Proper divisors

From Rosetta Code
Task
Proper divisors
You are encouraged to solve this task according to the task description, using any language you may know.

The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder.

For N > 1 they will always include 1,   but for N == 1 there are no proper divisors.


Examples

The proper divisors of     6     are   1, 2, and 3.
The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50.


Task
  1. Create a routine to generate all the proper divisors of a number.
  2. use it to show the proper divisors of the numbers 1 to 10 inclusive.
  3. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has.


Show all output here.


Related tasks



360 Assembly

Translation of: Rexx

This program uses two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. <lang 360asm>* Proper divisors 14/06/2016 PROPDIV CSECT

        USING  PROPDIV,R13        base register
        B      72(R15)            skip savearea
        DC     17F'0'             savearea
        STM    R14,R12,12(R13)    prolog
        ST     R13,4(R15)         "
        ST     R15,8(R13)         " 
        LR     R13,R15            "
        LA     R10,1              n=1

LOOPN1 C R10,=F'10' do n=1 to 10

        BH     ELOOPN1
        LR     R1,R10             n
        BAL    R14,PDIV           pdiv(n)
        ST     R0,NN              nn=pdiv(n)
        MVC    PG,PGT             init buffer 
        LA     R11,PG             pgi=0
        XDECO  R10,XDEC           edit n
        MVC    0(3,R11),XDEC+9    output n
        LA     R11,7(R11)         pgi=pgi+7
        L      R1,NN              nn
        XDECO  R1,XDEC            edit nn
        MVC    0(3,R11),XDEC+9    output nn
        LA     R11,20(R11)        pgi=pgi+20
        LA     R5,1               i=1

LOOPNI C R5,NN do i=1 to nn

        BH     ELOOPNI
        LR     R1,R5              i
        SLA    R1,2               *4
        L      R2,TDIV-4(R1)      tdiv(i)
        XDECO  R2,XDEC            edit tdiv(i)
        MVC    0(3,R11),XDEC+9    output tdiv(i)
        LA     R11,3(R11)         pgi=pgi+3
        LA     R5,1(R5)           i=i+1
        B      LOOPNI

ELOOPNI XPRNT PG,80 print buffer

        LA     R10,1(R10)         n=n+1
        B      LOOPN1

ELOOPN1 SR R0,R0 0

        ST     R0,M               m=0
        LA     R10,1              n=1

LOOPN2 C R10,=F'20000' do n=1 to 20000

        BH     ELOOPN2
        LR     R1,R10             n
        BAL    R14,PDIV           nn=pdiv(n)
        C      R0,M               if nn>m
        BNH    NNNHM
        ST     R10,II             ii=n
        ST     R0,M               m=nn

NNNHM LA R10,1(R10) n=n+1

        B      LOOPN2

ELOOPN2 MVC PG,PGR init buffer

        L      R1,II              ii
        XDECO  R1,XDEC            edit ii
        MVC    PG(5),XDEC+7       output ii
        L      R1,M               m
        XDECO  R1,XDEC            edit m
        MVC    PG+9(4),XDEC+8     output m
        XPRNT  PG,80              print buffer
        L      R13,4(0,R13)       epilog 
        LM     R14,R12,12(R13)    "
        XR     R15,R15            "
        BR     R14                exit
  • ------- pdiv --function(x)----->number of divisors---

PDIV ST R1,X x

        C      R1,=F'1'           if x=1
        BNE    NOTONE
        LA     R0,0               return(0)
        BR     R14

NOTONE LR R4,R1 x

        N      R4,=X'00000001'    mod(x,2)
        LA     R4,1(R4)           +1
        ST     R4,ODD             odd=mod(x,2)+1
        LA     R8,1               ia=1
        LA     R0,1               1
        ST     R0,TDIV            tdiv(1)=1
        SR     R9,R9              ib=0
        L      R7,ODD             odd
        LA     R7,1(R7)           j=odd+1

LOOPJ LR R5,R7 do j=odd+1 by odd

        MR     R4,R7              j*j
        C      R5,X               while j*j<x 
        BNL    ELOOPJ
        L      R4,X               x
        SRDA   R4,32              .
        DR     R4,R7              /j
        LTR    R4,R4              if mod(x,j)=0
        BNZ    ITERJ
        LA     R8,1(R8)           ia=ia+1
        LR     R1,R8              ia
        SLA    R1,2               *4 (F)
        ST     R7,TDIV-4(R1)      tdiv(ia)=j
        LA     R9,1(R9)           ib=ib+1
        L      R4,X               x
        SRDA   R4,32              .
        DR     R4,R7              j
        LR     R2,R9              ib
        SLA    R2,2               *4 (F)
        ST     R5,TDIVB-4(R2)     tdivb(ib)=x/j

ITERJ A R7,ODD j=j+odd

        B      LOOPJ

ELOOPJ LR R5,R7 j

        MR     R4,R7              j*j
        C      R5,X               if j*j=x
        BNE    JTJNEX
        LA     R8,1(R8)           ia=ia+1
        LR     R1,R8              ia
        SLA    R1,2               *4 (F)
        ST     R7,TDIV-4(R1)      tdiv(ia)=j

JTJNEX LA R1,TDIV(R1) @tdiv(ia+1)

        LA     R2,TDIVB-4(R2)     @tdivb(ib)
        LTR    R6,R9              do i=ib to 1 by -1
        BZ     ELOOPI

LOOPI MVC 0(4,R1),0(R2) tdiv(ia)=tdivb(i)

        LA     R8,1(R8)           ia=ia+1
        LA     R1,4(R1)           r1+=4
        SH     R2,=H'4'           r2-=4
        BCT    R6,LOOPI           i=i-1

ELOOPI LR R0,R8 return(ia)

        BR     R14                return to caller
  • ---- ----------------------------------------

TDIV DS 80F TDIVB DS 40F M DS F NN DS F II DS F X DS F ODD DS F PGT DC CL80'... has .. proper divisors:' PGR DC CL80'..... has ... proper divisors.' PG DC CL80' ' XDEC DS CL12

        YREGS
        END    PROPDIV</lang>
Output:
  1 has  0 proper divisors:
  2 has  1 proper divisors:  1
  3 has  1 proper divisors:  1
  4 has  2 proper divisors:  1  2
  5 has  1 proper divisors:  1
  6 has  3 proper divisors:  1  2  3
  7 has  1 proper divisors:  1
  8 has  3 proper divisors:  1  2  4
  9 has  2 proper divisors:  1  3
 10 has  3 proper divisors:  1  2  5
15120 has  79 proper divisors.

Ada

The first part of the task is to create a routine to generate a list of the proper divisors. To ease the re-use of this routine for other tasks, such as Abundant, Deficient and Perfect Number Classification [[1]], Abundant Odd Number [[2]], and Amicable Pairs [[3]], we define this routine as a function of a generic package:

<lang Ada>generic

  type Result_Type (<>) is limited private;
  None: Result_Type;
  with function One(X: Positive) return Result_Type;
  with function Add(X, Y: Result_Type) return Result_Type 
     is <>;

package Generic_Divisors is

 function Process
   (N: Positive; First: Positive := 1) return Result_Type is      
     (if First**2 > N or First = N then None 
     elsif (N mod First)=0 then 

(if First = 1 or First*First = N then Add(One(First), Process(N, First+1)) else Add(One(First), Add(One((N/First)), Process(N, First+1))))

     else Process(N, First+1));
     

end Generic_Divisors;</lang>

Now we instantiate the generic package to solve the other two parts of the task. Observe that there are two different instantiations of the package: one to generate a list of proper divisors, another one to count the number of proper divisors without actually generating such a list:

<lang Ada>with Ada.Text_IO, Ada.Containers.Generic_Array_Sort, Generic_Divisors;

procedure Proper_Divisors is

begin

  -- show the proper divisors of the numbers 1 to 10 inclusive.
  declare
     type Pos_Arr is array(Positive range <>) of Positive;
     subtype Single_Pos_Arr is Pos_Arr(1 .. 1);
     Empty: Pos_Arr(1 .. 0); 
     
     function Arr(P: Positive) return Single_Pos_Arr is ((others => P));
     
     package Divisor_List is new Generic_Divisors

(Result_Type => Pos_Arr, None => Empty, One => Arr, Add => "&");

     procedure Sort is new Ada.Containers.Generic_Array_Sort

(Positive, Positive, Pos_Arr);

  begin
     for I in 1 .. 10 loop

declare List: Pos_Arr := Divisor_List.Process(I); begin Ada.Text_IO.Put (Positive'Image(I) & " has" & Natural'Image(List'Length) & " proper divisors:"); Sort(List); for Item of List loop Ada.Text_IO.Put(Positive'Image(Item)); end loop; Ada.Text_IO.New_Line; end;

     end loop;
  end;
  
  -- find a number 1 .. 20,000 with the most proper divisors
  declare
     Number: Positive := 1;
     Number_Count: Natural := 0;
     Current_Count: Natural;
     
     function Cnt(P: Positive) return Positive is (1);
  
     package Divisor_Count is new Generic_Divisors

(Result_Type => Natural, None => 0, One => Cnt, Add => "+");

  begin
     for Current in 1 .. 20_000 loop

Current_Count := Divisor_Count.Process(Current); if Current_Count > Number_Count then Number := Current; Number_Count := Current_Count; end if;

     end loop;
     Ada.Text_IO.Put_Line

(Positive'Image(Number) & " has the maximum number of" & Natural'Image(Number_Count) & " proper divisors.");

  end;

end Proper_Divisors; </lang>

Output:
 1 has 0 proper divisors:
 2 has 1 proper divisors: 1
 3 has 1 proper divisors: 1
 4 has 2 proper divisors: 1 2
 5 has 1 proper divisors: 1
 6 has 3 proper divisors: 1 2 3
 7 has 1 proper divisors: 1
 8 has 3 proper divisors: 1 2 4
 9 has 2 proper divisors: 1 3
 10 has 3 proper divisors: 1 2 5

 15120 has the maximum number of 79 proper divisors.

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

<lang algol68># MODE to hold an element of a list of proper divisors # MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next );

  1. end of divisor list value #

REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL);

  1. resturns a DIVISORLIST containing the proper divisors of n #
  2. if n = 1, 0 or -1, we return no divisors #

PROC proper divisors = ( INT n )REF DIVISORLIST:

    BEGIN
        REF DIVISORLIST result   := nil divisor list;
        REF DIVISORLIST end list := result;
        INT abs n  = ABS n;
        IF abs n > 1 THEN
            # build the list of divisors backeards, so they are  #
            # returned in ascending order                        #
            INT root n = ENTIER sqrt( abs n );
            FOR d FROM root n BY -1 TO 2 DO
                IF abs n MOD d = 0 THEN
                    # found another divisor                      #
                    result := HEAP DIVISORLIST
                           := DIVISORLIST( d, result );
                    IF end list IS nil divisor list THEN
                        # first result                           #
                        end list := result
                    FI;
                    IF d * d /= n THEN
                        # add the other divisor to the end of    #
                        # the list                               #
                        next OF end list := HEAP DIVISORLIST
                                         := DIVISORLIST( abs n OVER d, nil divisor list );
                        end list         := next OF end list
                    FI
                FI
            OD;
            # 1 is always a proper divisor of numbers > 1        #
            result := HEAP DIVISORLIST
                   := DIVISORLIST( 1, result )
        FI;
        result
    END # proper divisors # ;
  1. returns the number of divisors in a DIVISORLIST #

PROC count divisors = ( REF DIVISORLIST list )INT:

    BEGIN
       INT result := 0;
       REF DIVISORLIST divisors := list;
       WHILE divisors ISNT nil divisor list DO
           result +:= 1;
           divisors := next OF divisors
       OD;
       result
    END # count divisors # ;
  1. find the proper divisors of 1 : 10 #

FOR n TO 10 DO

   REF DIVISORLIST divisors := proper divisors( n );
   print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) );
   WHILE divisors ISNT nil divisor list DO
       print( ( " ", whole( divisor OF divisors, 0 ) ) );
       divisors := next OF divisors
   OD;
   print( ( newline ) )

OD;

  1. find the first/only number in 1 : 20 000 with the most divisors #

INT max number = 20 000; INT max divisors  := 0; INT has max divisors  := 0; INT with max divisors := 0; FOR d TO max number DO

   INT divisor count = count divisors( proper divisors( d ) );
   IF divisor count > max divisors THEN
       # found a number with more divisors than the previous max  #
       max divisors       := divisor count;
       has max divisors   := d;
       with max divisors  := 1
   ELIF divisor count = max divisors THEN
       # found another number with that many divisors             #
       with max divisors +:= 1
   FI

OD; print( ( whole( has max divisors, 0 )

      , " is the "
      , IF with max divisors < 2 THEN "only" ELSE "first" FI
      , " number upto "
      , whole( max number, 0 )
      , " with "
      , whole( max divisors, 0 )
      , " divisors"
      , newline
      ) )</lang>
Output:
Proper divisors of:  1:
Proper divisors of:  2:  1
Proper divisors of:  3:  1
Proper divisors of:  4:  1 2
Proper divisors of:  5:  1
Proper divisors of:  6:  1 2 3
Proper divisors of:  7:  1
Proper divisors of:  8:  1 2 4
Proper divisors of:  9:  1 3
Proper divisors of: 10:  1 2 5
15120 is the first number upto 20000 with 79 divisors

Algol-M

Algol-M's maximum allowed integer value of 16,383 prevented searching up to 20,000 for the number with the most divisors, so the code here searches only up to 10,000. <lang algol> BEGIN

% COMPUTE P MOD Q % INTEGER FUNCTION MOD (P, Q); INTEGER P, Q; BEGIN

   MOD := P - Q * (P / Q);

END;

% COUNT, AND OPTIONALLY DISPLAY, PROPER DIVISORS OF N % INTEGER FUNCTION DIVISORS(N, DISPLAY); INTEGER N, DISPLAY; BEGIN

   INTEGER I, LIMIT, COUNT, START, DELTA;
   IF MOD(N, 2) = 0 THEN
     BEGIN
       START := 2;
       DELTA := 1;
     END
   ELSE  % ONLY NEED TO CHECK ODD DIVISORS %
     BEGIN
       START := 3;
       DELTA := 2;
     END;
   % 1 IS A DIVISOR OF ANY NUMBER > 1 %
   IF N > 1 THEN COUNT := 1 ELSE COUNT := 0;
   IF (DISPLAY <> 0) AND (COUNT <> 0) THEN WRITEON(1);
   % CHECK REMAINING POTENTIAL DIVISORS %
   I := START;
   LIMIT := N / START;
   WHILE I <= LIMIT DO
     BEGIN
       IF MOD(N, I) = 0 THEN
         BEGIN
           IF DISPLAY <> 0 THEN WRITEON(I);
           COUNT := COUNT + 1;
         END;
       I := I + DELTA;
       IF COUNT = 1 THEN LIMIT := N / I;
     END;
  DIVISORS := COUNT;

END;

COMMENT MAIN PROGRAM BEGINS HERE; INTEGER I, NDIV, TRUE, FALSE, HIGHDIV, HIGHNUM; TRUE := -1; FALSE := 0;

WRITE("PROPER DIVISORS OF FIRST TEN NUMBERS:"); FOR I := 1 STEP 1 UNTIL 10 DO

 BEGIN
    WRITE(I, " : ");
    NDIV := DIVISORS(I, TRUE);
 END;

WRITE("SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS ..."); HIGHDIV := 1; HIGHNUM := 1; FOR I := 1 STEP 1 UNTIL 10000 DO

 BEGIN
    NDIV := DIVISORS(I, FALSE);
    IF NDIV > HIGHDIV THEN 
      BEGIN
        HIGHDIV := NDIV;
        HIGHNUM := I;
      END;
 END;

WRITE("THE NUMBER IS:", HIGHNUM); WRITE("IT HAS", HIGHDIV, " DIVISORS");

END </lang>

Output:
PROPER DIVISORS OF FIRST TEN NUMBERS:
     1 :
     2 :      1
     3 :      1
     4 :      1     2
     5 :      1
     6 :      1     2     3
     7 :      1
     8 :      1     2     4
     9 :      1     3
    10 :      1     2     5
SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS:
THE NUMBER IS:  7560
IT HAS    63 DIVISORS

AppleScript

Translation of: JavaScript

<lang AppleScript>-- PROPER DIVISORS -----------------------------------------------------------

-- properDivisors :: Int -> [Int] on properDivisors(n)

   if n = 1 then
       {1}
   else
       set realRoot to n ^ (1 / 2)
       set intRoot to realRoot as integer
       set blnPerfectSquare to intRoot = realRoot
       
       -- isFactor :: Int -> Bool 
       script isFactor
           on |λ|(x)
               n mod x = 0
           end |λ|
       end script
       
       -- Factors up to square root of n,
       set lows to filter(isFactor, enumFromTo(1, intRoot))
       
       -- and quotients of these factors beyond the square root,
       
       -- integerQuotient :: Int -> Int
       script integerQuotient
           on |λ|(x)
               (n / x) as integer
           end |λ|
       end script
       
       -- excluding n itself (last item)
       items 1 thru -2 of (lows & map(integerQuotient, ¬
           items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows))
   end if

end properDivisors


-- TEST ---------------------------------------------------------------------- on run

   -- numberAndDivisors :: Int -> [Int]
   script numberAndDivisors
       on |λ|(n)
           {num:n, divisors:properDivisors(n)}
       end |λ|
   end script
   
