# Hofstadter Q sequence

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

The Hofstadter Q sequence is defined as:

{\displaystyle {\begin{aligned}Q(1)&=Q(2)=1,\\Q(n)&=Q{\big (}n-Q(n-1){\big )}+Q{\big (}n-Q(n-2){\big )},\quad n>2.\end{aligned}}}

It is defined like the Fibonacci sequence, but whereas the next term in the Fibonacci sequence is the sum of the previous two terms, in the Q sequence the previous two terms tell you how far to go back in the Q sequence to find the two numbers to sum to make the next term of the sequence.

• Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6
• Confirm and display that the 1000th term is:   502

Optional extra credit
• Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000th term.
• Ensure that the extra credit solution   safely   handles being initially asked for an nth term where   n   is large.

(This point is to ensure that caching and/or recursion limits, if it is a concern, is correctly handled).

## 11l

Translation of: C
V qseq = [0] * 100001
qseq[1] = 1
qseq[2] = 1

L(i) 3 .< qseq.len
qseq[i] = qseq[i - qseq[i-1]] + qseq[i - qseq[i-2]]

print(‘The first 10 terms are: ’qseq[1..10].map(q -> String(q)).join(‘, ’))
print(‘The 1000'th term is ’qseq[1000])

V less_than_preceding = 0
L(i) 2 .< qseq.len
I qseq[i] < qseq[i-1]
less_than_preceding++
print(‘Times a member of the sequence is less than its preceding term: ’less_than_preceding)
Output:
The first 10 terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6
The 1000'th term is 502
Times a member of the sequence is less than its preceding term: 49798


## 360 Assembly

Translation of: PL/I
*        Hofstrader q sequence for any n -   18/10/2015
USING  HOFSTRAD,R15       set base register
MVC    Q,=F'1'            q(1)=1
MVC    Q+4,=F'1'          q(2)=1
LA     R4,1               i=1
LOOPI    C      R4,N               do i=1 to n
BH     ELOOPI
C      R4,=F'3'           if i>=3 then
BL     NOTREC
LR     R1,R4              i
SLA    R1,2               i*4
L      R2,Q-8(R1)         q(i-1)
LR     R1,R4              i
SR     R1,R2              i-q(i-1)
SLA    R1,2               *4
L      R2,Q-4(R1)         r2=q(i-q(i-1))
LR     R1,R4              i
SLA    R1,2               i*4
L      R3,Q-12(R1)        q(i-2)
LR     R1,R4              i
SR     R1,R3              i-q(i-2)
SLA    R1,2               *4
L      R3,Q-4(R1)         r3=q(i-q(i-2))
AR     R2,R3              r2=r2+r3
LR     R1,R4              i
SLA    R1,2               i*4
ST     R2,Q-4(R1)         q(i)=q(i-q(i-1))+q(i-q(i-2))
NOTREC   C      R4,=F'10'          if i<=10
BNH    PRT
C      R4,N               or i=n then
BNE    NOPRT
PRT      XDECO  R4,XD              edit i
MVC    PG+2(4),XD+8       output i
LR     R1,R4              i
SLA    R1,2               i*4
L      R2,Q-4(R1)         q(i)
XDECO  R2,XD              edit q(i)
MVC    PG+10(4),XD+8      output q(i)
XPRNT  PG,80              print buffer
NOPRT    LA     R4,1(R4)           i=i+1
B      LOOPI
ELOOPI   XR     R15,R15            set return code
PG       DC     CL80'n=...., q=....'  buffer
XD       DS     CL12               temporary variable
LTORG                     insert literals for addressability
N        DC     F'1000'            n=1000
Q        DS     1000F              array q(1000)
YREGS
END    HOFSTRAD
Output:
n=   1, q=   1
n=   2, q=   1
n=   3, q=   2
n=   4, q=   3
n=   5, q=   3
n=   6, q=   4
n=   7, q=   5
n=   8, q=   5
n=   9, q=   6
n=  10, q=   6
n=1000, q= 502


