
    -[+>]<
]

[This program never terminates!                             ]
[This program isn't complete, (it only prints the hailstone ]
[sequence of a number until 1) but it may help other people ]
[to make complete versions.                                 ]
[                                                           ]
[This program only takes in up to 3 digit numbers as input  ]
[If you want to input 1 digit integers, add a 0 before. e.g ]
[04.                                                        ]
[                                                           ]
[Summary:                                                   ]
[This program takes 16 memory cells of space. Their data is ]
[presented below:                                           ]
[                                                           ]
[Cell 0: Temp cell.                                         ]
[Cell 1: Displays the current number. This changes based on ]
[Collatz' Conjecture.                                       ]
[Cell 14: Displays length of the hailstone sequence.        ]
[Cell 15: ASCII code for ",".                               ]
[Cell 16: ASCII code for " " (Space).                       ]
[Rest of the cells: Temp cells.                             ]

Brat

hailstone = { num |
  sequence = [num]
  while { num != 1 }
    { true? num % 2 == 0
      { num = num / 2 }
      { num = num * 3 + 1 }
      sequence << num
    }

  sequence
}

#Check sequence for 27
seq = hailstone 27
true? (seq[0,3] == [27 82 41 124] && seq[-1, -4] == [8 4 2 1])
  { p "Sequence for 27 is correct" }
  { p "Sequence for 27 is not correct!" }

#Find longest sequence for numbers < 100,000
longest = [number: 0 length: 0]

1.to 99999 { index |
    seq = hailstone index
    true? seq.length > longest[:length]
      { longest[:length] = seq.length
        longest[:number] = index
        p "Longest so far: #{index} @ #{longest[:length]} elements"
      }

    index = index + 1
  }

p "Longest was starting from #{longest[:number]} and was of length #{longest[:length]}"

Output:

Sequence for 27 is correct
Longest so far: 1 @ 1 elements
Longest so far: 2 @ 2 elements
Longest so far: 3 @ 8 elements
...
Longest so far: 52527 @ 340 elements
Longest so far: 77031 @ 351 elements
Longest was starting from 77031 and was of length 351

Bruijn

:import std/Combinator .
:import std/List .
:import std/Math M
:import std/Number/Binary .

# hailstone sequence using binary shifts
hailstone y [[(0 =? (+1b)) {}0 go]]
	go 0 : (=²?0 (1 /²0) (1 (↑¹0 + 0)))

# --- tests ---

seq-27 hailstone (+27b)

:test (∀seq-27) ((+112))
:test (take (+4) seq-27) ((+27b) : ((+82b) : ((+41b) : {}(+124b))))
:test (take (+4) <~>seq-27) ((+1b) : ((+2b) : ((+4b) : {}(+8b))))

below-100000 [0 : ∀(hailstone 0)] <$> seq
	seq take (+99999) (iterate ++‣ (+1b))

main [head (max-by (M.compare ⋔ tail) below-100000)]

Burlesque

blsq ) 27{^^^^2.%{3.*1.+}\/{2./}\/ie}{1!=}w!bx{\/+]}{\/isn!}w!L[
112

C

#include <stdio.h>
#include <stdlib.h>

int hailstone(int n, int *arry)
{
    int hs = 1;

    while (n!=1) {
        hs++;
        if (arry) *arry++ = n;
        n = (n&1) ? (3*n+1) : (n/2);
    }
    if (arry) *arry++ = n;
    return hs;
}

int main()
{
    int j, hmax = 0;
    int jatmax, n;
    int *arry;

    for (j=1; j<100000; j++) {
       n = hailstone(j, NULL);
       if (hmax < n) {
           hmax = n;
           jatmax = j;
       }
    }
    n = hailstone(27, NULL);
    arry = malloc(n*sizeof(int));
    n = hailstone(27, arry);

    printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
        arry[0],arry[1],arry[2],arry[3],
        arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
    printf("Max %d at j= %d\n", hmax, jatmax);
    free(arry);

    return 0;
}

Output:

[ 27, 82, 41, 124, ...., 8, 4, 2, 1] len= 112
Max 351 at j= 77031

With caching

Much faster if you want to go over a million or so.

#include <stdio.h>

#define N 10000000
#define CS N	/* cache size */

typedef unsigned long ulong;
ulong cache[CS] = {0};

ulong hailstone(ulong n)
{
	int x;
	if (n == 1) return 1;
	if (n < CS && cache[n]) return cache[n];

	x = 1 + hailstone((n & 1) ? 3 * n + 1 : n / 2);
	if (n < CS) cache[n] = x;
	return x;
}

int main()
{
	int i, l, max = 0, mi;
	for (i = 1; i < N; i++) {
		if ((l = hailstone(i)) > max) {
			max = l;
			mi = i;
		}
	}
	printf("max below %d: %d, length %d\n", N, mi, max);
	return 0;
}

C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Hailstone
{
    class Program
    {
        public static List<int> hs(int n,List<int> seq)
        {
            List<int> sequence = seq;
            sequence.Add(n);
            if (n == 1)
            {
                return sequence;
            }else{
                int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
                return hs(newn, sequence);
            }                        
        }

        static void Main(string[] args)
        {
            int n = 27;
            List<int> sequence = hs(n,new List<int>());
            Console.WriteLine(sequence.Count + " Elements");
            List<int> start = sequence.GetRange(0, 4);
            List<int> end = sequence.GetRange(sequence.Count - 4, 4);
            Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));            
            int number = 0, longest = 0;            
            for (int i = 1; i < 100000; i++)
            {
                int count = (hs(i, new List<int>())).Count;
                if (count > longest)
                {
                    longest = count;
                    number = i;
                }
            }
            Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
       }
    }
}

112 Elements
Starting with : 27,82,41,124 and ending with : 8,4,2,1
Number < 100000 with longest Hailstone seq.: 77031 with length of 351

With caching

As with the C example, much faster if you want to go over a million or so.

using System;
using System.Collections.Generic;

namespace ConsoleApplication1
{
    class Program
    {
        public static void Main()
        {
            int longestChain = 0, longestNumber = 0;

            var recursiveLengths = new Dictionary<int, int>();

            const int maxNumber = 100000;

            for (var i = 1; i <= maxNumber; i++)
            {
                var chainLength = Hailstone(i, recursiveLengths);
                if (longestChain >= chainLength) 
                    continue;

                longestChain = chainLength;
                longestNumber = i;
            }
            Console.WriteLine("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain);
        }

        private static int Hailstone(int num, Dictionary<int, int> lengths)
        {
            if (num == 1) 
                return 1;

            while (true)
            {
                if (lengths.ContainsKey(num))
                    return lengths[num];

                lengths[num] = 1 + ((num%2 == 0) ? Hailstone(num/2, lengths) : Hailstone((3*num) + 1, lengths));
            }
        }
    }
}

max below 100000: 77031 (351 steps)

C++

#include <iostream>
#include <vector>
#include <utility>

std::vector<int> hailstone(int i)
{
    std::vector<int> v;
    while(true){ 
        v.push_back(i);
        if (1 == i) break; 
        i = (i % 2) ? (3 * i + 1) : (i / 2);
    }
    return v;
}

std::pair<int,int> find_longest_hailstone_seq(int n)
{
    std::pair<int, int> maxseq(0, 0);
    int l; 
    for(int i = 1; i < n; ++i){
        l = hailstone(i).size(); 
        if (l > maxseq.second) maxseq = std::make_pair(i, l);
    }   
    return maxseq;
}

int main () {

// Use the routine to show that the hailstone sequence for the number 27 
    std::vector<int> h27;
    h27 = hailstone(27); 
// has 112 elements 
    int l = h27.size();
    std::cout << "length of hailstone(27) is " << l;
// starting with 27, 82, 41, 124 and 
    std::cout << " first four elements of hailstone(27) are ";
    std::cout << h27[0] << " " << h27[1] << " " 
              << h27[2] << " " << h27[3] << std::endl;
// ending with 8, 4, 2, 1
    std::cout << " last four elements of hailstone(27) are "
              << h27[l-4] << " " << h27[l-3] << " " 
              << h27[l-2] << " " << h27[l-1] << std::endl;

    std::pair<int,int> m = find_longest_hailstone_seq(100000); 

    std::cout << "the longest hailstone sequence under 100,000 is " << m.first 
              << " with " << m.second << " elements." <<std::endl;  

    return 0;
}

Output:

 length of hailstone(27) is 112 first four elements of hailstone(27) are 27 82 41 124
 last four elements of hailstone(27) are 8 4 2 1
 the longest hailstone sequence under 100,000 is 77031 with 351 elements.

Library: Qt

Uses: Qt

Templated solution works for all of Qt's sequential container classes (QLinkedList, QList, QVector).

#include <QDebug>
#include <QVector>

template <class T>
T hailstone(typename T::value_type n)
{
    T seq;
    for (seq << n; n != 1; seq << n) {
        n = (n&1) ? (3*n)+1 : n/2;
    }
    return seq;
}

template <class T>
T longest_hailstone_seq(typename T::value_type n)
{
    T maxSeq;
    for (; n > 0; --n) {
        const auto seq = hailstone<T>(n);
        if (seq.size() > maxSeq.size()) {
            maxSeq = seq;
        }
    }
    return maxSeq;
}

int main(int, char *[]) {
    const auto seq = hailstone<QVector<uint_fast16_t>>(27);
    qInfo() << "hailstone(27):";
    qInfo() << "  length:" << seq.size() << "elements";
    qInfo() << "  first 4 elements:" << seq.mid(0,4);
    qInfo() << "  last 4 elements:" << seq.mid(seq.size()-4);

    const auto max = longest_hailstone_seq<QVector<uint_fast32_t>>(100000);
    qInfo() << "longest sequence with starting element under 100000:";
    qInfo() << "  length:" << max.size() << "elements";
    qInfo() << "  starting element:" << max.first();
}

Output:

hailstone(27):
  length: 112 elements
  first 4 elements: QVector(27, 82, 41, 124)
  last 4 elements: QVector(8, 4, 2, 1)
longest sequence with starting element under 100000:
  length: 351 elements
  starting element: 77031

Ceylon

shared void run() {
	
	{Integer*} hailstone(variable Integer n) {
		variable [Integer*] stones = [n];
		while(n != 1) {
			n = if(n.even) then n / 2 else 3 * n + 1;
			stones = stones.append([n]);
		}
		return stones;
	}
	
	value hs27 = hailstone(27);
	print("hailstone sequence for 27 is ``hs27.take(3)``...``hs27.skip(hs27.size - 3).take(3)`` with length ``hs27.size``");
	
	variable value longest = hailstone(1);
	for(i in 2..100k - 1) {
		value current = hailstone(i);
		if(current.size > longest.size) {
			longest = current;
		}
	}
	print("the longest sequence under 100,000 starts with ``longest.first else "what?"`` and has length ``longest.size``");
}

CLIPS

(deftemplate longest
  (slot bound)             ; upper bound for the range of values to check
  (slot next (default 2))  ; next value that needs to be checked
  (slot start (default 1)) ; starting value of longest sequence
  (slot len (default 1))   ; length of longest sequence
)

(deffacts startup
  (query 27)
  (longest (bound 100000))
)

(deffunction hailstone-next
  (?n)
  (if (evenp ?n)
    then (div ?n 2)
    else (+ (* 3 ?n) 1)
  )
)

(defrule extend-sequence
  ?hail <- (hailstone $?sequence ?tail&:(> ?tail 1))
  =>
  (retract ?hail)
  (assert (hailstone ?sequence ?tail (hailstone-next ?tail)))
)

(defrule start-query
  (query ?num)
  =>
  (assert (hailstone ?num))
)

(defrule result-query
  (query ?num)
  (hailstone ?num $?sequence 1)
  =>
  (bind ?sequence (create$ ?num ?sequence 1))
  (printout t "Hailstone sequence starting with " ?num ":" crlf)
  (bind ?len (length ?sequence))
  (printout t "  Length: " ?len crlf)
  (printout t "  First four: " (implode$ (subseq$ ?sequence 1 4)) crlf)
  (printout t "  Last four: " (implode$ (subseq$ ?sequence (- ?len 3) ?len)) crlf)
  (printout t crlf)
)

(defrule longest-create-next-hailstone
  (longest (bound ?bound) (next ?next))
  (test (<= ?next ?bound))
  (not (hailstone ?next $?))
  =>
  (assert (hailstone ?next))
)

(defrule longest-check-next-hailstone
  ?longest <- (longest (bound ?bound) (next ?next) (start ?start) (len ?len))
  (test (<= ?next ?bound))
  ?hailstone <- (hailstone ?next $?sequence 1)
  =>
  (retract ?hailstone)
  (bind ?thislen (+ 2 (length ?sequence)))
  (if (> ?thislen ?len) then
    (modify ?longest (start ?next) (len ?thislen) (next (+ ?next 1)))
    else
    (modify ?longest (next (+ ?next 1)))
  )
)

(defrule longest-finished
  (longest (bound ?bound) (next ?next) (start ?start) (len ?len))
  (test (> ?next ?bound))
  =>
  (printout t "The number less than " ?bound " that has the largest hailstone" crlf)
  (printout t "sequence is " ?start " with a length of " ?len "." crlf)
  (printout t crlf)
)

Output:

The number less than 100000 that has the largest hailstone
sequence is 77031 with a length of 351.

Hailstone sequence starting with 27:
  Length: 112
  First four: 27 82 41 124
  Last four: 8 4 2 1

Clojure

(defn hailstone-seq [n]
  {:pre [(pos? n)]}
  (lazy-seq 
   (cond (= n 1)   '(1)
         (even? n) (cons n (hailstone-seq (/ n 2)))
         :else     (cons n (hailstone-seq (+ (* n 3) 1))))))

(let [hseq (hailstone-seq 27)]
  (->  hseq count      (= 112)            assert)
  (->> hseq (take 4)   (= [27 82 41 124]) assert)
  (->> hseq (drop 108) (= [8 4 2 1])      assert))

(let [{max-i :num, max-len :len}
      (reduce #(max-key :len %1 %2)
              (for [i (range 1 100000)]
                {:num i, :len (count (hailstone-seq i))}))]
  (println "Maximum length" max-len "was found for hailstone(" max-i ")."))

CLU

% Generate the hailstone sequence for a number
hailstone = iter (n: int) yields (int)
    while true do
        yield(n)
        if n=1 then break end
        if n//2 = 0 then
            n := n/2
        else
            n := 3*n + 1
        end
    end
end hailstone

% Make an array from an iterator
iter_array = proc [T,U: type] (i: itertype (U) yields (T), s: U) returns (array[T])
    arr: array[T] := array[T]$[]
    for item: T in i(s) do array[T]$addh(arr, item) end
    return(arr)
end iter_array

start_up = proc () 
    po: stream := stream$primary_output()
    
    % Generate the hailstone sequence for 27 
    h27: array[int] := iter_array[int,int](hailstone, 27)
    lo27: int := array[int]$low(h27)
    hi27: int := array[int]$high(h27)
    
    stream$putl(po, "The hailstone sequence for 27 has "
                 || int$unparse(array[int]$size(h27)) || " elements.")
    stream$puts(po, "The first 4 elements are:")
    for i: int in int$from_to(lo27, lo27+3) do
        stream$puts(po, " " || int$unparse(h27[i]))
    end
    stream$puts(po, ", and the last 4 elements are:")
    for i: int in int$from_to(hi27-3, hi27) do
        stream$puts(po, " " || int$unparse(h27[i]))
    end
    stream$putl(po, "")
    
    % Find whichever sequence < 100 000 has the longest sequence
    maxnum: int := 0
    maxlen: int := 0
    
    for i: int in int$from_to(1, 99999) do
        len: int := array[int]$size(iter_array[int,int](hailstone, i))
        if len > maxlen then
            maxnum, maxlen := i, len
        end
    end
    
    stream$putl(po, int$unparse(maxnum)
                 || " has the longest hailstone sequence < 100000: "
                 || int$unparse(maxlen))
end start_up

Output:

The hailstone sequence for 27 has 112 elements.
The first 4 elements are: 27 82 41 124, and the last 4 elements are: 8 4 2 1
77031 has the longest hailstone sequence < 100000: 351

COBOL

Testing with GnuCOBOL

       identification division.
       program-id. hailstones.
       remarks. cobc -x hailstones.cob.

       data division.
       working-storage section.
       01 most                 constant as 1000000.
       01 coverage             constant as 100000.       
       01 stones               usage binary-long.
       01 n                    usage binary-long.
       01 storm                usage binary-long.

       01 show-arg             pic 9(6).
       01 show-default         pic 99 value 27.
       01 show-sequence        usage binary-long.
       01 longest              usage binary-long occurs 2 times.

       01 filler.
          05 hail              usage binary-long
                               occurs 0 to most depending on stones.
       01 show                 pic z(10).
       01 low-range            usage binary-long.
       01 high-range           usage binary-long.
       01 range                usage binary-long.
        

       01 remain               usage binary-long.
       01 unused               usage binary-long.

       procedure division.
       accept show-arg from command-line
       if show-arg less than 1 or greater than coverage then
           move show-default to show-arg
       end-if
       move show-arg to show-sequence

       move 1 to longest(1)
       perform hailstone varying storm
                         from 1 by 1 until storm > coverage
       display "Longest at: " longest(2) " with " longest(1) " elements"
       goback.

      *> **************************************************************
       hailstone.
       move 0 to stones
       move storm to n
       perform until n equal 1
           if stones > most then
               display "too many hailstones" upon syserr
               stop run
           end-if

           add 1 to stones
           move n to hail(stones)
           divide n by 2 giving unused remainder remain
           if remain equal 0 then
               divide 2 into n
           else
               compute n = 3 * n + 1
           end-if
       end-perform
       add 1 to stones
       move n to hail(stones)

       if stones > longest(1) then
           move stones to longest(1)
           move storm to longest(2)
       end-if

       if storm equal show-sequence then
           display show-sequence ": " with no advancing
           perform varying range from 1 by 1 until range > stones
               move 5 to low-range
               compute high-range = stones - 4
               if range < low-range or range > high-range then
                   move hail(range) to show
                   display function trim(show) with no advancing
                   if range < stones then
                       display ", " with no advancing
                   end-if
               end-if
               if range = low-range and stones > 8 then
                   display "..., " with no advancing
               end-if
           end-perform
           display ": " stones " elements"
       end-if
       .

       end program hailstones.

Output:

prompt$ cobc -x hailstones.cob
prompt$ ./hailstones
+0000000027: 27, 82, 41, 124, ..., 8, 4, 2, 1: +0000000112 elements
Longest at: +0000077031 with +0000000351 elements
prompt$ ./hailstones 42
+0000000042: 42, 21, 64, 32, ..., 8, 4, 2, 1: +0000000009 elements
Longest at: +0000077031 with +0000000351 elements

CoffeeScript

Recursive version:

hailstone = (n) ->
  if n is 1
    [n]
 
  else if n % 2 is 0
    [n].concat hailstone n/2
 
  else
    [n].concat hailstone (3*n) + 1
 
h27 = hailstone 27
console.log "hailstone(27) = #{h27[0..3]} ... #{h27[-4..]} (length: #{h27.length})"
 
maxlength = 0
maxnums = []
 
for i in [1..100000]
  seq = hailstone i
 
  if seq.length is maxlength
    maxnums.push i
  else if seq.length > maxlength
    maxlength = seq.length
    maxnums = [i]
 
console.log "Max length: #{maxlength}; numbers generating sequences of this length: #{maxnums}"

hailstone(27) = 27,82,41,124 ... 8,4,2,1 (length: 112)
Max length: 351; numbers generating sequences of this length: 77031

Common Lisp

(defun hailstone (n)
  (cond ((= n 1) '(1))
	((evenp n) (cons n (hailstone (/ n 2))))
	(t (cons n (hailstone (+ (* 3 n) 1))))))

(defun longest (n)
  (let ((k 0) (l 0))
    (loop for i from 1 below n do
	 (let ((len (length (hailstone i))))
	   (when (> len l) (setq l len k i)))
	 finally (format t "Longest hailstone sequence under ~A for ~A, having length ~A." n k l))))

Sample session:

ROSETTA> (length (hailstone 27))
112
ROSETTA> (subseq (hailstone 27) 0 4)
(27 82 41 124)
ROSETTA> (last (hailstone 27) 4)
(8 4 2 1)
ROSETTA> (longest-hailstone 100000)
Longest hailstone sequence under 100000 for 77031, having length 351.
NIL

Cowgol

include "cowgol.coh";

# Generate the hailstone sequence for the given N and return the length.
# If a non-NULL pointer to a buffer is given, then store the sequence there.
sub hailstone(n: uint32, buf: [uint32]): (len: uint32) is
    len := 0;
    loop
        if buf != 0 as [uint32] then
            [buf] := n;
            buf := @next buf;
        end if;
        len := len + 1;
        if n == 1 then
            break;
        elseif n & 1 == 0 then
            n := n / 2;
        else
            n := 3*n + 1;
        end if;
    end loop;
end sub;

# Generate hailstone sequence for 27
var h27: uint32[113];
var h27len := hailstone(27, &h27[0]);

# Print information about it
print("The hailstone sequence for 27 has ");
print_i32(h27len);
print(" elements.\nThe first 4 elements are:");
var n: @indexof h27 := 0;
while n < 4 loop
    print_char(' ');
    print_i32(h27[n]);
    n := n + 1;
end loop;
print(", and the last 4 elements are:");
n := h27len as @indexof h27 - 4;
while n as uint32 < h27len loop
    print_char(' ');
    print_i32(h27[n]);
    n := n + 1;
end loop
print(".\n");

# Find longest hailstone sequence < 100,000
var i: uint32 := 1;
var max_i := i;
var len: uint32 := 0;
var max_len := len;
while i < 100000 loop
    len := hailstone(i, 0 as [uint32]);
    if len > max_len then
        max_i := i;
        max_len := len;
    end if;
    i := i + 1;
end loop;

print_i32(max_i);
print(" has the longest hailstone sequence < 100000: ");
print_i32(max_len);
print_nl();

Output:

The hailstone sequence for 27 has 112 elements.
The first 4 elements are: 27 82 41 124, and the last 4 elements are: 8 4 2 1.
77031 has the longest hailstone sequence < 100000: 351

Crystal

def hailstone(n)
    seq = [n]
    until n == 1
        n = n.even? ? n // 2 : n * 3 + 1
        seq << n
    end
    seq
end

max_len = (1...100_000).max_by{|n| hailstone(n).size }
max = hailstone(max_len)
puts ([max_len, max.size, max.max, max.first(4), max.last(4)])
# => [77031, 351, 21933016, [77031, 231094, 115547, 346642], [8, 4, 2, 1]]

twenty_seven = hailstone(27)
puts ([twenty_seven.size, twenty_seven.first(4), max.last(4)])
# => [112, [27, 82, 41, 124], [8, 4, 2, 1]]

D

Basic Version

import std.stdio, std.algorithm, std.range, std.typecons;

auto hailstone(uint n) pure nothrow {
  auto result = [n];
  while (n != 1) {
    n = (n & 1) ? (n * 3 + 1) : (n / 2);
    result ~= n;
  }
  return result;
}

void main() {
  enum M = 27;
  immutable h = M.hailstone;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.length, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Output:

hailstone(27)= [27, 82, 41, 124] ... [8, 4, 2, 1]
Length hailstone(27)= 112
Longest sequence in [1,100000]= 77031 with len 351

Lazy Version

Same output.

import std.stdio, std.algorithm, std.typecons, std.range;

auto hailstone(uint m) pure nothrow @nogc {
    return m
           .recurrence!q{ a[n - 1] & 1 ? a[n - 1] * 3 + 1 : a[n - 1]/2}
           .until!q{ a == 1 }(OpenRight.no);
}

void main() {
  enum M = 27;
  immutable h = M.hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Faster Lazy Version

Same output.

struct Hailstone {
  uint n;
  bool empty() const pure nothrow @nogc { return n == 0; }
  uint front() const pure nothrow @nogc { return n; }
  void popFront() pure nothrow @nogc {
    n = n == 1 ? 0 : (n & 1 ? (n * 3 + 1) : n / 2);
  }
}

void main() {
  import std.stdio, std.algorithm, std.range, std.typecons;

  enum M = 27;
  immutable h = M.Hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.Hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Lazy Version With Caching

Faster, same output.

import std.stdio, std.algorithm, std.range, std.typecons;

struct Hailstone(size_t cacheSize = 500_000) {
  size_t n;
  __gshared static size_t[cacheSize] cache;

  bool empty() const pure nothrow @nogc { return n == 0; }
  size_t front() const pure nothrow @nogc { return n; }

  void popFront() nothrow {
    if (n >= cacheSize) {
      n = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
    } else if (cache[n]) {
      n = cache[n];
    } else {
      immutable n2 = n == 1 ? 0 : (n & 1 ? n*3 + 1 : n/2);
      n = cache[n] = n2;
    }
  }
}

void main() {
  enum M = 27;
  const h = M.Hailstone!().array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.Hailstone!().walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Generator Range Version

import std.stdio, std.algorithm, std.range, std.typecons, std.concurrency;

auto hailstone(size_t n) {
    return new Generator!size_t({
        yield(n);
        while (n > 1) {
            n = (n & 1) ? (3 * n + 1) : (n / 2);
            yield(n);
        }
    });
}

void main() {
  enum M = 27;
  const h = M.hailstone.array;
  writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
  writeln("Length hailstone(", M, ")= ", h.length);

  enum N = 100_000;
  immutable p = iota(1, N)
                .map!(i => tuple(i.hailstone.walkLength, i))
                .reduce!max;
  writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}

Dart

import 'package:collection/collection.dart';
import 'dart:collection';
List<int> hailstone(int n) {
  if(n <= 0) {
    throw ArgumentError("start value must be >=1)");
  }
  var seq = Queue<int>();
  seq.add(n);
  while(n != 1) {
    n = n%2 == 0 ? n ~/ 2 : 3 * n + 1;
    seq.add(n);
  }
  return seq.toList();
}

main() {
  for(int i = 1; i <= 10; i++) {
    print("h($i) = ${hailstone(i)}");
  }
  var h27 = hailstone(27);
  var first4 = h27.take(4).toList();
  print("first 4 elements of h(27): $first4");
  assert(ListEquality().equals([27, 82, 41, 124], first4));

  var last4 = h27.skip(h27.length - 4).take(4).toList();
  print("last 4 elements of h(27): $last4");
  assert(ListEquality().equals([8, 4, 2, 1], last4));

  print("length of sequence h(27): ${h27.length}");
  assert(112 == h27.length);

  int seq = 0, max = 0;
  for(int i = 1; i <= 100000; i++) {
    var h = hailstone(i);
    if(h.length > max) {
      max = h.length;
      seq = i;
    }
  }
  print("up to 100000 the sequence h($seq) has the largest length ($max)");
}

Output:

h(1) = [1]
h(2) = [2, 1]
h(3) = [3, 10, 5, 16, 8, 4, 2, 1]
h(4) = [4, 2, 1]
h(5) = [5, 16, 8, 4, 2, 1]
h(6) = [6, 3, 10, 5, 16, 8, 4, 2, 1]
h(7) = [7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(8) = [8, 4, 2, 1]
h(9) = [9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
h(10) = [10, 5, 16, 8, 4, 2, 1]
first 4 elements of h(27): [27, 82, 41, 124]
last 4 elements of h(27): [8, 4, 2, 1]
length of sequence h(27): 112
up to 100000 the sequence h(77031) has the largest length (351)

Dc

Firstly, this code takes the value from the stack, computes and prints the corresponding Hailstone sequence, and the length of the sequence. The q procedure is for counting the length of the sequence. The e and o procedure is for even and odd number respectively. The x procedure is for overall control.

27
[[--: ]nzpq]sq
[d 2/ p]se
[d 3*1+ p]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx

Output:

82
41
124
62
(omitted)
8
4
2
1
--: 112

Then we could wrap the procedure x with a new procedure s, and call it with l which is loops the value of t from 1 to 100000, and cleaning up the stack after each time we finish up with a number. Register L for the length of the longest sequence and T for the corresponding number. Also, procedure q is slightly modified for storing L and T if needed, and all printouts in procedure e and o are muted.

0dsLsT1st
[dsLltsT]sM
[[zdlL<M q]sq
[d 2/]se
[d 3*1+ ]so
[d2% 0=e d1=q d2% 1=o d1=q lxx]dsxx]ss
[lt1+dstlsxc lt100000>l]dslx
lTn[:]nlLp

Output:

(Takes quite some time on a decent machine)

77031:351

DCL

$ n = f$integer( p1 )
$ i = 1
$ loop:
$  if p2 .nes. "QUIET" then $ s'i = n
$  if n .eq. 1 then $ goto done
$  i = i + 1
$  if .not. n
$  then
$   n = n / 2
$  else
$   if n .gt. 715827882 then $ exit  ! avoid overflowing
$   n = 3 * n + 1
$  endif
$  goto loop
$ done:
$ if p2 .nes. "QUIET"
$ then
$  penultimate_i = i - 1
$  antepenultimate_i = i - 2
$  preantepenultimate_i = i - 3
$  write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$ endif
$ sequence_length == i

Output:

$ @hailstone 27
sequence has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1

$ limit = f$integer( p1 )
$ i = 1
$ max_so_far = 0
$ loop:
$  call hailstone 'i quiet
$  if sequence_length .gt. max_so_far
$  then
$   max_so_far = sequence_length
$   current_record_holder = i
$  endif
$  i = i + 1
$  if i .lt. limit then $ goto loop
$ write sys$output current_record_holder, " is the number less than ", limit, " which has the longest hailstone sequence which is ", max_so_far, " in length"
$ exit
$
$ hailstone: subroutine
$ n = f$integer( p1 )
$ i = 1
$ loop:
$  if p2 .nes. "QUIET" then $ s'i = n
$  if n .eq. 1 then $ goto done
$  i = i + 1
$  if .not. n
$  then
$   n = n / 2
$  else
$   if n .gt. 715827882 then $ exit  ! avoid overflowing
$   n = 3 * n + 1
$  endif
$  goto loop
$ done:
$ if p2 .nes. "QUIET"
$ then
$  penultimate_i = i - 1
$  antepenultimate_i = i - 2
$  preantepenultimate_i = i - 3
$  write sys$output "sequence has ", i, " elements starting with ", s1, ", ", s2, ", ", s3, ", ", s4, " and ending with ", s'preantepenultimate_i, ", ", s'antepenultimate_i, ", ", s'penultimate_i, ", ", s'i
$ endif
$ sequence_length == I
$ exit
$ endsubroutine

Output:

$ @longest_hailstone 100000
77031 is the number less than 100000 which has the longest hailstone sequence which is 351 in length

Delphi

Using List<Integer>

program ShowHailstoneSequence;

{$APPTYPE CONSOLE}

uses SysUtils, Generics.Collections;

procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
  n: Integer;
begin
  aHailstoneList.Clear;
  aHailstoneList.Add(aStartingNumber);
  n := aStartingNumber;

  while n <> 1 do
  begin
    if Odd(n) then
      n := (3 * n) + 1
    else
      n := n div 2;
    aHailstoneList.Add(n);
  end;
end;

var
  i: Integer;
  lList: TList<Integer>;
  lMaxSequence: Integer;
  lMaxLength: Integer;
begin
  lList := TList<Integer>.Create;
  try
    GetHailstoneSequence(27, lList);
    Writeln(Format('27: %d elements', [lList.Count]));
    Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
      [lList[0], lList[1], lList[2], lList[3],
      lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
    Writeln;

    lMaxSequence := 0;
    lMaxLength := 0;
    for i := 1 to 100000 do
    begin
      GetHailstoneSequence(i, lList);
      if lList.Count > lMaxLength then
      begin
        lMaxSequence := i;
        lMaxLength := lList.Count;
      end;
    end;
    Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
  finally
    lList.Free;
  end;

  Readln;
end.