   -- maxDivisorCount :: Record -> Int -> Record
   script maxDivisorCount
       on |λ|(a, n)
           set intDivisors to length of properDivisors(n)
           
           if intDivisors ≥ divisors of a then
               {num:n, divisors:intDivisors}
           else
               a
           end if
       end |λ|
   end script
   
   {oneToTen:map(numberAndDivisors, ¬
       enumFromTo(1, 10)), mostDivisors:foldl(maxDivisorCount, ¬
       {num:0, divisors:0}, enumFromTo(1, 20000))} ¬
       

end run


-- GENERIC FUNCTIONS ---------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)

   if m > n then
       set d to -1
   else
       set d to 1
   end if
   set lst to {}
   repeat with i from m to n by d
       set end of lst to i
   end repeat
   return lst

end enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)

   tell mReturn(f)
       set lst to {}
       set lng to length of xs
       repeat with i from 1 to lng
           set v to item i of xs
           if |λ|(v, i, xs) then set end of lst to v
       end repeat
       return lst
   end tell

end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)

   tell mReturn(f)
       set v to startValue
       set lng to length of xs
       repeat with i from 1 to lng
           set v to |λ|(v, item i of xs, i, xs)
       end repeat
       return v
   end tell

end foldl

-- map :: (a -> b) -> [a] -> [b] on map(f, xs)

   tell mReturn(f)
       set lng to length of xs
       set lst to {}
       repeat with i from 1 to lng
           set end of lst to |λ|(item i of xs, i, xs)
       end repeat
       return lst
   end tell

end map

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)

   if class of f is script then
       f
   else
       script
           property |λ| : f
       end script
   end if

end mReturn</lang>

Output:

<lang AppleScript>{oneToTen:{{num:1, divisors:{1}}, {num:2, divisors:{1}}, {num:3, divisors:{1}}, {num:4, divisors:{1, 2}}, {num:5, divisors:{1}}, {num:6, divisors:{1, 2, 3}}, {num:7, divisors:{1}}, {num:8, divisors:{1, 2, 4}}, {num:9, divisors:{1, 3}}, {num:10, divisors:{1, 2, 5}}}, mostDivisors:{num:18480, divisors:79}}</lang>

Arc

<lang Arc>

Given num, return num and the list of its divisors

(= divisor (fn (num)

  (= dlist '())
  (when (is 1 num) (= dlist '(1 0)))
  (when (is 2 num) (= dlist '(2 1)))
  (unless (or (is 1 num) (is 2 num))
  (up i 1 (+ 1 (/ num 2))
    (if (is 0 (mod num i))
        (push i dlist)))
  (= dlist (cons num dlist)))
  dlist))
Find out what number has the most divisors between 2 and 20,000.
Print a list of the largest known number's divisors as it is found.

(= div-lists (fn (cnt (o show 0))

 (= tlist '()) (= clist tlist)
 (when (> show 0) (prn tlist))
 (up i 1 cnt
   (divisor i)
   (when (is 1 show) (prn dlist))
   (when (>= (len dlist) (len tlist))
       (= tlist dlist)
       (when (is show 2) (prn tlist))
       (let c (- (len dlist) 1)
       (push (list i c) clist))))
 (= many-divisors (list ((clist 0) 1)))
 (for n 0 (is ((clist n) 1) ((clist 0) 1)) (= n (+ 1 n))
   (push ((clist n) 0) many-divisors))
 (= many-divisors (rev many-divisors))
 (prn "The number with the most divisors under " cnt
      " has " (many-divisors 0) " divisors.")
 (prn "It is the number "
 (if (> 2 (len many-divisors)) (cut (many-divisors) 1)
     (many-divisors 1)) ".")
 (prn "There are " (- (len many-divisors) 1) " numbers"
      " with this trait, and they are "
      (map [many-divisors _] (range 1 (- (len many-divisors) 1))))
 (prn (map [divisor _] (cut many-divisors 1)))
 many-divisors))
Do the tasks

(div-lists 10 1) (div-lists 20000)

This took about 10 minutes on my machine.

</lang>

Output:

<lang Arc> (1 0) (2 1) (3 1) (4 2 1) (5 1) (6 3 2 1) (7 1) (8 4 2 1) (9 3 1) (10 5 2 1) The number with the most divisors under 10 has 3 divisors. It is the number 10. There are 3 numbers with this trait, and they are (10 8 6) ((10 5 2 1) (8 4 2 1) (6 3 2 1)) '(3 10 8 6)


The number with the most divisors under 20000 has 79 divisors. It is the number 18480. There are 2 numbers with this trait, and they are (18480 15120) </lang>

ARM Assembly

Works with: as version Raspberry Pi

<lang ARM Assembly> /* ARM assembly Raspberry PI */ /* program proFactor.s */

/* REMARK 1 : this program use routines in a include file

  see task Include a file language arm assembly 
  for the routine affichageMess conversion10 
  see at end of this program the instruction include */

/*******************************************/ /* Constantes */ /*******************************************/ .equ STDOUT, 1 @ Linux output console .equ EXIT, 1 @ Linux syscall .equ WRITE, 4 @ Linux syscall

/*******************************************/ /* Initialized data */ /*******************************************/ .data szMessStartPgm: .asciz "Program start \n" szMessEndPgm: .asciz "Program normal end.\n" szMessError: .asciz "\033[31mError Allocation !!!\n" szCarriageReturn: .asciz "\n"

/* datas message display */ szMessEntete: .ascii "Number :" sNumber: .space 12,' '

                        .asciz " Divisors :"

szMessResult: .ascii " " sValue: .space 12,' '

                        .asciz ""

szMessDivNumber: .ascii "\nnumber divisors :" sCounter: .space 12,' '

                        .asciz "\n"

szMessNumberMax: .ascii "Number :" sNumberMax: .space 12,' '

                        .ascii " has "

sDivMax: .space 12, ' '

                        .asciz " divisors\n"

/*******************************************/ /* UnInitialized data */ /*******************************************/ .bss /*******************************************/ /* code section */ /*******************************************/ .text .global main main: @ program start

   ldr r0,iAdrszMessStartPgm       @ display start message
   bl affichageMess
   mov r2,#1

1:

   mov r0,r2                       @  number
   ldr r1,iAdrsNumber              @ and convert ascii string
   bl conversion10
   ldr r0,iAdrszMessEntete         @ display result message
   bl affichageMess
   mov r0,r2                       @  number
   mov r1,#1                       @ display flag
   bl divisors                     @ display divisors
   ldr r1,iAdrsCounter              @ and convert ascii string
   bl conversion10
   ldr r0,iAdrszMessDivNumber      @ display result message
   bl affichageMess
   add r2,r2,#1
   cmp r2,#10
   ble 1b
   mov r2,#2
   mov r3,#0
   mov r4,#0
   ldr r5,iMaxi

2:

   mov r0,r2
   mov r1,#0                       @ display flag
   bl divisors                     @ display divisors
   cmp r0,r3
   movgt r3,r0
   movgt r4,r2
   add r2,r2,#1
   cmp r2,r5
   ble 2b
   mov r0,r4
   ldr r1,iAdrsNumberMax           @ and convert ascii string
   bl conversion10
   mov r0,r3
   ldr r1,iAdrsDivMax              @ and convert ascii string
   bl conversion10
   ldr r0,iAdrszMessNumberMax
   bl affichageMess


   ldr r0,iAdrszMessEndPgm         @ display end message
   bl affichageMess
   b 100f

99: @ display error message

   ldr r0,iAdrszMessError
   bl affichageMess

100: @ standard end of the program

   mov r0, #0                      @ return code
   mov r7, #EXIT                   @ request to exit program
   svc 0                           @ perform system call

iAdrszMessStartPgm: .int szMessStartPgm iAdrszMessEndPgm: .int szMessEndPgm iAdrszMessError: .int szMessError iAdrszCarriageReturn: .int szCarriageReturn iAdrszMessResult: .int szMessResult iAdrsValue: .int sValue iAdrszMessDivNumber: .int szMessDivNumber iAdrsCounter: .int sCounter iAdrszMessEntete: .int szMessEntete iAdrsNumber: .int sNumber iAdrszMessNumberMax: .int szMessNumberMax iAdrsDivMax: .int sDivMax iAdrsNumberMax: .int sNumberMax iMaxi: .int 20000 /******************************************************************/ /* divisors function */ /******************************************************************/ /* r0 contains the number */ /* r1 contains display flag (<>0: display, 0: no display ) /* r0 return divisors number */ divisors:

   push {r1-r8,lr}             @ save  registers 
   cmp r0,#1                   @ = 1 ?
   movle r0,#0
   ble 100f
   mov r7,r0
   mov r8,r1
   cmp r8,#0
   beq 1f
   mov r0,#1                   @ first divisor = 1
   ldr r1,iAdrsValue           @ and convert ascii string
   bl conversion10
   ldr r0,iAdrszMessResult     @ display result message
   bl affichageMess

1: @ begin loop

   lsr r4,r7,#1                @ Maxi
   mov r6,r4                   @ first divisor
   mov r5,#1                   @ Counter divisors

2:

   mov r0,r7                   @ dividende = number
   mov r1,r6                   @ divisor
   bl division
   cmp r3,#0                   @ remainder = 0 ?
   bne 3f
   add r5,r5,#1                @ increment counter
   cmp r8,#0                   @ display divisor ?
   beq 3f
   mov r0,r2                   @ divisor
   ldr r1,iAdrsValue           @ and convert ascii string
   bl conversion10
   ldr r0,iAdrszMessResult     @ display result message
   bl affichageMess

3:

   sub r6,r6,#1                @ decrement divisor
   cmp r6,#2                   @ End ?
   bge 2b                      @ no loop
   mov r0,r5                   @ return divisors number

100:

   pop {r1-r8,lr}              @ restaur registers
   bx lr                       @ return

/***************************************************/ /* ROUTINES INCLUDE */ /***************************************************/ .include "../affichage.inc" </lang>

Output:
Program start
Number :1            Divisors :
number divisors :0
Number :2            Divisors : 1            2
number divisors :2
Number :3            Divisors : 1            3
number divisors :2
Number :4            Divisors : 1            2
number divisors :2
Number :5            Divisors : 1
number divisors :1
Number :6            Divisors : 1            2            3
number divisors :3
Number :7            Divisors : 1
number divisors :1
Number :8            Divisors : 1            2            4
number divisors :3
Number :9            Divisors : 1            3
number divisors :2
Number :10           Divisors : 1            2            5
number divisors :3
Number :15120        has 79           divisors
Program normal end.

AutoHotkey

<lang AutoHotkey>proper_divisors(n) { Array := [] if n = 1 return Array Array[1] := true x := Floor(Sqrt(n)) loop, % x+1 if !Mod(n, i:=A_Index+1) && (floor(n/i) < n) Array[floor(n/i)] := true Loop % n/x if !Mod(n, i:=A_Index+1) && (i < n) Array[i] := true return Array }</lang> Examples:<lang AutoHotkey>output := "Number`tDivisors`tCount`n" loop, 10 { output .= A_Index "`t" for n, bool in x := proper_divisors(A_Index) output .= n " " output .= "`t" x.count() "`n" } maxDiv := 0, numDiv := [] loop, 20000 { Arr := proper_divisors(A_Index) numDiv[Arr.count()] := (numDiv[Arr.count()] ? numDiv[Arr.count()] ", " : "") A_Index maxDiv := maxDiv > Arr.count() ? maxDiv : Arr.count() } output .= "`nNumber(s) in the range 1 to 20,000 with the most proper divisors:`n" numDiv.Pop() " with " maxDiv " divisors" MsgBox % output return</lang>

Output:
---------------------------
Temp.ahk
---------------------------
Number	Divisors	Count
1			0
2	1 		1
3	1 		1
4	1 2 		2
5	1 		1
6	1 2 3 		3
7	1 		1
8	1 2 4 		3
9	1 3 		2
10	1 2 5 		3

Number(s) in the range 1 to 20,000 with the most proper divisors:
15120, 18480 with 79 divisors

AWK

<lang AWK>

  1. syntax: GAWK -f PROPER_DIVISORS.AWK

BEGIN {

   show = 0 # show divisors: 0=no, 1=yes
   print("    N  cnt  DIVISORS")
   for (i=1; i<=20000; i++) {
     divisors(i)
     if (i <= 10 || i == 100) { # including 100 as it was an example in task description
       printf("%5d  %3d  %s\n",i,Dcnt,Dstr)
     }
     if (Dcnt < max_cnt) {
       continue
     }
     if (Dcnt > max_cnt) {
       rec = ""
       max_cnt = Dcnt
     }
     rec = sprintf("%s%5d  %3d  %s\n",rec,i,Dcnt,show?Dstr:"divisors not shown")
   }
   printf("%s",rec)
   exit(0)

} function divisors(n, i) {

   if (n == 1) {
     Dcnt = 0
     Dstr = ""
     return
   }
   Dcnt = Dstr = 1
   for (i=2; i<n; i++) {
     if (n % i == 0) {
       Dcnt++
       Dstr = sprintf("%s %s",Dstr,i)
     }
   }
   return

} </lang>

output:

    N  cnt  DIVISORS
    1    0
    2    1  1
    3    1  1
    4    2  1 2
    5    1  1
    6    3  1 2 3
    7    1  1
    8    3  1 2 4
    9    2  1 3
   10    3  1 2 5
  100    8  1 2 4 5 10 20 25 50
15120   79  divisors not shown
18480   79  divisors not shown

BaCon

<lang qbasic> FUNCTION ProperDivisor(nr, show)

   LOCAL probe, total
   FOR probe = 1 TO nr-1
       IF MOD(nr, probe) = 0 THEN
           IF show THEN PRINT " ", probe;
           INCR total
       END IF
   NEXT
   RETURN total

END FUNCTION

FOR x = 1 TO 10

   PRINT x, ":";
   IF ProperDivisor(x, 1) = 0 THEN PRINT " 0";
   PRINT

NEXT

FOR x = 1 TO 20000

   DivisorCount = ProperDivisor(x, 0)
   IF DivisorCount > MaxDivisors THEN
       MaxDivisors = DivisorCount
       MagicNumber = x
   END IF

NEXT

PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors." </lang>

Output:
1: 0
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
Most proper divisors for number in the range 1-20000: 15120 with 79 divisors.

C

Brute Force

C has tedious boilerplate related to allocating memory for dynamic arrays, so we just skip the problem of storing values altogether. <lang c>

  1. include <stdio.h>
  2. include <stdbool.h>

int proper_divisors(const int n, bool print_flag) {

   int count = 0;
   for (int i = 1; i < n; ++i) {
       if (n % i == 0) {
           count++;
           if (print_flag)
               printf("%d ", i);
       }
   }
   if (print_flag)
       printf("\n");
   return count;

}

int main(void) {

   for (int i = 1; i <= 10; ++i) {
       printf("%d: ", i);
       proper_divisors(i, true);
   }
   int max = 0;
   int max_i = 1;
   for (int i = 1; i <= 20000; ++i) {
       int v = proper_divisors(i, false);
       if (v >= max) {
           max = v;
           max_i = i;
       }
   }
   printf("%d with %d divisors\n", max_i, max);
   return 0;

} </lang>

Output:
1: 
2: 1 
3: 1 
4: 1 2 
5: 1 
6: 1 2 3 
7: 1 
8: 1 2 4 
9: 1 3 
10: 1 2 5 
18480 with 79 divisors

Number Theoretic

There is no need to go through all the divisors if only the count is needed, this implementation refines the brute force approach by solving the second part of the task via a Number Theory formula. The running time is noticeably faster than the brute force method above. Output is same as the above. <lang C>

  1. include <stdio.h>
  2. include <stdbool.h>

int proper_divisors(const int n, bool print_flag) {

   int count = 0;

   for (int i = 1; i < n; ++i) {
       if (n % i == 0) {
           count++;
           if (print_flag)
               printf("%d ", i);
       }
   }

   if (print_flag)
       printf("\n");

   return count;

}

int countProperDivisors(int n){ int prod = 1,i,count=0;

while(n%2==0){ count++; n /= 2; }

prod *= (1+count);

for(i=3;i*i<=n;i+=2){ count = 0;

while(n%i==0){ count++; n /= i; }

prod *= (1+count); }

if(n>2) prod *= 2;

return prod - 1; }

int main(void) {

   for (int i = 1; i <= 10; ++i) {
       printf("%d: ", i);
       proper_divisors(i, true);
   }

   int max = 0;
   int max_i = 1;

   for (int i = 1; i <= 20000; ++i) {
       int v = countProperDivisors(i);
       if (v >= max) {
           max = v;
           max_i = i;
       }
   }

   printf("%d with %d divisors\n", max_i, max);
   return 0;

} </lang>

C#

<lang csharp>namespace RosettaCode.ProperDivisors {

   using System;
   using System.Collections.Generic;
   using System.Linq;
   internal static class Program
   {
       private static IEnumerable<int> ProperDivisors(int number)
       {
           return
               Enumerable.Range(1, number / 2)
                   .Where(divisor => number % divisor == 0);
       }
       private static void Main()
       {
           foreach (var number in Enumerable.Range(1, 10))
           {
               Console.WriteLine("{0}: {{{1}}}", number,
                   string.Join(", ", ProperDivisors(number)));
           }
           var record = Enumerable.Range(1, 20000).Select(number => new
           {
               Number = number,
               Count = ProperDivisors(number).Count()
           }).OrderByDescending(currentRecord => currentRecord.Count).First();
           Console.WriteLine("{0}: {1}", record.Number, record.Count);
       }
   }

}</lang>

Output:
1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79

C++

<lang cpp>#include <vector>

  1. include <iostream>
  2. include <algorithm>

std::vector<int> properDivisors ( int number ) {

  std::vector<int> divisors ;
  for ( int i = 1 ; i < number / 2 + 1 ; i++ )
     if ( number % i == 0 )

divisors.push_back( i ) ;

  return divisors ;

}

int main( ) {

  std::vector<int> divisors ;
  unsigned int maxdivisors = 0 ;
  int corresponding_number = 0 ;
  for ( int i = 1 ; i < 11 ; i++ ) {
     divisors =  properDivisors ( i ) ;
     std::cout << "Proper divisors of " << i << ":\n" ;
     for ( int number : divisors ) { 

std::cout << number << " " ;

     }
     std::cout << std::endl ;
     divisors.clear( ) ;
  }
  for ( int i = 11 ; i < 20001 ; i++ ) {
     divisors =  properDivisors ( i ) ;
     if ( divisors.size( ) > maxdivisors ) {

maxdivisors = divisors.size( ) ; corresponding_number = i ;

     }
     divisors.clear( ) ;
  }
  std::cout << "Most divisors has " << corresponding_number <<
     " , it has " << maxdivisors << " divisors!\n" ; 
  return 0 ;

} </lang>

Output:
Proper divisors of 1:

Proper divisors of 2:
1 
Proper divisors of 3:
1 
Proper divisors of 4:
1 2 
Proper divisors of 5:
1 
Proper divisors of 6:
1 2 3 
Proper divisors of 7:
1 
Proper divisors of 8:
1 2 4 
Proper divisors of 9:
1 3 
Proper divisors of 10:
1 2 5 
Most divisors has 15120 , it has 79 divisors!