## 8080 Assembly

puts:	equ	9	; CP/M call to print a string
org	100h
;;;	Generate the first 1000 members of the Q sequence
lxi	b,3	; Start at 3rd element (1 and 2 already defined)
genq:	dcx	b	; BC = N-1
call	q
mov	e,m	; DE = Q(N-1)
inx	h
mov	d,m
inx	b	; BC = (N-1)+1 = N
xchg		; HL = Q(N-1)
call	neg	; HL = -Q(N-1)
dad 	b	; HL = N-Q(N-1)
push	b	; Keep N
mov	b,h	; BC = N-Q(N-1)
mov	c,l
call	q	; HL = *Q(N-Q(N-1))
mov	e,m	; DE = Q(N-Q(N-1))
inx	h
mov	d,m
pop	b	; Restore N
push	d	; push Q(N-Q(N-1))
dcx	b	; BC = N-2
dcx	b
call	q	; DE = Q(N-2)
mov	e,m
inx	h
mov	d,m
inx	b	; BC = (N-2)+2 = N
inx	b
xchg		; HL = Q(N-2)
call	neg	; HL = -Q(N-2)
dad 	b	; HL = N-Q(N-2)
push	b	; Keep N
mov 	b,h	; BC = N-Q(N-2)
mov	c,l
call	q	; HL = *Q(N-Q(N-2))
mov	a,m	; HL = Q(N-Q(N-2))
inx	h
mov	h,m
mov	l,a
pop	b	; Restore N
pop	d	; pop Q(N-Q(N-1))
dad	d	; HL = Q(N-Q(N-1))+Q(N-Q(N-2))
xchg		; DE = Q(N-Q(N-1))+Q(N-Q(N-2))
call	q	; HL = *Q(N)
mov	m,e	; Store Q(N)
inx	h
mov	m,d
inx	b	; N = N+1
lxi	h,-1001
dad	b	; Are we there yet?
jnc	genq
;;;	Print first 10 terms
lxi	d,m10
mvi	c,puts
call	5
lxi	b,1	; Start at term 1
mvi	d,10	; 10 terms
p10:	push	b	; Save counters
push	d
call	prterm	; Print current term
pop	d	; Restore counters
pop 	b
inx	b	; Next term
dcr	d	; Repeat 10 times
jnz 	p10
;;;	Print 1000th term
lxi	d,m1000
mvi	c,puts
call	5
lxi	b,1000	; 1000th term
;;;	Print Q(BC)
prterm:	call	q 	; Load term into HL
mov	a,m
inx	h
mov	h,m
mov	l,a
lxi	b,num	; Push pointer to end of number buffer
push 	b
lxi	b,-10	; Divisor
dgt:	lxi	d,-1	; Quotient
divlp:	inx	d
jc	divlp
mvi	a,'0'+10
add	l	; Make ASCII digit
pop	h	; Get pointer
dcx	h
mov	m,a	; Store digit
push	h
xchg		; HL = next quotient
mov	a,h	; More digits?
ora	l
jnz	dgt
pop	d	; Print string
mvi	c,puts
jmp	5
;;;	Set HL = -HL
neg:	dcx	h
mov	a,h
cma
mov	h,a
mov	a,l
cma
mov	l,a
ret
;;;	Set HL to memory location of Q(BC)
q:	push	d	; Keep DE
mov	h,b	; HL = 2*(BC-1)
mov	l,c
dcx	h
lxi	d,qq	; Add to start of sequence
pop	d
ret
m10:	db	'The first 10 terms are: $' m1000: db 13,10,'The 1000th term is:$'
db	'*****'	; Placeholder for number
num:	db	' $' qq: dw 1,1 ; Q sequence stored here, starting with 1, 1 Output: The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 ## 8086 Assembly puts: equ 9 ; MS-DOS syscall to print a string cpu 8086 org 100h section .text ;;; Generate first 1000 elements of Q sequence mov dx,3 ; DX = N mov di,Q+4 ; DI = place to store elements mov cx,998 ; Generate 998 more terms genq: mov si,dx ; SI = N sub si,[di-2] ; SI -= Q[N-1] mov bp,dx ; BP = N sub bp,[di-4] ; BP -= Q[N-2] dec si ; SI = 2*(SI-1) (0-indexed, 2 bytes/term) shl si,1 dec bp ; Same for BP shl bp,1 mov ax,[si+Q] ; Load Q[n-Q[n-1]] add ax,[bp+Q] ; Add Q[n-Q[n-2]] stosw ; Store as Q[n] inc dx ; Increment N loop genq ;;; Print first 10 elements mov ah,puts mov dx,m10 int 21h mov cx,10 mov bx,1 p10: call prterm inc bx loop p10 ;;; Print 1000th element mov ah,puts mov dx,m1000 int 21h mov bx,1000 ;;; Print the term in BX prterm: push bx ; Save term dec bx shl bx,1 mov ax,[bx+Q] ; Load term into AX mov bp,10 ; Divisor mov bx,num ; Number buffer pointer .dgt: xor dx,dx div bp ; Divide number by 10 dec bx add dl,'0' ; DX = remainder, add '0' mov [bx],dl ; Stored digit test ax,ax ; Done yet? jnz .dgt ; If not, find next digit mov dx,bx ; Print the number mov ah,puts int 21h pop bx ; Restore term ret section .data m10: db 'First 10 terms are:$'
m1000:	db	13,10,'1000th term is: $' db '*****' ; Number placeholder num: db '$'
Q:	dw	1,1