Output:

27: 112 elements
[27 82 41 124 ... 8 4 2 1]

Longest sequence under 100,000: 77031 with 351 elements

Using Boost.Algorithm and TParallel.For

Library: System.SysUtils

Library: System.Types

Library: System.Threading

Library: System.SyncObjs

Library: Boost.Algorithm

Library: Boost.Int

[1]

Library: System.Diagnostics

program ShowHailstoneSequence;

{$APPTYPE CONSOLE}

uses
  System.SysUtils,
  System.Types,
  System.Threading,
  System.SyncObjs,
  Boost.Algorithm,
  Boost.Int,
  System.Diagnostics;

var
  lList: TIntegerDynArray;
  lMaxSequence, lMaxLength, i: Integer;
  StopWatch: TStopwatch;

begin
  lList := Hailstone(27);
  Writeln(Format('27: %d elements', [lList.Count]));
  Writeln(lList.toString(4), #10);

  lMaxSequence := 0;
  lMaxLength := 0;

  StopWatch := TStopwatch.Create;
  StopWatch.Start;

  TParallel.for (1, 1, 100000,
    procedure(idx: Integer)
    var
      lList: TIntegerDynArray;
    begin
      lList := Hailstone(idx);
      if lList.Count > lMaxLength then
      begin
        TInterlocked.Exchange(lMaxSequence, idx);
        TInterlocked.Exchange(lMaxLength, lList.Count);
      end;
    end);

  StopWatch.Stop;

  Write(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence,
    lMaxLength]));

  Writeln(Format(' in %d ms', [StopWatch.ElapsedMilliseconds]));

  Readln;
end.

Output:

27: 112 elements
[27, 82, 41, 124 ... 8, 4, 2, 1]

Longest sequence under 100,000: 77031 with 351 elements in 520 ms

Déjà Vu

local hailstone:
	swap [ over ]
	while < 1 dup:
		if % over 2:
			#odd
			++ * 3
		else:
			#even
			/ swap 2
		swap push-through rot dup
	drop
 
if = (name) :(main):
	local :h27 hailstone 27
	!. = 112 len h27
	!. = 27 h27! 0
	!. = 82 h27! 1
	!. = 41 h27! 2
	!. = 124 h27! 3
	!. = 8 h27! 108
	!. = 4 h27! 109
	!. = 2 h27! 110
	!. = 1 h27! 111
 
	local :max 0
	local :maxlen 0
	for i range 1 99999:
		dup len hailstone i
		if < maxlen:
			set :maxlen
			set :max i
		else:
			drop
	!print( "number: " to-str max ", length: " to-str maxlen )
else:
	@hailstone

Output:

true
true
true
true
true
true
true
true
true
number: 77031, length: 351

EasyLang

proc hailstone n . list[] .
   list[] = [ ]
   while n <> 1
      list[] &= n
      if n mod 2 = 0
         n = n / 2
      else
         n = 3 * n + 1
      .
   .
   list[] &= 1
.
hailstone 27 l[]
write "27 has length " & len l[] & " with "
for i to 4
   write l[i] & " "
.
write "... "
for i = len l[] - 3 to len l[]
   write l[i] & " "
.
print ""
for i = 1 to 100000
   hailstone i l[]
   if len l[] >= max_iter
      max_i = i
      max_iter = len l[]
   end
end
print max_i & " has length " & max_iter

EchoLisp

(lib 'hash)
(lib 'sequences)
(lib 'compile)

(define (hailstone n)
(when (> n 1)
	(if (even? n) (/ n 2) (1+ (* n 3)))))
	
(define H (make-hash))

;; (iterator/f seed f) returns seed, (f seed) (f(f seed)) ...

(define (hlength seed)
	(define collatz (iterator/f hailstone seed))
	(or
	   (hash-ref H seed) ;; known ?
	   (hash-set H seed
	      (for ((i (in-naturals)) (h collatz)) 
              ;; add length of subsequence if already known
	      #:break (hash-ref H h) => (+ i (hash-ref H h))
	      (1+ i)))))
	
(define (task (nmax 100000))
	(for ((n [1 .. nmax])) (hlength n)) ;; fill hash table

	(define hmaxlength (apply max (hash-values H)))
	(define hmaxseed (hash-get-key H hmaxlength))
	(writeln 'maxlength= hmaxlength 'for hmaxseed))

Output:

(define H27 (iterator/f hailstone 27))
(take H27 6)
   → (27 82 41 124 62 31)
(length H27)
   → 112
(list-tail (take H27 112) -6)
   → (5 16 8 4 2 1)

(task)
maxlength=     351     for     77031 

;; more ...
(lib 'bigint)
   
(task 200000)
    maxlength=     383     for     156159    
(task 300000)
    maxlength=     443     for     230631    
(task 400000)
    maxlength=     443     for     230631    
(task 500000)
    maxlength=     449     for     410011    
(task 600000)
    maxlength=     470     for     511935    
(task 700000)
    maxlength=     509     for     626331    
(task 800000)
    maxlength=     509     for     626331    
(task 900000)
    maxlength=     525     for     837799    
(task 1000000)
    maxlength=     525     for     837799

EDSAC order code

This program uses no optimization, and is best run on a fast simulator. Even with the storage-related code cut out, Part 2 of the task executes 182 million EDSAC orders. At 650 orders per second, the original EDSAC would have taken 78 hours.

 [Hailstone (or Collatz) task for Rosetta Code.
  EDSAC program, Initial Orders 2.]

 [This program shows how subroutines can be called via the
  phi, H, N, ..., V parameters, so that the code doesn't have
  to be changed if the subroutines are moved about in store.
  See Wilkes, Wheeler and Gill, 1951 edition, page 18.]

 [Library subroutine P7, prints long strictly positive integer;
  10 characters, right justified, padded left with spaces.
  Input: 0D = integer to be printed.
  Closed, even; 35 storage locations; working position 4D.]
            T   55 K  [call subroutine via V parameter]
            P   56 F  [address of subroutine]
            E   25 K
            T      V
   GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSFL4F
   T4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@

  [Subroutine to print a string placed after the subroutine call.
  One location per character, with character in top 5 bits.
  Last character flagged by having bit 0 set.
  17 locations, workspace 0F.]
            T   54 K  [call subroutine via C parameter]
            P   91 F  [address of subroutine]
            E   25 K
            T      C
   GKH16@A2FG4@A6@A2FT6@AFTFOFCFSFE3@A6@A3FT15@EFV2047F

 [************ Rosetta Code task ************
  Subroutine to generate and optionally store the hailstone
  (Collatz) sequence for the passed-in initial term n.
  Input:  4D = n, 35-bit positive integer
          6F = start address of sequence if stored;
               must be even; 0 = don't store
  Output: 7F = number of terms in sequence, or -1 if error
  Workspace: 0D (general), 8D (term of sequence)
  Must be loaded at an even address.]
            T   45 K  [call subroutine via H parameter]
            P  108 F  [address of subroutine]
            E   25 K
            T      H
            G      K
            A    3 F
            T   46 @
            H   54#@  [mult reg := 1 to test odd/even]
            A    4 D  [load n passed in by caller]
            T    8 D  [term := n]
            A   54 @  [load 1 (single)]
            T    7 F  [include initial term in count]
            A    6 F  [load address for store]
            S   56 @  [test for 0; allow for pre-inc]
            G   11 @  [skip next if storing not wanted]
            A   12 @  [make 'T addr D' order]
     [11]   T   21 @  [plant T order, or -ve value if not storing
                         (note that a T order is +ve as an integer)]
        [Loop: deal with current term in sequence
         First store it, if user requested that]
     [12]   T      D  [clear acc; also serves to make 'T addr D' order]
            A   21 @  [load T order to store term]
            G   22 @  [jump if caller doesn't want store]
            A   56 @  [pre-inc the address]
            U   21 @  [update T order]
            S   51 @  [check not gone beyond max EDSAC address]
            E   47 @  [error exit if it has]
            T      F  [clear acc]
            A    8 D  [load term]
     [21]   T      D  [store]
     [22]   T      F  [clear acc]
            A   54#@  [load 1 (double)]
            S    8 D  [1 - term]
            E   46 @  [if term = 1, jump out with acc = 0]
            T      F  [clear acc]
            C    8 D  [acc := term AND 1]
            S   54#@  [test whether 0 or 1]
            G   38 @  [jump if term is even]
         [Here if term is odd; acc = 0]
            A    8 D  [load term]
            S   52#@  [guard against numeric overflow]
            E   47 @  [jump if overflow]
            A   52#@  [restore term after test]
            L      D  [term*2]
            A    8 D  [term*3]
            A   54#@  [plus 1]
            E   41 @  [join common code]
         [Here if term is even]
     [38]   T      F  [clear acc]
            A    8 D  [load term]
            R      D  [term/2]
         [Common code, acc = new term]
     [41]   T    8 D  [store new term]
            A    7 F  [load count]
            A   54 @  [add 1]
            T    7 F  [update count]
            E   12 @  [loop back]
        [Here when sequence has reached 1
         Assume jump here with acc = 0]
     [46]   E      F  [return with acc = 0]
     [47]   T      F  [here on error]
            S   54 F  [acc := -1]
            T    7 F  [return that as count]
            E   46 @
   [Arrange the following to ensure even addresses for 35-bit values]
     [51]   T 1024 F  [for checking valid address]
     [52]   H  682 DT  682 D  [(2^34 - 1)/3]
     [54]   P      DP      F  [1]
     [56]   P    2 F  [to change addresses by 2]

        [Program to demonstrate Rosetta Code subroutine]
            T  180 K
            G      K
  [Double constants]
           [P 500 F  P F]  [maximum n = 1000"]
      [0]   & 848 F PF     [maximum n = 100000]
      [2]   P  13 D PF     [n = 27 as demo of sequence]
      [4]   P     D PF     [1]
  [Double variables]
      [6]   P  F P F  [n, start of Collatz sequence]
      [8]   P  F P F  [n with maximum count]
  [Single constants]
     [10]   P  400 F  [where to store sequence]
     [11]   P    2 F  [to change addresses by 2]
     [12]   @      F  [carriage return]
     [13]   &      F  [line feed]
     [14]   K 4096 F  [null char]
     [15]   A      D  [used for maiking 'A addr D' order]
     [16]   P    8 F  [ used for adding 8 to address]
  [Single variables]
     [17]   P      F  [maximum number of terms]
     [18]   P      F  [temporary store]
     [19]   P      F  [marks end of printing]

 [Subroutine to print 4 numbers starting at address in 6F.
  Prints new line (CR, LF) at end.]
     [20]   A    3 F  [plant link for return]
            T   40 @
            A    6 F  [load start address]
            A   15 @  [make 'A addr D' order]
            A   16 @  [inc address by 8 (4 double values)]
            U   19 @  [store as test for end]
            S   16 @  [restore 'A addr D' order for start]
     [27]   U   31 @  [plant 'A addr D' order in code]
            S   19 @  [test for end]
            E   38 @  [out if so]
            T      F  [clear acc]
     [31]   A      D  [load number]
            T      D  [to 0D for printing]
     [33]   A   33 @  [call print subroutine]
            G      V
            A   31 @  [load 'A addr D' order]
            A   11 @  [inc address to next double value]
            G   27 @  [loop back]
     [38]   O   12 @  [here when done, print CR LF]
            O   13 @
     [40]   E      F  [return]

         [Enter with acc = 0]
         [PART 1]
     [41]   A    2#@  [load demo value of n]
            T    4 D  [to 4D for subroutine]
            A   10 @  [address to store sequence]
            T    6 F  [to 6F for subroutine]
     [45]   A   45 @  [call subroutine to generate sequence]
            G      H
            A    7 F  [load length of sequence]
            G  198 @  [out if error]
            T   18 @
         [Print result]
     [50]   A   50 @  [print 'start' message]
            G      C
            K2048F SF TF AF RF TF !F !F #D
            A    2#@  [load demo value of n]
            T      D  [to 0D for printing]
     [63]   A   63 @  [print demo n]
            G      V
     [65]   A   65 @  [print 'length' string]
            G      C
            K2048F @F &F LF EF NF GF TF HF !F #D
            T      D  [ensure 1F and sandwich bit are 0]
            A   18 @  [load length]
            T      F  [to 0F (effectively 0D) for printing]
     [81]   A   81 @
            G      V
     [83]   A   83 @  [print 'first and last four' string]
            G      C
            K2048F @F &F FF IF RF SF TF !F AF NF DF !F LF AF SF TF !F FF OF UF RF @F &F #D
            A   18 @  [load length of sequence]
            L    1 F  [times 4]
            A    6 F  [make address of last 4]
            S   16 @
            T   18 @  [store address of last 4]
    [115]   A  115 @  [print first 4 terms]
            G   20 @
            A   18 @  [retrieve address of last 4]
            T    6 F  [pass as parameter]
    [119]   A  119 @  [print last 4 terms]
            G   20 @

         [PART 2]
            T      F
            T   17 @  [max count := 0]
            T    6#@  [n := 0]
         [Loop: update n, start new sequence]
    [124]   T      F  [clear acc]
            A    6#@  [load n]
            A    4#@  [add 1 (double)]
            U    6#@  [update n]
            T    4 D  [n to 4D for subroutine]
            T    6 F  [say no store]
    [130]   A  130 @  [call subroutine to generate sequence]
            G      H
            A    7 F  [load count returned by subroutine]
            G  198 @  [out if error]
            S   17 @  [compare with max count so far]
            G  140 @  [skip if less]
            A   17 @  [restore count after test]
            T   17 @  [update max count]
            A    6#@  [load n]
            T    8#@  [remember n that gave max count]
    [140]   T      F  [clear acc]
            A    6#@  [load n just done]
            S     #@  [compare with max(n)]
            G  124 @  [loop back if n < max(n)
                       else fall through with acc = 0]
         [Here whan reached maximum n. Print result.]
    [144]   A  144 @  [print 'max n' message]
            G      C
            K2048F MF AF XF !F NF !F !F #D
            A     #@  [load maximum n]
            T      D  [to 0D for printing]
    [157]   A  157 @  [call print subroutine]
            G      V
    [159]   A  159 @  [print 'max len' message]
            G      C
            K2048F @F &F MF AF XF !F LF EF NF #D
            T      D  [clear 1F and sandwich bit]
            A   17 @  [load max count (single)]
            T      F  [to 0F, effectively to 0D]
    [175]   A  175 @  [call print subroutine]
            G      V
    [177]   A  177 @  [print 'at n =' message]
            G      C
            K2048F @F &F AF TF !F NF !F #F VF !D
            A    8#@  [load n for which max count occurred]
            T      D  [to 0D for printing]
    [192]   A  192 @  [call print subroutine]
            G      V
    [194]   O   12 @  [print CR, LF]
            O   13 @
            O   14 @  [print null to flush teleprinter buffer]
            Z      F  [stop]
        [Here if term would overflow EDSAC 35-bit value.
         With a maximum n of 100,000 this doesn't happen.]
    [198]   A  198 @  [print 'overflow' message]
            G      C
            K2048F @F &F OF VF EF RF FF LF OF WD
            E  194 @  [jump to exit]

            E   41 Z  [define entry point]
            P      F  [acc = 0 on entry]

Output:

START          27
LENGTH        112
FIRST AND LAST FOUR
        27        82        41       124
         8         4         2         1
MAX N      100000
MAX LEN       351
AT N =      77031

Egel

import "prelude.eg"

namespace Hailstone (

    using System
    using List

    def even = [ N -> (N%2) == 0 ]

    def hailstone =
        [ 1 -> {1}
        | N -> if even N then cons N (hailstone (N/2))
               else cons N (hailstone (N * 3 + 1)) ]

    def hailpair =
        [ N -> (N, length (hailstone N)) ]

    def hailmax =
        [ (N, NMAX), (M, MMAX) -> if (NMAX < MMAX) then (M, MMAX) else (N, NMAX) ]

    def largest =
        [ 1 -> (1, 1)
        | N ->
            let M0 = hailpair N in
            let M1 = largest (N - 1) in
                hailmax M0 M1 ]
)

using System
using List
using Hailstone

def task0 = let H27 = hailstone 27 in length H27

def task1 = 
    let H27 = hailstone 27 in
    let L   = length H27 in
        (take 4 H27, drop (L - 4) H27)

def task2 = largest 100000

def main = (task0, task1, task2)

Eiffel

class
	APPLICATION

create
	make

feature

	make
		local
			test: LINKED_LIST [INTEGER]
			count, number, te: INTEGER
		do
			create test.make
			test := hailstone_sequence (27)
			io.put_string ("There are " + test.count.out + " elements in the sequence for the number 27.")
			io.put_string ("%NThe first 4 elements are: ")
			across
				1 |..| 4 as t
			loop
				io.put_string (test [t.item].out + "%T")
			end
			io.put_string ("%NThe last 4 elements are: ")
			across
				(test.count - 3) |..| test.count as t
			loop
				io.put_string (test [t.item].out + "%T")
			end
			across
				1 |..| 99999 as c
			loop
				test := hailstone_sequence (c.item)
				te := test.count
				if te > count then
					count := te
					number := c.item
				end
			end
			io.put_string ("%NThe longest sequence for numbers below 100000 is " + count.out + " for the number " + number.out + ".")
		end

	hailstone_sequence (n: INTEGER): LINKED_LIST [INTEGER]
			-- Members of the Hailstone Sequence starting from 'n'.
		require
			n_is_positive: n > 0
		local
			seq: INTEGER
		do
			create Result.make
			from
				seq := n
			until
				seq = 1
			loop
				Result.extend (seq)
				if seq \\ 2 = 0 then
					seq := seq // 2
				else
					seq := ((3 * seq) + 1)
				end
			end
			Result.extend (seq)
		ensure
			sequence_terminated: Result.last = 1
		end

end

Output:

There are 112 elements in the sequence for the number 27.
The first 4 elements are: 27    82    41    124
The last 4 elements are: 8    4    2    1
The longest sequence for numbers below 100000 is 351 for the number 77031.

Elena

ELENA 6.x :

import system'collections;
import extensions;
 
const int maxNumber = 100000;
 
Hailstone(int n,Map<int,int> lengths)
{
    if (n == 1)
    {
        ^ 1
    };
 
    while (true)
    {
        if (lengths.containsKey(n))
        {
            ^ lengths[n]
        }
        else
        {
            if (n.isEven())
            {
                lengths[n] := 1 + Hailstone(n/2, lengths)
            }
            else
            {
                lengths[n] := 1 + Hailstone(3*n + 1, lengths)
            }
        }
    }
}
 
public program()
{
    int longestChain := 0;
    int longestNumber := 0;
    auto recursiveLengths := new Map<int,int>(4096,4096);
 
    for(int i := 1; i < maxNumber; i+=1)
    {
        var chainLength := Hailstone(i, recursiveLengths);
        if (longestChain < chainLength)
        {
               longestChain := chainLength;
               longestNumber := i
        }
    };
 
    console.printFormatted("max below {0}: {1} ({2} steps)", maxNumber, longestNumber, longestChain)
}

Output:

max bellow 100000: 77031 (351 steps)

Elixir

defmodule Hailstone do
  require Integer
  
  def step(1)                        , do: 0
  def step(n) when Integer.is_even(n), do: div(n,2)
  def step(n)                        , do: n*3 + 1
  
  def sequence(n) do
    Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
  end
  
  def run do
    seq27 = sequence(27)
    len27 = length(seq27)
    repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
    IO.puts "Hailstone(27) has #{len27} elements: #{repr}"
    
    {len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
    IO.puts "Longest sequence starting under 100000 begins with #{start} and has #{len} elements."
  end
end

Hailstone.run

Output:

Hailstone(27) has 112 elements: [27, 82, 41, 124, ..., 8, 4, 2, 1]
Longest sequence starting under 100000 begins with 77031 and has 351 elements.

EMal

fun hailstone = List by int n
  List h = int[n]
  while n != 1 
    n = when((n % 2 == 0), n / 2, 3 * n + 1) 
    h.append(n)
  end
  return h
end
int NUMBER = 27
int LESS_THAN = 100000
List sequence = hailstone(NUMBER)
writeLine("The hailstone sequence for the number " + NUMBER +
  " has " + sequence.length + " elements")
writeLine("starting with " + 
  sequence.extractStart(4).join(", ") + " and ending with " + 
  sequence.extractEnd(4).join(", ") + ".")
int number = 0
sequence = int[]
for int i = 1; i < LESS_THAN; ++i
  List current = hailstone(i)
  if current.length > sequence.length
    sequence = current
    number = i
  end
end
writeLine("The number less than 100000 with longest hailstone sequence is " + 
  number + ", with length of " + sequence.length + ".")

Output:

The hailstone sequence for the number 27 has 112 elements
starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1.
The number less than 100000 with longest hailstone sequence is 77031, with length of 351.

Erlang

-module(hailstone).
-import(io).
-export([main/0]).

hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].

max_length(Start, Stop) ->
    F = fun (N) -> {length(hailstone(N)), N} end,
    Lengths = lists:map(F, lists:seq(Start, Stop)),
    lists:max(Lengths).

main() ->
    io:format("hailstone(4): ~w~n", [hailstone(4)]),
    Seq27 = hailstone(27),
    io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
    io:format("hailstone(27) first 4: ~w~n",
              [lists:sublist(Seq27, 4)]),
    io:format("hailstone(27) last 4: ~w~n",
              [lists:nthtail(length(Seq27) - 4, Seq27)]),
    io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
    {Length, N} = max_length(1, 100000),
    io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).

Output:

Eshell V5.8.4  (abort with ^G)
1> c(hailstone).
{ok,hailstone}
2> hailstone:main().
hailstone(4): [4,2,1]
hailstone(27) length: 112
hailstone(27) first 4: [27,82,41,124]
hailstone(27) last 4: [8,4,2,1]
finding maximum hailstone(N) length for 1 <= N <= 100000... done.
hailstone(77031) length: 351
ok

Erlang 2

This version has one collatz function for just calculating totals (just for fun) and the second generating lists.

-module(collatz).                                                            
-export([main/0,collatz/1,coll/1,max_atz_under/1]).                          
                                                                             
collatz(1) -> 1;                                                             
collatz(N) when N rem 2 == 0 -> 1 + collatz(N div 2);                        
collatz(N) when N rem 2 > 0 -> 1 + collatz(3 * N +1).                        
                                                                             
max_atz_under(N) ->                                                          
  F = fun (X) -> {collatz(X), X} end,                                        
  {_, Index} = lists:max(lists:map(F, lists:seq(1, N))),                     
  Index.                                                                     
                                                                             
coll(1) -> [1];                                                              
coll(N) when N rem 2 == 0 -> [N|coll(N div 2)];                              
coll(N) -> [N|coll(3 * N + 1)].                                              
                                                                             
main() ->                                                                    
    io:format("collatz(4) non-list total: ~w~n", [collatz(4)]),              
    io:format("coll(4) with lists ~w~n",  [coll(4)] ),                       
    Seq27 = coll(27),                                                        
    Seq1000 = coll(max_atz_under(100000)),                                   
    io:format("coll(27) length: ~B~n", [length(Seq27)]),                     
    io:format("coll(27) first 4: ~w~n", [lists:sublist(Seq27, 4)]),          
    io:format("collatz(27) last 4: ~w~n",                                    
              [lists:nthtail(length(Seq27) - 4, Seq27)]),                    
    io:format("maximum  N <= 100000..."),                             
    io:format("Max: ~w~n", [max_atz_under(100000)]),                   
    io:format("Total: ~w~n", [ length( Seq1000 ) ] ).

Output

64> collatz:main().
collatz(4) non-list total: 3
coll(4) with lists [4,2,1]
coll(27) length: 112
coll(27) first 4: [27,82,41,124]
collatz(27) last 4: [8,4,2,1]
maximum  N <= 100000...Max: 77031
Total: 351
ok

ERRE

In Italy it's known also as "Ulam conjecture".

PROGRAM ULAM

!$DOUBLE

PROCEDURE HAILSTONE(X,PRT%->COUNT)
   COUNT=1
   IF PRT% THEN PRINT(X,) END IF
   REPEAT
      IF X/2<>INT(X/2) THEN
          X=X*3+1
        ELSE
          X=X/2
      END IF
      IF PRT% THEN PRINT(X,) END IF
      COUNT=COUNT+1
   UNTIL X=1
   IF PRT% THEN PRINT END IF
END PROCEDURE

BEGIN
   HAILSTONE(27,TRUE->COUNT)
   PRINT("Sequence length for 27:";COUNT)
   MAX_COUNT=2
   NMAX=2
   FOR I=3 TO 100000 DO
      HAILSTONE(I,FALSE->COUNT)
      IF COUNT>MAX_COUNT THEN NMAX=I MAX_COUNT=COUNT END IF
   END FOR
   PRINT("Max. number is";NMAX;" with";MAX_COUNT;"elements")
END PROGRAM

Output:

        27        82        41       124        62
        31        94        47       142        71
       214       107       322       161       484
       242       121       364       182        91
       274       137       412       206       103
       310       155       466       233       700
       350       175       526       263       790
       395      1186       593      1780       890
       445      1336       668       334       167
       502       251       754       377      1132
       566       283       850       425      1276
       638       319       958       479      1438
       719      2158      1079      3238      1619
      4858      2429      7288      3644      1822
       911      2734      1367      4102      2051
      6154      3077      9232      4616      2308
      1154       577      1732       866       433
      1300       650       325       976       488
       244       122        61       184        92
        46        23        70        35       106
        53       160        80        40        20
        10         5        16         8         4
         2         1

Sequence length for 27: 112
Max. number is 77031 with 351 elements

Euler Math Toolbox

>function hailstone (n) ...
$  v=[n];
$  repeat
$    if mod(n,2) then n=3*n+1;
$    else n=n/2;
$    endif;
$    v=v|n;
$    until n==1;
$  end;
$  return v;
$  endfunction
>hailstone(27), length(%)
 [ 27  82  41  124  62  31  94  47  142  71  214  107  322  161  484  242
 121  364  182  91  274  137  412  206  103  310  155  466  233  700
 350  175  526  263  790  395  1186  593  1780  890  445  1336  668
 334  167  502  251  754  377  1132  566  283  850  425  1276  638  319
 958  479  1438  719  2158  1079  3238  1619  4858  2429  7288  3644
 1822  911  2734  1367  4102  2051  6154  3077  9232  4616  2308  1154
 577  1732  866  433  1300  650  325  976  488  244  122  61  184  92
 46  23  70  35  106  53  160  80  40  20  10  5  16  8  4  2  1 ]
 112
>function hailstonelength (n) ...
$  v=zeros(1,n);
$  v[1]=4; v[2]=2;
$  loop 3 to n;
$    count=1;
$    n=#;
$    repeat
$      if mod(n,2) then n=3*n+1;
$      else n=n/2;
$      endif;
$      if n<=cols(v) and v[n] then 
$        v[#]=v[n]+count;
$        break;
$      endif;
$      count=count+1;    
$    end;
$  end;
$  return v;
$  endfunction
>h=hailstonelength(100000);
>ex=extrema(h); ex[3], ex[4]
 351
 77031

Euphoria

function hailstone(atom n)
    sequence s
    s = {n}
    while n != 1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n + 1
        end if
        s &= n
    end while
    return s
end function

function hailstone_count(atom n)
    integer count
    count = 1
    while n != 1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n + 1
        end if
        count += 1
    end while
    return count
end function

sequence s
s = hailstone(27)
puts(1,"hailstone(27) =\n")
? s
printf(1,"len = %d\n\n",length(s))

integer max,imax,count
max = 0
for i = 2 to 1e5-1 do
    count = hailstone_count(i)
    if count > max then
        max = count
        imax = i
    end if
end for

printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",
    {imax,max})

Output:

hailstone(27) =
{27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,
91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,
1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,
850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,
7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,
577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,
106,53,160,80,40,20,10,5,16,8,4,2,1}
len = 112

The longest hailstone sequence under 100,000 is 77031 with 351 elements.

Excel

This example may be incorrect. Calculates the Hailstone sequence but might not complete everything from task description. Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself.

   In cell A1, place the starting number.
   In cell A2 enter this formula =IF(MOD(A1,2)=0,A1/2,A1*3+1)
   Drag and copy the formula down until 4, 2, 1

Ezhil

Ezhil is a Tamil programming language, see | Wikipedia entry.