Ceylon

<lang ceylon>shared void run() {

function divisors(Integer int) => if(int <= 1) then {} else (1..int / 2).filter((Integer element) => element.divides(int));

for(i in 1..10) { print("``i`` => ``divisors(i)``"); }

value start = 1; value end = 20k;

value mostDivisors = map {for(i in start..end) i->divisors(i).size} .inverse() .max(byKey(byIncreasing(Integer.magnitude)));

print("the number(s) with the most divisors between ``start`` and ``end`` is/are: ``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors"); }</lang>

Output:
1 => []
2 => { 1 }
3 => { 1 }
4 => { 1, 2 }
5 => { 1 }
6 => { 1, 2, 3 }
7 => { 1 }
8 => { 1, 2, 4 }
9 => { 1, 3 }
10 => { 1, 2, 5 }
the number(s) with the most divisors between 1 and 20000 is/are:
[15120, 18480] with 79 divisors

Clojure

<lang lisp>(ns properdivisors

 (:gen-class))

(defn proper-divisors [n]

 " Proper divisors of n"
 (if (= n 1)
   []
 (filter #(= 0 (rem n %)) (range 1 n))))
Property divisors of numbers 1 to 20,000 inclusive

(def data (for [n (range 1 (inc 20000))]

           [n (proper-divisors n)]))
Find Max

(defn maximal-key [k x & xs]

 " Normal max-key only finds one key that produces maximum, while this function finds them all "
 (reduce (fn [ys x]
           (let [c (compare (k x) (k (peek ys)))]
             (cond
               (pos? c) [x]
               (neg? c) ys
               :else    (conj ys x))))
         [x]
         xs))

(println "n\tcnt\tPROPER DIVISORS") (doseq [n (range 1 11)]

 (let [factors (proper-divisors n)]
   (println n "\t" (count factors) "\t" factors)))

(def max-data (apply maximal-key (fn i pd (count pd)) data))

(doseq [[n factors] max-data]

 (println n " has " (count factors) " divisors"))

</lang>

Output:
n	cnt	PROPER DIVISORS
1 	 0 	 []
2 	 1 	 (1)
3 	 1 	 (1)
4 	 2 	 (1 2)
5 	 1 	 (1)
6 	 3 	 (1 2 3)
7 	 1 	 (1)
8 	 3 	 (1 2 4)
9 	 2 	 (1 3)
10 	 3 	 (1 2 5)
15120  has  79  divisors
18480  has  79  divisors

Common Lisp

Ideally, the smallest-divisor function would only try prime numbers instead of odd numbers. <lang lisp>(defun proper-divisors-recursive (product &optional (results '(1)))

  "(int,list)->list::Function to find all proper divisors of a +ve integer."
 
  (defun smallest-divisor (x)
     "int->int::Find the smallest divisor of an integer > 1."
     (if (evenp x) 2
         (do ((lim (truncate (sqrt x)))
              (sd 3 (+ sd 2)))
             ((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd)))))
  
  (defun pd-rec (fac)
     "(int,int)->nil::Recursive function to find proper divisors of a +ve integer"
     (when (not (member fac results))
        (push fac results)
        (let ((hifac (/ fac (smallest-divisor fac))))
           (pd-rec hifac)
           (pd-rec (/ product hifac)))))
   
  (pd-rec product)
  (butlast (sort (copy-list results) #'<)))
 

(defun task (method &optional (n 1) (most-pds '(0)))

  (dotimes (i 19999)
     (let ((npds (length (funcall method (incf n))))
           (hiest (car most-pds)))
        (when (>= npds hiest)
           (if (> npds hiest)
               (setf most-pds (list npds (list n)))
               (setf most-pds (list npds (cons n (second most-pds))))))))
  most-pds)

(defun main ()

  (format t "Task 1:Proper Divisors of [1,10]:~%")
  (dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i))))
  (format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%"
          (task #'proper-divisors-recursive)))</lang>
Output:
CL-USER(10): (main)
Task 1:Proper Divisors of [1,10]:
1:NIL
2:(1)
3:(1)
4:(1 2)
5:(1)
6:(1 2 3)
7:(1)
8:(1 2 4)
9:(1 3)
10:(1 2 5)
Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:
(79 (18480 15120))
NIL

Component Pascal

<lang oberon2> MODULE RosettaProperDivisor; IMPORT StdLog;

PROCEDURE Pd*(n: LONGINT;OUT r: ARRAY OF LONGINT):LONGINT; VAR i,j: LONGINT; BEGIN i := 1;j := 0; IF n > 1 THEN WHILE (i < n) DO IF (n MOD i) = 0 THEN IF (j < LEN(r)) THEN r[j] := i END; INC(j) END; INC(i) END; END; RETURN j END Pd;

PROCEDURE Do*; VAR r: ARRAY 128 OF LONGINT; i,j,found,max,idxMx: LONGINT; mx: ARRAY 128 OF LONGINT; BEGIN FOR i := 1 TO 10 DO found := Pd(i,r); IF found > LEN(r) THEN (* Error. more pd than r can admit *) HALT(1) END; StdLog.Int(i);StdLog.String("[");StdLog.Int(found);StdLog.String("]:> "); FOR j := 0 TO found - 1 DO StdLog.Int(r[j]);StdLog.Char(' '); END; StdLog.Ln END;

max := 0;idxMx := 0;

 FOR i := 1 TO 20000 DO
 	found := Pd(i,r);
 	IF found > max THEN
   	idxMx:= 0;mx[idxMx] := i;max := found

ELSIF found = max THEN

   	INC(idxMx);mx[idxMx] := i
 	END;
 END;

StdLog.String("Found: ");StdLog.Int(idxMx + 1);

 StdLog.String(" Numbers with the longest proper divisors [");

StdLog.Int(max);StdLog.String("]: ");StdLog.Ln; FOR i := 0 TO idxMx DO

 	StdLog.Int(mx[i]);StdLog.Ln

END END Do;

END RosettaProperDivisor.

^Q RosettaProperDivisor.Do~ </lang>

Output:
 1[ 0]:> 
 2[ 1]:>  1 
 3[ 1]:>  1 
 4[ 2]:>  1  2 
 5[ 1]:>  1 
 6[ 3]:>  1  2  3 
 7[ 1]:>  1 
 8[ 3]:>  1  2  4 
 9[ 2]:>  1  3 
 10[ 3]:>  1  2  5 
Found:  2 Numbers with the longest proper divisors [ 79]: 
 15120
 18480

D

Translation of: Python

Currently the lambda of the filter allocates a closure on the GC-managed heap. <lang d>void main() /*@safe*/ {

   import std.stdio, std.algorithm, std.range, std.typecons;
   immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ =>
       iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x);
   iota(1, 11).map!properDivs.writeln;
   iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;

}</lang>

Output:
[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [1, 2, 4], [1, 3], [1, 2, 5]]
Tuple!(uint, int)(79, 18480)

The Run-time is about 0.67 seconds with the ldc2 compiler.

Delphi

Translation of: C#

<lang Delphi> program ProperDivisors;

{$APPTYPE CONSOLE}

{$R *.res}

uses

 System.SysUtils,
 System.Generics.Collections;

type

 TProperDivisors = TArray<Integer>;

function GetProperDivisors(const value: Integer): TProperDivisors; var

 i, count: Integer;

begin

 count := 0;
 for i := 1 to value div 2 do
 begin
   if value mod i = 0 then
   begin
     inc(count);
     SetLength(result, count);
     Result[count - 1] := i;
   end;
 end;

end;

procedure Println(values: TProperDivisors); var

 i: Integer;

begin

 Write('[');
 if Length(values) > 0 then
   for i := 0 to High(values) do
     Write(Format('%2d', [values[i]]));
 Writeln(']');

end;

var

 number, max_count, count, max_number: Integer;

begin

 for number := 1 to 10 do
 begin
   write(number, ': ');
   Println(GetProperDivisors(number));
 end;
 max_count := 0;
 for number := 1 to 20000 do
 begin
   count := length(GetProperDivisors(number));
   if count > max_count then
   begin
     max_count := count;
     max_number := number;
   end;
 end;
 Write(max_number, ': ', max_count);
 readln;

end. </lang>

Output:
1: []
2: [ 1]
3: [ 1]
4: [ 1 2]
5: [ 1]
6: [ 1 2 3]
7: [ 1]
8: [ 1 2 4]
9: [ 1 3]
10: [ 1 2 5]
15120: 79

Version with TParallel.For <lang Delphi> program ProperDivisors;

{$APPTYPE CONSOLE}

{$R *.res}

uses

 System.SysUtils,
 System.Threading,
 System.SyncObjs;

type

 TProperDivisors = array of Integer;

function GetProperDivisors(const value: Integer): TProperDivisors; var

 i, count: Integer;

begin

 count := 0;
 for i := 1 to value div 2 do
 begin
   if value mod i = 0 then
   begin
     inc(count);
     SetLength(result, count);
     Result[count - 1] := i;
   end;
 end;

end;

procedure Println(values: TProperDivisors); var

 i: Integer;

begin

 Write('[');
 if Length(values) > 0 then
   for i := 0 to High(values) do
     Write(Format('%2d', [values[i]]));
 Writeln(']');

end;

var

 number, max_count, count, max_number: Integer;

begin

 for number := 1 to 10 do
 begin
   write(number, ': ');
   Println(GetProperDivisors(number));
 end;
 max_count := 0;
 TParallel.for (1, 20000,
   procedure(I: Int64)
   begin
     count := length(GetProperDivisors(I));
     if count > max_count then
     begin
       TInterlocked.Exchange(max_count, count);
       TInterlocked.Exchange(max_number, I);
     end;
   end);
 Writeln(max_number, ': ', max_count);
 readln;

end. </lang>

Dyalect

Translation of: Swift

<lang dyalect>func properDivs(n) {

   if n == 1 {
       return
   }
   for x in 1..(n-1) {
       if n % x == 0 {
           yield x
       }
   }

}

for i in 1..10 {

   print("\(i): \(properDivs(i).toArray())")

}

var (num, max) = (0,0)

for i in 1..20000 {

   const count = properDivs(i).len()
   if count > max { 
       set (num, max) = (i, count) 
   }

}

print("\(num): \(max)")</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79

EchoLisp

<lang scheme> (lib 'list) ;; list-delete

let n = product p_i^a_i , p_i prime
number of divisors = product (a_i + 1) - 1

(define (numdivs n)

   (1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n))))))

(remember 'numdivs)

prime powers
input
a list g of grouped prime factors ( 3 3 3 ..)
returns (1 3 9 27 ...)

(define (ppows g (mult 1)) (for/fold (ppows '(1)) ((a g)) (set! mult (* mult a)) (cons mult ppows)))

proper divisors
decomp n into ((2 2 ..) ( 3 3 ..) ) prime factors groups
combines (1 2 4 8 ..) (1 3 9 ..) lists
remove n from the list

(define (divs n)

  (if (<= n 1) null 
   (list-delete
       (for/fold (divs'(1)) ((g (map  ppows (group (prime-factors n)))))

(for*/list ((a divs) (b g)) (* a b)))

   n )))
find number(s) with max # of proper divisors
returns list of (n . maxdivs) for n in range 2..N

(define (most-proper N)

   (define maxdivs 1)
   (define ndivs 0)
   (for/fold (most-proper null) ((n (in-range 2 N)))
      (set! ndivs (numdivs n))
       #:continue (< ndivs maxdivs)
       (when (> ndivs maxdivs)
       (set!-values (most-proper maxdivs) (values null ndivs)))
       (cons (cons n maxdivs) most-proper)))


</lang>

Output:

<lang scheme> (for ((i (in-range 1 11))) (writeln i (divs i))) 1 null 2 (1) 3 (1) 4 (2 1) 5 (1) 6 (2 3 1) 7 (1) 8 (4 2 1) 9 (3 1) 10 (2 5 1)


(most-proper 20000)

   → ((18480 . 79) (15120 . 79))

(most-proper 1_000_000)

   → ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))  

(lib 'bigint) (numdivs 95952222101012742144) → 666 ;; 🎩 </lang>

Eiffel

<lang Eiffel> class APPLICATION

create make

feature

make -- Test the feature proper_divisors. local list: LINKED_LIST [INTEGER] count, number: INTEGER do across 1 |..| 10 as c loop list := proper_divisors (c.item) io.put_string (c.item.out + ": ") across list as l loop io.put_string (l.item.out + " ") end io.new_line end across 1 |..| 20000 as c loop list := proper_divisors (c.item) if list.count > count then count := list.count number := c.item end end io.put_string (number.out + " has with " + count.out + " divisors the highest number of proper divisors.") end

proper_divisors (n: INTEGER): LINKED_LIST [INTEGER] -- Proper divisors of 'n'. do create Result.make across 1 |..| (n - 1) as c loop if n \\ c.item = 0 then Result.extend (c.item) end end end

end </lang>

Output:
1:
2: 1
3: 1
4: 1 2
5: 1 
6: 1 2 3 
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 has with 79 divisors the highest number of proper divisors.

Elixir

Translation of: Erlang

<lang elixir>defmodule Proper do

 def divisors(1), do: []
 def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort
 
 defp divisors(k,_n,q) when k>q, do: []
 defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)
 defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]
 defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]
 
 def most_divisors(limit) do
   {length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end)
                   |> Enum.max_by(fn {length,_nums} -> length end)
   IO.puts "With #{length}, Number #{inspect nums} has the most divisors"
 end

end

Enum.each(1..10, fn n ->

 IO.puts "#{n}: #{inspect Proper.divisors(n)}"

end) Proper.most_divisors(20000)</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
With 79, Number [18480, 15120] has the most divisors

Erlang

<lang erlang>-module(properdivs). -export([divs/1,sumdivs/1,longest/1]).

divs(0) -> []; divs(1) -> []; divs(N) -> lists:sort([1] ++ divisors(2,N,math:sqrt(N))).

divisors(K,_N,Q) when K > Q -> []; divisors(K,N,Q) when N rem K =/= 0 ->

   divisors(K+1,N,Q);

divisors(K,N,Q) when K * K == N ->

   [K] ++ divisors(K+1,N,Q);

divisors(K,N,Q) ->

   [K, N div K] ++ divisors(K+1,N,Q).

sumdivs(N) -> lists:sum(divs(N)).

longest(Limit) -> longest(Limit,0,0,1).

longest(L,Current,CurLeng,Acc) when Acc >= L ->

   io:format("With ~w, Number ~w has the most divisors~n", [CurLeng,Current]);

longest(L,Current,CurLeng,Acc) ->

   A = length(divs(Acc)),
   if A > CurLeng ->
       longest(L,Acc,A,Acc+1);
       true -> longest(L,Current,CurLeng,Acc+1)
   end.</lang>
Output:
1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) ||  N <- lists:seq(1,10)].                                      
X: 1, N: []
X: 2, N: [1]
X: 3, N: [1]
X: 4, N: [1,2]
X: 5, N: [1]
X: 6, N: [1,2,3]
X: 7, N: [1]
X: 8, N: [1,2,4]
X: 9, N: [1,3]
X: 10, N: [1,2,5]
[ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

2> properdivs:longest(20000).
With 79, Number 15120 has the most divisors

F#

<lang F#> let mutable a=0 let mutable b=0 let mutable c=0 let mutable d=0 let mutable e=0 let mutable f=0 for k=1 to 10 do

   b <- 0
   f <- k/2  
   printf "divisor "
   for l=1 to f do
       if k%l=0 then
          b <- b+1
          printf " %i," l
   printf "no of divisor %i" b
   printfn ""

for i=1 to 20000 do

   b <- 0
   f <- i/2
   for j=1 to f do
      if i%j=0 then
        b <- b+1
   if b=c then
        d <- 0
        d <- i
   if c<b then     
       c <- b
   

printfn "%i has %i divisor" d c

</lang>

A purely functional approach. <lang fsharp> // the simple function with the answer let propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0)

// to cache the result length; helpful for a long search let propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs

// UI: always the longest and messiest let show (n,count,divs) =

 let showCount = count |> function | 0-> "no proper divisors" | 1->"1 proper divisor" | _-> sprintf "%d proper divisors" count
 let showDiv = divs |> function | []->"" | x::[]->sprintf ": %d" x | _->divs |> Seq.map string |> String.concat "," |> sprintf ": %s"
 printfn "%d has %s%s" n showCount showDiv

// generate output [1..10] |> List.iter (propDivDat >> show)

// use a sequence: we don't really need to hold this data, just iterate over it Seq.init 20000 ( ((+) 1) >> propDivDat) |> Seq.fold (fun a b ->match a,b with | (_,c1,_),(_,c2,_) when c2 > c1 -> b | _-> a) (0,0,[]) |> fun (n,count,_) -> (n,count,[]) |> show </lang>

Output:
1 has no proper divisors
2 has 1 proper divisor: 1
3 has 1 proper divisor: 1
4 has 2 proper divisors: 1,2
5 has 1 proper divisor: 1
6 has 3 proper divisors: 1,2,3
7 has 1 proper divisor: 1
8 has 3 proper divisors: 1,2,4
9 has 2 proper divisors: 1,3
10 has 3 proper divisors: 1,2,5
15120 has 79 proper divisors

Factor

<lang factor>USING: formatting io kernel math math.functions math.primes.factors math.ranges prettyprint sequences ;

#divisors ( m -- n )
   dup sqrt >integer 1 + [1,b] [ divisor? ] with count dup +
   1 - ;

10 [1,b] [ dup pprint bl divisors but-last . ] each 20000 [1,b] [ #divisors ] supremum-by dup #divisors "%d with %d divisors.\n" printf</lang>

Output:
1 { }
2 { 1 }
3 { 1 }
4 { 1 2 }
5 { 1 }
6 { 1 2 3 }
7 { 1 }
8 { 1 2 4 }
9 { 1 3 }
10 { 1 2 5 }
15120 with 79 divisors.