Output:
First 10 terms are: 1 1 2 3 3 4 5 5 6 6
1000th term is: 502

## Action!

PROC Main()
DEFINE MAX="1000"
INT ARRAY q(MAX+1)
INT i

q(1)=1 q(2)=1
FOR i=3 TO MAX
DO
q(i)=q(i-q(i-1))+q(i-q(i-2))
OD

FOR i=1 TO 10
DO
PrintF("%I: %I%E",i,q(i))
OD
PrintF("%I: %I%E",MAX,q(MAX))
RETURN
Output:
1: 1
2: 1
3: 2
4: 3
5: 3
6: 4
7: 5
8: 5
9: 6
10: 6
1000: 502


with Ada.Text_IO;

type Callback is access procedure(N: Positive);

procedure Q(First, Last: Positive; Q_Proc: Callback) is
-- calls Q_Proc(Q(First)); Q_Proc(Q(First+1)); ... Q_Proc(Q(Last));
-- precondition: Last > 2

Q_Store: array(1 .. Last) of Natural := (1 => 1, 2 => 1, others => 0);
-- "global" array to store the Q(I)
-- if Q_Store(I)=0, we compute Q(I) and update Q_Store(I)
-- else we already know Q(I) = Q_Store(I)

function Q(N: Positive) return Positive is
begin
if Q_Store(N) = 0 then
Q_Store(N) := Q(N - Q(N-1)) + Q(N-Q(N-2));
end if;
return Q_Store(N);
end Q;

begin
for I in First .. Last loop
Q_Proc(Q(I));
end loop;
end Q;

procedure Print(P: Positive) is
begin
end Print;

Decrease_Counter: Natural := 0;
Previous_Value: Positive := 1;

procedure Decrease_Count(P: Positive) is
begin
if P < Previous_Value then
Decrease_Counter := Decrease_Counter + 1;
end if;
Previous_Value := P;
end Decrease_Count;

begin
Q(1, 10, Print'Access);
-- the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6

Q(1000, 1000,  Print'Access);
-- the 1000'th term is: 502

Q(2, 100_000, Decrease_Count'Access);
-- how many times a member of the sequence is less than its preceding term
-- for terms up to and including the 100,000'th term

Output:
 1 1 2 3 3 4 5 5 6 6
502
49798

## ALGOL 68

Translation of: C

Note: This specimen retains the original C coding style.

Works with: ALGOL 68 version Revision 1 - no extension to language used.
Works with: ALGOL 68G version Any - tested with release algol68g-2.3.5.

#!/usr/local/bin/a68g --script #

INT n = 100000;
main:
(
INT flip;
[n]INT q;

q[1] := q[2] := 1;

FOR i FROM 3 TO n DO
q[i] := q[i - q[i - 1]] + q[i - q[i - 2]] OD;

FOR i TO 10 DO
printf(($g(0)$, q[i], $b(l,x)$, i = 10)) OD;

printf(($g(0)l$, q[1000]));

flip := 0;
FOR i TO n-1 DO
flip +:= ABS (q[i] > q[i + 1]) OD;

printf(($"flips: "g(0)l$, flip))
)
Output:
1 1 2 3 3 4 5 5 6 6
502
flips: 49798


## ALGOL-M

begin
integer array Q[1:1000];
integer n;

Q[1] := Q[2] := 1;
for n := 3 step 1 until 1000 do
Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]];

write("The first 10 terms are:");
write("");
for n := 1 step 1 until 10 do writeon(Q[n]);

write("The 1000th term is:", Q[1000]);
end
Output:
The first 10 terms are:
1     1     2     3     3     4     5     5     6     6
The 1000th term is:   502