நிரல்பாகம்  hailstone ( எண் )
           பதிப்பி "=> ",எண் #hailstone seq
	    @( எண் == 1 )   ஆனால் 	        
	        பின்கொடு எண்
	    முடி

	      @( (எண்%2) == 1 )  ஆனால்
	      	   hailstone( 3*எண் + 1)
              இல்லை
	           hailstone( எண்/2 )
              முடி
முடி


எண்கள் = [5,17,19,23,37]
@(எண்கள் இல் இவ்வெண்) ஒவ்வொன்றாக
   பதிப்பி "****** calculating hailstone seq for ",இவ்வெண்," *********"
   hailstone( இவ்வெண் )
   பதிப்பி "**********************************************"
முடி

F#

let rec hailstone n = seq {
  match n with
  | 1                -> yield 1
  | n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
  | n                -> yield n; yield! hailstone (n * 3 + 1)
}

let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])

let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI

Output:

Maximum length 351 was found for hailstone(77031)

Factor

! rosetta/hailstone/hailstone.factor
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone

: hailstone ( n -- seq )
    [ 1vector ] keep
    [ dup 1 number= ]
    [
        dup even? [ 2 / ] [ 3 * 1 + ] if
        2dup swap push
    ] until
    drop ;

<PRIVATE
: main ( -- )
    27 hailstone dup dup
    "The hailstone sequence from 27:" print
    "  has length " write length .
    "  starts with " write 4 head [ unparse ] map ", " join print
    "  ends with " write 4 tail* [ unparse ] map ", " join print

    ! Maps n => { length n }, and reduces to longest Hailstone sequence.
    1 100000 [a,b)
    [ [ hailstone length ] keep 2array ]
    [ [ [ first ] bi@ > ] most ] map-reduce
    first2
    "The hailstone sequence from " write pprint
    " has length " write pprint "." print ;
PRIVATE>

MAIN: main

Output:

$ ./factor -run=rosetta.hailstone
Loading resource:work/rosetta/hailstone/hailstone.factor
The hailstone sequence from 27:
  has length 112
  starts with 27, 82, 41, 124
  ends with 8, 4, 2, 1
The hailstone sequence from 77031 has length 351.

FALSE

[$1&$[%3*1+0~]?~[2/]?]n:
[[$." "$1>][n;!]#%]s:
[1\[$1>][\1+\n;!]#%]c:
27s;! 27c;!."
"
0m:0f:
1[$100000\>][$c;!$m;>[m:$f:0]?%1+]#%
f;." has hailstone sequence length "m;.

Fermat

Array g[2]

Func Collatz(n, d) =
    {Runs the Collatz procedure for the number n and returns the number of steps.}
    {If d is nonzero, prints the terms in the sequence.}
    steps := 1;
    while n>1 do 
        if n|2=0 then n:=n/2 else n:=3n+1 fi; 
        if d then !!n fi;
        steps := steps + 1 
    od; 
    steps.

Function LongestTo(n) = 
    {Finds the number up to n for which the Collatz algorithm takes the most number of steps.}
    {The result is stored in the array [g]: g[1] is the number, g[2] is how many steps it takes.}
    champ:=0; 
    record:=0; 
    for i = 1, n do 
        q:=Collatz(i, 0); 
        if q > record then 
            champ:=i; record:=q; fi;
    od; 
    g[1]:=champ;
    g[2]:=record;
    .

Forth

: hail-next ( n -- n )
  dup 1 and if 3 * 1+ else 2/ then ;
: .hail ( n -- )
  begin dup . dup 1 > while hail-next repeat drop ;
: hail-len ( n -- n )
  1 begin over 1 > while swap hail-next swap 1+ repeat nip ;

27 hail-len . cr
27 .hail cr

: longest-hail ( max -- )
  0 0 rot 1+ 1 do    ( n length )
    i hail-len 2dup < if
      nip nip i swap
    else drop then
  loop
  swap . ." has hailstone sequence length " . ;

100000 longest-hail

Fortran

Works with: Fortran version 95 and later

program Hailstone
  implicit none

  integer :: i, maxn
  integer :: maxseqlen = 0, seqlen
  integer, allocatable :: seq(:)

  call hs(27, seqlen)
  allocate(seq(seqlen))
  call hs(27, seqlen, seq)
  write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
  write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ",  &
                                     seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen) 
  
  do i = 1, 99999
    call hs(i, seqlen)
    if (seqlen > maxseqlen) then
      maxseqlen = seqlen
      maxn = i
    end if
  end do
  write(*,*)
  write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"

  deallocate(seq)
  
contains

subroutine hs(number, length, seqArray)
  integer, intent(in)  :: number
  integer, intent(out) :: length  
  integer, optional, intent(inout) :: seqArray(:)
  integer :: n

  n = number
  length = 1
  if(present(seqArray)) seqArray(1) = n
  do while(n /= 1)
    if(mod(n,2) == 0) then
      n = n / 2
    else
      n = n * 3 + 1
    end if
    length = length + 1
    if(present(seqArray)) seqArray(length) = n
  end do
end subroutine

end program

Output:

Hailstone sequence for 27 has 112 elements
Sequence = 27, 82, 41, 124, ...., 8, 4, 2, 1

Longest sequence under 100000 is for 77031 with 351 elements

Frege

Translation of: Haskell

Works with: Frege version 3.21.586-g026e8d7

module Hailstone where

import Data.List (maximumBy)

hailstone :: Int -> [Int]
hailstone 1             = [1]
hailstone n | even n    = n : hailstone (n `div` 2)
            | otherwise = n : hailstone (n * 3 + 1)

withResult :: (t -> t1) -> t -> (t1, t)
withResult f x = (f x, x)

main :: IO ()
main = do
 let h27 = hailstone 27
 putStrLn $ show $ length h27
 let h4 = show $ take 4 h27
 let t4 = show $ drop (length h27 - 4) h27
 putStrLn ("hailstone 27: " ++ h4 ++ " ... " ++ t4)
 putStrLn $ show $ maximumBy (comparing fst) $ map (withResult (length . hailstone)) [1..100000]

Output:

112
hailstone 27: [27, 82, 41, 124] ... [8, 4, 2, 1]
(351, 77031)
runtime 0.969 wallclock seconds.

Frink

hailstone[n] :=
{
   results = new array

   while n != 1
   {
      results.push[n]
      if n mod 2 == 0    // n is even?
         n = n / 2
      else
         n = (3n + 1)
   }

   results.push[1]
   return results
}

longestLen = 0
longestN = 0
for n = 1 to 100000
{
   seq = hailstone[n]
   if length[seq] > longestLen
   {
      longestLen = length[seq]
      longestN = n
   }
}

println["$longestN has length $longestLen"]

FunL

def
  hailstone( 1 ) = [1]
  hailstone( n ) = n # hailstone( if 2|n then n/2 else n*3 + 1 )

if _name_ == '-main-'
  h27 = hailstone( 27 )
  assert( h27.length() == 112 and h27.startsWith([27, 82, 41, 124]) and h27.endsWith([8, 4, 2, 1]) )

  val (n, len) = maxBy( snd, [(i, hailstone( i ).length()) | i <- 1:100000] )

  println( n, len )

Output:

77031, 351

Futhark

fun hailstone_step(x: int): int =
  if (x % 2) == 0
  then x/2
  else (3*x) + 1

fun hailstone_seq(x: int): []int =
  let capacity = 100
  let i = 1
  let steps = replicate capacity (-1)
  let steps[0] = x
  loop ((capacity,i,steps,x)) = while x != 1 do
    let (steps, capacity) =
      if i == capacity then
        (concat steps (replicate capacity (-1)),
         capacity * 2)
      else (steps, capacity)
    let x = hailstone_step x
    let steps[i] = x
    in (capacity, i+1, steps, x)
  in #1 (split i steps)

fun hailstone_len(x: int): int =
  let i = 1
  loop ((i,x)) = while x != 1 do
    (i+1, hailstone_step x)
  in i

fun max (x: int) (y: int): int = if x < y then y else x

fun main (x: int) (n: int): ([]int, int) =
  (hailstone_seq x,
   reduce max 0 (map hailstone_len (map (1+) (iota (n-1)))))

FutureBasic

local fn Hailstone( n as NSInteger ) as NSInteger
  NSInteger count = 1
  
  while ( n != 1 )
    if ( n and 1 ) == 1
      n = n * 3 + 1
      count++
    end if
    n = n / 2
    count++
  wend
end fn = count



void local fn PrintHailstone( n as NSInteger )
  NSInteger count = 1, col = 1
  
  print "Sequence for number "; n; ":" : print
  print using "########"; n;
  
  col = 2
  while ( n != 1 )
    if ( n and 1 ) == 1
      n = n * 3 + 1
      count++
    else
      n = n / 2
      count++
    end if
    print using "########"; n;
    if col == 10 then print : col = 1 else col++
  wend
  
  print : print
  print "Sequence length = "; count
end fn

window 1, @"Hailstone Sequence", ( 0, 0, 620, 400 )

NSInteger n, seq_num, x, max_x, max_seq

seq_num = 27

print
fn PrintHailstone( seq_num )
print

for x = 1 to 100000
  n = fn Hailstone( x )
  if n > max_seq
    max_x = x
    max_seq = n
  end if
next

print  "The longest sequence is for "; max_x; ", it has a sequence length of "; max_seq; "."

HandleEvents

Output:

Sequence for number 27:

      27      82      41     124      62      31      94      47     142      71
     214     107     322     161     484     242     121     364     182      91
     274     137     412     206     103     310     155     466     233     700
     350     175     526     263     790     395    1186     593    1780     890
     445    1336     668     334     167     502     251     754     377    1132
     566     283     850     425    1276     638     319     958     479    1438
     719    2158    1079    3238    1619    4858    2429    7288    3644    1822
     911    2734    1367    4102    2051    6154    3077    9232    4616    2308
    1154     577    1732     866     433    1300     650     325     976     488
     244     122      61     184      92      46      23      70      35     106
      53     160      80      40      20      10       5      16       8       4
       2       1

Sequence length = 112

The longest sequence is for 77031, it has a sequence length of 351.

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Part 1. Create a routine to generate the hailstone sequence for a number

Part 2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with {27, 82, 41, 124} and ending with {8, 4, 2, 1}

Part 3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length

The sequence for the number 77,031 is the longest one, with 351 terms

Part 4. (Additional, not a requirement) Show the number less than 100,000 which has the highest value in its hailstone sequence together with that highest value

The sequence for the number 77,671 has the highest term: 1,570,824,736

GAP

CollatzSequence := function(n)
  local v;
  v := [ n ];
  while n > 1 do
    if IsEvenInt(n) then
      n := QuoInt(n, 2);
    else
      n := 3*n + 1;
    fi;
    Add(v, n);
  od;
  return v;
end;

CollatzLength := function(n)
  local m;
  m := 1;
  while n > 1 do
    if IsEvenInt(n) then
      n := QuoInt(n, 2);
    else
      n := 3*n + 1;
    fi;
    m := m + 1;
  od;
  return m;
end;

CollatzMax := function(a, b)
  local n, len, nmax, lmax;
  lmax := 0;
  for n in [a .. b] do
    len := CollatzLength(n);
    if len > lmax then
      nmax := n;
      lmax := len;
    fi;
  od;
  return [ nmax, lmax ];
end;

CollatzSequence(27);
# [ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 
#   103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 
#   251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 
#   7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 
#   650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 ]
CollatzLength(27);  
# 112

CollatzMax(1, 100);
# [ 97, 119 ]
CollatzMax(1, 1000);
# [ 871, 179 ]
CollatzMax(1, 10000);
# [ 6171, 262 ]
CollatzMax(1, 100000);
# [ 77031, 351 ]
CollatzMax(1, 1000000);
# [ 837799, 525 ]

Go

package main

import "fmt"

// 1st arg is the number to generate the sequence for.
// 2nd arg is a slice to recycle, to reduce garbage.
func hs(n int, recycle []int) []int {
    s := append(recycle[:0], n)
    for n > 1 {
        if n&1 == 0 {
            n = n / 2
        } else {
            n = 3*n + 1
        }
        s = append(s, n)
    }
    return s
}

func main() {
    seq := hs(27, nil)
    fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
        len(seq), seq[0], seq[1], seq[2], seq[3],
        seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])

    var maxN, maxLen int
    for n := 1; n < 100000; n++ {
        seq = hs(n, seq)
        if len(seq) > maxLen {
            maxN = n
            maxLen = len(seq)
        }
    }
    fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}

Output:

hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1]
hs(77031): 351 elements

Alternate solution (inspired both by recent news of a new proof submitted for publication and by recent chat on #rosettacode about generators.)

This solution interprets the wording of the task differently, and takes the word "generate" to mean use a generator. This has the advantage of not storing the whole sequence in memory at once. Elements are generated one at a time, counted and discarded. A time optimization added for task 3 is to store the sequence lengths computed so far.

Output is the same as version above.

package main

import "fmt"

// Task 1 implemented with a generator.  Calling newHg will "create
// a routine to generate the hailstone sequence for a number."
func newHg(n int) func() int {
    return func() (n0 int) {
        n0 = n
        if n&1 == 0 {
            n = n / 2
        } else {
            n = 3*n + 1
        }
        return
    }
}

func main() {
    // make generator for sequence starting at 27
    hg := newHg(27)
    // save first four elements for printing later
    s1, s2, s3, s4 := hg(), hg(), hg(), hg()
    // load next four elements in variables to use as shift register.
    e4, e3, e2, e1 := hg(), hg(), hg(), hg()
    // 4+4= 8 that we've generated so far
    ec := 8
    // until we get to 1, generate another value, shift, and increment.
    // note that intermediate elements--those shifted off--are not saved.
    for e1 > 1 {
        e4, e3, e2, e1 = e3, e2, e1, hg()
        ec++
    }
    // Complete task 2:
    fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
        ec, s1, s2, s3, s4, e4, e3, e2, e1)

    // Task 3:  strategy is to not store sequences, but just the length
    // of each sequence.  as soon as the sequence we're currently working on
    // dips into the range that we've already computed, we short-circuit
    // to the end by adding the that known length to whatever length
    // we've accumulated so far.

    var nMaxLen int // variable holds n with max length encounted so far
    // slice holds sequence length for each n as it is computed
    var computedLen [1e5]int
    computedLen[1] = 1
    for n := 2; n < 1e5; n++ {
        var ele, lSum int
        for hg := newHg(n); ; lSum++ {
            ele = hg()
            // as soon as we get an element in the range we have already
            // computed, we're done...
            if ele < n {
                break
            }
        }
        // just add the sequence length already computed from this point.
        lSum += computedLen[ele]
        // save the sequence length for this n
        computedLen[n] = lSum
        // and note if it's the maximum so far
        if lSum > computedLen[nMaxLen] {
            nMaxLen = n
        }
    }
    fmt.Printf("hs(%d): %d elements\n", nMaxLen, computedLen[nMaxLen])
}

Groovy

def hailstone = { long start ->
    def sequence = []
    while (start != 1) {
        sequence << start
        start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
    }
    sequence << start
}

Test Code

def sequence = hailstone(27)
assert sequence.size() == 112
assert sequence[0..3] == [27, 82, 41, 124]
assert sequence[-4..-1] == [8, 4, 2, 1]

def results = (1..100000).collect { [n:it, size:hailstone(it).size()] }.max { it.size }
println results

Output:

[n:77031, size:351]

Haskell

import Data.List (maximumBy)
import Data.Ord (comparing)

-------------------- HAILSTONE SEQUENCE ------------------

collatz :: Int -> Int
collatz n
  | even n = n `div` 2
  | otherwise = 1 + 3 * n

hailstone :: Int -> [Int]
hailstone = takeWhile (1 /=) . iterate collatz

longestChain :: Int
longestChain =
  fst $
    maximumBy (comparing snd) $
      (,) <*> (length . hailstone) <$> [1 .. 100000]

--------------------------- TEST -------------------------
main :: IO ()
main =
  mapM_
    putStrLn
    [ "Collatz sequence for 27: ",
      (show . hailstone) 27,
      "The number " <> show longestChain,
      "has the longest hailstone sequence",
      "for any number less then 100000. ",
      "The sequence has length: "
        <> (show . length . hailstone $ longestChain)
    ]

Output:

Collatz sequence for 27: 
[27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2]
The number 77031
has the longest hailstone sequence for any number less then 100000. 
The sequence has length: 350

The following is an older version, which some of the language examples on this page are translated from:

import Data.Ord (comparing)
import Data.List (maximumBy, intercalate)

hailstone :: Int -> [Int]
hailstone 1 = [1]
hailstone n
  | even n = n : hailstone (n `div` 2)
  | otherwise = n : hailstone (n * 3 + 1)

withResult :: (Int -> Int) -> Int -> (Int, Int)
withResult f x = (f x, x)

h27 :: [Int]
h27 = hailstone 27

main :: IO ()
main =
  mapM_
    putStrLn
    [ (show . length) h27
    , "hailstone 27: " ++
      intercalate " ... " (show <$> [take 4 h27, drop (length h27 - 4) h27])
    , show $
      maximumBy (comparing fst) $
      withResult (length . hailstone) <$> [1 .. 100000]
    ]

Output:

112
hailstone 27: [27,82,41,124] ... [8,4,2,1]
(351,77031)

Or, going back to basics, we can observe that the hailstone sequence is an 'anamorphism' – it builds up a list structure from a single integer value, which makes unfoldr the obvious first thing to reach for the first main task.
In turn, deriving the longest sequence for starting values below 100000 essentially involves a 'catamorphism' – it takes a list of hailstone sequences (or at least a list of their seed values and their lengths), and strips that structure down to a single (N, length) pair. This makes foldr the obvious recursion scheme to start with for the second main task.

One approach to using unfoldr and then foldr might be:

import Data.List (unfoldr)


-------------------- HAILSTONE SEQUENCE ------------------

hailStones :: Int -> [Int]
hailStones = (<> [1]) . unfoldr go
  where
    f x
      | even x = div x 2
      | otherwise = 1 + 3 * x
    go x
      | 2 > x = Nothing
      | otherwise = Just (x, f x)

mostStones :: Int -> (Int, Int)
mostStones = foldr go (0, 0) . enumFromTo 1
  where
    go x (m, ml)
      | l > ml = (x, l)
      | otherwise = (m, ml)
      where
        l = length (hailStones x)

------------------------- GENERIC ------------------------
lastN_ :: Int -> [Int] -> [Int]
lastN_ = (foldr (const (drop 1)) <*>) . drop

--------------------------- TEST -------------------------
h27, start27, end27 :: [Int]
[h27, start27, end27] = [id, take 4, lastN_ 4] <*> [hailStones 27]

maxNum, maxLen :: Int
(maxNum, maxLen) = mostStones 100000

main :: IO ()
main =
  mapM_
    putStrLn
    [ "Sequence 27 length:"
    , show $ length h27
    , "Sequence 27 start:"
    , show start27
    , "Sequence 27 end:"
    , show end27
    , ""
    , "N with longest sequence where N <= 100000"
    , show maxNum
    , "length of this sequence:"
    , show maxLen
    ]

Output:

Sequence 27 length:
112
Sequence 27 start:
[27,82,41,124]
Sequence 27 end:
[8,4,2,1]

N with longest sequence where N <= 100000
77031
length of this sequence:
351

HicEst

DIMENSION stones(1000)

H27 = hailstone(27)
ALIAS(stones,1, first4,4)
ALIAS(stones,H27-3,  last4,4)
WRITE(ClipBoard, Name) H27, first4, "...", last4

longest_sequence = 0
DO try = 1, 1E5
  elements = hailstone(try)
  IF(elements >= longest_sequence) THEN
      number = try
      longest_sequence = elements
      WRITE(StatusBar, Name) number, longest_sequence
  ENDIF
ENDDO
WRITE(ClipBoard, Name) number, longest_sequence
END

FUNCTION hailstone( n )
   USE : stones

   stones(1) = n
   DO i = 1, LEN(stones)
     IF(stones(i) == 1) THEN
         hailstone = i
         RETURN
     ELSEIF( MOD(stones(i),2) ) THEN
       stones(i+1) = 3*stones(i) + 1
     ELSE
       stones(i+1) = stones(i) / 2
     ENDIF
   ENDDO
END

H27=112; first4(1)=27; first4(2)=82; first4(3)=41; first4(4)=124; ...; last4(1)=8; last4(2)=4; last4(3)=2; last4(4)=1;
number=77031; longest_sequence=351;

Icon and Unicon

A simple solution that generates (in the Icon sense) the sequence is:

procedure hailstone(n)
    while n > 1 do {
        suspend n
        n := if n%2 = 0 then n/2 else 3*n+1
        }
    suspend 1
end

and a test program for this solution is:

procedure main(args)
    n := integer(!args) | 27
    every writes(" ",hailstone(n))
end

but this solution is computationally expensive when run repeatedly (task 3).

The following solution uses caching to improve performance on task 3 at the expense of space.

procedure hailstone(n)
    static cache
    initial {
        cache := table()
        cache[1] := [1]
        }
    /cache[n] := [n] ||| hailstone(if n%2 = 0 then n/2 else 3*n+1)
    return cache[n]
end

A test program is:

procedure main(args)
    n := integer(!args) | 27
    task2(n)
    write()
    task3()
end

procedure task2(n)
    count := 0
    every writes(" ",right(!(sequence := hailstone(n)),5)) do
        if (count +:= 1) % 15 = 0 then write()
    write()
    write(*sequence," value",(*sequence=1,"")|"s"," in the sequence.")
end

procedure task3()
    maxHS := 0
    every n := 1 to 100000 do {
        count := *hailstone(n)
        if maxHS <:= count then maxN := n
        }
    write(maxN," has a sequence of ",maxHS," values")
end

A sample run is:

->hs
    27    82    41   124    62    31    94    47   142    71   214   107   322   161   484
   242   121   364   182    91   274   137   412   206   103   310   155   466   233   700
   350   175   526   263   790   395  1186   593  1780   890   445  1336   668   334   167
   502   251   754   377  1132   566   283   850   425  1276   638   319   958   479  1438
   719  2158  1079  3238  1619  4858  2429  7288  3644  1822   911  2734  1367  4102  2051
  6154  3077  9232  4616  2308  1154   577  1732   866   433  1300   650   325   976   488
   244   122    61   184    92    46    23    70    35   106    53   160    80    40    20
    10     5    16     8     4     2     1
112 values in the sequence.

77031 has a sequence of 351 values
->

Inform 7

This solution uses a cache to speed up the length calculation for larger numbers.

Works with: Glulx virtual machine

Home is a room.

To decide which list of numbers is the hailstone sequence for (N - number):
	let result be a list of numbers;
	add N to result;
	while N is not 1:
		if N is even, let N be N / 2;
		otherwise let N be (3 * N) + 1;
		add N to result;
	decide on result.

Hailstone length cache relates various numbers to one number.

To decide which number is the hailstone sequence length for (N - number):
	let ON be N;
	let length so far be 0;
	while N is not 1:
		if N relates to a number by the hailstone length cache relation:
			let result be length so far plus the number to which N relates by the hailstone length cache relation;
			now the hailstone length cache relation relates ON to result;
			decide on result;
		if N is even, let N be N / 2;
		otherwise let N be (3 * N) + 1;
		increment length so far;
	let result be length so far plus 1;
	now the hailstone length cache relation relates ON to result;
	decide on result.

To say first and last (N - number) entry/entries in (L - list of values of kind K):
	let length be the number of entries in L;
	if length <= N * 2:
		say L;
	else:
		repeat with M running from 1 to N:
			if M > 1, say ", ";
			say entry M in L;
		say " ... ";
		repeat with M running from length - N + 1 to length:
			say entry M in L;
			if M < length, say ", ".

When play begins:
	let H27 be the hailstone sequence for 27;
	say "Hailstone sequence for 27 has [number of entries in H27] element[s]: [first and last 4 entries in H27].";
	let best length be 0;
	let best number be 0;
	repeat with N running from 1 to 99999:
		let L be the hailstone sequence length for N;
		if L > best length:
			let best length be L;
			let best number be N;
	say "The number under 100,000 with the longest hailstone sequence is [best number] with [best length] element[s].";
	end the story.

Output:

Hailstone sequence for 27 has 112 elements: 27, 82, 41, 124 ... 8, 4, 2, 1.
The number under 100,000 with the longest hailstone sequence is 77031 with 351 elements.

Io

Here is a simple, brute-force approach:

makeItHail := method(n,
  stones := list(n)
  while (n != 1,
    if(n isEven,
      n = n / 2,
      n = 3 * n + 1
    )
    stones append(n)
  )
  stones
)

out := makeItHail(27)
writeln("For the sequence beginning at 27, the number of elements generated is ", out size, ".")
write("The first four elements generated are ")
for(i, 0, 3,
  write(out at(i), " ")
)
writeln(".")

write("The last four elements generated are ")
for(i, out size - 4, out size - 1,
  write(out at(i), " ")
)
writeln(".")

numOfElems := 0
nn := 3
for(x, 3, 100000,
  out = makeItHail(x)
  if(out size > numOfElems,
    numOfElems = out size
    nn = x
  )
)

writeln("For numbers less than or equal to 100,000, ", nn,
" has the longest sequence of ", numOfElems, " elements.")

Output:

For the sequence beginning at 27, the number of elements generated is 112.
The first four elements generated are 27 82 41 124 .
The last four elements generated are 8 4 2 1 .
For numbers less than or equal to 100,000, 77031 has the longest sequence of 351 elements.

Ioke

This example may be incorrect. Calculates the Hailstone sequence but might not complete everything from task description. Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself.

collatz = method(n,
  n println
  unless(n <= 1,
    if(n even?, collatz(n / 2), collatz(n * 3 + 1)))
)

J

Solution:

hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0

Usage:

   # hailseq 27                 NB. sequence length
112
   4 _4 {."0 1 hailseq 27       NB. first & last 4 numbers in sequence
27 82 41 124
 8  4  2   1
   (>:@(i. >./) , >./) #@hailseq }.i. 1e5  NB. number < 100000 with max seq length & its seq length
77031 351

See also the Collatz Conjecture essay on the J wiki.

Java

Works with: Java version 1.5+

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Hailstone {

  public static List<Long> getHailstoneSequence(long n) {
    if (n <= 0)
      throw new IllegalArgumentException("Invalid starting sequence number");
    List<Long> list = new ArrayList<Long>();
    list.add(Long.valueOf(n));
    while (n != 1) {
      if ((n & 1) == 0)
        n = n / 2;
      else
        n = 3 * n + 1;
      list.add(Long.valueOf(n));
    }
    return list;
  }
  
  public static void main(String[] args) {
    List<Long> sequence27 = getHailstoneSequence(27);
    System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
    
    long MAX = 100000;
    // Simple way
    {
      long highestNumber = 1;
      int highestCount = 1;
      for (long i = 2; i < MAX; i++) {
        int count = getHailstoneSequence(i).size();
        if (count > highestCount) {
          highestCount = count;
          highestNumber = i;
        }
      }
      System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }
    
    // More memory efficient way
    {
      long highestNumber = 1;
      int highestCount = 1;
      for (long i = 2; i < MAX; i++) {
        int count = 1;
        long n = i;
        while (n != 1) {
          if ((n & 1) == 0)
            n = n / 2;
          else
            n = 3 * n + 1;
          count++;
        }
        if (count > highestCount) {
          highestCount = count;
          highestNumber = i;
        }
      }
      System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }
    
    // Efficient for analyzing all sequences
    {
      long highestNumber = 1;
      long highestCount = 1;
      Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
      sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
      
      List<Long> currentList = new ArrayList<Long>();
      for (long i = 2; i < MAX; i++) {
        currentList.clear();
        Long n = Long.valueOf(i);
        Integer count = null;
        while ((count = sequenceMap.get(n)) == null) {
          currentList.add(n);
          long nValue = n.longValue();
          if ((nValue & 1) == 0)
            n = Long.valueOf(nValue / 2);
          else
            n = Long.valueOf(3 * nValue + 1);
        }
        int curCount = count.intValue();
        for (int j = currentList.size() - 1; j >= 0; j--)
          sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
        if (curCount > highestCount) {
          highestCount = curCount;
          highestNumber = i;
        }
      }
      System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
    }
    return;
  }
}

Output:

Sequence for 27 has 112 elements: [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Method 1, number 77031 has the longest sequence, with a length of 351
Method 2, number 77031 has the longest sequence, with a length of 351
Method 3, number 77031 has the longest sequence, with a length of 351

JavaScript

ES5

Imperative

function hailstone (n) {
    var seq = [n];
    while (n > 1) {
        n = n % 2 ? 3 * n + 1 : n / 2;
        seq.push(n);
    }
    return seq;
}

// task 2: verify the sequence for n = 27
var h = hailstone(27), hLen = h.length;
print("sequence 27 is (" + h.slice(0, 4).join(", ") + " ... "
    + h.slice(hLen - 4, hLen).join(", ") + "). length: " + hLen);

// task 3: find the longest sequence for n < 100000
for (var n, max = 0, i = 100000; --i;) {
    var seq = hailstone(i), sLen = seq.length;
    if (sLen > max) {
        n = i;
        max = sLen;
    }
}
print("longest sequence: " + max + " numbers for starting point " + n);

Output:

sequence 27 is (27, 82, 41, 124 ... 8, 4, 2, 1). length: 112
longest sequence: 351 numbers for starting point 77031

Functional

This simple problem turns out to be a good test of the constraints on composing (ES5) JavaScript code in a functional style.

The first sub-problem falls easily within reach of a basic recursive definition (translating one of the Haskell solutions).