Fortran

Compiled using G95 compiler, run on x86 system under Puppy Linux <lang Fortran>

     function icntprop(num  )
     icnt=0
     do i=1 , num-1
         if (mod(num , i)  .eq. 0)  then
         icnt = icnt + 1
         if (num .lt. 11) print *,'    ',i
         end if
         end do
     icntprop =  icnt
     end function

     limit = 20000
     maxcnt = 0
     print *,'N   divisors'
     do j=1,limit,1
     if (j .lt. 11) print *,j
     icnt = icntprop(j)

     if (icnt .gt. maxcnt) then
     maxcnt = icnt
     maxj = j
     end if

     end do

     print *,' '
     print *,' from 1 to ',limit
     print *,maxj,' has max proper divisors: ',maxcnt
     end

</lang>

Output:

 N   divisors
 1
 2
      1
 3
      1
 4
      1
      2
 5
      1
 6
      1
      2
      3
 7
      1
 8
      1
      2
      4
 9
      1
      3
 10
      1
      2
      5
 
  from 1 to  20000
 15120  has max proper divisors:  79

FreeBASIC

<lang freebasic> ' FreeBASIC v1.05.0 win64

Sub ListProperDivisors(limit As Integer)

 If limit < 1 Then Return
 For i As Integer = 1 To limit
    Print Using "##"; i; 
    Print " ->";
    If i = 1 Then 
      Print " (None)"
      Continue For
    End if
    For j As Integer = 1 To i \ 2
      If i Mod j = 0 Then Print " "; j;
    Next j
    Print
 Next i

End Sub

Function CountProperDivisors(number As Integer) As Integer

 If number < 2 Then Return 0
 Dim count As Integer = 0
 For i As Integer = 1 To number \ 2
   If number Mod i = 0 Then count += 1
 Next
 Return count

End Function

Dim As Integer n, count, most = 1, maxCount = 0

Print "The proper divisors of the following numbers are :" Print ListProperDivisors(10)

For n As Integer = 2 To 20000

 count = CountProperDivisors(n)
 If count > maxCount Then
   maxCount = count
   most = n
 EndIf

Next

Print Print Str(most); " has the most proper divisors, namely"; maxCount Print Print "Press any key to exit the program" Sleep End </lang>

Output:
The proper divisors of the following numbers are :

 1 -> (None)
 2 ->  1
 3 ->  1
 4 ->  1  2
 5 ->  1
 6 ->  1  2  3
 7 ->  1
 8 ->  1  2  4
 9 ->  1  3
10 ->  1  2  5

15120 has the most proper divisors, namely 79

Frink

Frink's built-in factorization routines efficiently find factors of arbitrary-sized integers.

<lang frink> for n = 1 to 10

  println["$n\t" + join[" ", properDivisors[n]]]

println[]

d = new dict for n = 1 to 20000 {

  c = length[properDivisors[n]]
  d.addToList[c, n]

}

most = max[keys[d]] println[d@most + " have $most factors"]

properDivisors[n] := allFactors[n, true, false, true] </lang>

Output:
1
2       1
3       1
4       1 2
5       1
6       1 2 3
7       1
8       1 2 4
9       1 3
10      1 2 5

[15120, 18480] have 79 factors

GFA Basic

<lang> OPENW 1 CLEARW 1 ' ' Array f% is used to hold the divisors DIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed ' ' 1. Show proper divisors of 1 to 10, inclusive ' FOR i%=1 TO 10

 num%=@proper_divisors(i%)
 PRINT "Divisors for ";i%;":";
 FOR j%=1 TO num%
   PRINT " ";f%(j%);
 NEXT j%
 PRINT

NEXT i% ' ' 2. Find (smallest) number <= 20000 with largest number of proper divisors ' result%=1 ! largest so far number%=0 ! its number of divisors FOR i%=1 TO 20000

 num%=@proper_divisors(i%)
 IF num%>number%
   result%=i%
   number%=num%
 ENDIF

NEXT i% PRINT "Largest number of divisors is ";number%;" for ";result% ' ~INP(2) CLOSEW 1 ' ' find the proper divisors of n%, placing results in f% ' and return the number found ' FUNCTION proper_divisors(n%)

 LOCAL i%,root%,count%
 '
 ARRAYFILL f%(),0
 count%=1 ! index of next slot in f% to fill
 '
 IF n%>1
   f%(count%)=1
   count%=count%+1
   root%=SQR(n%)
   FOR i%=2 TO root%
     IF n% MOD i%=0
       f%(count%)=i%
       count%=count%+1
       IF i%*i%<>n% ! root% is an integer, so check if i% is actual squa- lists:seq(1,10)].                                      

X: 1, N: [] X: 2, N: [1] X: 3, N: [1] X: 4, N: [1,2] X: 5, N: [1] X: 6, N: [1,2,3] X: 7, N: [1] X: 8, N: [1,2,4] X: 9, N: [1,3] X: 10, N: [1,2,5] [ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

2> properdivs:longest(20000). With 79, Number 15120 has the most divisors re root of n%

         f%(count%)=n%/i%
         count%=count%+1
       ENDIF
     ENDIF
   NEXT i%
 ENDIF
 '
 RETURN count%-1

ENDFUNC </lang>

Output is:

Divisors for 1:
Divisors for 2: 1
Divisors for 3: 1
Divisors for 4: 1 2
Divisors for 5: 1
Divisors for 6: 1 2 3
Divisors for 7: 1
Divisors for 8: 1 2 4
Divisors for 9: 1 3
Divisors for 10: 1 2 5
Largest number of divisors is 79 for 15120

Go

Translation of: Kotlin

<lang go>package main

import (

   "fmt"
   "strconv"

)

func listProperDivisors(limit int) {

   if limit < 1 {
       return
   }
   width := len(strconv.Itoa(limit))
   for i := 1; i <= limit; i++ {
       fmt.Printf("%*d -> ", width, i)
       if i == 1 {
           fmt.Println("(None)")
           continue
       }
       for j := 1; j <= i/2; j++ {
           if i%j == 0 {
               fmt.Printf(" %d", j)
           }
       }
       fmt.Println()
   }

}

func countProperDivisors(n int) int {

   if n < 2 {
       return 0
   }
   count := 0
   for i := 1; i <= n/2; i++ {
       if n%i == 0 {
           count++
       }
   }
   return count

}

func main() {

   fmt.Println("The proper divisors of the following numbers are :\n")
   listProperDivisors(10)
   fmt.Println()
   maxCount := 0
   most := []int{1}
   for n := 2; n <= 20000; n++ {
       count := countProperDivisors(n)
       if count == maxCount {
           most = append(most, n)
       } else if count > maxCount {
           maxCount = count
           most = most[0:1]
           most[0] = n
       }
   }
   fmt.Print("The following number(s) <= 20000 have the most proper divisors, ")
   fmt.Println("namely", maxCount, "\b\n")
   for _, n := range most {
       fmt.Println(n)
   }

}</lang>

Output:
The proper divisors of the following numbers are :

 1 -> (None)
 2 ->  1
 3 ->  1
 4 ->  1 2
 5 ->  1
 6 ->  1 2 3
 7 ->  1
 8 ->  1 2 4
 9 ->  1 3
10 ->  1 2 5

The following number(s) <= 20000 have the most proper divisors, namely 79 

15120
18480

Haskell

<lang Haskell>import Data.Ord import Data.List

divisors :: (Integral a) => a -> [a] divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)]

main :: IO () main = do

 putStrLn "divisors of 1 to 10:"
 mapM_ (print . divisors) [1 .. 10]
 putStrLn "a number with the most divisors within 1 to 20000 (number, count):"
 print $ maximumBy (comparing snd)
   [(n, length $ divisors n) | n <- [1 .. 20000]]</lang>
Output:
divisors of 1 to 10:
[]
[1]
[1]
[1,2]
[1]
[1,2,3]
[1]
[1,2,4]
[1,3]
[1,2,5]
a number with the most divisors within 1 to 20000 (number, count):
(18480,79)

For a little more efficiency, we can filter only up to the root – deriving the higher proper divisors from the lower ones, as quotients:

<lang haskell>import Data.List (maximumBy) import Data.Ord (comparing) import Data.Bool (bool)

properDivisors

 :: Integral a
 => a -> [a]

properDivisors n =

 let root = (floor . sqrt . fromIntegral) n
     lows = filter ((0 ==) . rem n) [1 .. root]
 in init (lows ++ bool id tail (n == root * root) (reverse (quot n <$> lows)))

main :: IO () main = do

 putStrLn "Proper divisors of 1 to 10:"
 mapM_ (print . properDivisors) [1 .. 10]
 mapM_
   putStrLn
   [ ""
   , "A number in the range 1 to 20,000 with the most proper divisors,"
   , "as (number, count of proper divisors):"
   , ""
   ]
 print $
   maximumBy (comparing snd) $
   (,) <*> (length . properDivisors) <$> [1 .. 20000]</lang> 
Output:
Proper divisors of 1 to 10:
[]
[1]
[1]
[1,2]
[1]
[1,2,3]
[1]
[1,2,4]
[1,3]
[1,2,5]

A number in the range 1 to 20,000 with the most proper divisors,
as (number, count of proper divisors):

(18480,79)

and we can also define properDivisors in terms of primeFactors:

<lang haskell>import Data.Numbers.Primes (primeFactors) import Data.List (group, maximumBy, sort) import Data.Ord (comparing)

properDivisors :: Int -> [Int] properDivisors =

 init . sort . foldr (
   flip ((<*>) . fmap (*)) . scanl (*) 1
 ) [1] . group . primeFactors

TEST----------------------------

main :: IO () main = do

 putStrLn $
   fTable "Proper divisors of [1..10]:" show show properDivisors [1 .. 10]
 mapM_
   putStrLn
   [ ""
   , "A number in the range 1 to 20,000 with the most proper divisors,"
   , "as (number, count of proper divisors):"
   , ""
   ]
 print $
   maximumBy (comparing snd) $
   (,) <*> (length . properDivisors) <$> [1 .. 20000]

DISPLAY--------------------------

fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String fTable s xShow fxShow f xs =

 let rjust n c = (drop . length) <*> (replicate n c ++)
     w = maximum (length . xShow <$> xs)
 in unlines $
    s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs</lang>
Output:
Proper divisors of [1..10]:
 1 -> []
 2 -> [1]
 3 -> [1]
 4 -> [1,2]
 5 -> [1]
 6 -> [1,2,3]
 7 -> [1]
 8 -> [1,2,4]
 9 -> [1,3]
10 -> [1,2,5]


A number in the range 1 to 20,000 with the most proper divisors,
as (number, count of proper divisors):

(18480,79)

J

The proper divisors of an integer are the Factors of an integer without the integer itself.

So, borrowing from the J implementation of that related task:

<lang J>factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__ properDivisors=: factors -. ]</lang>

Proper divisors of numbers 1 through 10:

<lang J> (,&": ' -- ' ,&": properDivisors)&>1+i.10 1 -- 2 -- 1 3 -- 1 4 -- 1 2 5 -- 1 6 -- 1 2 3 7 -- 1 8 -- 1 2 4 9 -- 1 3 10 -- 1 2 5</lang>

Number(s) not exceeding 20000 with largest number of proper divisors (and the count of those divisors):

<lang J> (, #@properDivisors)&> 1+I.(= >./) #@properDivisors@> 1+i.20000 15120 79 18480 79</lang>

Note that it's a bit more efficient to simply count factors here, when selecting the candidate numbers.

<lang J> (, #@properDivisors)&> 1+I.(= >./) #@factors@> 1+i.20000 15120 79 18480 79</lang>

We could also arbitrarily toss either 15120 or 18480 (keeping the other number), if it were important that we produce only one result.

Java

Works with: Java version 1.5+

<lang java5>import java.util.Collections; import java.util.LinkedList; import java.util.List;

public class Proper{

   public static List<Integer> properDivs(int n){
       List<Integer> divs = new LinkedList<Integer>();
       if(n == 1) return divs;
       divs.add(1);
       for(int x = 2; x < n; x++){
           if(n % x == 0) divs.add(x);
       }
       
       Collections.sort(divs);
       
       return divs;
   }
   
   public static void main(String[] args){
       for(int x = 1; x <= 10; x++){
           System.out.println(x + ": " + properDivs(x));
       }
       
       int x = 0, count = 0;
       for(int n = 1; n <= 20000; n++){
           if(properDivs(n).size() > count){
               x = n;
               count = properDivs(n).size();
           }
       }
       System.out.println(x + ": " + count);
   }

}</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79

JavaScript

ES5

<lang JavaScript>(function () {

   // Proper divisors
   function properDivisors(n) {
       if (n < 2) return [];
       else {
           var rRoot = Math.sqrt(n),
               intRoot = Math.floor(rRoot),
               lows = range(1, intRoot).filter(function (x) {
                   return (n % x) === 0;
               });
           return lows.concat(lows.slice(1).map(function (x) {
               return n / x;
           }).reverse().slice((rRoot === intRoot) | 0));
       }
   }
   // [m..n]
   function range(m, n) {
       var a = Array(n - m + 1),
           i = n + 1;
       while (i--) a[i - 1] = i;
       return a;
   }
   var tblOneToTen = [
           ['Number', 'Proper Divisors', 'Count']
       ].concat(range(1, 10).map(function (x) {
           var ds = properDivisors(x);
           return [x, ds.join(', '), ds.length];
       })),
       dctMostBelow20k = range(1, 20000).reduce(function (a, x) {
           var lng = properDivisors(x).length;
           return lng > a.divisorCount ? {
               n: x,
               divisorCount: lng
           } : a;
       }, {
           n: 0,
           divisorCount: 0
       });


   // a -> bool -> s -> s
   function wikiTable(lstRows, blnHeaderRow, strStyle) {
       return '{| class="wikitable" ' + (
           strStyle ? 'style="' + strStyle + '"' : 
       ) + lstRows.map(function (lstRow, iRow) {
           var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');
           return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
               return typeof v === 'undefined' ? ' ' : v;
           }).join(' ' + strDelim + strDelim + ' ');
       }).join() + '\n|}';
   }
   return wikiTable(
       tblOneToTen,
       true
   ) + '\n\nMost proper divisors below 20,000:\n\n  ' + JSON.stringify(
       dctMostBelow20k
   );

})();</lang>

Output:
Number Proper Divisors Count
1 0
2 1 1
3 1 1
4 1, 2 2
5 1 1
6 1, 2, 3 3
7 1 1
8 1, 2, 4 3
9 1, 3 2
10 1, 2, 5 3

Most proper divisors below 20,000:

 {"n":15120,"divisorCount":79}

ES6

<lang JavaScript>(() => {

   'use strict';
   // properDivisors :: Int -> [Int]
   const properDivisors = n => {
       // The integer divisors of n, excluding n itself.
       const
           rRoot = Math.sqrt(n),
           intRoot = Math.floor(rRoot),
           blnPerfectSquare = rRoot === intRoot,
           lows = enumFromTo(1)(intRoot)
           .filter(x => 0 === (n % x));
       // For perfect squares, we can drop
       // the head of the 'highs' list
       return lows.concat(lows
               .map(x => n / x)
               .reverse()
               .slice(blnPerfectSquare | 0)
           )
           .slice(0, -1); // except n itself
   };