## ALGOL W

begin % find elements of the Hofstader Q sequence Q(1) = Q(2) = 1             %
% Q(n) = Q( n - Q( n - 1 ) ) + Q( n - Q( n - 2 ) ) for n > 2            %
integer MAX_Q;
max_Q := 100000;
begin
integer array Q ( 1 :: MAX_Q );
integer array xQ ( 1 :: 10 );
integer ltCount;
logical valuesOk;
% expected values of the first 10 elements                            %
xQ( 1 ) := xQ( 2 ) := 1;
xQ( 3 ) := 2; xQ( 4 ) := xQ( 5 ) := 3; xQ( 6 ) := 4;
xQ( 7 ) := xQ( 8 ) := 5; xQ( 9 ) := xQ( 10 ) := 6;
% calculate the sequence and count how often Q( n ) < Q( n - 1 )      %
ltCount := 0;
Q( 1 ) := Q( 2 ) := 1;
for n := 3 until MAX_Q do begin
Q( n ) := Q( n - Q( n - 1 ) ) + Q( n - Q( n - 2 ) );
if Q( n ) < Q( n - 1 ) then ltCount := ltCount + 1
end for_n ;
valuesOk := true;
write( "The first 10 terms of the Hofstader Q sequence:" );
for i := 1 until 10 do begin
writeon( i_w := 1, s_w := 0, " ", Q( i ) );
if Q( i ) not = xQ( i ) then begin
writeon( i_w := 1, s_w := 0, "-EXPECTED-", xQ( i ) );
valuesOk := false
end if_Q_i_ne_xQ_i
end for_i ;
write( i_w := 1, s_w := 0, "The 1000th term is: ", Q( 1000 ) );
if Q( 1000 ) not = 502 then begin
writeon( "-EXPECTED-502" );
valuesOk := false
end if_Q_100_ne_502 ;
if valuesOk then write( "    (Computed values are as expected)" )
else write( "Values NOT as expected" );
write( i_w := 1, s_w := 0, "Q(n) < Q(n-1) ", ltCount," times for n up to ", MAX_Q )
end
end.
Output:
The first 10 terms of the Hofstader Q sequence: 1 1 2 3 3 4 5 5 6 6
The 1000th term is: 502
(Computed values are as expected)
Q(n) < Q(n-1) 49798 times for n up to 100000


## APL

∇ Q_sequence;seq;size
size←100000
seq←{⍵,+/⍵[(1+⍴⍵)-¯2↑⍵]}⍣(size-2)⊢1 1

⎕←'The first 10 terms are:', seq[⍳10]
⎕←'The 1000th term is:', seq[1000]
⎕←(+/ 2>/seq),'terms were preceded by a larger term.'
∇

Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6
The 1000th term is: 502
49798 terms were preceded by a larger term.


## ARM Assembly

.text
.global _start
_start:	ldr	r6,=qs			@ R6 = base register for Q array
@@@	Write first 2 elements
mov	r0,#1			@ Q(1) and Q(2) are 1
strh	r0,[r6,#4]
strh	r0,[r6,#8]
@@@ 	Generate 100 thousand elements
mov	r1,#0x86A0
movt	r1,#1			@ 0x186A0 = 100.000
mov	r0,#3			@ Starting at element 3
1:	sub	r2,r0,#1		@ r2 = n-1
ldr	r2,[r6,r2,lsl#2]	@ r2 = Q[r2]
sub	r2,r0,r2		@ r2 = n-Q[r2]
ldr	r2,[r6,r2,lsl#2]	@ r2 = Q[r2]
sub	r3,r0,#2		@ r3 = n-2
ldr	r3,[r6,r3,lsl#2]	@ r3 = Q[r3]
sub	r3,r0,r3		@ r3 = n-Q[r3]
ldr	r3,[r6,r3,lsl#2]	@ r3 = Q[r3]
add	r2,r2,r3		@ r2 += r3
str	r2,[r6,r0,lsl#2]	@ Q[n] = r2
add	r0,r0,#1		@ n++