(function () {

  // Hailstone Sequence
  // n -> [n]
  function hailstone(n) {
    return n === 1 ? [1] : (
      [n].concat(
        hailstone(n % 2 ? n * 3 + 1 : n / 2)
      )
    )
  }

  var lstCollatz27 = hailstone(27);

  return {
    length: lstCollatz27.length,
    sequence: lstCollatz27
  };

})();

Output:

{"length":112,"sequence":[27,82,41,124,62,31,94,47,142,71,214,
107,322,161,484,242,121,364,182,91,274,137,412,206,103,310,155,466,233,700,350,
175,526, 263,790,395,1186,593,1780,890,445,1336,668,334,167,502,251,754,377,
1132,566,283,850,425,1276,638,319,958,479,1438,719,2158,1079,3238,1619,4858,
2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,
1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,
40,20,10,5,16,8,4,2,1]}

Attempting to fold that recursive function over an array of 100,000 elements, however, (to solve the second part of the problem) soon runs out of stack space, at least on the system used here.

The stack problem can be quickly fixed, as often, by simply applying a memoized function, which reuses previously calculated paths.

(function () {

  function memoizedHailstone() {
    var dctMemo = {};

    return function hailstone(n) {
      var value = dctMemo[n];

      if (typeof value === "undefined") {
        dctMemo[n] = value = (n === 1) ?
          [1] : ([n].concat(hailstone(n % 2 ? n * 3 + 1 : n / 2)));
      }
      return value;
    }
  }

  // Derived a memoized version of the function,
  // which can reuse previously calculated paths
  var fnCollatz = memoizedHailstone();

  // Iterative version of range
  // [m..n]
  function range(m, n) {
    var a = Array(n - m + 1),
      i = n + 1;
    while (i--) a[i - 1] = i;
    return a;
  }
  
  // Fold/reduce over an array to find the maximum length
  function longestBelow(n) {
    return range(1, n).reduce(
      function (a, x, i) {
        var lng = fnCollatz(x).length;

        return lng > a.l ? {
          n: i + 1,
          l: lng
        } : a

      }, {
        n: 0,
        l: 0
      }
    )
  }

  return longestBelow(100000);

})();

Output:

// Number, length of sequence
{"n":77031, "l":351}

For better time (as well as space) we can continue to memoize while falling back to a function which returns the sequence length alone, and is iteratively implemented. This also proves more scaleable, and we can still use a fold/reduce pattern over a list to find the longest collatz sequences for integers below one million, or ten million and beyond, without hitting the limits of system resources.

(function (n) {

  var dctMemo = {};

  // Length only of hailstone sequence
  // n -> n
  function collatzLength(n) {
    var i = 1,
      a = n,
      lng;

    while (a !== 1) {
      lng = dctMemo[a];
      if ('u' === (typeof lng)[0]) {
        a = (a % 2 ? 3 * a + 1 : a / 2);
        i++;
      } else return lng + i - 1;
    }
    return i;
  }

  // Iterative version of range
  // [m..n]
  function range(m, n) {
    var a = Array(n - m + 1),
      i = n + 1;
    while (i--) a[i - 1] = i;
    return a;
  }

  // Fold/reduce over an array to find the maximum length
  function longestBelow(n) {

    return range(1, n).reduce(
      function (a, x) {
        
        var lng = dctMemo[x] || (dctMemo[x] = collatzLength(x));

        return lng > a.l ? {
          n: x,
          l: lng
        } : a

      }, {
        n: 0,
        l: 0
      }
    )
  }

  return [100000, 1000000, 10000000].map(longestBelow);

})();

Output:

[
  {"n":77031, "l":351},   // 100,000
  {"n":837799, "l":525},  // 1,000,000
  {"n":8400511, "l":686}  // 10,000,000
]

longestBelow(100000000)
-> {"n":63728127, "l":950}

ES6

(() => {

    // hailstones :: Int -> [Int]
    const hailstones = x => {
        const collatz = memoized(n =>
            even(n) ? div(n, 2) : (3 * n) + 1);
        return reverse(until(
            xs => xs[0] === 1,
            xs => cons(collatz(xs[0]), xs), [x]
        ));
    };

    // collatzLength :: Int -> Int
    const collatzLength = n =>
        until(
            xi => xi[0] === 1,
            ([x, i]) => [(x % 2 ? 3 * x + 1 : x / 2), i + 1], //
            [n, 1]
        )[1];

    // GENERIC FUNCTIONS -----------------------------------------------------

    // comparing :: (a -> b) -> (a -> a -> Ordering)
    const comparing = f =>
        (x, y) => {
            const
                a = f(x),
                b = f(y);
            return a < b ? -1 : (a > b ? 1 : 0);
        };

    // cons :: a -> [a] -> [a]
    const cons = (x, xs) => [x].concat(xs);

    // div :: Int -> Int -> Int
    const div = (x, y) => Math.floor(x / y);

    // enumFromTo :: Int -> Int -> [Int]
    const enumFromTo = (m, n) =>
        Array.from({
            length: Math.floor(n - m) + 1
        }, (_, i) => m + i);

    // even :: Int -> Bool
    const even = n => n % 2 === 0;

    // fst :: (a, b) -> a
    const fst = pair => pair.length === 2 ? pair[0] : undefined;

    // map :: (a -> b) -> [a] -> [b]
    const map = (f, xs) => xs.map(f);

    // maximumBy :: (a -> a -> Ordering) -> [a] -> a
    const maximumBy = (f, xs) =>
        xs.length > 0 ? (
            xs.slice(1)
            .reduce((a, x) => f(x, a) > 0 ? x : a, xs[0])
        ) : undefined;

    // memoized :: (a -> b) -> (a -> b)
    const memoized = f => {
        const dctMemo = {};
        return x => {
            const v = dctMemo[x];
            return v !== undefined ? v : (dctMemo[x] = f(x));
        };
    };

    // reverse :: [a] -> [a]
    const reverse = xs =>
        xs.slice(0)
        .reverse();

    // unlines :: [String] -> String
    const unlines = xs => xs.join('\n');

    // until :: (a -> Bool) -> (a -> a) -> a -> a
    const until = (p, f, x) => {
        let v = x;
        while (!p(v)) v = f(v);
        return v;
    };

    // MAIN ------------------------------------------------------------------
    const
        // ceiling :: Int
        ceiling = 100000,

        // (maxLen, maxNum) :: (Int, Int)
        [maxLen, maxNum] =
        maximumBy(
            comparing(fst),
            map(i => [collatzLength(i), i], enumFromTo(1, ceiling))
        );
    return unlines([
        'Collatz sequence for 27: ',
        `${hailstones(27)}`,
        '',
        `The number ${maxNum} has the longest hailstone sequence`,
        `for any starting number under ${ceiling}.`,
        '',
        `The length of that sequence is ${maxLen}.`
    ]);
})();

Output:

(Run in the Atom editor, through the Script package)

Collatz sequence for 27: 
27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,
274,137,412,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,
1780,890,445,1336,668,334,167,502,251,754,377,1132,566,283,850,425,1276,
638,319,958,479,1438,719,2158,1079,3238,1619,4858,2429,7288,3644,1822,
911,2734,1367,4102,2051,6154,3077,9232,4616,2308,1154,577,1732,866,433,
1300,650,325,976,488,244,122,61,184,92,46,23,70,35,106,53,160,80,40,20,
10,5,16,8,4,2,1

The number 77031 has the longest hailstone sequence
for any starting number under 100000.

The length of that sequence is 351.

[Finished in 1.139s]

jq

Works with: jq version 1.4

# Generate the hailstone sequence as a stream to save space (and time) when counting
def hailstone:
  recurse( if . > 1 then
              if . % 2 == 0 then ./2|floor else 3*. + 1 end
           else empty
           end );

def count(g): reduce g as $i (0; .+1);

# return [i, length] for the first maximal-length hailstone sequence where i is in [1 .. n]
def max_hailstone(n):
  # state: [i, length]
  reduce range(1; n+1) as $i
    ([0,0]; 
     ($i | count(hailstone)) as $l
     | if $l > .[1] then [$i, $l] else . end);

Examples:

[27|hailstone] as $h
| "[27|hailstone]|length is \($h|length)",
  "The first four numbers: \($h[0:4])",
  "The last four numbers:  \($h|.[length-4:length])",
  "",
  max_hailstone(100000) as $m
  | "Maximum length for n|hailstone for n in 1..100000 is \($m[1]) (n == \($m[0]))"

Output:

$ jq -M -r -n -f hailstone.jq
[27|hailstone]|length is 112
The first four numbers: [27,82,41,124]
The last four numbers:  [8,4,2,1]

Maximum length for n|hailstone for n in 1..100000 is 351 (n == 77031)

Julia

Works with: Julia version 0.6 and 1.0+

Dynamic solution

function hailstonelength(n::Integer)
    len = 1
    while n > 1
        n = ifelse(iseven(n), n ÷ 2, 3n + 1)
        len += 1
    end
    return len
end

@show hailstonelength(27); nothing
@show findmax([hailstonelength(i) for i in 1:100_000]); nothing

Output:

hailstonelength(27) = 112
findmax((hailstonelength(i) for i = 1:100000)) = (351, 77031)

Solution with iterator

Julia 1.0

Works with: Julia version 1.0+

struct HailstoneSeq{T<:Integer}
    count::T
end

Base.eltype(::HailstoneSeq{T}) where T = T

function Base.iterate(h::HailstoneSeq, state=h.count)
    if state == 1
        (1, 0)
    elseif state < 1
        nothing
    elseif iseven(state)
        (state, state ÷ 2)
    elseif isodd(state)
        (state, 3state + 1)
    end
end

function Base.length(h::HailstoneSeq)
    len = 0
    for _ in h
        len += 1
    end
    return len
end

function Base.show(io::IO, h::HailstoneSeq)
    f5 = collect(Iterators.take(h, 5))
    print(io, "HailstoneSeq{", join(f5, ", "), "...}")
end

hs = HailstoneSeq(27)
println("Collection of the Hailstone sequence from 27: $hs")
cl = collect(hs)
println("First 5 elements: ", join(cl[1:5], ", "))
println("Last 5 elements: ", join(cl[end-4:end], ", "))
 
Base.isless(h::HailstoneSeq, s::HailstoneSeq) = length(h) < length(s)
println("The number with the longest sequence under 100,000 is: ", maximum(HailstoneSeq.(1:100_000)))

Output:

Collection of the Hailstone sequence from 27: HailstoneSeq{27, 82, 411, 124, 62...}
First 5 elements: 27, 82, 41, 124, 62
Last 5 elements: 16, 8, 4, 2, 1
The number with the longest sequence under 100,000 is: HailstoneSeq{777031, 231094, 115547, 346642, 173321...}

Julia 0.6

Works with: Julia version 0.6

struct HailstoneSeq{T<:Integer}
	start::T
end

Base.eltype(::HailstoneSeq{T}) where T = T

Base.start(hs::HailstoneSeq) = (-1, hs.start)
Base.done(::HailstoneSeq, state) = state == (1, 4)
function Base.next(::HailstoneSeq, state)
	_, s2 = state
	s1 = s2
	if iseven(s2)
		s2 = s2 ÷ 2
	else
		s2 = 3s2 + 1
	end
	return s1, (s1, s2)
end

function Base.length(hs::HailstoneSeq)
	r = 0
	for _ in hs
		r += 1
	end
	return r
end

function Base.show(io::IO, hs::HailstoneSeq)
	f5 = collect(Iterators.take(hs, 5))
	print(io, "HailstoneSeq(", join(f5, ", "), "...)")
end

hs = HailstoneSeq(27)
println("Collection of the Hailstone sequence from 27: $hs")
cl = collect(hs)
println("First 5 elements: ", join(cl[1:5], ", "))
println("Last 5 elements: ", join(cl[end-4:end], ", "))

Base.isless(h::HailstoneSeq, s::HailstoneSeq) = length(h) < length(s)
println("The number with the longest sequence under 100,000 is: ", maximum(HailstoneSeq.(1:100_000)))

Output:

Collection of the Hailstone sequence from 27: HailstoneSeq(27, 82, 41, 124, 62...)
First 5 elements: 27, 82, 41, 124, 62
Last 5 elements: 16, 8, 4, 2, 1
The number with the longest sequence under 100,000 is: HailstoneSeq(77031, 231094, 115547, 346642, 173321...)

K

  hail: (1<){:[x!2;1+3*x;_ x%2]}\
  seqn: hail 27

  #seqn
112
  4#seqn
27 82 41 124
  -4#seqn
8 4 2 1

  {m,x@s?m:|/s:{#hail x}'x}{x@&x!2}!:1e5
351 77031

Kotlin

fun hailstone(start: Int) = generateSequence(start) { n ->
    when {
        n == 1     -> null
        n % 2 == 0 -> n / 2
        else       -> n * 3 + 1
    }
}

fun main() {
    val hail27 = hailstone(27).toList()
    println("The hailstone sequence for 27 has ${hail27.size} elements:\n$hail27")

    val (n, length) = (1..100000).asSequence()
        .map { it to hailstone(it).count() }
        .maxBy { it.second }
    println("The number between 1 and 100000 with the longest hailstone sequence is $n, of length $length")
}

Alternative, doing it manually:

import java.util.ArrayDeque

fun hailstone(n: Int): ArrayDeque<Int> {
    val hails = when {
        n == 1 -> ArrayDeque<Int>()
        n % 2 == 0 -> hailstone(n / 2)
        else -> hailstone(3 * n + 1)
    }
    hails.addFirst(n)
    return hails
}

fun main(args: Array<String>) {
    val hail27 = hailstone(27)
    fun showSeq(s: List<Int>) = s.map { it.toString() }.reduce { a, b -> a + ", " + b }
    println("Hailstone sequence for 27 is " + showSeq(hail27.take(3)) + " ... "
            + showSeq(hail27.drop(hail27.size - 3)) + " with length ${hail27.size}.")

    var longestHail = hailstone(1)
    for (x in 1..99999)
        longestHail = arrayOf(hailstone(x), longestHail).maxBy { it.size } ?: longestHail
    println("${longestHail.first} is the number less than 100000 with " +
            "the longest sequence, having length ${longestHail.size}.")
}

Output:

Hailstone sequence for 27 is 27, 82, 41 ... 4, 2, 1 with length 112.
77031 is the number less than 100000 with the longest sequence, having length 351.

Lasso

[
	define_tag("hailstone", -required="n", -type="integer", -copy);
		local("sequence") = array(#n);
		while(#n != 1);
			((#n % 2) == 0) ? #n = (#n / 2) | #n = (#n * 3 + 1);
			#sequence->insert(#n);
		/while;
		return(#sequence);
	/define_tag;

	local("result");
	#result = hailstone(27);
	while(#result->size > 8);
		#result->remove(5);
	/while;
	#result->insert("...",5);

	"Hailstone sequence for n = 27 -> { " + #result->join(", ") + " }";

	local("longest_sequence") = 0;
	local("longest_index") = 0;
	loop(-from=1, -to=100000);
		local("length") = hailstone(loop_count)->size;
		if(#length > #longest_sequence);
			#longest_index = loop_count;
			#longest_sequence = #length;
		/if;
	/loop;

	"<br/>";
	"Number with the longest sequence under 100,000: " #longest_index + ", with " + #longest_sequence + " elements.";
]

Limbo

implement Hailstone;

include "sys.m"; sys: Sys;
include "draw.m";

Hailstone: module {
	init: fn(ctxt: ref Draw->Context, args: list of string);
};

init(nil: ref Draw->Context, nil: list of string)
{
	sys = load Sys Sys->PATH;

	seq := hailstone(big 27);
	l := len seq;

	sys->print("hailstone(27):  ");
	for(i := 0; i < 4; i++) {
		sys->print("%bd, ", hd seq);
		seq = tl seq;
	}
	sys->print("⋯");
	
	while(len seq > 4)
		seq = tl seq;

	while(seq != nil) {
		sys->print(", %bd", hd seq);
		seq = tl seq;
	}
	sys->print(" (length %d)\n", l);

	max := 1;
	maxn := big 1;
	for(n := big 2; n < big 100000; n++) {
		cur := len hailstone(n);
		if(cur > max) {
			max = cur;
			maxn = n;
		}
	}
	sys->print("hailstone(%bd) has length %d\n", maxn, max);
}

hailstone(i: big): list of big
{
	if(i == big 1)
		return big 1 :: nil;
	if(i % big 2 == big 0)
		return i :: hailstone(i / big 2);
	return i :: hailstone((big 3 * i) + big 1);
}

Output:

hailstone(27):  27, 82, 41, 124, ⋯, 8, 4, 2, 1 (length 112)
hailstone(77031) has length 351

Lingo

on hailstone (n, sequenceList)
  len = 1
  repeat while n<>1
    if listP(sequenceList) then sequenceList.add(n)
    if n mod 2 = 0 then
      n = n / 2
    else
      n = 3 * n + 1
    end if
    len = len + 1
  end repeat
  if listP(sequenceList) then sequenceList.add(n)
  return len
end

Usage:

sequenceList = []
hailstone(27, sequenceList)
put sequenceList
-- [27, 82, 41, 124, ... , 8, 4, 2, 1]

n = 0
maxLen = 0
repeat with i = 1 to 99999
  len = hailstone(i)
  if len>maxLen then
    n = i
    maxLen = len
  end if
end repeat
put n, maxLen
-- 77031 351

Logo

to hail.next :n
  output ifelse equal? 0 modulo :n 2 [:n/2] [3*:n + 1]
end

to hail.seq :n
  if :n = 1 [output [1]]
  output fput :n hail.seq hail.next :n
end

show hail.seq 27
show count hail.seq 27

to max.hail :n
  localmake "max.n 0
  localmake "max.length 0
  repeat :n [if greater? count hail.seq repcount  :max.length [
    make "max.n repcount
    make "max.length count hail.seq repcount
  ] ]
  (print :max.n [has hailstone sequence length] :max.length)
end

max.hail 100000

Logtalk

:- object(hailstone).

	:- public(generate_sequence/2).
	:- mode(generate_sequence(+natural, -list(natural)), zero_or_one).
	:- info(generate_sequence/2, [
		comment is 'Generates the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.',
		argnames is ['Start', 'Sequence']
	]).

	:- public(write_sequence/1).
	:- mode(write_sequence(+natural), zero_or_one).
	:- info(write_sequence/1, [
		comment is 'Writes to the standard output the Hailstone sequence that starts with its argument. Fails if the argument is not a natural number.',
		argnames is ['Start']
	]).

	:- public(sequence_length/2).
	:- mode(sequence_length(+natural, -natural), zero_or_one).
	:- info(sequence_length/2, [
		comment is 'Calculates the length of the Hailstone sequence that starts with its first argument. Fails if the argument is not a natural number.',
		argnames is ['Start', 'Length']
	]).

	:- public(longest_sequence/4).
	:- mode(longest_sequence(+natural, +natural, -natural, -natural), zero_or_one).
	:- info(longest_sequence/4, [
		comment is 'Calculates the longest Hailstone sequence in the interval [Start, End]. Fails if the interval is not valid.',
		argnames is ['Start', 'End', 'N', 'Length']
	]).

	generate_sequence(Start, Sequence) :-
		integer(Start),
		Start >= 1,
		sequence(Start, Sequence).

	sequence(1, [1]) :-
		!. 
	sequence(N, [N| Sequence]) :-
		(	N mod 2 =:= 0 ->
			M is N // 2
		;	M is (3 * N) + 1
		),
		sequence(M, Sequence).

	write_sequence(Start) :-
		integer(Start),
		Start >= 1,
		sequence(Start).

	sequence(1) :-
		!,
		write(1), nl. 
	sequence(N) :-
		write(N), write(' '),
		(	N mod 2 =:= 0 ->
			M is N // 2
		;	M is (3 * N) + 1
		),
		sequence(M).

	sequence_length(Start, Length) :-
		integer(Start),
		Start >= 1,
		sequence_length(Start, 1, Length).

	sequence_length(1, Length, Length) :-
		!.
	sequence_length(N, Length0, Length) :-
		Length1 is Length0 + 1,
		(	N mod 2 =:= 0 ->
			M is N // 2
		;	M is (3 * N) + 1
		),
		sequence_length(M, Length1, Length).

	longest_sequence(Start, End, N, Length) :-
		integer(Start),
		integer(End),
		Start >= 1,
		Start =< End,
		longest_sequence(Start, End, 1, N, 1, Length).

	longest_sequence(Current, End, N, N, Length, Length) :-
		Current > End,
		!.
	longest_sequence(Current, End, N0, N, Length0, Length) :-
		sequence_length(Current, 1, CurrentLength),
		Next is Current + 1,
		(	CurrentLength > Length0 ->
			longest_sequence(Next, End, Current, N, CurrentLength, Length)
		;	longest_sequence(Next, End, N0, N, Length0, Length)
		).

:- end_object.

Testing:

| ?- hailstone::write_sequence(27).
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
true

| ?- hailstone::sequence_length(27, Length).
Length = 112
true

| ?- hailstone::longest_sequence(1, 100000, N, Length).
N = 77031, Length = 351
true

LOLCODE

There is presently no way to query a BUKKIT for the existence of a given key, thus making memoization infeasible. This solution takes advantage of prior knowledge to run in reasonable time.

HAI 1.3

HOW IZ I hailin YR stone
    I HAS A sequence ITZ A BUKKIT
    sequence HAS A length ITZ 1
    sequence HAS A SRS 0 ITZ stone

    IM IN YR stoner
        BOTH SAEM stone AN 1, O RLY?
            YA RLY, FOUND YR sequence
        OIC

        MOD OF stone AN 2, O RLY?
            YA RLY, stone R SUM OF PRODUKT OF stone AN 3 AN 1
            NO WAI, stone R QUOSHUNT OF stone AN 2
        OIC

        sequence HAS A SRS sequence'Z length ITZ stone
        sequence'Z length R SUM OF sequence'Z length AN 1
    IM OUTTA YR stoner
IF U SAY SO

I HAS A hail27 ITZ I IZ hailin YR 27 MKAY
VISIBLE "hail(27) = "!

IM IN YR first4 UPPIN YR i TIL BOTH SAEM i AN 4
    VISIBLE hail27'Z SRS i " "!
IM OUTTA YR first4
VISIBLE "..."!

IM IN YR last4 UPPIN YR i TIL BOTH SAEM i AN 4
    VISIBLE " " hail27'Z SRS SUM OF 108 AN i!
IM OUTTA YR last4
VISIBLE ", length = " hail27'Z length

I HAS A max, I HAS A len ITZ 0

BTW, DIS IZ RLY NOT FAST SO WE ONLY CHEK N IN [75000, 80000)
IM IN YR maxer UPPIN YR n TIL BOTH SAEM n AN 5000
    I HAS A n ITZ SUM OF n AN 75000
    I HAS A seq ITZ I IZ hailin YR n MKAY
    BOTH SAEM len AN SMALLR OF len AN seq'Z length, O RLY?
        YA RLY, max R n, len R seq'Z length
    OIC
IM OUTTA YR maxer

VISIBLE "len(hail(" max ")) = " len

KTHXBYE

Output:

hail(27) = 27 82 41 124 ... 8 4 2 1, length = 112
len(hail(77031)) = 351

Lua

function hailstone( n, print_numbers )
    local n_iter = 1

    while n ~= 1 do
        if print_numbers then print( n ) end
        if n % 2 == 0 then 
            n = n / 2
        else
            n = 3 * n + 1
        end    
        
        n_iter = n_iter + 1
    end
    if print_numbers then print( n ) end
    
    return n_iter;
end

hailstone( 27, true )

max_i, max_iter = 0, 0
for i = 1, 100000 do
    num = hailstone( i, false )
    if num >= max_iter then
        max_i = i
        max_iter = num
    end
end

print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )

M2000 Interpreter

Use of two versions of Hailstone, one which return each n, and another one which return only the length of sequence.

Also we use current stack as FIFO to get the last 4 numbers

Module hailstone.Task {
      hailstone=lambda  (n as long)->{
            =lambda n  (&val) ->{
                  if n=1 then =false: exit
                  =true
                  if n mod 2=0 then n/=2 : val=n: exit
                  n*=3 : n++: val=n
            }
      }
      Count=Lambda (n) ->{
            m=lambda n ->{
                  if n=1 then =false: exit
                  =true :if n mod 2=0 then n/=2 :exit
                  n*=3 : n++
            }
            c=1
            While m() {c++}
            =c
            
      }
      k=Hailstone(27)
      counter=1
      x=0
      Print 27,
      While k(&x) {
            counter++
            Print x,
            if counter=4 then exit
      }
      Print
      Flush  ' empty current stack
      While k(&x) {
            counter++
            data x   ' send to end of stack -used as FIFO
            if stack.size>4 then drop
      }
      \\ [] return a stack object and leave empty current stack
      \\ Print use automatic iterator to print all values in columns.
      Print []
      Print "counter:";counter
      m=0
      For i=2 to 99999 {
            m1=max.data(count(i), m)
            if m1<>m then m=m1: im=i
      }
      Print Format$("Number {0} has then longest hailstone sequence of length {1}", im, m)
}
hailstone.Task

Output:

      27      82      41     124
       8       4       2       1
counter:112
Number 77031 has then longest hailstone sequence of length 351

Maple

Define the procedure:

hailstone := proc( N )
    local n := N, HS := Array([n]);
    while n > 1 do
        if type(n,even) then
            n := n/2;
        else
            n := 3*n+1;
        end if;
        HS(numelems(HS)+1) := n;
    end do;
    HS;
end proc;

Run the command and show the appropriate portion of the result;

> r := hailstone(27):
                              [ 1..112 1-D Array     ]
                         r := [ Data Type: anything  ]
                              [ Storage: rectangular ]
                              [ Order: Fortran_order ]
> r(1..4) ... r(-4..);
                       [27, 82, 41, 124] .. [8, 4, 2, 1]

Compute the first 100000 sequences:

longest := 0; n := 0;
for i from 1 to 100000 do
    len := numelems(hailstone(i));
    if len > longest then
        longest := len;
        n := i;
    end if;
od:
printf("The longest Hailstone sequence in the first 100k is n=%d, with %d terms\n",n,longest);

Output:

The longest Hailstone sequence in the first 100k is n=77031, with 351 terms

Mathematica / Wolfram Language

Here are four ways to generate the sequence.

Nested function call formulation

HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]

This is probably the most readable and shortest implementation.

Fixed-Point formulation

HailstoneFP[n_] := Most@FixedPointList[Switch[#, 1, 1, _?OddQ , 3# + 1, _, #/2] &, n]

Recursive formulation

HailstoneR[1] = {1}
HailstoneR[n_?OddQ] := Prepend[HailstoneR[3 n + 1], n]
HailstoneR[n_] := Prepend[HailstoneR[n/2], n]

Procedural implementation

HailstoneP[n_] := Module[{x = {n}, s = n}, 
 While[s > 1, x = {x, s = If[OddQ@s, 3 s + 1, s/2]}]; Flatten@x]

Validation

I use this version to do the validation:

Hailstone[n_] := 
 NestWhileList[If[Mod[#, 2] == 0, #/2, ( 3*# + 1) ] &, n, # != 1 &];


c27 = Hailstone@27;
Print["Hailstone sequence for n = 27: [", c27[[;; 4]], "...", c27[[-4 ;;]], "]"]
Print["Length Hailstone[27] = ", Length@c27]

longest = -1; comp = 0;
Do[temp = Length@Hailstone@i;
 If[comp < temp, comp = temp; longest = i],
 {i, 100000}
 ]
Print["Longest Hailstone sequence at n = ", longest, "\nwith length = ", comp];

Output:

Hailstone sequence for n = 27: [{27,82,41,124}...{8,4,2,1}]
Length Hailstone[27] = 112
Longest Hailstone sequence at n = 77031
with length = 351

I think the fixed-point and the recursive piece-wise function formulations are more idiomatic for Mathematica

Sequence 27

With[{seq = HailstoneFP[27]}, { Length[seq], Take[seq, 4], Take[seq, -4]}]

Output:

{112, {27, 82, 41, 124}, {8, 4, 2, 1}}

Alternatively,

Short[HailstoneFP[27],0.45]

Output:

{27, 82, 41, 124, <<104>>, 8, 4, 2, 1}

Longest sequence length

MaximalBy[Table[{i, Length[HailstoneFP[i]]}, {i, 100000}], Last]

Output:

{{77031, 351}}

MATLAB / Octave

Hailstone Sequence For N

function x = hailstone(n)
  x = n;
  while n > 1
       % faster than mod(n, 2)
    if n ~= floor(n / 2) * 2
      n = n * 3 + 1;
    else
      n = n / 2;
    end
    x(end + 1) = n; %#ok
  end

Show sequence of hailstone(27) and number of elements:

x = hailstone(27);
fprintf('hailstone(27): %d %d %d %d ... %d %d %d %d\nnumber of elements: %d\n', x(1:4), x(end-3:end), numel(x))

Output:

hailstone(27): 27 82 41 124 ... 8 4 2 1
number of elements: 112

Longest Hailstone Sequence Under N

Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length:

Basic Version (use the above routine)

N = 1e5;
maxLen = 0;
for k = 1:N
  kLen = numel(hailstone(k));
  if kLen > maxLen
    maxLen = kLen;
    n = k;
  end
end

Output:

n = 77031
maxLen = 351

Faster Version

function [n, maxLen] = longestHailstone(N)
  maxLen = 0;
  for k = 1:N
    a = k;
    kLen = 1;
    while a > 1
      if a ~= floor(a / 2) * 2
        a = a * 3 + 1;
      else
        a = a / 2;
      end
      kLen = kLen + 1;
    end
    if kLen > maxLen
      maxLen = kLen;
      n = k;
    end
  end

Output:

>> [n, maxLen] = longestHailstone(1e5)
n = 77031
maxLen = 351

Much Faster Version With Caching

function [n, maxLen] = longestHailstone(N)
  lenList(N) = 0;
  lenList(1) = 1;
  maxLen = 0;
  for k = 2:N
    a = k;
    kLen = 0;
    while a >= k
      if a == floor(a / 2) * 2
        a = a / 2;
      else
        a = a * 3 + 1;
      end
      kLen = kLen + 1;
    end
    kLen = kLen + lenList(a);
    lenList(k) = kLen;
    if kLen > maxLen
      maxLen = kLen;
      n = k;
    end
  end

Output:

>> [n, maxLen] = longestHailstone(1e5)
n = 77031
maxLen = 351

Maxima

collatz(n) := block([L], L: [n], while n > 1 do
(n: if evenp(n) then n/2 else 3*n + 1, L: endcons(n, L)), L)$

collatz_length(n) := block([m], m: 1, while n > 1 do
(n: if evenp(n) then n/2 else 3*n + 1, m: m + 1), m)$

collatz_max(n) := block([j, m, p], m: 0,
for i from 1 thru n do
   (p: collatz_length(i), if p > m then (m: p, j: i)),
[j, m])$

collatz(27);           /* [27, 82, 41, ..., 4, 2, 1] */
length(%);             /* 112 */
collatz_length(27);    /* 112 */
collatz_max(100000);   /* [77031, 351] */

Mercury

The actual calculation (including module ceremony) providing both a function and a predicate implementation:

:- module hailstone.