   // ------------------------TESTS-----------------------
   // main :: IO ()
   const main = () =>
       console.log([
           fTable('Proper divisors of [1..10]:')(str)(
               JSON.stringify
           )(properDivisors)(enumFromTo(1)(10)),
           ,
           'Example of maximum divisor count in the range [1..20000]:',
           '    ' + maximumBy(comparing(snd))(
               enumFromTo(1)(20000).map(
                   n => [n, properDivisors(n).length]
               )
           ).join(' has ') + ' proper divisors.'
       ].join('\n'));


   // -----------------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);
       };
   // enumFromTo :: Int -> Int -> [Int]
   const enumFromTo = m => n =>
       Array.from({
           length: 1 + n - m
       }, (_, i) => m + i);
   // fTable :: String -> (a -> String) -> (b -> String)
   //                      -> (a -> b) -> [a] -> String
   const fTable = s => xShow => fxShow => f => xs => {
       // Heading -> x display function ->
       //           fx display function ->
       //    f -> values -> tabular string
       const
           ys = xs.map(xShow),
           w = Math.max(...ys.map(x => x.length));
       return s + '\n' + zipWith(
           a => b => a.padStart(w, ' ') + ' -> ' + b
       )(ys)(
           xs.map(x => fxShow(f(x)))
       ).join('\n');
   };
   // maximumBy :: (a -> a -> Ordering) -> [a] -> a
   const maximumBy = f => xs =>
       0 < xs.length ? (
           xs.slice(1)
           .reduce((a, x) => 0 < f(x)(a) ? x : a, xs[0])
       ) : undefined;
   // snd :: (a, b) -> b
   const snd = tpl => tpl[1];
   // str :: a -> String
   const str = x => x.toString();
   // until :: (a -> Bool) -> (a -> a) -> a -> a
   const until = p => f => x => {
       let v = x;
       while (!p(v)) v = f(v);
       return v;
   };
   // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
   const zipWith = f => xs => ys => {
       const
           lng = Math.min(xs.length, xs.length),
           as = xs.slice(0, lng),
           bs = ys.slice(0, lng);
       return Array.from({
           length: lng
       }, (_, i) => f(as[i])(
           bs[i]
       ));
   };
   // MAIN ---
   return main();

})();</lang>

Output:
Proper divisors of [1..10]:
 1 -> []
 2 -> [1]
 3 -> [1]
 4 -> [1,2]
 5 -> [1]
 6 -> [1,2,3]
 7 -> [1]
 8 -> [1,2,4]
 9 -> [1,3]
10 -> [1,2,5]

Example of maximum divisor count in the range [1..20000]:
    15120 has 79 proper divisors.

jq

Works with: jq version 1.4

In the following, proper_divisors returns a stream. In order to count the number of items in the stream economically, we first define "count(stream)": <lang jq>def count(stream): reduce stream as $i (0; . + 1);

  1. unordered

def proper_divisors:

 . as $n
 | if $n > 1 then 1,
     ( range(2; 1 + (sqrt|floor)) as $i
       | if ($n % $i) == 0 then $i,
           (($n / $i) | if . == $i then empty else . end)
        else empty

end)

   else empty
   end;
  1. The first integer in 1 .. n inclusive
  2. with the maximal number of proper divisors in that range:

def most_proper_divisors(n):

 reduce range(1; n+1) as $i
   ( [null, 0];
     count( $i | proper_divisors ) as $count
     | if $count > .[1] then [$i, $count] else . end);</lang>

The tasks: <lang jq>"The proper divisors of the numbers 1 to 10 inclusive are:", (range(1;11) as $i | "\($i): \( [ $i | proper_divisors] )"), "", "The pair consisting of the least number in the range 1 to 20,000 with", "the maximal number proper divisors together with the corresponding", "count of proper divisors is:", most_proper_divisors(20000) </lang>

Output:

<lang sh>$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jq The proper divisors of the numbers 1 to 10 inclusive are: 1: [] 2: [1] 3: [1] 4: [1,2] 5: [1] 6: [1,2,3] 7: [1] 8: [1,2,4] 9: [1,3] 10: [1,2,5]

The pair consisting of the least number in the range 1 to 20,000 with the maximal number proper divisors together with the corresponding count of proper divisors is: [15120,79]</lang>

Julia

Use factor to obtain the prime factorization of the target number. I adopted the argument handling style of factor in my properdivisors function. <lang Julia> function properdivisors{T<:Integer}(n::T)

   0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got $n"))
   1 < n || return T[]
   !isprime(n) || return T[one(T), n]
   f = factor(n)
   d = T[one(T)]
   for (k, v) in f
       c = T[k^i for i in 0:v]
       d = d*c'
       d = reshape(d, length(d))
   end
   sort!(d)
   return d[1:end-1]

end

lo = 1 hi = 10 println("List the proper divisors for ", lo, " through ", hi, ".") for i in lo:hi

   println(@sprintf("%4d", i), " ", properdivisors(i))

end

hi = 2*10^4 println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.")

maxdiv = 0 nlst = Int[]

for i in lo:hi

   ndiv = length(properdivisors(i))
   if ndiv > maxdiv
       maxdiv = ndiv
       nlst = [i]
   elseif ndiv == maxdiv
       push!(nlst, i)
   end

end

println(nlst, " have the maximum proper divisor count of ", maxdiv, ".") </lang>

Output:
List the proper divisors for 1 through 10.
   1 []
   2 [1,2]
   3 [1,3]
   4 [1,2]
   5 [1,5]
   6 [1,2,3]
   7 [1,7]
   8 [1,2,4]
   9 [1,3]
  10 [1,2,5]

Find the numbers within [1,20000] having the most divisors.
[15120,18480] have the maximum proper divisor count of 79.

Kotlin

<lang scala>// version 1.0.5-2

fun listProperDivisors(limit: Int) {

   if (limit < 1) return
   for(i in 1..limit) {
       print(i.toString().padStart(2) + " -> ")
       if (i == 1) {
           println("(None)")
           continue
       }
       (1..i/2).filter{ i % it == 0 }.forEach { print(" $it") }
       println()
   }

}

fun countProperDivisors(n: Int): Int {

   if (n < 2) return 0
   return (1..n/2).count { (n % it) == 0 }

}

fun main(args: Array<String>) {

   println("The proper divisors of the following numbers are :\n")
   listProperDivisors(10)
   println()
   var count: Int
   var maxCount = 0
   val most: MutableList<Int> = mutableListOf(1)
   for (n in 2..20000) {
       count = countProperDivisors(n)
       if (count == maxCount)
           most.add(n)
       else if (count > maxCount) {
           maxCount = count
           most.clear()
           most.add(n)
       }
   }
   println("The following number(s) have the most proper divisors, namely " + maxCount + "\n")
   for (n in most) println(n)

}</lang>

Output:
The proper divisors of the following numbers are :

 1 -> (None)
 2 ->  1
 3 ->  1
 4 ->  1 2
 5 ->  1
 6 ->  1 2 3
 7 ->  1
 8 ->  1 2 4
 9 ->  1 3
10 ->  1 2 5

The following number(s) have the most proper divisors, namely 79

15120
18480

Lua

<lang Lua>-- Return a table of the proper divisors of n function propDivs (n)

   if n < 2 then return {} end
   local divs, sqr = {1}, math.sqrt(n)
   for d = 2, sqr do
       if n % d == 0 then
           table.insert(divs, d)
           if d ~= sqr then table.insert(divs, n/d) end
       end
   end
   table.sort(divs)
   return divs

end

-- Show n followed by all values in t function show (n, t)

   io.write(n .. ":\t")
   for _, v in pairs(t) do io.write(v .. " ") end
   print()

end

-- Main procedure local mostDivs, numDivs, answer = 0 for i = 1, 10 do show(i, propDivs(i)) end for i = 1, 20000 do

   numDivs = #propDivs(i)
   if numDivs > mostDivs then
       mostDivs = numDivs
       answer = i
   end

end print(answer .. " has " .. mostDivs .. " proper divisors.")</lang>

Output:
1:
2:      1
3:      1
4:      1 2
5:      1
6:      1 2 3
7:      1
8:      1 2 4
9:      1 3
10:     1 2 5
15120 has 79 proper divisors.

Mathematica / Wolfram Language

A Function that yields the proper divisors of an integer n: <lang Mathematica>ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;</lang>

Proper divisors of n from 1 to 10: <lang Mathematica>Grid@Table[{n, ProperDivisors[n]}, {n, 1, 10}]</lang>

Output:
1	{}
2	{1}
3	{1}
4	{1,2}
5	{1}
6	{1,2,3}
7	{1}
8	{1,2,4}
9	{1,3}
10	{1,2,5}

The number with the most divisors between 1 and 20,000: <lang Mathematica>Fold[

Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &,
{0, 0},
Range[20000]]</lang>
Output:
{18480, 79}

An alternate way to find the number with the most divisors between 1 and 20,000: <lang Mathematica>Last@SortBy[

 Table[
   {n, Length@ProperDivisors[n]},
   {n, 1, 20000}],
 Last]</lang>
Output:
{15120, 79}

MATLAB

<lang matlab>function D=pd(N) K=1:ceil(N/2); D=K(~(rem(N, K)));</lang>

Output:
for I=1:10
   disp([num2str(I) ' : ' num2str(pd(I))])
end
1 : 1
2 : 1
3 : 1
4 : 1  2
5 : 1
6 : 1  2  3
7 : 1
8 : 1  2  4
9 : 1  3
10 : 1  2  5

maxL=0; maxI=0;
for I=1:20000
   L=length(pd(I));
   if L>maxL
      maxL=L; maxI=I;
   end
end
maxI

maxI =

       15120

maxL

maxL =

    79

Modula-2

<lang modula2>MODULE ProperDivisors; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;

PROCEDURE WriteInt(n : INTEGER); VAR buf : ARRAY[0..15] OF CHAR; BEGIN

   FormatString("%i", buf, n);
   WriteString(buf)

END WriteInt;

PROCEDURE proper_divisors(n : INTEGER; print_flag : BOOLEAN) : INTEGER; VAR count,i : INTEGER; BEGIN

   count := 0;
   FOR i:=1 TO n-1 DO
       IF n MOD i = 0 THEN
           INC(count);
           IF print_flag THEN
               WriteInt(i);
               WriteString(" ")
           END
       END
   END;
   IF print_flag THEN WriteLn END;
   RETURN count;

END proper_divisors;

VAR

   buf : ARRAY[0..63] OF CHAR;
   i,max,max_i,v : INTEGER;

BEGIN

   FOR i:=1 TO 10 DO
       WriteInt(i);
       WriteString(": ");
       proper_divisors(i, TRUE)
   END;
   max := 0;
   max_i := 1;
   FOR i:=1 TO 20000 DO
       v := proper_divisors(i, FALSE);
       IF v>= max THEN
           max := v;
           max_i := i
       END
   END;
   FormatString("%i with %i divisors\n", buf, max_i, max);
   WriteString(buf);
   ReadChar

END ProperDivisors.</lang>

Nim

Translation of: C

<lang nim>import strformat

proc properDivisors(n: int) =

 var count = 0
 for i in 1..<n:
   if n mod i == 0:
     inc count
     write(stdout, fmt"{i} ")
 write(stdout, "\n")

proc countProperDivisors(n: int): int =

 var nn = n
 var prod = 1
 var count = 0
 while nn mod 2 == 0:
   inc count
   nn = nn div 2
 prod *= (1 + count)
 for i in countup(3, n, 2):
   count = 0
   while nn mod i == 0:
     inc count
     nn = nn div i
   prod *= (1 + count)
 if nn > 2:
   prod *= 2
 prod - 1

for i in 1..10:

 write(stdout, fmt"{i:2}: ")
 properDivisors(i)

var max = 0 var maxI = 1

for i in 1..20000:

 var v = countProperDivisors(i)
 if v >= max:
   max = v
   maxI = i

echo fmt"{maxI} with {max} divisors"</lang>

Output:
 1: 
 2: 1 
 3: 1 
 4: 1 2 
 5: 1 
 6: 1 2 3 
 7: 1 
 8: 1 2 4 
 9: 1 3 
10: 1 2 5 
18480 with 79 divisors

Oberon-2

<lang oberon2> MODULE ProperDivisors; IMPORT

 Out;

CONST

   initialSize = 128;

TYPE

 Result* = POINTER TO ResultDesc;
 ResultDesc = RECORD 
   found-: LONGINT; (* number of slots in pd *)
   pd-: POINTER TO ARRAY OF LONGINT;
   cap: LONGINT;   (* Capacity *)
 END;
 

VAR

 i,found,max,idxMx: LONGINT;
 mx: ARRAY 32 OF LONGINT;
 rs: Result;
 PROCEDURE (r: Result) Init(size: LONGINT);
 BEGIN 
   r.found := 0;
   r.cap := size;
   NEW(r.pd,r.cap);
 END Init;
 PROCEDURE (r: Result) Add(n: LONGINT);
 BEGIN
   (* Out.String("--->");Out.LongInt(n,0);Out.String(" At: ");Out.LongInt(r.found,0);Out.Ln; *)
   IF (r.found < LEN(r.pd^) - 1) THEN 
     r.pd[r.found] := n;
   ELSE
     (* expand pd for more room *)
   END;
   INC(r.found);
 END Add;
 PROCEDURE (r:Result) Show();
 VAR
   i: LONGINT;
 BEGIN
     Out.String("(Result:");Out.LongInt(r.found + 1,0);(* Out.String("/");Out.LongInt(r.cap,0);*)
     Out.String("-");
     IF r.found > 0 THEN
       FOR i:= 0 TO r.found - 1 DO
         Out.LongInt(r.pd[i],0);
         IF i = r.found - 1 THEN Out.Char(')') ELSE Out.Char(',') END
       END
     END;
     Out.Ln
 END Show;
 PROCEDURE (r:Result) Reset();
 BEGIN 
   r.found := 0;
 END Reset;
 PROCEDURE GetFor(n: LONGINT;VAR rs: Result);
 VAR
   i: LONGINT;
 BEGIN
   IF n > 1 THEN 
     rs.Add(1);i := 2;
     WHILE (i < n) DO
       IF (n MOD i) = 0 THEN rs.Add(i) END;
       INC(i)
     END
   END;
 END GetFor;
 

BEGIN

 NEW(rs);rs.Init(initialSize);
 FOR i := 1 TO 10 DO 
   Out.LongInt(i,4);Out.Char(':');
   GetFor(i,rs);
   rs.Show();
   rs.Reset();
 END;
 Out.LongInt(100,4);Out.Char(':');GetFor(100,rs);rs.Show();rs.Reset();
 max := 0;idxMx := 0;found := 0;
 FOR i := 1 TO 20000 DO
   GetFor(i,rs);
   IF rs.found > max THEN
     idxMx:= 0;mx[idxMx] := i;max := rs.found
   ELSIF rs.found = max THEN
     INC(idxMx);mx[idxMx] := i
   END;
   rs.Reset()
 END;
 Out.String("Found: ");Out.LongInt(idxMx + 1,0);
 Out.String(" Numbers with most proper divisors ");
 Out.LongInt(max,0);Out.String(": ");Out.Ln;
 FOR i := 0 TO idxMx DO
   Out.LongInt(mx[i],0);Out.Ln
 END

END ProperDivisors. </lang>

Output:
   1:(Result:1-
   2:(Result:2-1)
   3:(Result:2-1)
   4:(Result:3-1,2)
   5:(Result:2-1)
   6:(Result:4-1,2,3)
   7:(Result:2-1)
   8:(Result:4-1,2,4)
   9:(Result:3-1,3)
  10:(Result:4-1,2,5)
 100:(Result:9-1,2,4,5,10,20,25,50)
Found: 2 Numbers with most proper divisors 79: 
15120
18480

Objeck

<lang objeck>use Collection;

class Proper{

 function : Main(args : String[]) ~ Nil {
   for(x := 1; x <= 10; x++;) {
     Print(x, ProperDivs(x));
   };
       
   x := 0;
   count := 0;
   
   for(n := 1; n <= 20000; n++;) {
     if(ProperDivs(n)->Size() > count) {
       x := n;
       count := ProperDivs(n)->Size();
     };
   };
   "{$x}: {$count}"->PrintLine();
 }
 function : ProperDivs(n : Int) ~ IntVector {
   divs := IntVector->New();
       
   if(n = 1) {
     return divs;
   };
   divs->AddBack(1);
       
   for(x := 2; x < n; x++;) {
     if(n % x = 0) { 
       divs->AddBack(x);
     };
   };
   divs->Sort();
       
   return divs;
 }
   
 function : Print(x : Int, result : IntVector) ~ Nil {
   "{$x}: "->Print();
   result->ToArray()->ToString()->PrintLine();
 }

} </lang>

Output:

1: []
2: [1]
3: [1]
4: [1,2]
5: [1]
6: [1,2,3]
7: [1]
8: [1,2,4]
9: [1,3]
10: [1,2,5]
15120: 79

Oforth

<lang Oforth>Integer method: properDivs self 2 / seq filter(#[ self swap mod 0 == ]) }

10 seq apply(#[ dup print " : " print properDivs println ]) 20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println</lang>

Output:
1 : []
2 : [1]
3 : [1]
4 : [1, 2]
5 : [1]
6 : [1, 2, 3]
7 : [1]
8 : [1, 2, 4]
9 : [1, 3]
10 : [1, 2, 5]
[79, 15120]

PARI/GP

<lang parigp>proper(n)=if(n==1, [], my(d=divisors(n)); d[2..#d]); apply(proper, [1..10]) r=at=0; for(n=1,20000, t=numdiv(n); if(t>r, r=t; at=n)); [at, numdiv(t)-1]</lang>