cmp	r0,r1
bls	1b			@ If r0<=r1, generate next
@@@	Print first 10 elements
ldr 	r1,=f10m
bl	pstr
mov	r8,#1			@ Start at element 1
1:	ldr	r0,[r6,r8,lsl#2]	@ Grab current element
bl	pnum			@ Print it
ldr	r1,=space		@ Print a space
bl 	pstr
cmp	r8,#10			@ Keep going until 10 elements printed
bls	1b
ldr 	r1,=nl 			@ Print newline
bl	pstr
@@@	Print 1000th element
ldr	r1,=f1000m
bl	pstr
mov	r8,#1000		@ Grab 1000th element
ldr	r0,[r6,r8,lsl#2]
bl	pnum
ldr	r1,=nl 			@ Print newline
bl	pstr
@@@	Find how many times a member is less than its preceding term
mov	r0,#0 			@ counter
mov	r1,#0x86A0		@ max element
movt	r1,#1
mov	r2,#1			@ value of previous element
mov	r3,#2			@ number of current element
2:	ldr	r4,[r6,r3,lsl#2]	@ get value of current element
cmp	r2,r4			@ if previous more than current
addhi	r0,r0,#1		@ then increment counter
mov	r2,r4			@ current el is now prevous el
add	r3,r3,#1		@ increment element index
cmp	r3,r1			@ are we there yet?
bls 	2b			@ if not, keep going
bl	pnum			@ otherwise, print the number
ldr	r1,=ltermm		@ and the corresponding message
bl	pstr
mov	r0,#0			@ and then exit
mov	r7,#1
swi	#0
@@@	Print a length-prefixed string (in r1)
pstr:	push	{r7,lr}			@ Save syscall and link registers
mov	r0,#1			@ 1 = stdout
ldrb	r2,[r1],#1		@ Get length and advance r1
mov	r7,#4			@ Write
swi	#0
pop	{r7,pc}
@@@	Print unsigned number in r0 using Linux
pnum:	push	{r7,lr}			@ Save syscall and link registers
ldr	r7,=qs			@ May as well use R7 as buffer pointer
1:	mov	r1,#10			@ Div-mod by 10
bl	divmod
add	r1,r1,#'0		@ This makes an ASCII digit
strb	r1,[r7,#-1]!		@ Store it in the buffer
tst	r0,r0 			@ Are there more digits?
bne	1b			@ If so, calculate them
mov	r0,#1			@ 1 = stdout
mov	r1,r7			@ Start of number in R1
ldr	r2,=qs			@ Calculate length
sub	r2,r2,r1
mov	r7,#4			@ 4 = write
swi	#0
pop	{r7,pc}
@@@	Division routine: r0=r0/r1, r1=r0%r1
divmod:	mov	r2,#0			@ R2 = counter
1:	cmp	r1,r0			@ Double R1 until R1>R0
lslls	r1,r1,#1
bls	1b
mov	r3,#0
2:	lsl	r3,r3,#1
subs	r0,r0,r1		@ Trial subtraction
addhs	r3,r3,#1		@ If it worked, mark
addlo	r0,r0,r1		@ If it didn't, undo
lsr	r1,r1,#1		@ Halve R1
subs	r2,r2,#1		@ Decrement counter
bhs	2b			@ Keep going until zero
mov	r1,r0			@ R1 = modulus
mov	r0,r3			@ R0 = quotient
bx	lr
.data
space:	.ascii	"\x1 "
nl:	.ascii 	"\x1\n"
f10m:	.ascii	"\x18The first 10 terms are: "
f1000m:	.ascii	"\x14The 1000th term is: "
ltermm:	.ascii	"' terms were preceded by a larger term.\n"
.bss
.align  4
.space	8			@ Buffer for number output
qs:     .space  4 * 100001 		@ One word per term

Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6
The 1000th term is: 502
49798 terms were preceded by a larger term.

## Arturo

q: new [1 1]
n: 2
while [n<1001][
'q ++ (get q n-q$n-1]) + get q n-q\[n-2] n: n+1 ] print ["First ten items:" first.n: 10 q] print ["1000th item:" q\[999]]  Output: First ten items: [1 1 2 3 3 4 5 5 6 6] 1000th item: 502 ## AutoHotkey SetBatchLines, -1 Q := HofsQSeq(100000) Loop, 10 Out .= Q[A_Index] ", " MsgBox, % "First ten:t" Out "n" . "1000th:tt" Q[1000] "n" . "Flips:tt" Q.flips HofsQSeq(n) { Q := {1: 1, 2: 1, "flips": 0} Loop, % n - 2 { i := A_Index + 2 , Q[i] := Q[i - Q[i - 1]] + Q[i - Q[A_Index]] if (Q[i] < Q[i - 1]) Q.flips++ } return Q }  Output: First ten: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 1000th: 502 Flips: 49798 ## AWK #!/usr/bin/awk -f BEGIN { N = 100000 print "Q-sequence(1..10) : " Qsequence(10) Qsequence(N,Q) print "1000th number of Q sequence : " Q[1000] for (n=2; n<=N; n++) { if (Q[n]<Q[n-1]) NN++ } print "number of Q(n)<Q(n+1) for n<=100000 : " NN } function Qsequence(N,Q) { Q[1] = 1 Q[2] = 1 seq = "1 1" for (n=3; n<=N; n++) { Q[n] = Q[n-Q[n-1]]+Q[n-Q[n-2]] seq = seq" "Q[n] } return seq }  Q-sequence(1..10) : 1 1 2 3 3 4 5 5 6 6 1000th number of Q sequence : 502 number of Q(n)<Q(n+1) for n<=100000 : 49798 ## BASIC ### BASIC256 Translation of: FreeBASIC limite = 100000 dim Q[limite+1] cont = 0 Q[1] = 1 Q[2] = 1 for i = 3 to limite Q[i] = Q[i-Q[i-1]] + Q[i-Q[i-2]] if Q[i] < Q[i-1] then cont += 1 next i print "Primeros 10 términos: "; for i = 1 to 10 print Q[i] + " "; next i print "Término número 1000: "; Q[1000] print "Términos menores que los anteriores: "; cont Output: Igual que la entrada de FreeBASIC.  ### BBC BASIC  PRINT "First 10 terms of Q = " ; FOR i% = 1 TO 10 : PRINT ;FNq(i%, c%) " "; : NEXT : PRINT PRINT "1000th term = " ; FNq(1000, c%) PRINT "100000th term = " ; FNq(100000, c%) PRINT "Term is less than preceding term " ; c% " times" END DEF FNq(n%, RETURN c%) LOCAL i%,q%() IF n% < 3 THEN = 1 ELSE IF n% = 3 THEN = 2 DIM q%(n%) q%(1) = 1 : q%(2) = 1 : q%(3) = 2 c% = 0 FOR i% = 3 TO n% q%(i%) = q%(i% - q%(i%-1)) + q%(i% - q%(i%-2)) IF q%(i%) < q%(i%-1) THEN c% += 1 NEXT = q%(n%)  Output: First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6 1000th term = 502 100000th term = 48157 Term is less than preceding term 49798 times  ### QBasic Works with: QBasic version 1.1 Works with: QuickBasic version 4.5 CONST limite = 10000 DIM Q(limite) Q(1) = 1 Q(2) = 1 cont = 0 FOR i = 3 TO limite Q(i) = Q(i - Q(i - 1)) + Q(i - Q(i - 2)) IF Q(i) < Q(i-1) THEN cont = cont + 1 NEXT i PRINT "First 10 terms: "; FOR i = 1 TO 10 PRINT Q(i); " "; NEXT i PRINT PRINT "Term 1000: "; Q(1000) PRINT "Terms less than preceding in first 100k: "; cont  ### True BASIC LET limite = 100000 DIM q(0) MAT REDIM q(limite) LET q(1) = 1 LET q(2) = 1 LET count = 0 FOR i = 3 TO limite LET q(i) = q(i-q(i-1))+q(i-q(i-2)) IF q(i) < q(i-1) THEN LET count = count + 1 END IF NEXT i PRINT "First 10 terms: "; FOR i = 1 TO 10 PRINT q(i); NEXT i PRINT PRINT "Term 1000: "; q(1000) PRINT "Terms less than preceding in first 100k: "; count END  Output: Same as FreeBASIC entry. ### XBasic Works with: Windows XBasic PROGRAM "Hofstadter Q sequence" VERSION "0.0000" DECLARE FUNCTION Entry () FUNCTION Entry () limite = 1e5 DIM q[limite] q[1] = 1 q[2] = 1 count = 0 FOR i = 3 TO limite q[i] = q[i-q[i-1]] + q[i-q[i-2]] IF q[i] < q[i-1] THEN INC count END IF NEXT i PRINT "First 10 terms: "; FOR i = 1 TO 10 PRINT q[i]; NEXT i PRINT "\nTerm 1000: "; q[1000] PRINT "Terms less than preceding in first 100k: "; count END FUNCTION END PROGRAM  Output: Same as FreeBASIC entry. ### Yabasic limite = 1e5 dim q(limite) q(1) = 1 q(2) = 1 count = 0 for i = 3 to limite q(i) = q(i-q(i-1)) + q(i-q(i-2)) if q(i) < q(i-1) count = count + 1 next i print "First 10 terms: "; for i = 1 to 10 print q(i), " "; next i print "\nTerm 1000: ", q(1000) print "Terms less than preceding in first 100k: ", count end  Output: Same as FreeBASIC entry. ## Bracmat ( 0:?memocells & tbl(memo,!memocells+1) { allocate array } & ( Q = . !arg:(1|2)&1 | !arg:>2 & ( !arg:>!memocells:?memocells { Array is too small. } & tbl(memo,!memocells+1) { Let array grow to needed size. } | { Array is not too small. } ) & ( !