:- interface.

:- import_module int, list.

:- func hailstone(int) = list(int).
:- pred hailstone(int::in, list(int)::out) is det.

:- implementation.

hailstone(N) = S :- hailstone(N, S).

hailstone(N, [N|S]) :-
  ( N = 1 ->       S = []
  ; N mod 2 = 0 -> hailstone(N/2, S)
  ;                hailstone(3 * N + 1, S) ).

:- end_module hailstone.

The mainline test driver (making use of unification for more succinct tests):

:- module test_hailstone.

:- interface.

:- import_module io.

:- pred main(io.state::di, io.state::uo) is det.

:- implementation.

:- import_module int, list.
:- import_module hailstone.

:- pred longest(int::in, int::out, int::out) is det.
:- pred longest(int::in, int::in, int::in, int::out, int::out) is det.

longest(M, N, L) :- longest(M, 0, 0, N, L).

longest(N, CN, CL, MN, ML) :-
  ( N > 1 ->
      L = list.length(hailstone(N)),
      ( L > CL -> longest(N - 1, N,  L,  MN, ML)
      ;           longest(N - 1, CN, CL, MN, ML) )
  ;   MN = CN, ML = CL ).


main(!IO) :-
  S = hailstone(27),
  ( list.length(S) = 112,
    list.append([27, 82, 41, 124], _, S),
    list.remove_suffix(S, [8, 4, 2, 1], _),
    longest(100000, 77031, 351) ->
      io.write_string("All tests succeeded.\n", !IO)
  ;   io.write_string("At least one test failed.\n", !IO) ).

:- end_module test_hailstone.

Output:

of running this program is

All tests succeeded.

For those unused to logic programming languages it seems that nothing has been proved in terms of confirming anything, but if you look at the predicate declaration for longest/3 …

:- pred longest(int::in, int::out, int::out) is det.

… you see that the second and third parameters are output parameters. This by calling longest(100000, 77031, 351) you prove, through unification, that the longest sequence is with the number 77031 and that it is 351 cycles long.

Similarly, using list.append([27, 82, 41, 124], _, S) automatically proves that the generated sequence begins with the provided sequence, etc. Thus we know that the correct sequences and values were generated without bothering to print them out.

MiniScript

Non-cached version

Calculates sequence without using previous calculated sequences.

getSequence = function(n)
	results = [n]
	while n > 1
		if n % 2 then
			n = 3 * n + 1
		else
			n = n / 2
		end if
		results.push n
	end while
	return results
end function

h = getSequence(27)
print "The hailstone sequence for 27 has 112 elements starting with"
print h[:4]
print "and ending with"
print h[-4:]

maxSeqLen = 0
maxSeqVal = 0
for i in range(1,100000)
	h = getSequence(i)
	if h.len > maxSeqLen then
		maxSeqLen = h.len
		maxSeqVal = i
	end if
end for
print
print "The number < 100,000 which has the longest hailstone sequence is " + maxSeqVal + "."
print "This sequence has " + maxSeqLen + " elements."

Output:

The hailstone sequence for 27 has 112 elements starting with
[27, 82, 41, 124]
and ending with
[8, 4, 2, 1]

The number < 100,000 which has the longest hailstone sequence is 77031.
This sequence has 351 elements.

Cached version

Calculations are stored for used in later calculations.

cache = {}
calc = function(n)
	if cache.hasIndex(n) then return
	items = [n]
	origNum = n
	while n > 1 and not cache.hasIndex(n)
		if n % 2 then
			n = 3 * n + 1
		else
			n = n /2 
		end if
		items.push n
	end while
	cache[origNum] = {"len": items.len,"items":items}
end function

getLen = function(n)
	if not cache.hasIndex(n) then calc n
	if n == 1 then return 1
	return cache[n].len + getLen(cache[n].items[-1]) - 1
end function

getSequence = function(n)
	if not cache.hasIndex(n) then calc n
	if n == 1 then return [1]
	return cache[n].items[:-1] + getSequence(cache[n].items[-1])
end function

h = getSequence(27)
print "The hailstone sequence for 27 has " + h.len + " elements starting with"
print h[:4]
print "and ending with"
print h[-4:]

longSeq = 0
longSeqVal =0
for i in range(2, 100000)
	seq = getLen(i)
	if longSeq < seq then
		longSeq = seq
		longSeqVal = i
	end if
end for
print "The number < 100,000 which has the longest hailstone sequence is " + longSeqVal + "."
print "This sequence has " + longSeq + " elements."

Output:

Output is the same as the non-cached version above.

The hailstone sequence for 27 has 112 elements starting with
[27, 82, 41, 124]
and ending with
[8, 4, 2, 1]

The number < 100,000 which has the longest hailstone sequence is 77031.
This sequence has 351 elements.

ML

MLite

fun hail (x = 1) = [1]
       | (x rem 2 = 0) = x :: hail (x div 2)
       | x = x :: hail (x * 3 + 1)

fun hailstorm
		([], i, largest, largest_at) = (largest_at, largest)
	| 	(x :: xs, i, largest, largest_at) =
		let 
			val k = len (hail x)
		in 
			if k > largest then
				hailstorm (xs, i + 1, k, i)
			else
				hailstorm (xs, i + 1, largest, largest_at)
			end
	| 	(x :: xs) = hailstorm (x :: xs, 1, 0, 0)
 
 ;

val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";

val biggest = hailstorm ` iota (100000 - 1);

print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);

Output:

hailstone sequence for the number 27 has 112 elements starting with [27, 82, 41, 124] and ending with [8, 4, 2, 1].
The number less than 100,000 which has the longest hailstone sequence is at element 77031 and is of length 351

Modula-2

MODULE hailst;

IMPORT  InOut;

CONST   maxCard         = MAX (CARDINAL) DIV 3;
TYPE    action          = (List, Count, Max);
VAR     a               : CARDINAL;

PROCEDURE HailStone (start  : CARDINAL;  type  : action) : CARDINAL;

VAR     n, max, count           : CARDINAL;

BEGIN
  count := 1;
  n := start;
  max := n;
  LOOP
    IF  type = List  THEN
      InOut.WriteCard (n, 12);
      IF  count MOD 6 = 0  THEN  InOut.WriteLn  END
    END;
    IF  n = 1  THEN  EXIT  END;
    IF  ODD (n)  THEN
      IF  n < maxCard  THEN
        n := 3 * n + 1;
        IF   n > max  THEN  max := n  END
      ELSE
        InOut.WriteString ("Exceeding max value for type CARDINAL at count = ");
        InOut.WriteCard (count, 10);
        InOut.WriteString (" for intermediate value ");
        InOut.WriteCard (n, 10);
        InOut.WriteString (". Aborting.");
        HALT
      END
    ELSE
      n := n DIV 2
    END;
    INC (count)
  END;
  IF  type = Max  THEN  RETURN  max  ELSE  RETURN  count  END
END HailStone;

PROCEDURE FindMax (num   : CARDINAL);

VAR     val, maxCount, maxVal, cnt      : CARDINAL;

BEGIN
  maxCount := 0;
  maxVal := 0;
  FOR  val := 2 TO num  DO
   cnt := HailStone (val, Count);
    IF  cnt > maxCount  THEN
      maxVal := val;
      maxCount := cnt
    END
  END;
  InOut.WriteString ("Longest sequence below ");        InOut.WriteCard (num, 1);
  InOut.WriteString (" is ");           InOut.WriteCard (HailStone (maxVal, Count), 1);
  InOut.WriteString (" for n = ");      InOut.WriteCard (maxVal, 1);
  InOut.WriteString (" with an intermediate maximum of ");
  InOut.WriteCard (HailStone (maxVal, Max), 1);
  InOut.WriteLn
END FindMax;

BEGIN
  a := HailStone (27, List);
  InOut.WriteLn;
  InOut.WriteString ("Iterations total = ");    InOut.WriteCard (HailStone (27, Count), 12);
  InOut.WriteString (" max value = ");          InOut.WriteCard (HailStone (27, Max)  , 12);
  InOut.WriteLn;
  FindMax (100000);
  InOut.WriteString ("Done.");          InOut.WriteLn
END hailst.

Producing:

jan@Beryllium:~/modula/rosetta$ hailst
          27          82          41         124          62          31
          94          47         142          71         214         107
         322         161         484         242         121         364
         182          91         274         137         412         206
         103         310         155         466         233         700
         350         175         526         263         790         395
        1186         593        1780         890         445        1336
         668         334         167         502         251         754
         377        1132         566         283         850         425
        1276         638         319         958         479        1438
         719        2158        1079        3238        1619        4858
        2429        7288        3644        1822         911        2734
        1367        4102        2051        6154        3077        9232
        4616        2308        1154         577        1732         866
         433        1300         650         325         976         488
         244         122          61         184          92          46
          23          70          35         106          53         160
          80          40          20          10           5          16
           8           4           2           1
Iterations total =          112 max value =         9232
Longest sequence below 100000 is 351 for n = 77031 with an intermediate maximum of 21933016
Done.

When trying the same for all values below 1 million:

Exceeding max value for type CARDINAL at n = 159487 , count = 60 and intermediate value 1699000271. Aborting.

MUMPS

hailstone(n)	;
	If n=1 Quit n
	If n#2 Quit n_" "_$$hailstone(3*n+1)
	Quit n_" "_$$hailstone(n\2)
Set x=$$hailstone(27) Write !,$Length(x," ")," terms in ",x,!
112 terms in 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

Nanoquery

def hailstone(n)
	seq = list()
 
	while (n > 1)
		append seq n 
		if (n % 2)=0
			n = int(n / 2)
		else
			n = int((3 * n) + 1)
		end
	end
	append seq n
	return seq
end
 
h = hailstone(27)
println "hailstone(27)"
println "total elements: " + len(hailstone(27))
print   h[0] + ", " + h[1] + ", " + h[2] + ", " + h[3] + ", ..., " 
println h[-4] + ", " + h[-3] + ", " + h[-2] + ", " + h[-1]
 
max = 0
maxLoc = 0
for i in range(1,99999)
	result = len(hailstone(i))
	if (result > max)
		max = result
		maxLoc = i
	end
end
print   "\nThe number less than 100,000 with the longest sequence is "
println maxLoc + " with a length of " + max

Output:

hailstone(27)
total elements: 112
27, 82, 41, 124, ..., 8, 4, 2, 1

The number less than 100,000 with the longest sequence is 77031 with a length of 351

NetRexx

/* NetRexx */

options replace format comments java crossref savelog symbols binary

do
  start = 27
  hs = hailstone(start)
  hsCount = hs.words
  say 'The number' start 'has a hailstone sequence comprising' hsCount 'elements'
  say '  its first four elements are:' hs.subword(1, 4)
  say '   and last four elements are:' hs.subword(hsCount - 3)

  hsMax = 0
  hsCountMax = 0
  llimit = 100000
  loop x_ = 1 to llimit - 1
    hs = hailstone(x_)
    hsCount = hs.words
    if hsCount > hsCountMax then do
      hsMax = x_
      hsCountMax = hsCount
      end
    end x_

  say 'The number' hsMax 'has the longest hailstone sequence in the range 1 to' llimit - 1 'with a sequence length of' hsCountMax
catch ex = Exception
  ex.printStackTrace
end

return

method hailstone(hn = long) public static returns Rexx signals IllegalArgumentException

  hs = Rexx('')
  if hn <= 0 then signal IllegalArgumentException('Invalid start point.  Must be a positive integer greater than 0')

  loop label n_ while hn > 1
    hs = hs' 'hn
    if hn // 2 \= 0 then hn = hn * 3 + 1
                    else hn = hn % 2
    end n_
  hs = hs' 'hn

  return hs.strip

Output:

The number 27 has a hailstone sequence comprising 112 elements
  its first four elements are: 27 82 41 124
   and last four elements are: 8 4 2 1
The number 77031 has the longest hailstone sequence in the range 1 to 99999 with a sequence length of 351

Nim

proc hailstone(n: int): seq[int] =
  result = @[n]
  var n = n
  while n > 1:
    if (n and 1) == 1:
      n = 3 * n + 1
    else:
      n = n div 2
    result.add n


when isMainModule:
  import strformat, strutils
  let h = hailstone(27)
  echo &"Hailstone sequence for number 27 has {h.len} elements."
  let first = h[0..3].join(", ")
  let last = h[^4..^1].join(", ")
  echo &"This sequence begins with {first} and ends with {last}."

  var m, mi = 0
  for i in 1..<100_000:
    let n = hailstone(i).len
    if n > m:
      m = n
      mi = i
  echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."

Output:

Hailstone sequence for number 27 has 112 elements.
This sequence begins with 27, 82, 41, 124 and ends with 8, 4, 2, 1.

For numbers < 100_000, maximum length 351 was found for Hailstone(77031).

Oberon-2

MODULE hailst;

IMPORT  Out;

CONST   maxCard         = MAX (INTEGER) DIV 3;
        List            = 1;
        Count           = 2;
        Max             = 3;

VAR     a               : INTEGER;

PROCEDURE HailStone (start, type  : INTEGER) : INTEGER;

VAR     n, max, count           : INTEGER;

BEGIN
  count := 1;
  n := start;
  max := n;
  LOOP
    IF  type = List  THEN
      Out.Int (n, 12);
      IF  count MOD 6 = 0  THEN  Out.Ln  END
    END;
    IF  n = 1  THEN  EXIT  END;
    IF  ODD (n)  THEN
      IF  n < maxCard  THEN
        n := 3 * n + 1;
        IF   n > max  THEN  max := n  END
      ELSE
        Out.String ("Exceeding max value for type INTEGER at: ");
        Out.String (" n = ");           Out.Int (start, 12);
        Out.String (" , count = ");     Out.Int (count, 12);
        Out.String (" and intermediate value ");
        Out.Int (n, 1);
        Out.String (". Aborting.");
        Out.Ln;
        HALT (2)
      END
    ELSE
      n := n DIV 2
    END;
    INC (count)
  END;
  IF  type = Max  THEN  RETURN  max  ELSE  RETURN  count  END
END HailStone;


PROCEDURE FindMax (num   : INTEGER);

VAR     val, maxCount, maxVal, cnt      : INTEGER;

BEGIN
  maxCount := 0;
  maxVal := 0;
  FOR  val := 2 TO num  DO
   cnt := HailStone (val, Count);
    IF  cnt > maxCount  THEN
      maxVal := val;
      maxCount := cnt
    END
  END;
  Out.String ("Longest sequence below ");       Out.Int (num, 1);
  Out.String (" is ");                          Out.Int (HailStone (maxVal, Count), 1);
  Out.String (" for n = ");                     Out.Int (maxVal, 1);
  Out.String (" with an intermediate maximum of ");
  Out.Int (HailStone (maxVal, Max), 1);
  Out.Ln
END FindMax;

BEGIN
  a := HailStone (27, List);
  Out.Ln;
  Out.String ("Iterations total = ");   Out.Int (HailStone (27, Count), 12);
  Out.String (" max value = ");         Out.Int (HailStone (27, Max)  , 12);
  Out.Ln;
  FindMax (1000000);
  Out.String ("Done.");
  Out.Ln
END hailst.

Producing

          27          82          41         124          62          31
          94          47         142          71         214         107
         322         161         484         242         121         364
         182          91         274         137         412         206
         103         310         155         466         233         700
         350         175         526         263         790         395
        1186         593        1780         890         445        1336
         668         334         167         502         251         754
         377        1132         566         283         850         425
        1276         638         319         958         479        1438
         719        2158        1079        3238        1619        4858
        2429        7288        3644        1822         911        2734
        1367        4102        2051        6154        3077        9232
        4616        2308        1154         577        1732         866
         433        1300         650         325         976         488
         244         122          61         184          92          46
          23          70          35         106          53         160
          80          40          20          10           5          16
           8           4           2           1

Iterations total = 112 max value =  9232

Exceeding max value for type INTEGER at:  n = 113383 , count = 120 and intermediate value 827370449. Aborting.

OCaml

#load "nums.cma";;
open Num;;

(* generate Hailstone sequence *)
let hailstone n =
  let one = Int 1
  and two = Int 2
  and three = Int 3 in
  let rec g s x =
    if x =/ one
    then x::s
    else g (x::s) (if mod_num x two =/ one
                   then three */ x +/ one
                   else x // two)
  in
  g [] (Int n)
;;

(* compute only sequence length *)
let haillen n =
  let one = Int 1
  and two = Int 2
  and three = Int 3 in
  let rec g s x =
    if x =/ one
    then s+1
    else g (s+1) (if mod_num x two =/ one
                  then three */ x +/ one
                  else x // two)
  in
  g 0 (Int n)
;;

(* max length for starting values in 1..n *)
let hailmax =
  let rec g idx len = function
  | 0 -> (idx, len)
  | i -> 
      let a = haillen i in
      if a > len
      then g i a (i-1)
      else g idx len (i-1)
  in
  g 0 0
;;

hailmax 100000 ;;
(* - : int * int = (77031, 351) *)

List.rev_map string_of_num (hailstone 27) ;;

(* - : string list =
["27"; "82"; "41"; "124"; "62"; "31"; "94"; "47"; "142"; "71"; "214"; "107";
 "322"; "161"; "484"; "242"; "121"; "364"; "182"; "91"; "274"; "137"; "412";
 "206"; "103"; "310"; "155"; "466"; "233"; "700"; "350"; "175"; "526"; "263";
 "790"; "395"; "1186"; "593"; "1780"; "890"; "445"; "1336"; "668"; "334";
 "167"; "502"; "251"; "754"; "377"; "1132"; "566"; "283"; "850"; "425";
 "1276"; "638"; "319"; "958"; "479"; "1438"; "719"; "2158"; "1079"; "3238";
 "1619"; "4858"; "2429"; "7288"; "3644"; "1822"; "911"; "2734"; "1367";
 "4102"; "2051"; "6154"; "3077"; "9232"; "4616"; "2308"; "1154"; "577";
 "1732"; "866"; "433"; "1300"; "650"; "325"; "976"; "488"; "244"; "122";
 "61"; "184"; "92"; "46"; "23"; "70"; "35"; "106"; "53"; "160"; "80"; "40";
 "20"; "10"; "5"; "16"; "8"; "4"; "2"; "1"] *)

Oforth

: hailstone   // n -- [n]
| l |
   ListBuffer new ->l
   while(dup 1 <>) [ dup l add dup isEven ifTrue: [ 2 / ] else: [ 3 * 1+ ] ]
   l add l dup freeze ;

hailstone(27) dup size println dup left(4) println right(4) println
100000 seq map(#[ dup hailstone size swap Pair new ]) reduce(#maxKey) println

Output:

112
[27, 82, 41, 124]
[8, 4, 2, 1]
[351, 77031]

Onyx (wasm)

use core { * }

hailstone :: (n: u32) -> [..]u32 {
    seq: [..]u32;
    array.push(&seq, n);
    while n > 1 {
        n = n/2 if n%2 == 0 else (n*3)+1;
        array.push(&seq, n);
    }
    return seq;
}

Longest :: struct { num, len : u32; }

main :: () {
    
    // -------
    // task 1: 
    // -------
    // "Create a routine to generate the hailstone 
    // sequence for a number."
    
    i := 27;
    seq := hailstone(i);
    printf("Task 1:\n{}: {}\n\n", 
        i,
        seq
    );
    
    // -------
    // task 2:
    // -------
    // "Use the routine to show that the hailstone 
    // sequence for the number 27 has 
    // 112 elements starting with 
    // 27, 82, 41, 124 and ending with 8, 4, 2, 1"
    
    slice_size := 4;
    len := seq.length;
    slice_first := seq[0..slice_size];
    slice_last := seq[seq.length-slice_size..seq.length];
    printf("Task 2:\nlength: {}, first: {}, last: {}\n\n", 
        len, 
        slice_first, 
        slice_last
    );
    
    // -------
    // task 3:
    // -------
    // "Show the number less than 100,000 
    // which has the longest hailstone sequence 
    // together with that sequences length."
    
    l : Longest;
    for i in 1..100000 {
        seq := hailstone(i);
        if l.len < seq.length { l = .{num = i, len = seq.length}; }
    }
    printf("Task 3:\nLongest Num: {}, Sequence Length: {}\n", l.num, l.len);
}

ooRexx

sequence = hailstone(27)
say "Hailstone sequence for 27 has" sequence~items "elements and is ["sequence~toString('l', ", ")"]"

highestNumber = 1
highestCount = 1

loop i = 2 to 100000
    sequence = hailstone(i)
    count = sequence~items
    if count > highestCount then do
        highestNumber = i
        highestCount = count
    end
end
say "Number" highestNumber "has the longest sequence with" highestCount "elements"

-- short routine to generate a hailstone sequence
::routine hailstone
  use arg n

  sequence = .array~of(n)
  loop while n \= 1
      if n // 2 == 0 then n = n / 2
      else n = 3 * n + 1
      sequence~append(n)
  end
  return sequence

Output:

Hailstone sequence for 27 has 112 elements and is [27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 77, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 102, 051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 0, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Number 77031 has the longest sequence with 351 elements

Order

To display the length, and first and last elements, of the hailstone sequence for 27, we could do this:

#include <order/interpreter.h>

#define ORDER_PP_DEF_8hailstone ORDER_PP_FN(                  \
8fn(8N,                                                       \
    8cond((8equal(8N, 1), 8seq(1))                            \
          (8is_0(8remainder(8N, 2)),                          \
           8seq_push_front(8N, 8hailstone(8quotient(8N, 2)))) \
          (8else,                                             \
           8seq_push_front(8N, 8hailstone(8inc(8times(8N, 3))))))) )

ORDER_PP(
  8lets((8H, 8seq_map(8to_lit, 8hailstone(27)))
        (8S, 8seq_size(8H)),
        8print(8(h(27) - length:) 8to_lit(8S) 8comma 8space
               8(starts with:) 8seq_take(4, 8H) 8comma 8space
               8(ends with:) 8seq_drop(8minus(8S, 4), 8H))
        ) )

Output:

h(27) - length:112, starts with:(27)(82)(41)(124), ends with:(8)(4)(2)(1)

Unfortunately, the C preprocessor not really being designed with large amounts of garbage collection in mind, trying to compute the hailstone sequences up to 100000 is almost guaranteed to run out of memory (and take a very, very long time). If we wanted to try, we could add this to the program, which in most languages would use relatively little memory:

#define ORDER_PP_DEF_8h_longest ORDER_PP_FN( \
8fn(8M, 8P, \
    8if(8is_0(8M), \
        8P, \
        8let((8L, 8seq_size(8hailstone(8M))), \
             8h_longest(8dec(8M), \
                        8if(8greater(8L, 8tuple_at_1(8P)), \
                            8pair(8M, 8L), 8P))))) )

ORDER_PP(
  8let((8P, 8h_longest(8nat(1,0,0,0,0,0), 8pair(0, 0))),
       8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P))))
)

...or even this "more elegant" version, which will run out of memory very quickly indeed (but in practice seems to work better for smaller ranges):

ORDER_PP(
  8let((8P,
        8seq_head(
          8seq_sort(8fn(8P, 8Q, 8greater(8tuple_at_1(8P),
                                         8tuple_at_1(8Q))),
                    8seq_map(8fn(8N,
                                 8pair(8N, 8seq_size(8hailstone(8N)))),
                             8seq_iota(1, 8nat(1,0,0,0,0,0)))))),
       8pair(8to_lit(8tuple_at_0(8P)), 8to_lit(8tuple_at_1(8P)))) )

Notice that large numbers (>100) must be entered as digit sequences with 8nat. 8to_lit converts a digit sequence back to a readable number.

Oz

declare
  fun {HailstoneSeq N}
     N > 0 = true %% assert
     if N == 1 then         [1]
     elseif {IsEven N} then N|{HailstoneSeq N div 2}
     else                   N|{HailstoneSeq 3*N+1}
     end
  end

  HSeq27 = {HailstoneSeq 27}
  {Length HSeq27} = 112
  {List.take HSeq27 4} = [27 82 41 124]
  {List.drop HSeq27 108} = [8 4 2 1]

  fun {MaxBy2nd A=A1#A2 B=B1#B2}
     if B2 > A2 then B else A end
  end

  Pairs = {Map {List.number 1 99999 1}
           fun {$ I} I#{Length {HailstoneSeq I}} end}

  MaxI#MaxLen = {List.foldL Pairs MaxBy2nd 0#0}
  {System.showInfo
   "Maximum length "#MaxLen#" was found for hailstone("#MaxI#")"}

Output:

Maximum length 351 was found for hailstone(77031)

PARI/GP

Version #1.

show(n)={
  my(t=1);
  while(n>1,
    print1(n",");
    n=if(n%2,
      3*n+1
    ,
      n/2
    );
    t++
  );
  print(1);
  t
};

len(n)={
  my(t=1);
  while(n>1,
    if(n%2,
      t+=2;
      n+=(n>>1)+1
    ,
      t++;
      n>>=1
    )
  );
  t
};

show(27)
r=0;for(n=1,1e5,t=len(n);if(t>r,r=t;ra=n));print(ra"\t"r)

Output:

27,82,41,124,62,31,94,47,142,71,214,107,322,161,484,242,121,364,182,91,274,137,4
12,206,103,310,155,466,233,700,350,175,526,263,790,395,1186,593,1780,890,445,133
6,668,334,167,502,251,754,377,1132,566,283,850,425,1276,638,319,958,479,1438,719
,2158,1079,3238,1619,4858,2429,7288,3644,1822,911,2734,1367,4102,2051,6154,3077,
9232,4616,2308,1154,577,1732,866,433,1300,650,325,976,488,244,122,61,184,92,46,2
3,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1

and

77031   351

Version #2.

Works with: PARI/GP version 2.7.4 and above

Different kind of PARI scripts for Collatz sequences you can find in OEIS, e.g.: A070165

\\ Get vector with Collatz sequence for the specified starting number.
\\ Limit vector to the lim length, or less, if 1 (one) term is reached (when lim=0).
\\ 3/26/2016 aev
Collatz(n,lim=0)={
my(c=n,e=0,L=List(n)); if(lim==0, e=1; lim=n*10^6); 
for(i=1,lim, if(c%2==0, c=c/2, c=3*c+1); listput(L,c); if(e&&c==1, break));
return(Vec(L)); } 
Collatzmax(ns,nf)={
my(V,vn,mxn=1,mx,im=1);
print("Search range: ",ns,"..",nf);
for(i=ns,nf, V=Collatz(i); vn=#V; if(vn>mxn, mxn=vn; im=i); kill(V)); 
print("Hailstone/Collatz(",im,") has the longest length = ",mxn);
} 

{
\\ Required tests:
print("Required tests:");
my(Vr,vrn);
Vr=Collatz(27); vrn=#Vr;
print("Hailstone/Collatz(27): ",Vr[1..4]," ... ",Vr[vrn-3..vrn],"; length = ",vrn);
Collatzmax(1,100000);
}

Output:

Required tests:
Hailstone/Collatz(27): [27, 82, 41, 124] ... [8, 4, 2, 1]; length = 112
Search range: 1..100000
Hailstone/Collatz(77031) has the longest length = 351

(15:32) gp > ##
  ***   last result computed in 15,735 ms.

Pascal

See Delphi or try this transformed Delphi version without generics.Use of a static array.

program ShowHailstoneSequence;
{$IFDEF FPC}
  {$MODE delphi} //or objfpc
{$Else}
  {$Apptype Console} // for delphi
{$ENDIF}
uses
  SysUtils;// format
const
  maxN = 10*1000*1000;// for output 1000*1000*1000

type
  tiaArr = array[0..1000] of Uint64;
  tIntArr = record
               iaMaxPos : integer;
               iaArr    : tiaArr
            end;
  tpiaArr = ^tiaArr;

function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
  result := 0;
  //ensure n to be odd
  while not(ODD(n)) do
  Begin
    inc(result);
    n := n shr 1;
  end;

  IF n > 1 then
  repeat
    //now n == odd -> so two steps in one can be made
    repeat
      n := (3*n+1) SHR 1;inc(result,2);
    until NOT(Odd(n));
    //now n == even -> so only one step can be made
    repeat
      n := n shr 1;      inc(result);
    until odd(n);
  until n = 1;
end;

procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
  maxPos: NativeInt;
  n: UInt64;
  pArr : tpiaArr;
begin
  with aHailstoneList do
  begin
    maxPos := 0;
    pArr := @iaArr;
  end;
  n  := aStartingNumber;
  pArr^[maxPos] := n;
  while n <> 1 do
  begin
    if odd(n) then
      n := (3*n+1)
    else
      n := n shr 1;
    inc(maxPos);
    pArr^[maxPos] := n;
  end;
  aHailstoneList.iaMaxPos  := maxPos;
end;

var
  i,Limit: NativeInt;
  lList: tIntArr;
  lAverageLength:Uint64;
  lMaxSequence: NativeInt;
  lMaxLength,lgth: NativeInt;
begin
  lList.iaMaxPos := 0;
  GetHailstoneSequence(27, lList);//319804831
  with lList do
  begin
    Limit := iaMaxPos;
    writeln(Format('sequence of %d has %d  elements',[iaArr[0],Limit+1]));
    write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
    For i := iaMaxPos-3 to iaMaxPos-1 do
       write(iaArr[i],',');
    writeln(iaArr[iaMaxPos]);
  end;
  Writeln;

  lMaxSequence := 0;
  lMaxLength := 0;
  i := 1;
  limit := 10*i;
  writeln(' Limit      : number with max length | average length');
  repeat
    lAverageLength:= 0;
    repeat
      lgth:= HailstoneSeqCnt(i);
      inc(lAverageLength, lgth);
      if lgth >= lMaxLength then
      begin
        lMaxSequence := i;
        lMaxLength := lgth+1;
      end;
      inc(i);
    until i = Limit;
    Writeln(Format(' %10d : %9d    |  %4d   |      %7.3f',
                   [limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
    limit := limit*10;
  until Limit > maxN;
end.