Output:
%1 = [[], [2], [3], [2, 4], [5], [2, 3, 6], [7], [2, 4, 8], [3, 9], [2, 5, 10]]
%2 = [15120, 7]

Pascal

Works with: Free Pascal

Using prime factorisation <lang pascal>{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} uses

 sysutils;

const

 MAXPROPERDIVS = 1920;

type

 tRes = array[0..MAXPROPERDIVS] of LongWord;
 tPot = record
          potPrim,
          potMax :LongWord;
        end;
 tprimeFac = record
                pfPrims : array[1..10] of tPot;
                pfCnt,
                pfNum   : LongWord;
              end;
 tSmallPrimes = array[0..6541] of longWord;

var

 SmallPrimes: tSmallPrimes;

procedure InitSmallPrimes; var

 pr,testPr,j,maxprimidx: Longword;
 isPrime : boolean;

Begin

 maxprimidx := 0;
 SmallPrimes[0] := 2;
 pr := 3;
 repeat
   isprime := true;
   j := 0;
   repeat
     testPr := SmallPrimes[j];
     IF testPr*testPr > pr then
       break;
     If pr mod testPr = 0 then
     Begin
       isprime := false;
       break;
     end;
     inc(j);
   until false;
   if isprime then
   Begin
     inc(maxprimidx);
     SmallPrimes[maxprimidx]:= pr;
   end;
   inc(pr,2);
 until pr > 1 shl 16 -1;

end;

procedure PrimeFacOut(primeDecomp:tprimeFac); var

 i : LongWord;

begin

 with primeDecomp do
 Begin
   write(pfNum,' = ');
   For i := 1 to pfCnt-1 do
     with pfPrims[i] do
       If potMax = 1 then
         write(potPrim,'*')
       else
         write(potPrim,'^',potMax,'*');
   with pfPrims[pfCnt] do
     If potMax = 1 then
       write(potPrim)
     else
       write(potPrim,'^',potMax);
 end;

end;

procedure PrimeDecomposition(n:LongWord;var res:tprimeFac); var

 i,pr,cnt,quot{to minimize divisions} : LongWord;

Begin

 res.pfNum := n;
 res.pfCnt:= 0;
 i := 0;
 cnt := 0;
 repeat
   pr := SmallPrimes[i];
   IF pr*pr>n then
     Break;
   quot := n div pr;
   IF pr*quot = n then
     with res do
     Begin
       inc(pfCnt);
       with pfPrims[pfCnt] do
       Begin
         potPrim := pr;
         potMax := 0;
         repeat
           n := quot;
           quot := quot div pr;
           inc(potMax);
         until pr*quot <> n;
       end;
     end;
    inc(i);
 until false;
 //a big prime left over?
 IF n <> 1 then
   with res do
   Begin
     inc(pfCnt);
     with pfPrims[pfCnt] do
     Begin
       potPrim := n;
       potMax := 1;
     end;
   end;

end;

function CntProperDivs(const primeDecomp:tprimeFac):LongWord; //count of proper divisors var

  i: LongWord;

begin

 result := 1;
 with primeDecomp do
   For i := 1 to pfCnt do
     result := result*(pfPrims[i].potMax+1);
 //remove
 dec(result);

end;

function findProperdivs(n:LongWord;var res:TRes):LongWord; //simple trial division to get a sorted list of all proper divisors var

 i,j: LongWord;

Begin

 result := 0;
 i := 1;
 j := n;
 while j>i do
 begin
   j := n DIV i;
   IF i*j = n then
   Begin
     //smaller factor part at the beginning upwards
     res[result]:= i;
     IF i <> j then
       //bigger factor at the end downwards
       res[MAXPROPERDIVS-result]:= j
     else
       //n is square number
       res[MAXPROPERDIVS-result]:= 0;
     inc(result);
   end;
   inc(i);
 end;
 If result>0 then
 Begin
   //move close together
   i := result;
   j := MAXPROPERDIVS-result+1;
   result := 2*result-1;
   repeat
     res[i] := res[j];
     inc(j);
     inc(i);
   until i > result;
   if res[result-1] = 0 then
     dec(result);
 end;

end;

procedure AllFacsOut(n: Longword); var

 res:TRes;
 i,k,j:LongInt;

Begin

  j := findProperdivs(n,res);
  write(n:5,' : ');
  For k := 0 to j-2 do write(res[k],',');
  IF j>=1 then
    write(res[j-1]);
  writeln;

end;

var

 primeDecomp: tprimeFac;
 rs : tRes;
 i,j,max,maxcnt: LongWord;

BEGIN

 InitSmallPrimes;
 For i := 1 to 10 do
   AllFacsOut(i);
 writeln;
 max    := 0;
 maxCnt := 0;
 For i := 1 to 20*1000 do
 Begin
   PrimeDecomposition(i,primeDecomp);
   j := CntProperDivs(primeDecomp);
   IF j> maxCnt then
   Begin
     maxcnt := j;
     max := i;
   end;
 end;
 PrimeDecomposition(max,primeDecomp);
 j := CntProperDivs(primeDecomp);
 PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
 //https://en.wikipedia.org/wiki/Highly_composite_number <= HCN
 //http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN
 max := 3491888400;
 PrimeDecomposition(max,primeDecomp);
 j := CntProperDivs(primeDecomp);
 PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
END.</lang>
Output:

   1 :
   2 : 1
   3 : 1
   4 : 1,2
   5 : 1
   6 : 1,2,3
   7 : 1
   8 : 1,2,4
   9 : 1,3
  10 : 1,2,5

15120 = 2^4*3^3*5*7 79 factors

3491888400 = 2^4*3^3*5^2*7*11*13*17*19 1919 factors

real 0m0.004s

Perl

Using a module for divisors

Library: ntheory

<lang perl>use ntheory qw/divisors/; sub proper_divisors {

 my $n = shift;
 # Like Pari/GP, divisors(0) = (0,1) and divisors(1) = ()
 return 1 if $n == 0;
 my @d = divisors($n);
 pop @d;  # divisors are in sorted order, so last entry is the input
 @d;

} say "$_: ", join " ", proper_divisors($_) for 1..10;

  1. 1. For the max, we can do a traditional loop.

my($max,$ind) = (0,0); for (1..20000) {

 my $nd = scalar proper_divisors($_);
($max,$ind) = ($nd,$_) if $nd > $max;

} say "$max $ind";

  1. 2. Or we can use List::Util's max with decoration (this exploits its implementation)

{

 use List::Util qw/max/;
 no warnings 'numeric';
 say max(map { scalar(proper_divisors($_)) . " $_" } 1..20000);

}</lang>

Output:
1: 
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
79 15120
79 18480

Note that the first code will choose the first max, while the second chooses the last.

Phix

The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is <lang Phix>global function factors(atom n, integer include1=0) -- -- returns a list of all integer factors of n -- if include1 is 0 (the default), result does not contain either 1 or n -- if include1 is 1 the result contains 1 and n -- if include1 is -1 the result contains 1 but not n --

   sequence lfactors = {}, hfactors = {}
   atom hfactor
   integer p = 2,
           lim = floor(sqrt(n))
   if include1!=0 then
       lfactors = {1}
       if n!=1 and include1=1 then
           hfactors = {n}
       end if
   end if
   while p<=lim do
       if remainder(n,p)=0 then
           lfactors = append(lfactors,p)
           hfactor = n/p
           if hfactor=p then exit end if
           hfactors = prepend(hfactors,hfactor)
       end if
       p += 1
   end while 
   return lfactors & hfactors

end function</lang> The compiler knows where to find that, but note the above does not permit 0, so the main program is: <lang Phix>function proper_divisors(integer n)

   if n=0 then return {} end if
   return factors(n,-1) -- (-1 means "1 but not n")

end function

for i=0 to 10 do

   printf(1,"%d: %v\n",{i,proper_divisors(i)})

end for

sequence candidates = {} integer maxd = 0 for i=1 to 20000 do

   integer k = length(factors(i,-1))
   if k>=maxd then
       if k=maxd then
           candidates &= i
       else
           candidates = {i}
           maxd = k
       end if
   end if

end for printf(1,"%d divisors: %v\n", {maxd,candidates})</lang>

Output:
0: {}
1: {1}
2: {1}
3: {1}
4: {1,2}
5: {1}
6: {1,2,3}
7: {1}
8: {1,2,4}
9: {1,3}
10: {1,2,5}
79 divisors: {15120,18480}

PHP

<lang php><?php function ProperDivisors($n) {

 yield 1;
 $large_divisors = [];
 for ($i = 2; $i <= sqrt($n); $i++) {
   if ($n % $i == 0) {
     yield $i;
     if ($i*$i != $n) {
       $large_divisors[] = $n / $i;
     }
   }
 }
 foreach (array_reverse($large_divisors) as $i) {
   yield $i;
 }

}

assert([1, 2, 4, 5, 10, 20, 25, 50] ==

       iterator_to_array(ProperDivisors(100)));

foreach (range(1, 10) as $n) {

 echo "$n =>";
 foreach (ProperDivisors($n) as $divisor) {
   echo " $divisor";
 }
 echo "\n";

}

$divisorsCount = []; for ($i = 1; $i < 20000; $i++) {

 $divisorsCount[sizeof(iterator_to_array(ProperDivisors($i)))][] = $i;

} ksort($divisorsCount);

echo "Numbers with most divisors: ", implode(", ", end($divisorsCount)), ".\n"; echo "They have ", key($divisorsCount), " divisors.\n";

</lang>

Outputs:

1 => 1
2 => 1
3 => 1
4 => 1 2
5 => 1
6 => 1 2 3
7 => 1
8 => 1 2 4
9 => 1 3
10 => 1 2 5
Numbers with most divisors: 15120, 18480.
They have 79 divisors.

PicoLisp

<lang PicoLisp># Generate all proper divisors. (de propdiv (N)

  (head -1 (filter
     '((X) (=0 (% N X)))
     (range 1 N) )) )
  1. Obtaining the values from 1 to 10 inclusive.

(mapcar propdiv (range 1 10))

  1. Output:
  2. (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))</lang>

Brute-force

<lang PicoLisp>(de propdiv (N)

  (cdr
     (rot
        (make
           (for I N
              (and (=0 (% N I)) (link I)) ) ) ) ) )

(de countdiv (N)

  (let C -1 
     (for I N
        (and (=0 (% N I)) (inc 'C)) )
     C ) )

(let F (-5 -8)

  (tab F "N" "LIST")
  (for I 10
     (tab F
        I
        (glue " + " (propdiv I)) ) ) )

(println

  (maxi
     countdiv
     (range 1 20000) ) )</lang>

Factorization

<lang PicoLisp>(de accu1 (Var Key)

  (if (assoc Key (val Var))
     (con @ (inc (cdr @)))
     (push Var (cons Key 2)) )
  Key )

(de factor (N)

  (let
     (R NIL
        D 2
        L (1 2 2 . (4 2 4 2 4 6 2 6 .))
        M (sqrt N) )
     (while (>= M D)
        (if (=0 (% N D))
           (setq M
              (sqrt (setq N (/ N (accu1 'R D)))) )
           (inc 'D (pop 'L)) ) )
     (accu1 'R N)
     (dec (apply * (mapcar cdr R))) ) )

(bench

  (println
     (maxi
        factor
        (range 1 20000) ) 
     @@ ) )</lang>

Output:

15120 79
0.081 sec

PL/I

<lang pli>*process source xref;

(subrg):
cpd: Proc Options(main);
p9a=time();
Dcl (p9a,p9b) Pic'(9)9';
Dcl cnt(3) Bin Fixed(31) Init((3)0);
Dcl x Bin Fixed(31);
Dcl pd(300) Bin Fixed(31);
Dcl sumpd   Bin Fixed(31);
Dcl npd     Bin Fixed(31);
Dcl hi      Bin Fixed(31) Init(0);
Dcl (xl(10),xi) Bin Fixed(31);
Dcl i       Bin Fixed(31);
Do x=1 To 10;
  Call proper_divisors(x,pd,npd);
  Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2)));
  End;
xi=0;
Do x=1 To 20000;
  Call proper_divisors(x,pd,npd);
  Select;
    When(npd>hi) Do;
      xi=1;
      xl(1)=x;
      hi=npd;
      End;
    When(npd=hi) Do;
      xi+=1;
      xl(xi)=x;
      End;
    Otherwise;
    End;
  End;
Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));
x= 166320; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));
x=1441440; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));


p9b=time();
Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a);
Return;
proper_divisors: Proc(n,pd,npd);
Dcl (n,pd(300),npd) Bin Fixed(31);
Dcl (d,delta)       Bin Fixed(31);
npd=0;
If n>1 Then Do;
  If mod(n,2)=1 Then  /* odd number  */
    delta=2;
  Else                /* even number */
    delta=1;
  Do d=1 To n/2 By delta;
    If mod(n,d)=0 Then Do;
      npd+=1;
      pd(npd)=d;
      End;
    End;
  End;
End;
End;</lang>
Output:
 1 ->
 2 ->  1
 3 ->  1
 4 ->  1 2
 5 ->  1
 6 ->  1 2 3
 7 ->  1
 8 ->  1 2 4
 9 ->  1 3
10 ->  1 2 5
 79 ->  15120 18480
  166320 ->  159
 1441440 ->  287
 0.530 seconds elapsed

PowerShell

version 1

<lang PowerShell> function proper-divisor ($n) {

   if($n -ge 2) {
       $lim = [Math]::Floor([Math]::Sqrt($n))
       $less, $greater = @(1), @()
       for($i = 2; $i -lt $lim; $i++){
           if($n%$i -eq 0) {
               $less += @($i)
               $greater = @($n/$i) + $greater
           }
       }
       if(($lim -ne 1) -and ($n%$lim -eq 0)) {$less += @($lim)}
       $($less + $greater)
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>

version 2

<lang PowerShell> function proper-divisor ($n) {

   if($n -ge 2) {
       $lim = [Math]::Floor($n/2)+1
       $proper = @(1)
       for($i = 2; $i -lt $lim; $i++){
           if($n%$i -eq 0) {
               $proper += @($i)
           }
       }
       $proper
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>

version 3

<lang PowerShell> function eratosthenes ($n) {

   if($n -gt 1){
       $prime = @(0..$n| foreach{$true})
       $m = [Math]::Floor([Math]::Sqrt($n))
       function multiple($i) {
           for($j = $i*$i; $j -le $n; $j += $i) {
               $prime[$j] = $false
           }
       }
       multiple 2
       for($i = 3; $i -le $m; $i += 2) {
           if($prime[$i]) {multiple $i}
       }
       2
       for($i = 3; $i -le $n; $i += 2) {
           if($prime[$i]) {$i}
       }
       
   } else {
       Write-Error "$n is not greater than 1"
   }

} function prime-decomposition ($n) {

   $array = eratosthenes $n
   $prime = @()
   foreach($p in $array) {
       while($n%$p -eq 0) {
           $n /= $p
           $prime += @($p)
       }
   }
   $prime

} function proper-divisor ($n) {

   if($n -ge 2) {
       $array = prime-decomposition $n
       $lim = $array.Count
       function state($res, $i){  
           if($i -lt $lim) {
               state ($res) ($i + 1)
               state ($res*$array[$i]) ($i + 1)   
           } elseif($res -lt $n) {$res}
       }
       state 1 0 | sort -Unique
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang> Output:

1 2 4 5 10 20 25 50
1 2 4 8 16 31 62 124 248
1 2 4 8 16 32 64 128 256 512 1024

Prolog

Works with: SWI-Prolog 7

Taking a cue from an SO answer:

<lang prolog>divisor(N, Divisor) :-

   UpperBound is round(sqrt(N)),
   between(1, UpperBound, D),
   0 is N mod D,
   (
       Divisor = D
    ;
       LargerDivisor is N/D,
       LargerDivisor =\= D,
       Divisor = LargerDivisor
   ).

proper_divisor(N, D) :-

   divisor(N, D),
   D =\= N.


%% Task 1 %

proper_divisors(N, Ds) :-

   setof(D, proper_divisor(N, D), Ds).


%% Task 2 %

show_proper_divisors_of_range(Low, High) :-

   findall( N:Ds,
            ( between(Low, High, N),
              proper_divisors(N, Ds) ),
            Results ),
   maplist(writeln, Results).


%% Task 3 %

proper_divisor_count(N, Count) :-

   proper_divisors(N, Ds),
   length(Ds, Count).

find_most_proper_divisors_in_range(Low, High, Result) :-

   aggregate_all( max(Count, N),
                  ( between(Low, High, N),
                    proper_divisor_count(N, Count) ),
                  max(MaxCount, Num) ),
   Result = (num(Num)-divisor_count(MaxCount)).</lang>

Output:

<lang prolog>?- show_proper_divisors_of_range(1,10). 2:[1] 3:[1] 4:[1,2] 5:[1] 6:[1,2,3] 7:[1] 8:[1,2,4] 9:[1,3] 10:[1,2,5] true.