(!argmemo):>0 { Set index to !arg. Return value at index if > 0 } | Q(!arg+-1*Q(!arg+-1))+Q(!arg+-1*Q(!arg+-2)) : ?(!arg?memo) { Set index to !arg. Store value just found. } ) ) & 0:?i & whl ' (1+!i:~>10:?i&put(str(Q!i " "))) & put\n & whl'(1+!i:~>1000:?i&Q!i) & out(Q1000) & 0:?previous:?lessThan:?i & whl ' ( 1+!i:~>100000:?i & Q!i : ( <!previous&1+!lessThan:?lessThan | ? ) : ?previous ) & out!lessThan ); Output: 1 1 2 3 3 4 5 5 6 6 502 49798 ## BCPL get "libhdr" let start() be ( let Q = vec 1000 Q!1 := 1 Q!2 := 1 for n = 3 to 1000 do Q!n := Q!(n-Q!(n-1)) + Q!(n-Q!(n-2)) writes("The first 10 terms are:") for n = 1 to 10 do writef(" %N", Q!n) writef("*NThe 1000th term is: %N*N", Q!1000) ) Output: The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 ## C #include <stdio.h> #include <stdlib.h> #define N 100000 int main() { int i, flip, *q = (int*)malloc(sizeof(int) * N) - 1; q[1] = q[2] = 1; for (i = 3; i <= N; i++) q[i] = q[i - q[i - 1]] + q[i - q[i - 2]]; for (i = 1; i <= 10; i++) printf("%d%c", q[i], i == 10 ? '\n' : ' '); printf("%d\n", q[1000]); for (flip = 0, i = 1; i < N; i++) flip += q[i] > q[i + 1]; printf("flips: %d\n", flip); return 0; }  Output: 1 1 2 3 3 4 5 5 6 6 502 flips: 49798  ## C# using System; using System.Collections.Generic; namespace HofstadterQSequence { class Program { // Initialize the dictionary with the first two indices filled. private static readonly Dictionary<int, int> QList = new Dictionary<int, int> { {1, 1}, {2, 1} }; private static void Main() { int lessThanLast = 0; /* Initialize our variable that holds the number of times * a member of the sequence was less than its preceding term. */ for (int n = 1; n <= 100000; n++) { int q = Q(n); // Get Q(n). if (n > 1 && QList[n - 1] > q) // If Q(n) is less than Q(n - 1), lessThanLast++; // then add to the counter. if (n > 10 && n != 1000) continue; /* If n is greater than 10 and not 1000, * the rest of the code in the loop does not apply, * and it will be skipped. */ if (!Confirm(n, q)) // Confirm Q(n) is correct. throw new Exception(string.Format("Invalid result: Q({0}) != {1}", n, q)); Console.WriteLine("Q({0}) = {1}", n, q); // Write Q(n) to the console. } Console.WriteLine("Number of times a member of the sequence was less than its preceding term: {0}.", lessThanLast); } private static bool Confirm(int n, int value) { if (n <= 10) return new[] {1, 1, 2, 3, 3, 4, 5, 5, 6, 6}[n - 1] == value; if (n == 1000) return 502 == value; throw new ArgumentException("Invalid index.", "n"); } private static int Q(int n) { int q; if (!QList.TryGetValue(n, out q)) // Try to get Q(n) from the dictionary. { q = Q(n - Q(n - 1)) + Q(n - Q(n - 2)); // If it's not available, then calculate it. QList.Add(n, q); // Add it to the dictionary. } return q; } } } Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 Number of times a member of the sequence was less than its preceding term: 49798. ## C++ solution modeled after Perl solution #include <iostream> int main() { const int size = 100000; int hofstadters[size] = { 1, 1 }; for (int i = 3 ; i < size; i++) hofstadters[ i - 1 ] = hofstadters[ i - 1 - hofstadters[ i - 1 - 1 ]] + hofstadters[ i - 1 - hofstadters[ i - 2 - 1 ]]; std::cout << "The first 10 numbers are: "; for (int i = 0; i < 10; i++) std::cout << hofstadters[ i ] << ' '; std::cout << std::endl << "The 1000'th term is " << hofstadters[ 999 ] << " !" << std::endl; int less_than_preceding = 0; for (int i = 0; i < size - 1; i++) if (hofstadters[ i + 1 ] < hofstadters[ i ]) less_than_preceding++; std::cout << "In array of size: " << size << ", "; std::cout << less_than_preceding << " times a number was preceded by a greater number!" << std::endl; return 0; }  Output: The first 10 numbers are: 1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502 ! In array of size: 100000, 49798 times a number was preceded by a greater number! ## Clojure The qs function, given the initial subsequence of Q of length n, produces the initial subsequence of length n+1. The subsequences are vectors for efficient indexing. qfirst iterates qs so the nth iteration is Q{1..n]. (defn qs [q] (let [n (count q)] (condp = n 0 [1] 1 [1 1] (conj q (+ (q (- n (q (- n 1)))) (q (- n (q (- n 2))))))))) (defn qfirst [n] (-> (iterate qs []) (nth n))) (println "first 10:" (qfirst 