Output:

sequence of 27 has 112  elements
27,82,41,124..8,4,2,1

 Limit      : number with max length | average length
         10 :         9    |    20   |        5.490
        100 :        97    |   119   |       27.504
       1000 :       871    |   179   |       50.683
      10000 :      6171    |   262   |       71.119
     100000 :     77031    |   351   |       89.137
    1000000 :    837799    |   525   |      108.613
   10000000 :   8400511    |   686   |      127.916
  100000000 :  63728127    |   950   |      147.337
 1000000000 : 670617279    |   987   |      166.780

real  6m22.968s // 32-bit compiled
real  3m56.573s // 64-bit compiled

PascalABC.NET

function hailstone(n: Integer): sequence of integer;
begin
  result := seq(n);
  while n > 1 do
  begin
    n := n.IsEven ? n div 2 : 3 * n + 1;
    result := result + seq(n);
  end;
end;

begin
  var a := hailstone(27);
  println(a.Count, a.Take(4), a.TakeLast(4));
  var max := 0; 
  var maxi := 0;
  for var i := 1 to 100_000 do 
    if hailstone(i).Count > max then
    begin
      max := hailstone(i).Count;
      maxi := i;
    end;
  println(maxi, max);
end.

Output:

112 [27,82,41,124] [8,4,2,1]
77031 351

Perl

Straightforward

#!/usr/bin/perl

use warnings;
use strict;

my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";

my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
    @h = hailstone($x);
    if (scalar @h > $max) {
        ($max, $n) = (scalar @h, $x);
    }
}

print "Max length $max was found for hailstone($n) for numbers < 100_000\n";


sub hailstone {
    my ($n) = @_;

    my @sequence = ($n);

    while ($n > 1) {
        if ($n % 2 == 0) {
            $n = int($n / 2);
        } else {
            $n = $n * 3 + 1;
        }

        push @sequence, $n;
    }

    return @sequence;
}

Output:

Length of hailstone(27) = 112
[27, 82, 41, 124, ..., 8, 4, 2, 1]
Max length 351 was found for hailstone(77031) for numbers < 100_000

Compact

A more compact version:

#!/usr/bin/perl
use strict;

sub hailstone {
    @_ = local $_ = shift;
    push @_, $_ = $_ % 2 ? 3 * $_ + 1 : $_ / 2 while $_ > 1;
    @_;
}

my @h = hailstone($_ = 27);
print "$_: @h[0 .. 3] ... @h[-4 .. -1] (".@h.")\n";

@h = ();
for (1 .. 99_999) { @h = ($_, $h[2]) if ($h[2] = hailstone($_)) > $h[1] }
printf "%d: (%d)\n", @h;

Output:

27: 27 82 41 124 ... 8 4 2 1 (112)
77031: (351)

Phix

Copy of Euphoria

with javascript_semantics
function hailstone(atom n)
    sequence s = {n}
    while n!=1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n+1
        end if
        s &= n
    end while
    return s
end function
 
function hailstone_count(atom n)
    integer count = 1
    while n!=1 do
        if remainder(n,2)=0 then
            n /= 2
        else
            n = 3*n+1
        end if
        count += 1
    end while
    return count
end function
 
sequence s = hailstone(27)
printf(1,"hailstone(27) = %v\n",{shorten(s,"numbers",4)})
 
integer hmax = 1, imax = 1,count
for i=2 to 1e5-1 do
    count = hailstone_count(i)
    if count>hmax then
        hmax = count
        imax = i
    end if
end for
 
printf(1,"The longest hailstone sequence under 100,000 is %d with %d elements.\n",{imax,hmax})

Output:

hailstone(27) = {27,82,41,124,"...",8,4,2,1," (112 numbers)"}
The longest hailstone sequence under 100,000 is 77031 with 351 elements.

PHP

function hailstone($n,$seq=array()){
	$sequence = $seq;
	$sequence[] = $n;
	if($n == 1){
		return $sequence;
	}else{
		$n = ($n%2==0) ? $n/2 : (3*$n)+1;
		return hailstone($n, $sequence);
	}
}

$result = hailstone(27);
echo count($result) . ' Elements.<br>';
echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>';

$maxResult = array(0);

for($i=1;$i<=100000;$i++){
		$result = count(hailstone($i));
		if($result > max($maxResult)){
			$maxResult = array($i=>$result);
		}
}
foreach($maxResult as $key => $val){
echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val;
}

112 Elements.
Starting with : 27,82,41,124 and ending with : 8,4,2,1
Number < 100000 with longest Hailstone seq.: 77031 with length of 351

Picat

import util.

go =>
   println("H27:"),
   H27 = hailstoneseq(27),
   H27Len = H27.len,
   println(len=H27.len),   
   println(take(H27,4)++['...']++drop(H27,H27Len-4)),
   nl,

   println("Longest sequence < 100_000:"),
   longest_seq(99_999),

   nl.

% The Hailstone value of a number
hailstone(N) = N // 2, N mod 2 == 0 => true.
hailstone(N) = 3*N+1, N mod 2 == 1 => true.

% Sequence for a number
hailstoneseq(N) = Seq =>
   Seq := [N],
   while (N > 1)
      N := hailstone(N),
      Seq := Seq ++ [N]
   end.

%
% Use a map to cache the lengths.
% Here we don't care about the actual sequence.
%
longest_seq(Limit) =>
   Lens = new_map(), % caching the lengths
   MaxLen = 0,
   MaxN = 1,

   foreach(N in 1..Limit-1) 
      M = N,
      CLen = 1,
      while (M > 1) 
         if Lens.has_key(M) then
            CLen := CLen + Lens.get(M) - 1,
            M := 1
         else
            M := hailstone(M), % call the 
            CLen := CLen + 1
         end
      end,
      Lens.put(N, CLen),
      if CLen > MaxLen then
         MaxLen := CLen,
         MaxN := N
      end
   end,
   println([maxLen=MaxLen, maxN=MaxN]),
   nl.

Output:

H27:
len = 112
[27,82,41,124,...,8,4,2,1]

Longest sequence < 100_000:
[maxLen = 351,maxN = 77031]

Mode-directed tabling

If we just want to get the length of the longest sequence - and are not forced to use the same Hailstone function as for the H27 task - then this version using model-directed tabling is faster than longest_seq/1: 0.055s vs 0.127s. (Original idea by Neng-Fa Zhou.)

go2 =>
   time(max_chain(MaxN,MaxLen)),
   printf("MaxN=%w,MaxLen=%w%n",MaxN,MaxLen).

table (-,max)
max_chain(N,Len) =>
    between(2,99_999,N),
    gen(N,Len).

table (+,-)
gen(1,Len) => Len=1.
gen(N,Len), N mod 2 == 0 => 
    gen(N div 2,Len1),
    Len=Len1+1.
gen(N,Len) =>
    gen(3*N+1,Len1),
    Len=Len1+1.

PicoLisp

(de hailstone (N)
   (make
      (until (= 1 (link N))
         (setq N
            (if (bit? 1 N)
               (inc (* N 3))
               (/ N 2) ) ) ) ) )

(let L (hailstone 27)
   (println 27 (length L) (head 4 L) '- (tail 4 L)) )

(let N (maxi '((N) (length (hailstone N))) (range 1 100000))
   (println N (length (hailstone N))) )

Output:

27 112 (27 82 41 124) - (8 4 2 1)
77031 351

Pike

#!/usr/bin/env pike

int next(int n)
{
    if (n==1)
        return 0;
    if (n%2)
        return 3*n+1;
    else
        return n/2;
}

array(int) hailstone(int n)
{
    array seq = ({ n });
    while (n=next(n))
        seq += ({ n });
    return seq;
}

void main()
{
    array(int) two = hailstone(27);
    if (equal(two[0..3], ({ 27, 82, 41, 124 })) && equal(two[<3..], ({ 8,4,2,1 })))
        write("sizeof(({ %{%d, %}, ... %{%d, %} }) == %d\n", two[0..3], two[<3..], sizeof(two));

    mapping longest = ([ "length":0, "start":0 ]);

    foreach(allocate(100000); int start; )
    {
        int length = sizeof(hailstone(start));
        if (length > longest->length)
        {
            longest->length = length;
            longest->start = start;
        }
    }
    write("longest sequence starting at %d has %d elements\n", longest->start, longest->length);
}

Output:

 sizeof(({ 27, 82, 41, 124, , ... 8, 4, 2, 1,  }) == 112
 longest sequence starting at 77031 has 351 elements

PL/I

test: proc options (main);
   declare (longest, n) fixed (15);
   declare flag bit (1);
   declare (i, value) fixed (15);

   /* Task 1: */
   flag = '1'b;
   put skip list ('The sequence for 27 is');
   i = hailstones(27);

   /* Task 2: */
   flag = '0'b;
   longest = 0;
   do i = 1 to 99999;
      if longest < hailstones(i) then
         do; longest = hailstones(i); value = i; end;
   end;
   put skip edit (value, ' has the longest sequence of ', longest) (a);

hailstones: procedure (n) returns ( fixed (15));
   declare n fixed (15) nonassignable;
   declare (m, p) fixed (15);

   m = n;
   p = 1;
   if flag then put skip list (m);
   do p = 1 by 1 while (m > 1);
      if iand(m, 1) = 0 then
         m = m/2;
      else
         m = 3*m + 1;
      if flag then put skip list (m);
   end;
   if flag then put skip list ('The hailstone sequence has length' || p);
   return (p);
end hailstones;

end test;

Output:

The sequence for 27 is 
                27 
                82 
                41 
               124 
                62 
                31 
                94 
                47 
               142 
                71 
               214 
               107 
               322 
               161 
               484 
               242 
               121 
               364 
               182 
                91 
               274 
               137 
               412 
               206 
               103 
               310 
               155 
               466 
               233 
               700 
               350 
               175 
               526 
               263 
               790 
               395 
              1186 
               593 
              1780 
               890 
               445 
              1336 
               668 
               334 
               167 
               502 
               251 
               754 
               377 
              1132 
               566 
               283 
               850 
               425 
              1276 
               638 
               319 
               958 
               479 
              1438 
               719 
              2158 
              1079 
              3238 
              1619 
              4858 
              2429 
              7288 
              3644 
              1822 
               911 
              2734 
              1367 
              4102 
              2051 
              6154 
              3077 
              9232 
              4616 
              2308 
              1154 
               577 
              1732 
               866 
               433 
              1300 
               650 
               325 
               976 
               488 
               244 
               122 
                61 
               184 
                92 
                46 
                23 
                70 
                35 
               106 
                53 
               160 
                80 
                40 
                20 
                10 
                 5 
                16 
                 8 
                 4 
                 2 
                 1 
The hailstone sequence has length               112 
             77031 has the longest sequence of                351

PL/I-80

hailstone_demo: proc options (main);
    %replace
        true by '1'b,
        false by '0'b;
    dcl
        (slen, longest) fixed bin(15),
        (n, n_longest,limit) fixed decimal(12),
        answer char(1);
    put skip list ('Display hailstone sequence for what number? ');
    get list (n);
    slen = hailstone(n, true);
    put skip list ('Sequence length = ', slen);
    put skip(2) list ('Search for longest sequence (y/n)? ');
    get list (answer);
    if ((answer ^= 'y') & (answer ^= 'Y')) then stop;
    put list ('Search to what limit? ');
    get list (limit);
    longest = 1;
    n = 2;
    do while (n < limit);
        slen = hailstone(n, false);
        if slen > longest then
            do;
               longest = slen;
               n_longest = n;
            end;
        n = n + 1;
    end;
    put skip edit ('Longest sequence =',longest,' for n =',n_longest)
        (a,f(4),a,f(6));


/* compute, and optionally display, hailstone sequence for n */
hailstone: 
    procedure(n, show) returns (fixed binary);
    dcl
        (len, col) fixed binary,
        (n, k) fixed decimal(12),
        show bit(1);
    /* make local copy since n is passed by reference */
    k = n; 
    col = 1;
    len = 1;
    do while ((k ^= 1) & (k > 0));
      if (show) then   /* print 8 columns across */ 
          do;
              put edit (k) (f(8));
              col = col + 1;
              if col > 8 then
                  do;
                      put skip;
                      col = 1;
                  end;
          end;
      if (mod(k,2) = 0) then
          k = k / 2;
      else
          k = k * 3 + 1;
      len = len + 1;
    end;
    if (show) then put edit (k) (f(8));
    return (len);
    end hailstone;

end hailstone_demo;

Output:

Display hailstone sequence for what number? 27            
      27      82      41     124      62      31      94      47
     142      71     214     107     322      161    484     242
     121     364     182      91     274      137    412     206
     103     310     155     466     233      700    350     175
     526     263     790     395    1186      593   1780     890
     445    1136     668     334     167      502    251     754
     377    1132     566     283     850      425   1276     638
     319     958     479    1438     719     2158   1079    3238
    1619    4858    2429    7288    3644     1822    911    2734
    1367    4102    2051    6154    3077     9232   4616    2308
    1154     577    1732     866     433     1300    650     325
     976     488     244     122      61      184     92      46
      23      70      35     106      53      160     80      40
      20      10       5      16       8        4      2       1
Sequence length =        112

Search for longest sequence (y/n)? y
Search to what limit? 100000
Longest sequence = 351 for n = 77031

Plain TeX

The following code works with any TeX engine.

\newif\ifprint
\newcount\itercount
\newcount\currentnum
\def\hailstone#1{\itercount=0 \currentnum=#1 \hailstoneaux}
\def\hailstoneaux{%
	\advance\itercount1
	\ifprint\number\currentnum\space\space\fi
	\ifnum\currentnum>1
		\ifodd\currentnum
			\multiply\currentnum3 \advance\currentnum1
		\else
			\divide\currentnum2
		\fi
		\expandafter\hailstoneaux
	\fi
}

\parindent=0pt
\printtrue\hailstone{27}
Length = \number\itercount
\bigbreak

\newcount\ii \ii=1
\printfalse
\def\lenmax{0}
\def\seed{0}
\loop
	\ifnum\ii<100000
		\hailstone\ii
		\ifnum\itercount>\lenmax\relax
			\edef\lenmax{\number\itercount}%
			\edef\seed{\number\ii}%
		\fi
		\advance\ii1
\repeat
Seed max = \seed, length = \lenmax
\bye

pdf or dvi output:

27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206
103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167
502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619
4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732
866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16
8 4 2 1 Length = 112

Seed max = 77031, length = 351

Pointless

output =
  println(format(fmt,
    [seqLength, initSeq, tailSeq] ++ toList(longestPair)
  ))

fmt = """getSeq(27) (length): {}
getSeq(27) (first 4): {}
getSeq(27) (last 4): {}
max length {} for n = {}"""

-----------------------------------------------------------
 
seq       = getSeq(27)
seqLength = length(seq)
initSeq   = take(4, seq)
tailSeq   = drop(seqLength - 4, seq)
 
-----------------------------------------------------------
 
longestPair =
  range(1, 99999)
  |> map(n => (length(getSeq(n)), n))
  |> argmax(at(0))
 
-----------------------------------------------------------
-- generate full sequence
 
getSeq(n) =
  iterate(step, n)
  |> takeUntil(eq(1))
 
-----------------------------------------------------------
-- get the next number in a sequence
 
step(n) =
  if n % 2 == 0 then round(n / 2) else n * 3 + 1

Output:

getSeq(27) (length): 112
getSeq(27) (first 4): [27, 82, 41, 124]
getSeq(27) (last 4): [8, 4, 2, 1]
max length 351 for n = 77031

PowerShell

Works with: PowerShell version 3.0+

function Get-HailStone {
    param($n)
           
    switch($n) {
        1              {$n;return}
        {$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
        {$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}      
    }
}

function Get-HailStoneBelowLimit {
    param($UpperLimit)

    for ($i = 1; $i -lt $UpperLimit; $i++) { 
        [pscustomobject]@{
            'Number' = $i
            'Count' = (Get-HailStone $i).count
        } 
    }
}

Output:

PS C:\> Get-HailStone 27
27
82
41
...
8
4
2
1

PS C:\> (Get-HailStone 27).count
112

PS C:\> Get-HailStoneBelowLimit 100000 | Sort Count -Descending | Select -first 1
Number         Count
------         -----
 77031           351

Prolog

1. Create a routine to generate the hailstone sequence for a number.

hailstone(1,[1]) :- !.
hailstone(N,[N|S]) :- 0 is N mod 2, N1 is N / 2, hailstone(N1,S).
hailstone(N,[N|S]) :- 1 is N mod 2, N1 is (3 * N) + 1, hailstone(N1, S).

2. Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1.

The following query performs the test.

hailstone(27,X),
length(X,112),
append([27, 82, 41, 124], _, X),
append(_, [8, 4, 2, 1], X).

3. Show the number less than 100,000 which has the longest hailstone sequence together with that sequences length.

longestHailstoneSequence(M, Seq, Len) :- longesthailstone(M, 1, 1, Seq, Len).
longesthailstone(1, Cn, Cl, Mn, Ml):- Mn = Cn,
	                               Ml = Cl.
longesthailstone(N, _, Cl, Mn, Ml) :- hailstone(N, X),
                                       length(X, L),
                                       Cl < L,
                                       N1 is N-1,
                                       longesthailstone(N1, N, L, Mn, Ml).
longesthailstone(N, Cn, Cl, Mn, Ml) :- N1 is N-1,
                                       longesthailstone(N1, Cn, Cl, Mn, Ml).

run this query.

longestHailstoneSequence(100000, Seq, Len).

to get the following result

Seq = 77031,
Len = 351 

Constraint Handling Rules

CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker

:- use_module(library(chr)).
:- chr_option(debug, off).
:- chr_option(optimize, full).

:- chr_constraint collatz/2, hailstone/1, clean/0.

% to remove all constraints hailstone/1 after computation
clean @ clean \ hailstone(_) <=> true.
clean @ clean <=> true.

% compute Collatz number
init @ collatz(1,X) <=>  X = 1 | true.
collatz @ collatz(N, C) <=> (N mod 2 =:= 0 -> C is N / 2; C is 3 * N + 1).

% Hailstone loop
hailstone(1) ==> true.
hailstone(N) ==> N \= 1 | collatz(N, H), hailstone(H).

Code for task one :

task1 :-
	hailstone(27),
	findall(X, find_chr_constraint(hailstone(X)), L),
	clean,
	% check the requirements
	(   (length(L, 112), append([27, 82, 41, 124 | _], [8,4,2,1], L)) -> writeln(ok); writeln(ko)).

Output:

 ?- task1.
ok
true.

Code for task two :

longest_sequence :-
	seq(2, 100000, 1-[1], Len-V),
	format('For ~w sequence has ~w len ! ~n', [V, Len]).


% walk through 2 to 100000 and compute the length of the sequences
% memorize the longest
seq(N, Max, Len-V, Len-V) :- N is Max + 1, !.
seq(N, Max, CLen - CV, FLen - FV) :-
	len_seq(N, Len - N),
	(   Len > CLen -> Len1 = Len, V1 = [N]
	;   Len = CLen -> Len1 = Len, V1 = [N | CV]
	;   Len1 = CLen, V1 = CV),
	N1 is N+1,
	seq(N1, Max, Len1 - V1, FLen - FV).

% compute the len of the Hailstone sequence for a number
len_seq(N, Len - N) :-
	hailstone(N),
	findall(hailstone(X), find_chr_constraint(hailstone(X)), L),
	length(L, Len),
	clean.

Output:

 ?- longest_sequence.
For [77031] sequence has 351 len ! 
true.

Pure

// 1. Create a routine to generate the hailstone sequence for a number. 
type odd x::int = x mod 2;
type even x::int = ~odd x;
odd x = typep odd x;
even x = typep even x;

hailstone 1       = [1];
hailstone n::even = n:hailstone (n div 2);
hailstone n::odd  = n:hailstone (3*n + 1);

// 2. Use the routine to show that the hailstone sequence for the number 27 
//    has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
n = 27;
hs = hailstone n;
l = # hs;
using system;

printf 
    ("the hailstone sequence for the number %d has %d elements " + 
     "starting with %s and ending with %s\n") 
    (n, l, __str__ (hs!!(0..3)), __str__ ( hs!!((l-4)..l)));

// 3. Show the number less than 100,000 which has the longest hailstone 
//    sequence together with that sequences length.
printf ("the number under 100,000 with the longest sequence is %d " + 
        "with a sequence length of %d\n")
       (foldr (\ (a,b) (c,d) -> if (b > d) then (a,b) else (c,d)) 
             (0,0) 
             (map (\ x -> (x, # hailstone x)) (1..100000)));

Output:

the hailstone sequence for the number 27 has 112 elements starting with [27,82,41,124] and ending with [8,4,2,1]
the number under 100,000 with the longest sequence is 77031 with a sequence length of 351

Python

Procedural

def hailstone(n):
    seq = [n]
    while n > 1:
        n = 3 * n + 1 if n & 1 else n // 2
        seq.append(n)
    return seq


if __name__ == '__main__':
    h = hailstone(27)
    assert (len(h) == 112
            and h[:4] == [27, 82, 41, 124]
            and h[-4:] == [8, 4, 2, 1])
    max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
    print(f"Maximum length {max_length} was found for hailstone({n}) "
          f"for numbers <100,000")

Output:

Maximum length 351 was found for hailstone(77031) for numbers <100,000

Using a generator

from itertools import islice


def hailstone(n):
    yield n
    while n > 1:
        n = 3 * n + 1 if n & 1 else n // 2
        yield n


if __name__ == '__main__':
    h = hailstone(27)
    assert list(islice(h, 4)) == [27, 82, 41, 124]
    for _ in range(112 - 4 * 2):
        next(h)
    assert list(islice(h, 4)) == [8, 4, 2, 1]
    max_length, n = max((sum(1 for _ in hailstone(i)), i)
                        for i in range(1, 100_000))
    print(f"Maximum length {max_length} was found for hailstone({n}) "
          f"for numbers <100,000")

Output:

Maximum length 351 was found for hailstone(77031) for numbers <100,000

Composition of pure functions

Works with: Python version 3.7

'''Hailstone sequences'''

from itertools import (islice, takewhile)


# hailstone :: Int -> [Int]
def hailstone(x):
    '''Hailstone sequence starting with x.'''
    def p(n):
        return 1 != n
    return list(takewhile(p, iterate(collatz)(x))) + [1]


# collatz :: Int -> Int
def collatz(n):
    '''Next integer in the hailstone sequence.'''
    return 3 * n + 1 if 1 & n else n // 2


# ------------------------- TEST -------------------------
# main :: IO ()
def main():
    '''Tests.'''

    n = 27
    xs = hailstone(n)
    print(unlines([
        f'The hailstone sequence for {n} has {len(xs)} elements,',
        f'starting with {take(4)(xs)},',
        f'and ending with {drop(len(xs) - 4)(xs)}.\n'
    ]))

    (a, b) = (1, 99999)
    (i, x) = max(
        enumerate(
            map(compose(len)(hailstone), enumFromTo(a)(b))
        ),
        key=snd
    )
    print(unlines([
        f'The number in the range {a}..{b} '
        f'which produces the longest sequence is {1 + i},',
        f'generating a hailstone sequence of {x} integers.'
    ]))


# ----------------------- GENERIC ------------------------

# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
    '''Function composition.'''
    return lambda f: lambda x: g(f(x))


# drop :: Int -> [a] -> [a]
# drop :: Int -> String -> String
def drop(n):
    '''The sublist of xs beginning at
       (zero-based) index n.
    '''
    def go(xs):
        if isinstance(xs, (list, tuple, str)):
            return xs[n:]
        else:
            take(n)(xs)
            return xs
    return go


# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
    '''Integer enumeration from m to n.'''
    return lambda n: range(m, 1 + n)


# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
    '''An infinite list of repeated
       applications of f to x.
    '''
    def go(x):
        v = x
        while True:
            yield v
            v = f(v)
    return go


# snd :: (a, b) -> b
def snd(tpl):
    '''Second component of a tuple.'''
    return tpl[1]


# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
    '''The prefix of xs of length n,
       or xs itself if n > length xs.
    '''
    def go(xs):
        return (
            xs[0:n]
            if isinstance(xs, (list, tuple))
            else list(islice(xs, n))
        )
    return go


# unlines :: [String] -> String
def unlines(xs):
    '''A single newline-delimited string derived
       from a list of strings.'''
    return '\n'.join(xs)


if __name__ == '__main__':
    main()

Output:

The hailstone sequence for 27 has 112 elements,
starting with [27, 82, 41, 124],
and ending with [8, 4, 2, 1].

The number in the range 1..99999 which produces the longest sequence is 77031,
generating a hailstone sequence of 351 integers.

Quackery

[ 1 & ]                                 is odd      ( n --> b )
  
[ []
  [ over join swap
    dup 1 > while
    dup odd iff
      [ 3 * 1 + ]
    else
      [ 2 / ]
    swap again ]
   drop ]                              is hailstone ( n --> [ )

 [ stack ]                             is longest   (   --> s )

 [ stack ]                             is length    (   --> s )

27 hailstone
say "The hailstone sequence for 27 has "
dup size echo say " elements." cr
say "It starts with"
dup 4 split drop witheach [ sp echo ]
say " and ends with"
-4 split nip witheach [ sp echo ]
say "." cr cr 

0 longest put  0 length put
99999 times 
  [ i^ 1+ hailstone size
    dup length share > if
      [ dup length replace
        i^ 1+ longest replace ] 
    drop ]
longest take echo
say " has the longest sequence of any number less than 100000."
cr say "It is " length take echo say " elements long." cr

Output:

The hailstone sequence for 27 has 112 elements.
It starts with 27 82 41 124 and ends with 8 4 2 1.

77031 has the longest sequence of any number less than 100000.
It is 351 elements long.

R

Iterative solution

### PART 1:
makeHailstone <- function(n){
  hseq <- n  
  while (hseq[length(hseq)] > 1){
    current.value <- hseq[length(hseq)]
    if (current.value %% 2 == 0){
      next.value <- current.value / 2
    } else {
      next.value <- (3 * current.value) + 1
    }
    hseq <- append(hseq, next.value)
  }
  return(list(hseq=hseq, seq.length=length(hseq)))
}

### PART 2:
twenty.seven <- makeHailstone(27)
twenty.seven$hseq
twenty.seven$seq.length

### PART 3:
max.length <- 0;  lower.bound <- 1;  upper.bound <- 100000

for (index in lower.bound:upper.bound){
  current.hseq <- makeHailstone(index)  
  if (current.hseq$seq.length > max.length){
    max.length <- current.hseq$seq.length
    max.index  <- index
  }
}

cat("Between ", lower.bound, " and ", upper.bound, ", the input of ", 
    max.index, " gives the longest hailstone sequence, which has length ", 
    max.length, ". \n", sep="")

Output:

> twenty.seven$hseq
  [1]   27   82   41  124   62   31   94   47  142   71  214  107  322  161  484
 [16]  242  121  364  182   91  274  137  412  206  103  310  155  466  233  700
 [31]  350  175  526  263  790  395 1186  593 1780  890  445 1336  668  334  167
 [46]  502  251  754  377 1132  566  283  850  425 1276  638  319  958  479 1438
 [61]  719 2158 1079 3238 1619 4858 2429 7288 3644 1822  911 2734 1367 4102 2051
 [76] 6154 3077 9232 4616 2308 1154  577 1732  866  433 1300  650  325  976  488
 [91]  244  122   61  184   92   46   23   70   35  106   53  160   80   40   20
[106]   10    5   16    8    4    2    1

> twenty.seven$seq.length
[1] 112

Between 1 and 1e+05, the input of 77031 gives the longest hailstone sequence, 
which has length 351.

Vectorization solution

The previous solution is entirely satisfactory and may be more efficient than the following solution. However, problems like these are a great chance to show off the strength of R's vectorization. Also, this lets us show off how the <- syntax can do multiple variable assignments in one line. Observe how short the following code is:

###Task 1:
collatz <- function(n)
{
  lastIndex <- 1
  output <- lastEntry <- n
  while(lastEntry != 1)
  {
    #Each branch updates lastEntry, lastIndex, and appends a new element to the end of output.
    #Note that the return value of lastIndex <- lastIndex + 1 is lastIndex + 1.
    #You may be surprised that output can be appended to despite starting as just a single number.
    #If so, recall that R's numerics are vectors, meaning that output<-n created a vector of length 1.
    #It's ugly, but efficient.
    if(lastEntry %% 2) lastEntry <- output[lastIndex <- lastIndex + 1] <- 3 * lastEntry + 1
    else lastEntry <- output[lastIndex <- lastIndex + 1] <- lastEntry %/% 2
  }
  output
}

###Task 2:
#Notice how easy it is to access the required elements:
twentySeven <- collatz(27)
cat("The first four elements are:", twentySeven[1:4], "and the last four are:", twentySeven[length(twentySeven) - 3:0], "\n")

###Task 3:
#Notice how a several line long loop can be avoided with R's sapply or Vectorize:
seqLenghts <- sapply(seq_len(99999), function(x) length(collatz(x)))
longest <- which.max(seqLenghts)
cat("The longest sequence before the 100000th is found at n =", longest, "and it has length", seqLenghts[longest], "\n")
#Equivalently, line 1 could have been: seqLenghts <- sapply(Vectorize(collatz)(1:99999), length).
#Another good option would be seqLenghts <- lengths(Vectorize(collatz)(1:99999)).