?- find_most_proper_divisors_in_range(1,20000,Result). Result = num(15120)-divisor_count(79). </lang>

PureBasic

<lang PureBasic> EnableExplicit

Procedure ListProperDivisors(Number, List Lst())

 If Number < 2 : ProcedureReturn : EndIf
 Protected i
 For i = 1 To Number / 2
   If Number % i = 0
     AddElement(Lst())
     Lst() = i
   EndIf
 Next

EndProcedure

Procedure.i CountProperDivisors(Number)

 If Number < 2 : ProcedureReturn 0 : EndIf
 Protected i, count = 0
 For i = 1 To Number / 2
   If Number % i = 0
     count + 1
   EndIf
 Next
 ProcedureReturn count

EndProcedure

Define n, count, most = 1, maxCount = 0 If OpenConsole()

 PrintN("The proper divisors of the following numbers are : ")
 PrintN("")
 NewList lst()
 For n = 1 To 10
   ListProperDivisors(n, lst())
   Print(RSet(Str(n), 3) + " -> ")
   If ListSize(lst()) = 0
     Print("(None)")
   Else  
     ForEach lst()
       Print(Str(lst()) + " ")
     Next
   EndIf
   ClearList(lst())
   PrintN("")
 Next
 For n = 2 To 20000
   count = CountProperDivisors(n)
   If count > maxCount
     maxCount = count
     most = n
   EndIf
 Next
 PrintN("")
 PrintN(Str(most) + " has the most proper divisors, namely " + Str(maxCount))
 PrintN("")
 PrintN("Press any key to close the console")
 Repeat: Delay(10) : Until Inkey() <> ""
 CloseConsole()

EndIf </lang>

Output:
The proper divisors of the following numbers are :

  1 -> (None)
  2 -> 1
  3 -> 1
  4 -> 1 2
  5 -> 1
  6 -> 1 2 3
  7 -> 1
  8 -> 1 2 4
  9 -> 1 3
 10 -> 1 2 5

15120 has the most proper divisors, namely 79

Python

Python: Literal

A very literal interpretation <lang python>>>> def proper_divs2(n): ... return {x for x in range(1, (n + 1) // 2 + 1) if n % x == 0 and n != x} ... >>> [proper_divs2(n) for n in range(1, 11)] [set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}] >>> >>> n, length = max(((n, len(proper_divs2(n))) for n in range(1, 20001)), key=lambda pd: pd[1]) >>> n 15120 >>> length 79 >>> </lang>


Python: From prime factors

I found a reference on how to generate factors from all the prime factors and the number of times each prime factor goes into N - its multiplicity.

For example, given N having prime factors P(i) with associated multiplicity M(i}) then the factors are given by:

for m[0] in range(M(0) + 1):
    for m[1] in range(M[1] + 1):
        ...
                for m[i - 1] in range(M[i - 1] + 1):
                    mult = 1
                    for j in range(i):
                        mult *= P[j] ** m[j]
                    yield mult

This version is over an order of magnitude faster for generating the proper divisors of the first 20,000 integers; at the expense of simplicity. <lang python>from math import sqrt from functools import lru_cache, reduce from collections import Counter from itertools import product


MUL = int.__mul__


def prime_factors(n):

   'Map prime factors to their multiplicity for n'
   d = _divs(n)
   d = [] if d == [n] else (d[:-1] if d[-1] == d else d)
   pf = Counter(d)
   return dict(pf)

@lru_cache(maxsize=None) def _divs(n):

   'Memoized recursive function returning prime factors of n as a list'
   for i in range(2, int(sqrt(n)+1)):
       d, m  = divmod(n, i)
       if not m:
           return [i] + _divs(d)
   return [n]


def proper_divs(n):

   Return the set of proper divisors of n.
   pf = prime_factors(n)
   pfactors, occurrences = pf.keys(), pf.values()
   multiplicities = product(*(range(oc + 1) for oc in occurrences))
   divs = {reduce(MUL, (pf**m for pf, m in zip(pfactors, multis)), 1)
           for multis in multiplicities}
   try:
       divs.remove(n)
   except KeyError:
       pass
   return divs or ({1} if n != 1 else set())


if __name__ == '__main__':

   rangemax = 20000
   
   print([proper_divs(n) for n in range(1, 11)])
   print(*max(((n, len(proper_divs(n))) for n in range(1, 20001)), key=lambda pd: pd[1]))</lang>
Output:
[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
15120 79


Python: Functional

Defining a list of proper divisors in terms of the prime factorization:

Works with: Python version 3.7

<lang python>Proper divisors

from itertools import accumulate, chain, groupby, product from functools import reduce from math import floor, sqrt from operator import mul


  1. properDivisors :: Int -> [Int]

def properDivisors(n):

   The ordered divisors of n, excluding n itself.
   
   def go(a, group):
       return [x * y for x, y in product(
           a,
           accumulate(chain([1], group), mul)
       )]
   return sorted(
       reduce(go, [
           list(g) for _, g in groupby(primeFactors(n))
       ], [1])
   )[:-1] if 1 < n else []


  1. --------------------------TEST---------------------------
  2. main :: IO ()

def main():

   Tests
   print(
       fTable('Proper divisors of [1..10]:')(str)(str)(
           properDivisors
       )(range(1, 1 + 10))
   )
   print('\nExample of maximum divisor count in the range [1..20000]:')
   print(
       max(
           [(n, len(properDivisors(n))) for n in range(1, 1 + 20000)],
           key=snd
       )
   )


  1. -------------------------DISPLAY-------------------------
  1. fTable :: String -> (a -> String) ->
  2. (b -> String) -> (a -> b) -> [a] -> String

def fTable(s):

   Heading -> x display function -> fx display function ->
      f -> xs -> tabular string.
   
   def go(xShow, fxShow, f, xs):
       ys = [xShow(x) for x in xs]
       w = max(map(len, ys))
       return s + '\n' + '\n'.join(map(
           lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
           xs, ys
       ))
   return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
       xShow, fxShow, f, xs
   )


  1. -------------------------GENERIC-------------------------
  1. primeFactors :: Int -> [Int]

def primeFactors(n):

   A list of the prime factors of n.
   
   def f(qr):
       r = qr[1]
       return step(r), 1 + r
   def step(x):
       return 1 + (x << 2) - ((x >> 1) << 1)
   def go(x):
       root = floor(sqrt(x))
       def p(qr):
           q = qr[0]
           return root < q or 0 == (x % q)
       q = until(p)(f)(
           (2 if 0 == x % 2 else 3, 1)
       )[0]
       return [x] if q > root else [q] + go(x // q)
   return go(n)


  1. snd :: (a, b) -> b

def snd(tpl):

   Second member of a pair.
   return tpl[1]


  1. until :: (a -> Bool) -> (a -> a) -> a -> a

def until(p):

   The result of repeatedly applying f until p holds.
      The initial seed value is x.
   
   def go(f, x):
       v = x
       while not p(v):
           v = f(v)
       return v
   return lambda f: lambda x: go(f, x)


  1. MAIN ---

if __name__ == '__main__':

   main()</lang>
Output:
Proper divisors of [1..10]:
 1 -> []
 2 -> [1]
 3 -> [1]
 4 -> [1, 2]
 5 -> [1]
 6 -> [1, 2, 3]
 7 -> [1]
 8 -> [1, 2, 4]
 9 -> [1, 3]
10 -> [1, 2, 5]

Example of maximum divisor count in the range [1..20000]:
(15120, 79)

R

Package solution

Works with: R version 3.3.2 and above

<lang r># Proper divisors. 12/10/16 aev require(numbers); V <- sapply(1:20000, Sigma, k = 0, proper = TRUE); ind <- which(V==max(V)); cat(" *** max number of divisors:", max(V), "\n"," *** for the following indices:",ind, "\n");</lang>

Output:
Loading required package: numbers
  *** max number of divisors: 79 
  *** for the following indices: 15120 18480 

Filter solution

<lang r>#Task 1

  1. Has no input error checking.

properDivisors<-function(n) {

 #Returning NULL seems like bad code, but task 2 demands some output for n=1, which has no proper divisors.
 if(n==1) NULL else Filter(function(x) n %% x == 0, 1:(n%/%2))

}

  1. Task 2
  2. The output could be put in to a cleaner form than a list, but this is the idiomatic way.

Vectorize(properDivisors)(1:10)

  1. Task 3
  2. Although there are two, the task only asks for one suitable number so that is all we give.
  3. Similarly, we have seen no need to make sure that "divisors" is only a plural when it should be.
  4. Be aware that this solution uses both length and lengths. It would not work if the index of the
  5. desired number was not also the number itself. However, this is always the case.

mostProperDivisors<-function(N) {

 divisorList<-Vectorize(properDivisors)(1:N)
 numberWithMostDivisors<-which.max(lengths(divisorList))
 return(paste0("The number with the most proper divisors between 1 and ",N,
               " is ",numberWithMostDivisors,
               ". It has ",length(divisorListnumberWithMostDivisors)," proper divisors.")) 

} mostProperDivisors(20000)</lang>

Output:
#Task 2
> Vectorize(properDivisors)(1:10)
[[1]]
NULL

[[2]]
[1] 1

[[3]]
[1] 1

[[4]]
[1] 1 2

[[5]]
[1] 1

[[6]]
[1] 1 2 3

[[7]]
[1] 1

[[8]]
[1] 1 2 4

[[9]]
[1] 1 3

[[10]]
[1] 1 2 5

#Task 3
> mostProperDivisors(20000)
[1] "The number with the most proper divisors between 1 and 20000 is 15120. It has 79 proper divisors."

Racket

Short version

<lang racket>#lang racket (require math) (define (proper-divisors n) (drop-right (divisors n) 1)) (for ([n (in-range 1 (add1 10))])

 (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))

(define most-under-20000

 (for/fold ([best '(1)]) ([n (in-range 2 (add1 20000))])
   (define divs (proper-divisors n))
   (if (< (length (cdr best)) (length divs)) (cons n divs) best)))

(printf "~a has ~a proper divisors\n"

       (car most-under-20000) (length (cdr most-under-20000)))</lang>
Output:
proper divisors of: 1	()
proper divisors of: 2	(1)
proper divisors of: 3	(1)
proper divisors of: 4	(1 2)
proper divisors of: 5	(1)
proper divisors of: 6	(1 2 3)
proper divisors of: 7	(1)
proper divisors of: 8	(1 2 4)
proper divisors of: 9	(1 3)
proper divisors of: 10	(1 2 5)
15120 has 79 proper divisors


Long version

The main module will only be executed when this file is executed. When used as a library, it will not be used. <lang racket>#lang racket/base (provide fold-divisors ; name as per "Abundant..."

        proper-divisors)

(define (fold-divisors v n k0 kons)

 (define n1 (add1 n))
 (cond
   [(>= n1 (vector-length v))
    (define rv (make-vector n1 k0))
    (for* ([n (in-range 1 n1)] [m (in-range (* 2 n) n1 n)])
      (vector-set! rv m (kons n (vector-ref rv m))))
    rv]
   [else v]))

(define proper-divisors

 (let ([p.d-v (vector)])
   (λ (n)
     (set! p.d-v (reverse (fold-divisors p.d-v n null cons)))
     (vector-ref p.d-v n))))

(module+ main

 (for ([n (in-range 1 (add1 10))])
   (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))
 (define count-proper-divisors
   (let ([p.d-v (vector)])
     (λ(n) (set! p.d-v (fold-divisors p.d-v n 0 (λ (d n) (add1 n))))
           (vector-ref p.d-v n))))
 (void (count-proper-divisors 20000))
 (define-values [C I]
   (for*/fold ([C 0] [I 1])
              ([i (in-range 1 (add1 20000))]
               [c (in-value (count-proper-divisors i))]
               #:when [> c C])
     (values c i)))
 (printf "~a has ~a proper divisors\n" I C))</lang>

The output is the same as the short version above.

Raku

(formerly Perl 6)

Works with: rakudo version 2018.10

Once your threshold is over 1000, the maximum proper divisors will always include 2, 3 and 5 as divisors, so only bother to check multiples of 2, 3 and 5.

There really isn't any point in using concurrency for a limit of 20_000. The setup and bookkeeping drowns out any benefit. Really doesn't start to pay off until the limit is 50_000 and higher. Try swapping in the commented out race map iterator line below for comparison. <lang perl6>sub propdiv (\x) {

   my @l = 1 if x > 1;
   (2 .. x.sqrt.floor).map: -> \d {
       unless x % d { @l.push: d; my \y = x div d; @l.push: y if y != d }
   }
   @l

}

put "$_ [{propdiv($_)}]" for 1..10;

my @candidates; loop (my int $c = 30; $c <= 20_000; $c += 30) {

  1. (30, *+30 …^ * > 500_000).race.map: -> $c {
   my \mx = +propdiv($c);
   next if mx < @candidates - 1;
   @candidates[mx].push: $c

}

say "max = {@candidates - 1}, candidates = {@candidates.tail}";</lang>

Output:
1 []
2 [1]
3 [1]
4 [1 2]
5 [1]
6 [1 2 3]
7 [1]
8 [1 2 4]
9 [1 3]
10 [1 2 5]
max = 79, candidates = 15120 18480

REXX

version 1

<lang rexx>Call time 'R' Do x=1 To 10

 Say x '->' proper_divisors(x)
 End

hi=1 Do x=1 To 20000

 /* If x//1000=0 Then Say x */
 npd=count_proper_divisors(x)
 Select
   When npd>hi Then Do
     list.npd=x
     hi=npd
     End
   When npd=hi Then
     list.hi=list.hi x
   Otherwise
     Nop
   End
 End

Say hi '->' list.hi

Say ' 166320 ->' count_proper_divisors(166320) Say '1441440 ->' count_proper_divisors(1441440)

Say time('E') 'seconds elapsed' Exit

proper_divisors: Procedure Parse Arg n If n=1 Then Return pd= /* Optimization reduces 37 seconds to 28 seconds */ If n//2=1 Then /* odd number */

 delta=2

Else /* even number */

 delta=1

Do d=1 To n%2 By delta

 If n//d=0 Then
   pd=pd d
 End

Return space(pd)

count_proper_divisors: Procedure Parse Arg n Return words(proper_divisors(n))</lang>

Output:
1 ->
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5
79 -> 15120 18480
 166320 -> 159
1441440 -> 287
28.342000 seconds elapsed

version 2

The following REXX version is an adaptation of the   optimized   version for the REXX language example for   Factors of an integer.

This REXX version handles all integers   (negative, zero, positive)   and automatically adjusts the precision (decimal digits).
It also allows the specification of the ranges (for display and for finding the maximum),   and allows for extra numbers to be
specified.

With the (function) optimization, it's over   20   times faster. <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*obtain optional arguments from the CL*/ if bot== | bot=="," then bot= 1 /*Not specified? Then use the default.*/ if top== | top=="," then top= 10 /* " " " " " " */ if inc== | inc=="," then inc= 1 /* " " " " " " */ if range== | range=="," then range= 20000 /* " " " " " " */ w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/ numeric digits max(9, w + 1) /*have enough digits for // operator.*/ @.= 'and' /*a literal used to separate #s in list*/

     do n=bot  to top  by inc                   /*process the first range specified.   */
     q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/
     if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */
     say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q
     end   /*n*/

m=0 /*M ≡ maximum number of Pdivs (so far).*/

     do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */
     #= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/
     if #<m  then iterate                       /*Less then max?   Then ignore this #. */
     @.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/
     end   /*r*/                                /* [↑]   process 2nd range of integers.*/

say say m ' is the highest number of proper divisors in range 1──►'range,

      ", and it's for: "       subword(@.m, 3)

say /* [↓] handle any given extra numbers.*/

     do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/
     w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/
     numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/
     q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/
     say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."
     end   /*i*/                                /* [↑] support extra specified integers*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x,b; x= abs(x); if x==1 then return /*unity?*/

      odd= x // 2;                            if x==0  then return '∞'         /*zero ?*/
      a= 1                                      /* [↓]  use all, or only odd #s.    ___*/
          do j=2+odd  by 1+odd  while j*j < x   /*divide by some integers up to    √ X */
          if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
                           end
          end   /*j*/                           /* [↑]  %  is the REXX integer division*/
                                                /* [↓]  adjust for a square.        ___*/
      if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
                       return  a   b            /*return the divisors  (both lists).   */</lang>
output   when using the following input:   0   10   1       20000       166320   1441440   11796480000
                   0 has  ∞   proper divisors:  ∞
                   1 has  0   proper divisors:
                   2 has  1   proper divisors:  1
                   3 has  1   proper divisors:  1
                   4 has  2   proper divisors:  1 2
                   5 has  1   proper divisors:  1
                   6 has  3   proper divisors:  1 2 3
                   7 has  1   proper divisors:  1
                   8 has  3   proper divisors:  1 2 4
                   9 has  2   proper divisors:  1 3
                  10 has  3   proper divisors:  1 2 5

79 is the highest number of proper divisors in range 1──►20000, and it's for:  15120 and 18480

              166320 has 159  proper divisors.
             1441440 has 287  proper divisors.
         11796480000 has 329  proper divisors.

version 3

When factoring          20,000 integers,   this REXX version is about   10%   faster than the REXX version 2.
When factoring        200,000 integers,   this REXX version is about   30%   faster.
When factoring     2,000,000 integers,   this REXX version is about   40%   faster.
When factoring   20,000,000 integers,   this REXX version is about   38%   faster.

(The apparent slowdown for the last example above is probably due to a shortage of real storage, causing more page faults.)