Output:

The first four elements are: 27 82 41 124 and the last four are: 8 4 2 1 
The longest sequence before the 100000th is found at n = 77031 and it has length 351

Racket

#lang racket

(define hailstone
  (let ([t (make-hasheq)])
    (hash-set! t 1 '(1))
    (λ(n) (hash-ref! t n
            (λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))

(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
        `(,@(take h27 4) ... ,@(take-right h27 4))
        (length h27))

(define N 100000)
(define longest
  (for/fold ([m #f]) ([i (in-range 1 (add1 N))])
    (define h (hailstone i))
    (if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
        N (car longest) (cdr longest))

Output:

h(27) = (27 82 41 124 ... 8 4 2 1), 112 items
for x<=100000, 77031 has the longest sequence with 351 items

Raku

(formerly Perl 6)

sub hailstone($n) { $n, { $_ %% 2 ?? $_ div 2 !! $_ * 3 + 1 } ... 1 }

my @h = hailstone(27);
say "Length of hailstone(27) = {+@h}";
say ~@h;

my $m = max ( (1..99_999).race.map: { +hailstone($_) => $_ } );
say "Max length {$m.key} was found for hailstone({$m.value}) for numbers < 100_000";

Output:

Length of hailstone(27) = 112
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
Max length 351 was found for hailstone(77031) for numbers < 100_000

REBOL

hail: func [
	"Returns the hailstone sequence for n"
	n [integer!]
	/local seq
] [
	seq: copy reduce [n]
	while [n <> 1] [
		append seq n: either n % 2 == 0 [n / 2] [3 * n + 1]
	]
	seq
]

hs27: hail 27
print [
	"the hail sequence of 27 has length" length? hs27
	"and has the form " copy/part hs27 3 "..."
	back back back tail hs27
]

maxN: maxLen: 0
repeat n 99999 [
	if (len: length? hail n) > maxLen [
		maxN: n
		maxLen: len
	]
]

print [
	"the number less than 100000 with the longest hail sequence is"
	maxN "with length" maxLen
]

Output:

the hail sequence of 27 has length 112 and has the form  27 82 41 ... 4 2 1
the number less than 100000 with the longest hail sequence is 77031 with length 351

Refal

$ENTRY Go {
    = <ShowHailstone 27>
      <ShowLongest 100000>;
}

Hailstone {
    1 = 1;
    s.N, <Mod s.N 2>: {
        0 = s.N <Hailstone <Div s.N 2>>;
        1 = s.N <Hailstone <+ 1 <* 3 s.N>>>;
    };
};

ShowHailstone {
    s.N, <Hailstone s.N>: e.Seq,
         <Lenw e.Seq>: s.Len s.1 s.2 s.3 s.4 e.X s.D4 s.D3 s.D2 s.D1
        = <Prout 'The hailstone sequence for the number '
             <Symb s.N> ' has ' <Symb s.Len> ' elements,\n'
             'starting with ' s.1 s.2 s.3 s.4
             'and ending with ' s.D4 s.D3 s.D2 <Symb s.D1>'.'>;
}

FindLongest {
    s.Max = <FindLongest s.Max 1 1 1>;
    s.Max s.Max s.Long s.Len = s.Long s.Len;
    s.Max s.Cur s.Long s.Len,
        <Hailstone s.Cur>: e.CurSeq,
        <Lenw e.CurSeq>: s.CurLen e.X,
        <+ s.Cur 1>: s.Next,
        <Compare s.CurLen s.Len>: {
            '+' = <FindLongest s.Max s.Next s.Cur s.CurLen>;
            s.X = <FindLongest s.Max s.Next s.Long s.Len>;
        };
};

ShowLongest {
    s.Max, <FindLongest s.Max>: s.Long s.Len
        = <Prout 'The number < ' <Symb s.Max> ' which has the longest'
                 ' hailstone sequence is ' <Symb s.Long> '.\n'
                 'The length of its Hailstone sequence is ' 
                 <Symb s.Len> '.'>;
};

Output:

The hailstone sequence for the number 27 has 112 elements,
starting with 27 82 41 124 and ending with 8 4 2 1.
The number < 100000 which has the longest hailstone sequence is 77031.
The length of its Hailstone sequence is 351.

REXX

non-optimized

/*REXX program tests a  number  and also a  range for  hailstone  (Collatz)  sequences. */
numeric digits 20                                /*be able to handle gihugeic numbers.  */
parse arg x y .                                  /*get optional arguments from the C.L. */
if x=='' | x==","   then x=     27               /*No  1st  argument?  Then use default.*/
if y=='' | y==","   then y= 100000 - 1           /* "  2nd      "        "   "     "    */
$= hailstone(x)     /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say  x   ' has a hailstone sequence of '      words($)
say      '    and starts with: '              subword($, 1, 4)    " ∙∙∙"
say      '    and  ends  with:  ∙∙∙'          subword($, max(5, words($)-3))
if y==0  then exit  /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
w= 0;         do j=1  for y;  call hailstone j   /*traipse through the range of numbers.*/
              if #hs<=w  then iterate            /*Not big 'nuff?   Then keep traipsing.*/
              bigJ= j;   w= #hs                  /*remember what # has biggest hailstone*/
              end   /*j*/
say '(between 1 ──►'   y") "       bigJ      ' has the longest hailstone sequence: '   w
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
hailstone: procedure expose #hs; parse arg n 1 s /*N and S: are set to the 1st argument.*/
                     do #hs=1   while  n\==1     /*keep loop while   N   isn't  unity.  */
                     if n//2  then n= n * 3  + 1 /*N is odd ?   Then calculate  3*n + 1 */
                              else n= n % 2      /*"  " even?   Then calculate  fast ÷  */
                     s= s n                      /* [↑]  %   is REXX integer division.  */
                     end   /*#hs*/               /* [↑]  append  N  to the sequence list*/
           return s                              /*return the  S  string to the invoker.*/

output   when using the default inputs:

27  has a hailstone sequence of  112
    and starts with:  27 82 41 124  ∙∙∙
    and  ends  with:  ∙∙∙ 8 4 2 1

(between 1 ──► 99999)  77031  has the longest hailstone sequence:  351

optimized

This version is about   7   times faster than the previous (unoptimized) version.

It makes use of:

•   previously calculated Collatz sequences (memoization)
•   a faster method of determining if an integer is even

/*REXX program tests a  number  and also a  range for  hailstone  (Collatz)  sequences. */
!.=0;     !.0=1;  !.2=1;  !.4=1;  !.6=1;  !.8=1  /*assign even numerals to be  "true".  */
numeric digits 20;  @.= 0                        /*handle big numbers; initialize array.*/
parse arg x y z .;  !.h= y                       /*get optional arguments from the C.L. */
if x=='' | x==","   then x=     27               /*No  1st  argument?  Then use default.*/
if y=='' | y==","   then y= 100000 - 1           /* "  2nd      "        "   "     "    */
if z=='' | z==","   then z=     12               /*head/tail number?     "   "     "    */
hm= max(y, 500000)                               /*use memoization (maximum num for  @.)*/
$= hailstone(x)     /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 1▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say  x   ' has a hailstone sequence of '       words($)
say      '    and starts with: '               subword($, 1, z)    " ∙∙∙"
say      '    and  ends  with:  ∙∙∙'           subword($, max(z+1, words($)-z+1))
if y==0  then exit  /*▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒task 2▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒*/
say
w= 0;        do j=1  for y;  $= hailstone(j)     /*traipse through the range of numbers.*/
             #hs= words($)                       /*find the length of the hailstone seq.*/
             if #hs<=w  then iterate             /*Not big enough?  Then keep traipsing.*/
             bigJ= j;   w= #hs                   /*remember what # has biggest hailstone*/
             end   /*j*/
say '(between 1 ──►'   y") "      bigJ     ' has the longest hailstone sequence: '   w
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
hailstone: procedure expose @. !. hm;  parse arg n 1 s 1 o,@.1  /*N,S,O: are the 1st arg*/
                    do  while @.n==0             /*loop while the residual is unknown.  */
                    parse var  n  ''  -1  L      /*extract the last decimal digit of  N.*/
                    if !.L  then n= n % 2        /*N is even?   Then calculate  fast ÷  */
                            else n= n * 3  +  1  /*"  " odd ?     "      "      3*n + 1 */
                    s= s n                       /* [↑]  %: is the REXX integer division*/
                    end   /*while*/              /* [↑]  append  N  to the sequence list*/
           s= s @.n                              /*append the number to a sequence list.*/
           @.o= subword(s, 2);   parse var s _ r /*use memoization for this hailstone #.*/
              do  while r\=='';  parse var r _ r /*obtain the next  hailstone sequence. */
              if @._\==0  then leave             /*Was number already found?  Return  S.*/
              if _>hm     then iterate           /*Is  number  out of range?  Ignore it.*/
              @._= r                             /*assign subsequence number to array.  */
              end   /*while*/;         return s

output   when using the default inputs:

27  has a hailstone sequence of  112
    and starts with:  27 82 41 124 62 31 94 47 142 71 214 107  ∙∙∙
    and  ends  with:  ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1

(between 1 ──► 99999)  77031  has the longest hailstone sequence:  351

output   when using the inputs:     ,   1000000

27  has a hailstone sequence of  112
    and starts with:  27 82 41 124 62 31 94 47 142 71 214 107  ∙∙∙
    and  ends  with:  ∙∙∙ 53 160 80 40 20 10 5 16 8 4 2 1

(between 1 ──► 1000000)  837799  has the longest hailstone sequence:  525

Ring

size = 27
aList = []
hailstone(size)

func hailstone n 
     add(aList,n)
     while n != 1 
           if n % 2 = 0  n = n / 2
           else n = 3 * n + 1 ok  
           add(aList, n)              
     end   
     see "first 4 elements : "     
     for i = 1 to 4
         see "" + aList[i]  + " "
     next
     see nl
     see "last 4 elements : "
     for i = len(aList) - 3 to len(aList)
         see "" + aList[i] + " "
     next

RPL

HP-28 emulator's timedog preventing from longlasting code execution, the third item of the task is achieved by calculating the 1-100,000 sequence in increments of 5,000 numbers, with a derivative of the generator that only provides the length of the sequence to increase execution speed.

Works with: Halcyon Calc version 4.2.7

Code	Comments
≪ DUP 1 →LIST SWAP WHILE DUP 1 ≠ REPEAT IF DUP 2 MOD THEN 3 * 1 + ELSE 2 / END SWAP OVER + SWAP END DROP ≫ 'SYRAQ' STO ≪ 1 SWAP WHILE DUP 1 ≠ REPEAT IF DUP 2 MOD THEN 3 * 1 + ELSE 2 / END SWAP 1 + SWAP END DROP ≫ 'SYRAF' STO ≪ SyMax SYRAF SWAP DUP 5000 + DUP 4 ROLLD FOR n IF DUP n SYRAF < THEN DROP n SYRAF n 'SyMax' STO END NEXT DROP ≫ 'SYR5K' STO	( n -- { sequence } ) initialize sequence with n Calculate next element Store it into sequence Forget n ( n -- sequence_length ) Initialize counter Calculate next element Increment counter Forget n ( n -- n+10000 ) Initialize loop If sequence length greater than previous ones... ... memorize n

The following instructions deliver what is required:

27 SYRAQ SIZE
1 SYR5K … SYR5K [20 times] DROP SyMax RCL

Output:

2: 127
1: 77031

Ruby

This program uses new methods (Integer#even? and Enumerable#max_by) from Ruby 1.8.7.

Works with: Ruby version 1.8.7

def hailstone n
  seq = [n]
  until n == 1
    n = (n.even?) ? (n / 2) : (3 * n + 1)
    seq << n
  end
  seq
end

puts "for n = 27, show sequence length and first and last 4 elements"
hs27 = hailstone 27
p [hs27.length, hs27[0..3], hs27[-4..-1]]

# find the longest sequence among n less than 100,000
n = (1 ... 100_000).max_by{|n| hailstone(n).length}
puts "#{n} has a hailstone sequence length of #{hailstone(n).length}"
puts "the largest number in that sequence is #{hailstone(n).max}"

Output:

for n = 27, show sequence length and first and last 4 elements
[112, [27, 82, 41, 124], [8, 4, 2, 1]]
77031 has a hailstone sequence length of 351
the largest number in that sequence is 21933016

With shared structure

This version builds some linked lists with shared structure. Hailstone::ListNode is an adaptation of ListNode from Singly-linked list/Element definition#Ruby. When two sequences contain the same value, those two lists share a tail. This avoids recomputing the end of the sequence.

Works with: Ruby version 1.8.7

module Hailstone
  ListNode = Struct.new(:value, :size, :succ) do
    def each
      node = self
      while node
        yield node.value
        node = node.succ
      end
    end
  end
  
  @@sequence = {1 => ListNode[1,1]}
  
  module_function
  
  def sequence(n)
    unless @@sequence[n]
      m, ary = n, []
      until succ = @@sequence[m]
        ary << m
        m = m.even? ? (m / 2) : (3 * m + 1)
      end
      ary.reverse_each do |m|
        @@sequence[m] = succ = ListNode[m, succ.size + 1, succ]
      end
    end
    @@sequence[n]
  end
end

puts "for n = 27, show sequence length and first and last 4 elements"
hs27 = Hailstone.sequence(27).entries
p [hs27.size, hs27[0..3], hs27[-4..-1]]

# find the longest sequence among n less than 100,000
n = (1 ... 100_000).max_by{|n| Hailstone.sequence(n).size}
puts "#{n} has a hailstone sequence length of #{Hailstone.sequence(n).size}"
puts "the largest number in that sequence is #{Hailstone.sequence(n).max}"

output is the same as the above.

Rust

fn hailstone(start : u32) -> Vec<u32> {
    let mut res = Vec::new();
    let mut next = start;

    res.push(start);

    while next != 1  {
        next = if next % 2 == 0 { next/2 } else { 3*next+1 };
        res.push(next);
    }
    res
}

 
fn main() {
    let test_num = 27;
    let test_hailseq = hailstone(test_num);

    println!("For {} number of elements is {} ", test_num, test_hailseq.len());

    let fst_slice = test_hailseq[0..4].iter()
                        .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " });
    let last_slice = test_hailseq[test_hailseq.len()-4..].iter()
                        .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " });
    
    println!("  hailstone starting with {} ending with {} ", fst_slice, last_slice);

    let max_range = 100000;
    let mut max_len = 0;
    let mut max_seed = 0;
    for i_seed in 1..max_range {
        let i_len = hailstone(i_seed).len();

        if i_len > max_len {
            max_len = i_len;
            max_seed = i_seed;
        }
    }
    println!("Longest sequence is {} element long for seed {}", max_len, max_seed);
}

Output:

For 27 number of elements is 112 
  hailstone starting with 27, 82, 41, 124,  ending with 8, 4, 2, 1,  
Longest sequence is 351 element long for seed 77031

S-lang

% lst=1, return list of elements; lst=0 just return length
define hailstone(n, lst)
{
  variable l;
  if (lst) l = {n};
  else l = 1;

  while (n > 1) {
    if (n mod 2)
      n = 3 * n + 1;
    else
      n /= 2;
    if (lst)
      list_append(l, n);
    else
      l++;
    % if (prn) () = printf("%d, ", n);
  }
  % if (prn) () = printf("\n");
  return l;
}

variable har = list_to_array(hailstone(27, 1)), more = 0;
() = printf("Hailstone(27) has %d elements starting with:\n\t", length(har));

foreach $1 (har[[0:3]])
  () = printf("%d, ", $1);

() = printf("\nand ending with:\n\t");
foreach $1 (har[[length(har)-4:]]) {
  if (more) () = printf(", ");
  more = printf("%d", $1);
}

() = printf("\ncalculating...\r");
variable longest, longlen = 0, h;
_for $1 (2, 99999, 1) {
  $2 = hailstone($1, 0);
  if ($2 > longlen) {
    longest = $1;
    longlen = $2;
    () = printf("longest sequence started w/%d and had %d elements  \r", longest, longlen);
  }
}
() = printf("\n");

Output:

Hailstone(27) has 112 elements starting with:
        27, 82, 41, 124,
and ending with:
        8, 4, 2, 1
longest sequence started w/77031 and had 351 elements

SAS

* Create a routine to generate the hailstone sequence for one number;
%macro gen_seq(n);
   data hailstone;
      array hs_seq(100000);
      n=&n;
      do until (n=1);
         seq_size + 1;
         hs_seq(seq_size) = n;
         if mod(n,2)=0 then n=n/2;
         else n=(3*n)+1; 
      end;
	  seq_size + 1;
      hs_seq(seq_size)=n;
	  call symputx('seq_length',seq_size);
   run;

   proc sql;
      title "First and last elements of Hailstone Sequence for number &n";
	  select seq_size as sequence_length, hs_seq1, hs_seq2, hs_seq3, hs_seq4
			%do i=&seq_length-3 %to &seq_length;
				, hs_seq&i
			%end; 
		from hailstone;
	quit;
%mend;

* Use the routine to output the first and last four numbers in the sequence for 27;
%gen_seq(27);

* Show the number less than 100,000 which has the longest hailstone sequence, and what that length is ;
%macro longest_hailstone(start_num, end_num);
	data hailstone_analysis;
	  do start=&start_num to &end_num;
	    n=start;
		length_of_sequence=1;
		do while (n>1);
		  length_of_sequence+1;
		  if mod(n,2)=0 then n=n/2;
		  else n=(3*n) + 1;
		end;
		output;
	  end;
	run;

	proc sort data=hailstone_analysis;
	  by descending length_of_sequence;
	run;

	proc print data=hailstone_analysis (obs=1) noobs;
	  title "Number from &start_num to &end_num with longest Hailstone sequence";
	  var start length_of_sequence;
	run;
%mend;
%longest_hailstone(1,99999);

Output:

                   First and last elements of Hailstone Sequence for number 27
    sequence_
       length   hs_seq1   hs_seq2   hs_seq3   hs_seq4  hs_seq109  hs_seq110  hs_seq111  hs_seq112
-------------------------------------------------------------------------------------------------
          112        27        82        41       124          8          4          2          1

                      Number from 1 to 99999 with longest Hailstone sequence
                                            length_of_
                                    start     sequence
                                    77031        351

S-BASIC

comment
  Compute and display "hailstone" (i.e., Collatz) sequence
  for a given number and find the longest sequence in the
  range permitted by S-BASIC's 16-bit integer data type.
end

$lines

$constant false = 0
$constant true = FFFFH

rem - compute p mod q 
function mod(p, q = integer) = integer
end = p - q * (p/q)

comment
  Compute, and optionally display, hailstone sequence for n.
  Return length of sequence or zero on overflow
end
function hailstone(n, display = integer) = integer
  var length = integer
  length = 1
  while (n <> 1) and (n > 0) do
    begin
      if display then print using "#####   ", n;
      if mod(n,2) = 0 then
        n = n / 2
      else
        n = (n * 3) + 1
     length = length + 1
    end
  if display then print using "#####   ", n
  rem - return 0 on overflow
  if n < 0 then length = 0
end = length

var n, limit, slen, longest, n_longest = integer

input "Display hailstone sequence for what number"; n
slen = hailstone(n, true)
print "Sequence length = "; slen

rem - find longest sequence before overflow
n = 2
longest = 1
slen = 1
limit = 1000;
print "Searching for longest sequence up to N =", limit," ..."
while (n < limit) and (slen <> 0) do
  begin
    slen = hailstone(n, false)
    if slen > longest then
      begin
        longest = slen
        n_longest = n
      end
    n = n + 1
  end
if slen = 0 then print "Search terminated with overflow at";n-1
print "Maximum sequence length =";longest;" for N =";n_longest

end

Output:

Display hailstone sequence for what number? 27
   27     82     41    124     62     31     94     47    142     71
  214    107    322    161    484    242    121    364    182     91
  274    137    412    206    103    310    155    466    233    700
  350    175    526    263    790    395   1186    593   1780    890
  445   1336    668    334    167    502    251    754    377   1132
  566    283    850    425   1276    638    319    958    479   1438
  719   2158   1079   3238   1619   4858   2429   7288   3644   1822
  911   2734   1367   4102   2051   6154   3077   9232   4616   2308
 1154    577   1732    866    433   1300    650    325    976    488
  244    122     61    184     92     46     23     70     35    106
   53    160     80     40     20     10      5     16      8      4
    2      1
Sequence length = 112
Searching for longest sequence up to N = 1000 ...
Search terminated with overflow at 447
Maximum sequence length = 144 for N = 327

Scala

Library: Scala

Works with: Scala version 2.10.2

object HailstoneSequence extends App {
  def hailstone(n: Int): Stream[Int] =
    n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1))

  val nr = args.headOption.map(_.toInt).getOrElse(27)
  val collatz = hailstone(nr)
  println(s"Use the routine to show that the hailstone sequence for the number: $nr.")
  println(collatz.toList)
  println(s"It has ${collatz.length} elements.")
  println
  println(
    "Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.")
  val (n, len) = (1 until 100000).map(n => (n, hailstone(n).length)).maxBy(_._2)
  println(s"Longest hailstone sequence length= $len occurring with number $n.")
}

Output:

Use the routine to show that the hailstone sequence for the number: 27.
List(27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1)
It has 112 elements.

Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.
Longest hailstone sequence length= 351 occurring with number 77031.

Scheme

(define (collatz n)
(if (= n 1) '(1)
(cons n (collatz (if (even? n) (/ n 2) (+ 1 (* 3 n)))))))

(define (collatz-length n)
(let aux ((n n) (r 1)) (if (= n 1) r
(aux (if (even? n) (/ n 2) (+ 1 (* 3 n))) (+ r 1)))))

(define (collatz-max a b)
(let aux ((i a) (j 0) (k 0))
(if (> i b) (list j k)
(let ((h (collatz-length i)))
(if (> h k) (aux (+ i 1) i h) (aux (+ i 1) j k))))))

(collatz 27)
; (27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182
; 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395
; 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283
; 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429
; 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154
; 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35
; 106 53 160 80 40 20 10 5 16 8 4 2 1)

(collatz-length 27)
; 112

(collatz-max 1 100000)
; (77031 351)

Scilab

Translation of: MATLAB

function x=hailstone(n)
    // iterative definition
    // usage: global verbose; verbose=%T; hailstone(27)
    global verbose
    x=0; loop=%T
    while(loop) 
        x=x+1
        if verbose then
            printf('%i ',n)
        end
        if n==1 then
            loop=%F
        elseif modulo(n,2)==1 then
            n=3*n+1
        else 
            n=n/2
        end
    end
endfunction

global verbose;
verbose=1;
N=hailstone(27);
printf('\n\n%i\n',N);
  
global verbose;
verbose=0;
N=100000;
M=zeros(N,1);
for k=1:N
  M(k)=hailstone(k);
end;
[maxLength,n]=max(M)

Output:

27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 
112
 n  =      77031.  
 maxLength  =      351.  

Seed7

$ include "seed7_05.s7i";

const func array integer: hailstone (in var integer: n) is func
  result
    var array integer: hSequence is 0 times 0;
  begin
    while n <> 1 do
      hSequence &:= n;
      if odd(n) then
        n := 3 * n + 1;
      else
        n := n div 2;
      end if;
    end while;
    hSequence &:= n;
  end func;

const func integer: hailstoneSequenceLength (in var integer: n) is func
  result
    var integer: sequenceLength is 1;
  begin
    while n <> 1 do
      incr(sequenceLength);
      if odd(n) then
        n := 3 * n + 1;
      else
        n := n div 2;
      end if;
    end while;
  end func;

const proc: main is func
  local
    var integer: number is 0;
    var integer: length is 0;
    var integer: maxLength is 0;
    var integer: numberOfMaxLength is 0;
    var array integer: h27 is 0 times 0;
  begin
    for number range 1 to 99999 do
      length := hailstoneSequenceLength(number);
      if length > maxLength then
        maxLength := length;
        numberOfMaxLength := number;
      end if;
    end for;
    h27 := hailstone(27);
    writeln("hailstone(27):");
    for number range 1 to 4 do
      write(h27[number] <& ", ");
    end for;
    write("....");
    for number range length(h27) -3 to length(h27) do
      write(", " <& h27[number]);
    end for;
    writeln("  length=" <& length(h27));
    writeln("Maximum length " <& maxLength <& " at number=" <& numberOfMaxLength);
  end func;

Output:

hailstone(27):
27, 82, 41, 124, ...., 8, 4, 2, 1  length=112
Maximum length 351 at number=77031

SETL

program hailstone_sequence;
    hail27 := hailstone(27);
    print("The hailstone sequence for the number 27 has", #hail27, "elements,");
    print("starting with", hail27(..4), "and ending with", hail27(#hail27-3..));

    sizes := [#hailstone(n) : n in [1..99999]];
    maxsize := max/sizes;
    maxelem := [n : n in [1..#sizes] | sizes(n) = maxsize](1);

    print("The number < 100,000 with the longest hailstone sequence is",maxelem);
    print("The length of its sequence is",sizes(maxelem));

    proc hailstone(n);
        seq := [];
        loop doing seq with:= n; while n/=1 do
            if even n then
                n div:= 2;
            else
                n := 3*n + 1;
            end if;
        end loop;
        return seq;
    end proc;
end program;

Output:

The hailstone sequence for the number 27 has 112 elements,
starting with [27 82 41 124] and ending with [8 4 2 1]
The number < 100,000 with the longest hailstone sequence is 77031
The length of its sequence is 351

Sidef

func hailstone (n) {
    var sequence = [n]
    while (n > 1) {
        sequence << (
            n.is_even ? n.div!(2)
                      : n.mul!(3).add!(1)
        )
    }
    return(sequence)
}
 
# The hailstone sequence for the number 27
var arr = hailstone(var nr = 27)
say "#{nr}: #{arr.first(4)} ... #{arr.last(4)} (#{arr.len})"
 
# The longest hailstone sequence for a number less than 100,000
var h = [0, 0]
for i (1 .. 99_999) {
    (var l = hailstone(i).len) > h[1] && (
        h = [i, l]
    )
}
 
printf("%d: (%d)\n", h...)

Smalltalk

Works with: GNU Smalltalk

Object subclass: Sequences [
  Sequences class >> hailstone: n [
      |seq| 
      seq := OrderedCollection new.
      seq add: n.
      (n = 1) ifTrue: [ ^seq ].
      (n even) ifTrue: [ seq addAll: (Sequences hailstone: (n / 2)) ]
               ifFalse: [ seq addAll: (Sequences hailstone: ( (3*n) + 1 ) ) ].
      ^seq.
  ]

  Sequences class >> hailstoneCount: n [
      ^ (Sequences hailstoneCount: n num: 1)
  ]

  "this 'version' avoids storing the sequence, it just counts
   its length - no memoization anyway"
  Sequences class >> hailstoneCount: n num: m [
      (n = 1) ifTrue: [ ^m ].
      (n even) ifTrue: [ ^ Sequences hailstoneCount: (n / 2) num: (m + 1) ]
               ifFalse: [ ^ Sequences hailstoneCount: ( (3*n) + 1) num: (m + 1) ].
  ]
].

|r|
r := Sequences hailstone: 27.  "hailstone 'from' 27"
(r size) displayNl.            "its length"

"test 'head' ..."
( (r first: 4) = #( 27  82  41  124 ) asOrderedCollection ) displayNl.

"... and 'tail'"
( ( (r last: 4 ) ) = #( 8 4 2 1 ) asOrderedCollection) displayNl.

|longest|
longest := OrderedCollection from: #( 1 1 ).
2 to: 100000 do: [ :c |
  |l|
  l := Sequences hailstoneCount: c.
  (l > (longest at: 2) ) ifTrue: [ longest replaceFrom: 1 to: 2 with: { c . l }  ].
].

('Sequence generator %1, sequence length %2' % { (longest at: 1) . (longest at: 2) })
   displayNl.

SNUSP

   /@+@@@+++# 27
   |    halve odd   /===count<<\    /recurse\    #/?\ zero
$>@/===!/===-?\==>?!/-<+++\    \!/=!\@\>?!\@/<@\.!\-/
 /+<-\!>\?-<+>/++++<\?>+++/*6+4  |    |   \=/  \=itoa=@@@+@+++++#
 \=>?/<=!=\   |                  |    !     /+ !/+ !/+ !/+   \    mod10
        |//!==/========\         |    /<+> -\!?-\!?-\!?-\!?-\!
 /=>?\<=/\<+>!\->+>+<<?/>>=print@/\ln \?!\-?!\-?!\-?!\-?!\-?/\    div10
 \+<-/!<     ----------.++++++++++/      #  +/! +/! +/! +/! +/

Swift

func hailstone(var n:Int) -> [Int] {

    var arr = [n]
    
    while n != 1 {

        if n % 2 == 0 {
            n /= 2
        } else {
            n = (3 * n) + 1
        }    

        arr.append(n)
    }
    
    return arr
}

let n = hailstone(27)

println("hailstone(27): \(n[0...3]) ... \(n[n.count-4...n.count-1]) for a count of \(n.count).")

var longest = (n: 1, len: 1)

for i in 1...100_000 {

    let new = hailstone(i)
    
    if new.count > longest.len {
        longest = (i, new.count)
    }
}

println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")

Output:

hailstone(27): [27, 82, 41, 124] ... [8, 4, 2, 1] for a count of 112
Longest sequence for numbers under 100,000 is with 77031. Which has 351 items.

Tcl

The core looping structure is an example of an n-plus-one-half loop, except the loop is officially infinite here.

proc hailstone n {
    while 1 {
	lappend seq $n
	if {$n == 1} {return $seq}
	set n [expr {$n & 1 ? $n*3+1 : $n/2}]
    }
}

set h27 [hailstone 27]
puts "h27 len=[llength $h27]"
puts "head4 = [lrange $h27 0 3]"
puts "tail4 = [lrange $h27 end-3 end]"

set maxlen [set max 0]
for {set i 1} {$i<100000} {incr i} {
    set l [llength [hailstone $i]]
    if {$l>$maxlen} {set maxlen $l;set max $i}
}
puts "max is $max, with length $maxlen"

Output:

h27 len=112
head4 = 27 82 41 124
tail4 = 8 4 2 1
max is 77031, with length 351

TI-83 BASIC

Task 1

prompt N
N→M: 0→X: 1→L
While L=1
X+1→X
Disp M
If M=1 
Then: 0→L
Else
If remainder(M,2)=1
Then: 3*M+1→M
Else: M/2→M
End
End
End
{N,X}

Output:

        10
         5
        16
         8
         4
         2
         1
{27,112}

Task 2

As the calculator is quite slow, so the output is for N=200

prompt N
0→A:0→B
for(I,1,N)
I→M: 0→X: 1→L
While L=1
X+1→X
If M=1
Then: 0→L
Else
If remainder(M,2)=1
Then: 3*M+1→M
Else: M/2→M
End
End
End
If X>B: Then
I→A:X→B
End
Disp {I,X}
End
{A,B}

Output:

{171,125}

Transd

Translation of: Python

#lang transd

MainModule: {
    hailstone: (λ n Int()
        (with seq Vector<Int>([n])
            (while (> n 1)
                (= n (if (mod n 2) (+ (* 3 n) 1)
                      else (/ n 2)))
                (append seq n)
            )
            (ret seq)
        )
    ),

    _start: (λ 
        (with h (hailstone 27) l 0 n 0 t 0
            (lout "Length of (27): " (size h))
            (lout "First 4 of (27): " Range(in: h 0 4))
            (lout "Last 4 of (27): " Range(in: h -4 -0))
        (for i in Range(100000) do
            (= t (size (hailstone (to-Int i)))) (if (> t l) (= l t) (= n i))
        )
        (lout "For n < 100.000 the max. sequence length is " l " for " n)
        )
    )
}

Output:

Length of (27): 112
First 4 of (27): [27, 82, 41, 124]
Last 4 of (27): [8, 4, 2, 1]
For n < 100.000 the max. sequence length is 351 for 77031

TXR

@(do (defun hailstone (n)
       (cons n 
             (gen (not (eq n 1))
                  (set n (if (evenp n)
                           (trunc n 2)
                           (+ (* 3 n) 1)))))))
@(next :list @(mapcar* (fun tostring) (hailstone 27)))
27
82
41
124
@(skip)
8
4
2
1
@(eof)
@(do (let ((max 0) maxi)
       (each* ((i (range 1 99999))
               (h (mapcar* (fun hailstone) i))
               (len (mapcar* (fun length) h)))
         (if (> len max)
           (progn
             (set max len)
             (set maxi i))))
       (format t "longest sequence is ~a for n = ~a\n" max maxi)))

$ txr -l hailstone.txr
longest sequence is 351 for n = 77031

uBasic/4tH

Translation of: FreeBASIC

' ------=< MAIN >=------

m = 0
Proc _hailstone_print(27)
Print

For x = 1 To 10000
    n = Func(_hailstone(x))
    If n > m Then
        t = x
        m = n
    EndIf
Next

Print  "The longest sequence is for "; t; ", it has a sequence length of "; m

End

_hailstone_print Param (1)
    ' print the number and sequence

    Local (1)
    b@ = 1

    Print "sequence for number "; a@
    Print Using "________"; a@;   'starting number

    Do While a@ # 1
        If (a@ % 2 ) = 1 Then
            a@ = a@ * 3 + 1   ' n * 3 + 1
        Else
            a@ = a@ / 2       ' n / 2
        EndIf

        b@ = b@ + 1
        Print Using "________"; a@;

        If (b@ % 10) = 0 Then Print
    Loop

    Print : Print
    Print "sequence length = "; b@
    Print

    For b@ = 0 To 79
      Print "-";
    Next

    Print
Return
 
_hailstone Param (1)
    ' normal version
    ' only counts the sequence

    Local (1)
    b@ = 1

    Do While a@ # 1
        If (a@ % 2) = 1 Then
            a@ = a@ * 3 + 1  ' n * 3 + 1
        Else
            a@ = a@ / 2      ' divide number by 2
        EndIf

        b@ = b@ + 1
    Loop

Return (b@)

uBasic is an interpreted language. Doing a sequence up to 100,000 would take over an hour, so we did up to 10,000 here.

Output:

sequence for number 27
     27      82      41     124      62      31      94      47     142      71
    214     107     322     161     484     242     121     364     182      91
    274     137     412     206     103     310     155     466     233     700
    350     175     526     263     790     395    1186     593    1780     890
    445    1336     668     334     167     502     251     754     377    1132
    566     283     850     425    1276     638     319     958     479    1438
    719    2158    1079    3238    1619    4858    2429    7288    3644    1822
    911    2734    1367    4102    2051    6154    3077    9232    4616    2308
   1154     577    1732     866     433    1300     650     325     976     488
    244     122      61     184      92      46      23      70      35     106
     53     160      80      40      20      10       5      16       8       4
      2       1

sequence length = 112
--------------------------------------------------------------------------------
The longest sequence is for 6171, it has a sequence length of 262

UNIX Shell

The best way is to use a shell with built-in arrays and arithmetic, such as Bash.

Works with: Bash

#!/bin/bash
# seq is the array genereated by hailstone
# index is used for seq
declare -a seq
declare -i index

# Create a routine to generate the hailstone sequence for a number
hailstone () {
  unset seq index
  seq[$((index++))]=$((n=$1))
  while [ $n -ne 1 ]; do
    [ $((n % 2)) -eq 1 ] && ((n=n*3+1)) || ((n=n/2))
    seq[$((index++))]=$n
  done
}

# Use the routine to show that the hailstone sequence for the number 27
# has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1
i=27
hailstone $i
echo "$i: ${#seq[@]}"
echo "${seq[@]:0:4} ... ${seq[@]:(-4):4}"

# Show the number less than 100,000 which has the longest hailstone
# sequence together with that sequences length.
# (But don't show the actual sequence)!
max=0
maxlen=0
for ((i=1;i<100000;i++)); do
  hailstone $i
  if [ $((len=${#seq[@]})) -gt $maxlen ]; then
    max=$i
    maxlen=$len
  fi
done

echo "${max} has a hailstone sequence length of ${maxlen}"

Output:

27: 112
27 82 41 124 ... 8 4 2 1
77031 has a hailstone sequence of 351

Bourne Shell

This script follows tradition for the Bourne Shell; its hailstone() function writes the sequence to standard output, so the shell can capture or pipe this output. This script is very slow because it forks many processes. Each `command substitution` forks a subshell, and each expr(1) command forks a process.

• Therefore, this script only examines sequences from 1 to 1000, not 100000. A fast computer might run this script in 45 to 120 seconds, using most time to run system calls in kernel mode. If the script went to 100000, it would need several hours.

Works with: Bourne Shell

# Outputs a hailstone sequence from $1, with one element per line.
# Clobbers $n.
hailstone() {
	n=`expr "$1" + 0`
	eval "test $? -lt 2 || return $?"  # $n must be integer.

	echo $n
	while test $n -ne 1; do
		if expr $n % 2 >/dev/null; then
			n=`expr 3 \* $n + 1`
		else
			n=`expr $n / 2`
		fi
		echo $n
	done
}

set -- `hailstone 27`
echo "Hailstone sequence from 27 has $# elements:"
first="$1, $2, $3, $4"
shift `expr $# - 4`
echo "  $first, ..., $1, $2, $3, $4"

i=1 max=0 maxlen=0
while test $i -lt 1000; do
	len=`hailstone $i | wc -l | tr -d ' '`
	test $len -gt $maxlen && max=$i maxlen=$len
	i=`expr $i + 1`
done
echo "Hailstone sequence from $max has $maxlen elements."

C Shell

This script is several times faster than the previous Bourne Shell script, because it uses C Shell expressions, not the expr(1) command. This script is slow, but it can reach 100000, and a fast computer might run it in less than 15 minutes.

# Outputs a hailstone sequence from !:1, with one element per line.
# Clobbers $n.
alias hailstone eval \''@ n = \!:1:q		\\
	echo $n					\\
	while ( $n != 1 )			\\
		if ( $n % 2 ) then		\\
			@ n = 3 * $n + 1	\\
		else				\\
			@ n /= 2		\\
		endif				\\
		echo $n				\\
	end					\\
'\'

set sequence=(`hailstone 27`)
echo "Hailstone sequence from 27 has $#sequence elements:"
@ i = $#sequence - 3
echo "  $sequence[1-4] ... $sequence[$i-]"

# hailstone-length $i
#   acts like
# @ len = `hailstone $i | wc -l | tr -d ' '`
#   but without forking any subshells.
alias hailstone-length eval \''@ n = \!:1:q	\\
	@ len = 1				\\
	while ( $n != 1 )			\\
		if ( $n % 2 ) then		\\
			@ n = 3 * $n + 1	\\
		else				\\
			@ n /= 2		\\
		endif				\\
		@ len += 1			\\
	end					\\
'\'

@ i = 1
@ max = 0
@ maxlen = 0
while ($i < 100000)
	# XXX - I must run hailstone-length in a subshell, because my
	# C Shell has a bug when it runs hailstone-length inside this
	# while ($i < 1000) loop: it forgets about this loop, and
	# reports an error <<end: Not in while/foreach.>>
	@ len = `hailstone-length $i; echo $len`
	if ($len > $maxlen) then
		@ max = $i
		@ maxlen = $len
	endif
	@ i += 1
end
echo "Hailstone sequence from $max has $maxlen elements."

Output:

$ csh -f hailstone.csh
Hailstone sequence from 27 has 112 elements:
  27 82 41 124 ... 8 4 2 1
Hailstone sequence from 77031 has 351 elements.

Ursa

Implementation

hailstone.u

import "math"

def hailstone (int n)
	decl int<> seq
	while (> n 1)
		append n seq
		if (= (mod n 2) 0)
			set n (floor (/ n 2))
		else
			set n (int (+ (* 3 n) 1))
		end if
	end while
	append n seq
	return seq
end hailstone

Usage

Output:

> import "hailstone.u"
> out (hailstone 27) endl console
class java.lang.Integer<27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1>
> out (size (hailstone 27)) endl console
112
> decl int i max maxLoc
> for (set i 1) (< i 100000) (inc i) 
..	decl int result
..	set result (size (hailstone i))
..
..	if (> result max)
..		set max result
..		set maxLoc i
..	end if
..end for
> out "hailstone(" maxLoc ")= " max endl console
hailstone(77031)= 351
> _

Ursala

#import std
#import nat

hail = @iNC ~&h~=1->x ^C\~& @h ~&h?\~&t successor+ sum@iNiCX

#show+

main =

<
   ^T(@ixX take/$4; %nLP~~lrxPX; ^|TL/~& :/'...',' has length '--@h+ %nP+ length) hail 27,
   ^|TL(~&,:/' has sequence length ') %nP~~ nleq$^&r ^(~&,length+ hail)* nrange/1 100000>

The hail function computes the sequence as follows.

• Given a number as an argument, @iNC makes a list containing only that number before passing it to the rest of the function. The i in the expression stands for the identity function, N for the constant null function, and C for the cons operator.
• The iteration combinator (->) is used with a predicate of ~&h~=l which tests the condition that the head (~&h) of its argument is not equal (~=) to 1. Iteration of the rest of the function continues while this predicate holds.
• The x suffix says to return the reversal of the list after the iteration finishes.
• The function being iterated builds a list using the cons operator (^C) with the identity function (~&) of the argument for the tail, and the result of the rest of the line for the head.
• The @h operator says that the function following will be applied to the head of the list.
• The conditional operator (?) has the head function (~&h) as its predicate, which tests whether the head of its argument is non-null.
• In this case, the argument is a natural number, but naturals are represented as lists of booleans, so taking the head of a number is the same as testing the least significant bit.
• If the condition is not met, the number has a 0 least significant bit, and therefore is even. In this case, the conditional predicate calls for taking its tail (~&t), effectively dividing it by 2 using a bit shift.
• If the condition is met, the number is odd, so the rest of the function computes the successor of the number multiplied by three.
• Rather than multiplying the hard way, the function sum@iNiCX computes the sum of the pair (X) of numbers given by the identity function (i) of the argument, and the doubling of the argument (NiC), also obtained by a bit shift, with a zero bit (N) consed (C) with the identity (i).

Most of the main expression pertains to less interesting printing and formatting, but the part that searches for the longest sequence in the range is nleq$^&r ^(~&,length+ hail)* nrange/1 100000.

• The expression nrange/1 100000 evaluates to the list of the first 100000 positive integers.
• The map operator (*) causes a list to be made of the results of its operand applied to each number.
• The operand to the map operator, applied to an individual number in the list, constructs a pair (^) with the identity function (~&) of the number on the left, and the length of the hail sequence on the right.
• The maximizing operator ($^) with respect to the natural less or equal relation (nleq) applied to the right sides (&r) of its pair of arguments extracts the number with the maximum length sequence.

Output:

<27,82,41,124>...<8,4,2,1> has length 112
77031 has sequence length 351

Ursalang

let hailstone = fn(n_) {
  var n = n_
  let steps = [n]
  loop {
    if n < 2 { break steps }
    n := (if n % 2 == 0 { n / 2 } else { n * 3 + 1 })
    steps.push(n)
  }
}

let twentySeven = hailstone(27)
print("The Hailstone sequence for 27 starts with:")
print(twentySeven.slice(0, 4))
print("and ends with:")
print(twentySeven.slice(-4))
print("and its length is:")
print(twentySeven.len())

print("Checking sequences up to 100,000")
var longest = 0
var longestLen = 0
for i in range(100000) {
  let len = hailstone(i).len()
  if len > longestLen {
    longest := i
    longestLen := len
  }
  if i % 10000 == 0 {print(i)}
}
print("The longest sequence under 100,000 is for:")
print(longest)
print("and has length:")
print(longestLen)

Output:

The Hailstone sequence for 27 starts with:
[ 27, 82, 41, 124 ]
and ends with:
[ 8, 4, 2, 1 ]
and its length is:
112
Checking sequences up to 100,000
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
The longest sequence under 100,000 is for:
77031
and has length:
351

VBA

Translation of: Phix

Private Function hailstone(ByVal n As Long) As Collection
    Dim s As New Collection
    s.Add CStr(n), CStr(n)
    i = 0
    Do While n <> 1
        If n Mod 2 = 0 Then
            n = n / 2
        Else
            n = 3 * n + 1
        End If
        s.Add CStr(n), CStr(n)
    Loop
    Set hailstone = s
End Function
 
Private Function hailstone_count(ByVal n As Long)
    Dim count As Long: count = 1
    Do While n <> 1
        If n Mod 2 = 0 Then
            n = n / 2
        Else
            n = 3 * n + 1
        End If
        count = count + 1
    Loop
    hailstone_count = count
End Function
 
Public Sub rosetta()
    Dim s As Collection, i As Long
    Set s = hailstone(27)
    Dim ls As Integer: ls = s.count
    Debug.Print "hailstone(27) = ";
    For i = 1 To 4
        Debug.Print s(i); ", ";
    Next i
    Debug.Print "... ";
    For i = s.count - 4 To s.count - 1
        Debug.Print s(i); ", ";
    Next i
    Debug.Print s(s.count)
    Debug.Print "length ="; ls
    Dim hmax As Long: hmax = 1
    Dim imax As Long: imax = 1
    Dim count As Integer
    For i = 2 To 100000# - 1
        count = hailstone_count(i)
        If count > hmax Then
            hmax = count
            imax = i
        End If
    Next i
    Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub

Output:

hailstone(27) = 27, 82, 41, 124, ... 16, 8, 4, 2, 1
length = 112 

The longest hailstone sequence under 100,000 is 77031 with 351 elements.

VBScript

'function arguments: "num" is the number to sequence and "return" is the value to return - "s" for the sequence or 
'"e" for the number elements.
Function hailstone_sequence(num,return)
    n = num
	sequence = num
	elements = 1
	Do Until n = 1
		If n Mod 2 = 0 Then
			n = n / 2
		Else
			n = (3 * n) + 1
		End If
		sequence = sequence & " " & n
		elements = elements + 1	
	Loop
	Select Case return
		Case "s"
			hailstone_sequence = sequence
		Case "e"
			hailstone_sequence = elements
	End Select
End Function

'test driving.
'show sequence for 27
WScript.StdOut.WriteLine "Sequence for 27: " & hailstone_sequence(27,"s")
WScript.StdOut.WriteLine "Number of Elements: " & hailstone_sequence(27,"e")
WScript.StdOut.WriteBlankLines(1)
'show the number less than 100k with the longest sequence
count = 1
n_elements = 0
n_longest = ""
Do While count < 100000
	current_n_elements = hailstone_sequence(count,"e")
	If current_n_elements > n_elements Then
		n_elements = current_n_elements
		n_longest = "Number: " & count & " Length: " & n_elements
	End If
	count = count + 1
Loop
WScript.StdOut.WriteLine "Number less than 100k with the longest sequence: "
WScript.StdOut.WriteLine n_longest

Output:

Sequence for 27: 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
Number of Elements: 112

Number less than 100k with the longest sequence: 
Number: 77031 Length: 351

Visual Basic

Translation of: PL/I

Works with: Visual Basic version VB6 Standard

Option Explicit
Dim flag As Boolean ' true to print values
Sub main()
    Dim longest As Long, n As Long
    Dim i As Long, value As Long
    ' Task 1:
    flag = True
    i = 27
    Debug.Print "The hailstone sequence has length of "; i; " is "; hailstones(i)
    ' Task 2:
    flag = False
    longest = 0
    For i = 1 To 99999
        If longest < hailstones(i) Then
            longest = hailstones(i)
            value = i
        End If
    Next i
    Debug.Print value; " has the longest sequence of "; longest
End Sub 'main
Function hailstones(n As Long) As Long
    Dim m As Long, p As Long
    Dim m1 As Long, m2 As Long, m3 As Long, m4 As Long
    If flag Then Debug.Print "The sequence for"; n; "is: ";
    p = 1
    m = n
    If flag Then Debug.Print m;
    While m > 1
        p = p + 1
        If (m Mod 2) = 0 Then
            m = m / 2
        Else
            m = 3 * m + 1
        End If
        If p <= 4 Then If flag Then Debug.Print m;
        m4 = m3
        m3 = m2
        m2 = m1
        m1 = m
    Wend
    If flag Then
        If p <= 4 Then
            Debug.Print
        ElseIf p = 5 Then
            Debug.Print m1
        ElseIf p = 6 Then
            Debug.Print m2; m1
        ElseIf p = 7 Then
            Debug.Print m3; m2; m1
        ElseIf p = 8 Then
            Debug.Print m4; m3; m2; m1
        Else
            Debug.Print "..."; m4; m3; m2; m1
        End If
    End If
    hailstones = p
End Function 'hailstones

Output:

The sequence for 27 is:  27  82  41  124 ... 8  4  2  1 
The hailstone sequence has length of  27  is  112 
 77031  has the longest sequence of  351 

Visual Basic .NET

Works with: Visual Basic .NET version 2005+

Module HailstoneSequence
    Sub Main()
        ' Checking sequence of 27.

        Dim l As List(Of Long) = HailstoneSequence(27)
        Console.WriteLine("27 has {0} elements in sequence:", l.Count())

        For i As Integer = 0 To 3 : Console.Write("{0}, ", l(i)) : Next
        Console.Write("... ")
        For i As Integer = l.Count - 4 To l.Count - 1 : Console.Write(", {0}", l(i)) : Next

        Console.WriteLine()

        ' Finding longest sequence for numbers below 100000.

        Dim max As Integer = 0
        Dim maxCount As Integer = 0

        For i = 1 To 99999
            l = HailstoneSequence(i)
            If l.Count > maxCount Then
                max = i
                maxCount = l.Count
            End If
        Next
        Console.WriteLine("Max elements in sequence for number below 100k: {0} with {1} elements.", max, maxCount)
        Console.ReadLine()
    End Sub

    Private Function HailstoneSequence(ByVal n As Long) As List(Of Long)
        Dim valList As New List(Of Long)()
        valList.Add(n)

        Do Until n = 1
            n = IIf(n Mod 2 = 0, n / 2, (3 * n) + 1)
            valList.Add(n)
        Loop

        Return valList
    End Function

End Module

Output:

27 has 112 elements in sequence:
27, 82, 41, 124, ... , 8, 4, 2, 1
Max elements in sequence for number below 100k: 77031 with 351 elements.

V (Vlang)

Translation of: go

// 1st arg is the number to generate the sequence for.
// 2nd arg is a slice to recycle, to reduce garbage.
fn hs(nn int, recycle []int) []int {
    mut n := nn
    mut s := recycle[..0]
    s << n
    for n > 1 {
        if n&1 == 0 {
            n /= 2
        } else {
            n = 3*n + 1
        }
        s << n
    }
    return s
}
 
fn main() {
    mut seq := hs(27, [])
    println("hs(27): $seq.len elements: [${seq[0]} ${seq[1]} ${seq[2]} ${seq[3]} ... ${seq[seq.len-4]} ${seq[seq.len-3]} ${seq[seq.len-2]} ${seq[seq.len-1]}]")
 
    mut max_n, mut max_len := 0,0
    for n in 1..100000 {
        seq = hs(n, seq)
        if seq.len > max_len {
            max_n = n
            max_len = seq.len
        }
    }
    println("hs($max_n): $max_len elements")
}

Output:

hs(27): 112 elements: [27 82 41 124 ... 8 4 2 1]
hs(77031): 351 elements

Wren

var hailstone = Fn.new { |n|
    if (n < 1) Fiber.abort("Parameter must be a positive integer.")
    var h = [n]
    while (n != 1) { 
        n = (n%2 == 0) ? (n/2).floor : 3*n + 1 
        h.add(n)
    }
    return h
}

var h = hailstone.call(27)
System.print("For the Hailstone sequence starting with n = 27:")
System.print("   Number of elements  = %(h.count)")
System.print("   First four elements = %(h[0..3])")
System.print("   Final four elements = %(h[-4..-1])")

System.print("\nThe Hailstone sequence for n < 100,000 with the longest length is:")
var longest = 0
var longlen = 0
for (n in 1..99999) {
    var h = hailstone.call(n)
    var c = h.count
    if (c > longlen) {
        longest = n
        longlen = c
    }
}
System.print("   Longest = %(longest)")
System.print("   Length  = %(longlen)")

Output:

For the Hailstone sequence starting with n = 27:
   Number of elements  = 112
   First four elements = [27, 82, 41, 124]
   Final four elements = [8, 4, 2, 1]

The Hailstone sequence for n < 100,000 with the longest length is:
   Longest = 77031
   Length  = 351

XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations
int Seq(1000);          \more than enough for longest sequence

func Hailstone(N);      \Return length of Hailstone sequence starting at N
int  N;                 \ also fills Seq array with sequence
int  I;
[I:= 0;
loop [Seq(I):= N;  I:= I+1;
     if N=1 then return I;
     N:= if N&1 then N*3+1 else N/2;
     ];
];

int N, SN, Len, MaxLen;
[Len:= Hailstone(27);
Text(0, "27's Hailstone length = ");  IntOut(0, Len);  CrLf(0);

Text(0, "Sequence = ");
for N:= 0 to 3 do [IntOut(0, Seq(N));  ChOut(0, ^ )];
Text(0, "... ");
for N:= Len-4 to Len-1 do [IntOut(0, Seq(N));  ChOut(0, ^ )];
CrLf(0);

MaxLen:= 0;
for N:= 1 to 100_000-1 do
    [Len:= Hailstone(N);
    if Len > MaxLen then [MaxLen:= Len;  SN:= N];       \save N with max length
    ];
IntOut(0, SN);  Text(0, "'s Hailstone length = ");  IntOut(0, MaxLen);
]

Output:

27's Hailstone length = 112
Sequence = 27 82 41 124 ... 8 4 2 1 
77031's Hailstone length = 351

Z80 Assembly

This task will be split into two parts, in order to fit all the output of each on one Amstrad CPC screen.

Show The Sequence with n=27

Output is in hexadecimal but is otherwise correct.

;;;;;;;;;;;;;;;;;;; HEADER   ;;;;;;;;;;;;;;;;;;;
read "\SrcCPC\winape_macros.asm"
read "\SrcCPC\MemoryMap.asm"
read "\SrcALL\winapeBuildCompat.asm"
;;;;;;;;;;;;;;;;;;; PROGRAM  ;;;;;;;;;;;;;;;;;;;

org &8000
ld de,27
call doHailstone	
;returns length of sequence, and writes each entry in the sequence
;	to RAM

;print the sequence length (in hex)
ld a,h
call ShowHex	
ld a,l
ld (memdump_smc),a
;just to prove I didn't need to know the sequence length at
;	compile time, I'll store the calculated length as the operand
;	of "doMemDump" which normally takes a constant embedded after
;	it as the number of bytes to display.

;      If that doesn't make sense, don't worry. 
;      This has nothing to do with calculating the hailstone sequence, just showing the results.
call ShowHex

call NewLine		;prints CRLF
call NewLine

call doMemDump
memdump_smc:
byte 0			;operand of "doMemDump" (gets overwritten with the sequence length)
word HailstoneBuffer	;operand of "doMemDump"


ret

;;;;;;;;;;;;;;;;;;; LIBRARY  ;;;;;;;;;;;;;;;;;;;
read "\SrcCPC\winape_stringop.asm"
read "\SrcCPC\winape_showhex.asm"


doHailstone:
;you need the proper input for the function "hailstone"
;returns addr. of last element in IX.
call hailstone
ld de,HailstoneBuffer
or a	;clear carry
push ix
pop hl		;returns element count in HL.
sbc hl,de	;subtract the two to get the length of the array.

SRL H
RR L		;divide array size by 2, since each entry is 2 bytes.
INC L
ret nz		;if no carry, don't increment H.
INC H
ret


hailstone:
;input - de = n
ld ix,HailstoneBuffer
ld a,d
or e
ret z	;zero is not allowed.
loop_hailstone:
ld (IX+0),e
ld (IX+1),d
ld a,e
cp 1
jr nz,continue_hailstone
ld a,d
or a
ret z	;if de = 1, stop.
continue_hailstone:
bit 0,e
jr z,DE_IS_EVEN
;de is odd
push de
pop hl		;ld hl,de

SLA E
RL D
add hl,de	;hl = de*3

ld de,1
add hl,de

push hl	
pop de		;ld de,hl

inc ix
inc ix
jr loop_hailstone

DE_IS_EVEN:
SRL D		;A/2
RR E
inc ix
inc ix
jr loop_hailstone


doMemDump:
;show the hailstone sequence to the screen. This is just needed to display the data, if you don't care about that
;you can stop reading here.
pop hl		;get PC
ld b,(hl)	;get byte count
inc hl
ld e,(hl)	;get low byte of start addr.
inc hl
ld d,(hl)	;get high byte of start addr.
inc hl
push hl		;now when we return we'll skip the data block.
ex de,hl


call NewLine

;we'll dump 8 words per line.

ld c,8
loop_doMemDump:
inc hl
ld a,(hl)
call ShowHex
dec hl
ld a,(hl)
call ShowHex
ld a,' '
call PrintChar
inc hl
inc hl
dec c
ld a,c
and %00001111
jr nz,continueMemdump
ld c,8
continueMemdump:
djnz loop_doMemDump
ret



HailstoneBuffer:
ds 512,0

Output:

call &8000
0070

001B 0052 0029 007C 003E 001F 005E 002F
008E 0047 00D6 006B 0142 00A1 01E4 00F2
0079 016C 00B6 005B 0112 0089 019C 00CE
0067 0136 009B 01D2 00E9 02BC 015E 00AF
01BD 0538 029C 014E 00A7 01F6 00FB 02F2
0179 046C 0236 011B 0352 01A9 04FC 027E
013F 03BE 01DF 059E 02CF 086E 0437 0CA6
0653 12FA 097D 1C78 0E3C 071E 038F 0AAE
0557 1006 0803 180A 0C05 2410 1208 0904
0482 0241 06C4 0362 01B1 0514 028A 0145
03D0 01E8 00F4 007A 003D 00B8 005C 002E
0017 0046 0023 006A 0035 00A0 0050 0028
0014 000A 0005 0010 0008 0004 0002 0001

zkl

fcn collatz(n,z=L()){ z.append(n); if(n==1) return(z);
   if(n.isEven) return(self.fcn(n/2,z)); return(self.fcn(n*3+1,z)) }

This uses tail recursion and thus is stack efficient.

Output:

var n=collatz(27)
n.len()
112
n[0,4]
L(27,82,41,124)
n[-4,*]
L(8,4,2,1)

Rather than write a function that calculates the length, just roll through all 100,000 sequences and save the largest (length,sequence start) pair. Creating all those Collatz lists isn't quick. This works by using a [mutable] list to hold state as the pump does the basic looping.

[2..0d100_000].pump(Void,  // loop n from 2 to 100,000
   collatz,              // generate Collatz sequence(n)
   fcn(c,n){           // if new longest sequence, save length/C, return longest
      if(c.len()>n[0]) n.clear(c.len(),c[0]); n}.fp1(L(0,0)))

Output:

L(351,77031)  // length, hailstone

ZX Spectrum Basic

Translation of: BBC_BASIC

10 LET n=27: LET s=1
20 GO SUB 1000
30 PRINT '"Sequence length = ";seqlen
40 LET maxlen=0: LET s=0
50 FOR m=2 TO 100000
60 LET n=m
70 GO SUB 1000
80 IF seqlen>maxlen THEN LET maxlen=seqlen: LET maxnum=m
90 NEXT m
100 PRINT "The number with the longest hailstone sequence is ";maxnum
110 PRINT "Its sequence length is ";maxlen
120 STOP 
1000 REM Hailstone
1010 LET l=0
1020 IF s THEN PRINT n;"  ";
1030 IF n=1 THEN LET seqlen=l+1: RETURN 
1040 IF FN m(n,2)=0 THEN LET n=INT (n/2): GO TO 1060
1050 LET n=3*n+1
1060 LET l=l+1
1070 GO TO 1020
2000 DEF FN m(a,b)=a-INT (a/b)*b