It accomplishes a faster speed by incorporating the calculation of an   integer square root   of an integer   (without using any floating point arithmetic). <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*obtain optional arguments from the CL*/ if bot== | bot=="," then bot= 1 /*Not specified? Then use the default.*/ if top== | top=="," then top= 10 /* " " " " " " */ if inc== | inc=="," then inc= 1 /* " " " " " " */ if range== | range=="," then range=20000 /* " " " " " " */ w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/ numeric digits max(9, w + 1) /*have enough digits for // operator.*/ @.= 'and' /*a literal used to separate #s in list*/

     do n=bot  to top  by inc                   /*process the first range specified.   */
     q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/
     if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */
     say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q
     end   /*n*/

m=0 /*M ≡ maximum number of Pdivs (so far).*/

     do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */
     #= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/
     if #<m  then iterate                       /*Less then max?   Then ignore this #. */
     @.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/
     end   /*r*/                                /* [↑]   process 2nd range of integers.*/

say say m ' is the highest number of proper divisors in range 1──►'range,

      ", and it's for: "       subword(@.m, 3)

say /* [↓] handle any given extra numbers.*/

     do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/
     w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/
     numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/
     q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/
     say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."
     end   /*i*/                                /* [↑] support extra specified integers*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x 1 z,b; x= abs(x); if x==1 then return /*unity?*/

      odd= x // 2;                                if x==0  then return '∞'     /*zero ?*/
      r= 0;         q= 1                        /* [↓] ══integer square root══     ___ */
           do while q<=z; q=q*4; end            /*R:  an integer which will be    √ X  */
           do while q>1;  q=q%4; _= z-r-q;  r=r%2;  if _>=0  then  do;  z=_;  r=r+q;  end
           end   /*while q>1*/                  /* [↑]  compute the integer sqrt of  X.*/
      a=1                                       /* [↓]  use all, or only odd #s.   ___ */
          do j=2 +odd  by 1 +odd to r -(r*r==x) /*divide by some integers up to   √ X  */
          if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
                           end
          end   /*j*/                           /* [↑]  %  is the REXX integer division*/
                                                /* [↓]  adjust for a square.        ___*/
      if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
                       return  a   b            /*return the divisors  (both lists).   */</lang>
output   is identical to the 2nd REXX version when using the same inputs.


Ring

<lang ring>

  1. Project : Proper divisors

limit = 10 for n=1 to limit

   if n=1
      see "" + 1 + " -> (None)" + nl
      loop
   ok
   see "" + n + " -> "
   for m=1 to n-1
       if n%m = 0
          see " " + m 
       ok
   next
   see nl

next </lang> Output:

1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

Ruby

<lang ruby>require "prime"

class Integer

 def proper_divisors
   return [] if self == 1
   primes = prime_division.flat_map{|prime, freq| [prime] * freq}
   (1...primes.size).each_with_object([1]) do |n, res|
     primes.combination(n).map{|combi| res << combi.inject(:*)}
   end.flatten.uniq
 end

end

(1..10).map{|n| puts "#{n}: #{n.proper_divisors}"}

size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.max select.each do |n|

 puts "#{n} has #{size} divisors"

end</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120 has 79 divisors
18480 has 79 divisors

An Alternative Approach

<lang ruby>#Determine the integer within a range of integers that has the most proper divisors

  1. Nigel Galloway: December 23rd., 2014

require "prime" n, g = 0 (1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g[1]+1)}

                   n, g = e, i if e > n}

puts "#{g} has #{n-1} proper divisors"</lang>

Output:

In the range 1..200000

15120 has 79 proper divisors

and in the ranges 1..2000000 & 1..20000000

166320 has 159 proper divisors
1441440 has 287 proper divisors

Rust

<lang rust>trait ProperDivisors {

   fn proper_divisors(&self) -> Option<Vec<u64>>;

}

impl ProperDivisors for u64 {

   fn proper_divisors(&self) -> Option<Vec<u64>> {
       if self.le(&1) {
           return None;
       }
       let mut divisors: Vec<u64> = Vec::new();
       for i in 1..*self {
           if *self % i == 0 {
               divisors.push(i);
           }
       }
       Option::from(divisors)
   }

}

fn main() {

   for i in 1..11 {
       println!("Proper divisors of {:2}: {:?}", i,
                i.proper_divisors().unwrap_or(vec![]));
   }
   let mut most_idx: u64 = 0;
   let mut most_divisors: Vec<u64> = Vec::new();
   for i in 1..20_001 {
       let divs = i.proper_divisors().unwrap_or(vec![]);
       if divs.len() > most_divisors.len() {
           most_divisors = divs;
           most_idx = i;
       }
   }
   println!("In 1 to 20000, {} has the most proper divisors at {}", most_idx,
            most_divisors.len());

} </lang>

Output:
Proper divisors of  1: []
Proper divisors of  2: [1]
Proper divisors of  3: [1]
Proper divisors of  4: [1, 2]
Proper divisors of  5: [1]
Proper divisors of  6: [1, 2, 3]
Proper divisors of  7: [1]
Proper divisors of  8: [1, 2, 4]
Proper divisors of  9: [1, 3]
Proper divisors of 10: [1, 2, 5]
In 1 to 20000, 15120 has the most proper divisors at 79

S-Basic

<lang Basic> $constant false = 0 $constant true = FFFFH

rem - compute p mod q function mod(p, q = integer) = integer end = p - q * (p/q)

rem - count, and optionally display, proper divisors of n function divisors(n, display = integer) = integer

 var i, limit, count, start, delta = integer
 if mod(n, 2) = 0 then
   begin
     start = 2
     delta = 1
   end
 else
   begin
     start = 3
     delta = 2
   end
 if n < 2 then count = 0 else count = 1
 if display and (count = 1) then print using "#####"; 1;
 i = start
 limit = n / start
 while i <= limit do
   begin
     if mod(n, i) = 0 then
       begin
         if display then print using "#####"; i;
         count = count + 1
       end
     i = i + delta
     if count = 1 then limit = n / i
   end
 if display then print

end = count

rem - main program begins here var i, ndiv, highdiv, highnum = integer

print "Proper divisors of first 10 numbers:" for i = 1 to 10

 print using "### : "; i;
 ndiv = divisors(i, true)

next i

print "Searching for number with most divisors ..." highdiv = 1 highnum = 1 for i = 1 to 20000

 ndiv = divisors(i, false)
 if ndiv > highdiv then
   begin
     highdiv = ndiv
     highnum = i
   end

next i print "Searched up to"; i print highnum; " has the most divisors: "; highdiv

end </lang>

Output:
Proper divisors of first 10 numbers:
  1 :
  2 :     1
  3 :     1
  4 :     1     2
  5 :     1
  6 :     1     2    3
  7 :     1
  8 :     1     2    4
  9 :     1     3
 10 :     1     2    5
Searching for number with most divisors ...
Searched up to 20000
 15120 has the most divisors:  79

Scala

Simple proper divisors

<lang Scala>def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) def format(i: Int, divisors: Seq[Int]) = f"$i%5d ${divisors.length}%2d ${divisors mkString " "}"

println(f" n cnt PROPER DIVISORS") val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) =>

   val divisors = properDivisors(i)
   if (i <= 10 || i == 100) println( format(i, divisors) )
   if (max._1 < divisors.length) (divisors.length, List(i))
   else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i))
   else max

}

list.foreach( number => println(f"$number%5d ${properDivisors(number).length}") )</lang>

Output:
    n   cnt   PROPER DIVISORS
    1     0   
    2     1   1
    3     1   1
    4     2   1 2
    5     1   1
    6     3   1 2 3
    7     1   1
    8     3   1 2 4
    9     2   1 3
   10     3   1 2 5
  100     8   1 2 4 5 10 20 25 50
15120    79
18480    79

Proper divisors for integers for big integers

If Longs are enough to you you can replace every BigInt with Long and the one BigInt(1) with 1L

<lang Scala>import scala.annotation.tailrec

def factorize(x: BigInt): List[BigInt] = {

 @tailrec
 def foo(x: BigInt, a: BigInt = 2, list: List[BigInt] = Nil): List[BigInt] = a * a > x match {
   case false if x % a == 0 => foo(x / a, a, a :: list)
   case false => foo(x, a + 1, list)
   case true => x :: list
 }
 foo(x)

}

def properDivisors(n: BigInt): List[BigInt] = {

 val factors = factorize(n)
 val products = (1 until factors.length).flatMap(i => factors.combinations(i).map(_.product).toList).toList
 (BigInt(1) :: products).filter(_ < n)

}</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: writeProperDivisors (in integer: n) is func

 local
   var integer: i is 0;
 begin
   for i range 1 to n div 2 do
     if n rem i = 0 then
       write(i <& " ");
     end if;
   end for;
   writeln;
 end func;

const func integer: countProperDivisors (in integer: n) is func

 result
   var integer: count is 0;
 local
   var integer: i is 0;
 begin
   for i range 1 to n div 2 step succ(n rem 2) do
     if n rem i = 0 then
       incr(count);
     end if;
   end for;
 end func;

const proc: main is func

 local
   var integer: i is 0;
   var integer: v is 0;
   var integer: max is 0;
   var integer: max_i is 1;
 begin
   for i range 1 to 10 do
     write(i <& ": ");
     writeProperDivisors(i);
   end for;
   for i range 1 to 20000 do
     v := countProperDivisors(i);
     if v > max then
       max := v;
       max_i := i;
     end if;
   end for;
   writeln(max_i <& " with " <& max <& " divisors");
 end func;</lang>
Output:
1: 
2: 1 
3: 1 
4: 1 2 
5: 1 
6: 1 2 3 
7: 1 
8: 1 2 4 
9: 1 3 
10: 1 2 5 
15120 with 79 divisors

Sidef

Translation of: Raku

<lang ruby>func propdiv (n) {

   n.divisors.slice(0, -2)

}

var max = 0 var candidates = [] for i in (1..20_000) { var divs = propdiv(i).len if (divs > max) { candidates = [] max = divs } candidates << i if (divs == max) } say "max = #{max}, candidates = #{candidates}"</lang>
Output:
 1: []
 2: [1]
 3: [1]
 4: [1, 2]
 5: [1]
 6: [1, 2, 3]
 7: [1]
 8: [1, 2, 4]
 9: [1, 3]
10: [1, 2, 5]
max = 79, candidates = [15120, 18480]

Swift

Simple function: <lang Swift>func properDivs1(n: Int) -> [Int] {

   return filter (1 ..< n) { n % $0 == 0 }

}</lang> More efficient function: <lang Swift>import func Darwin.sqrt

func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func properDivs(n: Int) -> [Int] {

   if n == 1 { return [] }
   
   var result = [Int]()
   
   for div in filter (1 ... sqrt(n), { n % $0 == 0 }) {
       
       result.append(div)
       if n/div != div && n/div != n { result.append(n/div) }
   }
   
   return sorted(result)
   

}</lang> Rest of the task: <lang Swift>for i in 1...10 {

   println("\(i): \(properDivs(i))")

}

var (num, max) = (0,0)

for i in 1...20_000 {

   let count = properDivs(i).count
   if (count > max) { (num, max) = (i, count) }

}

println("\(num): \(max)")</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79

tbas

<lang vb> dim _proper_divisors(100)

sub proper_divisors(n) dim i dim _proper_divisors_count = 0 if n <> 1 then for i = 1 to (n \ 2) if n %% i = 0 then _proper_divisors_count = _proper_divisors_count + 1 _proper_divisors(_proper_divisors_count) = i end if next end if return _proper_divisors_count end sub

sub show_proper_divisors(n, tabbed) dim cnt = proper_divisors(n) print str$(n) + ":"; tab(4);"(" + str$(cnt) + " items) "; dim j for j = 1 to cnt if tabbed then print str$(_proper_divisors(j)), else print str$(_proper_divisors(j)); end if if (j < cnt) then print ","; next print end sub

dim i for i = 1 to 10

   show_proper_divisors(i, false)

next

dim c dim maxindex = 0 dim maxlength = 0 for t = 1 to 20000 c = proper_divisors(t) if c > maxlength then maxindex = t maxlength = c end if next

print "A maximum at "; show_proper_divisors(maxindex, false) </lang>

>tbas proper_divisors.bas
1:  (0 items)
2:  (1 items) 1
3:  (1 items) 1
4:  (2 items) 1,2
5:  (1 items) 1
6:  (3 items) 1,2,3
7:  (1 items) 1
8:  (3 items) 1,2,4
9:  (2 items) 1,3
10: (3 items) 1,2,5
A maximum at 15120:(79 items) 1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28,30,
35,36,40,42,45,48,54,56,60,63,70,72,80,84,90,105,108,112,120,126,135,
140,144,168,180,189,210,216,240,252,270,280,315,336,360,378,420,432,
504,540,560,630,720,756,840,945,1008,1080,1260,1512,1680,1890,2160,
2520,3024,3780,5040,7560

Tcl

Note that if a number, , greater than 1 divides exactly, both and are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.) <lang tcl>proc properDivisors {n} {

   if {$n == 1} return
   set divs 1
   for {set i 2} {$i*$i <= $n} {incr i} {

if {!($n % $i)} { lappend divs $i if {$i*$i < $n} { lappend divs [expr {$n / $i}] } }

   }
   return $divs

}

for {set i 1} {$i <= 10} {incr i} {

   puts "$i => {[join [lsort -int [properDivisors $i]] ,]}"

} set maxI [set maxC 0] for {set i 1} {$i <= 20000} {incr i} {

   set c [llength [properDivisors $i]]
   if {$c > $maxC} {

set maxI $i set maxC $c

   }

} puts "max: $maxI => (...$maxC…)"</lang>

Output:
1 => {}
2 => {1}
3 => {1}
4 => {1,2}
5 => {1}
6 => {1,2,3}
7 => {1}
8 => {1,2,4}
9 => {1,3}
10 => {1,2,5}
max: 15120 => (...79...)

VBA

<lang vb>Public Sub Proper_Divisor() Dim t() As Long, i As Long, l As Long, j As Long, c As Long

   For i = 1 To 10
       Debug.Print "Proper divisor of " & i & " : " & Join(S(i), ", ")
   Next
   For i = 2 To 20000
       l = UBound(S(i)) + 1
       If l > c Then c = l: j = i
   Next
   Debug.Print "Number in the range 1 to 20,000 with the most proper divisors is : " & j
   Debug.Print j & " count " & c & " proper divisors"

End Sub

Private Function S(n As Long) As String() 'returns the proper divisors of n Dim j As Long, t() As String, c As Long

   't = list of proper divisor of n
   If n > 1 Then
       For j = 1 To n \ 2
           If n Mod j = 0 Then
               ReDim Preserve t(c)
               t(c) = j
               c = c + 1
           End If
       Next
   End If
   S = t

End Function</lang>

Output:
Proper divisor of 1 : 
Proper divisor of 2 : 1
Proper divisor of 3 : 1
Proper divisor of 4 : 1, 2
Proper divisor of 5 : 1
Proper divisor of 6 : 1, 2, 3
Proper divisor of 7 : 1
Proper divisor of 8 : 1, 2, 4
Proper divisor of 9 : 1, 3
Proper divisor of 10 : 1, 2, 5
Number in the range 1 to 20,000 with the most proper divisors is : 15120
15120 count 79 proper divisors

Visual Basic .NET

Translation of: C#

<lang vbnet>Module Module1

   Function ProperDivisors(number As Integer) As IEnumerable(Of Integer)
       Return Enumerable.Range(1, number / 2).Where(Function(divisor As Integer) number Mod divisor = 0)
   End Function
   Sub Main()
       For Each number In Enumerable.Range(1, 10)
           Console.WriteLine("{0}: {{{1}}}", number, String.Join(", ", ProperDivisors(number)))
       Next
       Dim record = Enumerable.Range(1, 20000).Select(Function(number) New With {.Number = number, .Count = ProperDivisors(number).Count()}).OrderByDescending(Function(currentRecord) currentRecord.Count).First()
       Console.WriteLine("{0}: {1}", record.Number, record.Count)
   End Sub

End Module</lang>

Output:
1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79

Wren

Library: Wren-fmt
Library: Wren-math

<lang ecmascript>import "/fmt" for Fmt import "/math" for Int

for (i in 1..10) System.print("%(Fmt.d(2, i)) -> %(Int.properDivisors(i))")

System.print("\nThe number in the range [1, 20000] with the most proper divisors is:") var number = 1 var maxDivs = 0 for (i in 2..20000) {

   var divs = Int.properDivisors(i).count
   if (divs > maxDivs) {
       number = i
       maxDivs = divs
   }

} System.print("%(number) which has %(maxDivs) proper divisors.")</lang>

Output:
 1 -> []
 2 -> [1]
 3 -> [1]
 4 -> [1, 2]
 5 -> [1]
 6 -> [1, 2, 3]
 7 -> [1]
 8 -> [1, 2, 4]
 9 -> [1, 3]
10 -> [1, 2, 5]

The number in the range [1, 20000] with the most proper divisors is:
15120 which has 79 proper divisors.

zkl

Translation of: D

This is the simple version : <lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) }</lang> This version is MUCH faster (the output isn't ordered however): <lang zkl>fcn properDivs(n){

  if(n==1) return(T);
  ( pd:=[1..(n).toFloat().sqrt()].filter('wrap(x){ n%x==0 }) )
  .pump(pd,'wrap(pd){ if(pd!=1 and (y:=n/pd)!=pd ) y else Void.Skip })

}</lang>

<lang zkl>[1..10].apply(properDivs).println(); [1..20_001].apply('wrap(n){ T(properDivs(n).len(),n) })

  .reduce(fcn([(a,_)]ab, [(c,_)]cd){ a>c and ab or cd },T(0,0))
  .println();</lang>
Output:
L(L(),L(1),L(1),L(1,2),L(1),L(1,2,3),L(1),L(1,2,4),L(1,3),L(1,2,5))
L(79,18480)