10)) (println "1000th:" (last (qfirst 1000))) (println "extra credit:" (->> (qfirst 100000) (partition 2 1) (filter #(apply > %)) count))  Output: first 10: [1 1 2 3 3 4 5 5 6 6] 1000th: 502 extra credit: 49798  ## CLU q_seq = proc (n: int) returns (sequence[int]) q: array[int] := array[int][1,1] for i: int in intfrom_to(3,n) do array[int]addh(q, q[i-q[i-1]] + q[i-q[i-2]]) end return(sequence[int]a2s(q)) end q_seq start_up = proc () po: stream := streamprimary_output() q: sequence[int] := q_seq(100000) streamputs(po, "First 10 terms:") for i: int in intfrom_to(1,10) do streamputs(po, " " || intunparse(q[i])) end streamputs(po, "\n1000th term: " || intunparse(q[1000])) flips: int := 0 for i: int in intfrom_to(2, sequence[int]size(q)) do if q[i-1]>q[i] then flips := flips + 1 end end streamputl(po, "\nflips: " || intunparse(flips)) end start_up Output: First 10 terms: 1 1 2 3 3 4 5 5 6 6 1000th term: 502 flips: 49798 ## CoffeeScript Translation of: JavaScript hofstadterQ = do -> memo = [ 1 ,1, 1] Q = (n) -> result = memo[n] if typeof result != 'number' result = memo[n] = Q(n - Q(n - 1)) + Q(n - Q(n - 2)) result # some results: console.log 'Q(' + i + ') = ' + hofstadterQ(i) for i in [1..10] console.log 'Q(1000) = ' + hofstadterQ(1000)  Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 ## COBOL  IDENTIFICATION DIVISION. PROGRAM-ID. Q-SEQ. DATA DIVISION. WORKING-STORAGE SECTION. 01 SEQ. 02 Q PIC 9(3) OCCURS 1000 TIMES. 02 Q-TMP1 PIC 9(3). 02 Q-TMP2 PIC 9(3). 02 N PIC 9(4). 01 DISPLAYING. 02 ITEM PIC Z(3). 02 IX PIC Z(4). PROCEDURE DIVISION. MAIN-PROGRAM. PERFORM GENERATE-SEQUENCE. PERFORM SHOW-ITEM VARYING N FROM 1 BY 1 UNTIL N IS GREATER THAN 10. SET N TO 1000. PERFORM SHOW-ITEM. STOP RUN. GENERATE-SEQUENCE. SET Q(1) TO 1. SET Q(2) TO 1. PERFORM GENERATE-ITEM VARYING N FROM 3 BY 1 UNTIL N IS GREATER THAN 1000. GENERATE-ITEM. COMPUTE Q-TMP1 = N - Q(N - 1). COMPUTE Q-TMP2 = N - Q(N - 2). COMPUTE Q(N) = Q(Q-TMP1) + Q(Q-TMP2). SHOW-ITEM. MOVE N TO IX. MOVE Q(N) TO ITEM. DISPLAY 'Q(' IX ') = ' ITEM.  Output: Q( 1) = 1 Q( 2) = 1 Q( 3) = 2 Q( 4) = 3 Q( 5) = 3 Q( 6) = 4 Q( 7) = 5 Q( 8) = 5 Q( 9) = 6 Q( 10) = 6 Q(1000) = 502 ## Common Lisp (defparameter *mm* (make-hash-table :test #'equal)) ;;; generic memoization macro (defmacro defun-memoize (f (&rest args) &body body) (defmacro hash () (gethash (cons ',f (list ,@args)) *mm*)) (let ((h (gensym))) (defun ,f (,@args) (let ((,h (hash))) (if ,h ,h (setf (hash) (progn ,@body))))))) ;;; def q (defun-memoize q (n) (if (<= n 2) 1 (+ (q (- n (q (- n 1)))) (q (- n (q (- n 2))))))) ;;; test (format t "First of Q: ~a~%Q(1000): ~a~%Bumps up to 100000: ~a~%" (loop for i from 1 to 10 collect (q i)) (q 1000) (loop with c = 0 with last-q = (q 1) for i from 2 to 100000 do (let ((next-q (q i))) (if (< next-q last-q) (incf c)) (setf last-q next-q)) finally (return c)))  Output: First of Q: (1 1 2 3 3 4 5 5 6 6) Q(1000): 502 Bumps up to 100000: 49798 Although the above definition of q is more general, for this specific problem the following is faster: (let ((cc (make-array 3 :element-type 'integer :initial-element 1 :adjustable t :fill-pointer 3))) (defun q (n) (when (>= n (length cc)) (loop for i from (length cc) below n do (q i)) (vector-push-extend (+ (aref cc (- n (aref cc (- n 1)))) (aref cc (- n (aref cc (- n 2))))) cc)) (aref cc n)))  ## Cowgol include "cowgol.coh"; # Generate 1000 terms of the Q sequence var Q: uint16[1001]; Q[1] := 1; Q[2] := 1; var n: @indexof Q := 3; while n <= 1000 loop Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]]; n := n + 1; end loop; # Print first 10 terms print("The first 10 terms are: "); n := 1; while n <= 10 loop print_i16(Q[n]); print_char(' '); n := n + 1; end loop; print_nl(); # Print 1000th term print("The 1000th term is: "); print_i16(Q[1000]); print_nl(); Output: The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 ## D import std.stdio, std.algorithm, std.functional, std.range; int Q(in int n) nothrow in { assert(n > 0); } body { alias mQ = memoize!Q; if (n == 1 || n == 2) return 1; else return mQ(n - mQ(n - 1)) + mQ(n - mQ(n - 2)); } void main() { writeln("Q(n) for n = [1..10] is: ", iota(1, 11).map!Q); writeln("Q(1000) = ", Q(1000)); writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.", iota(2, 100_001).count!(i => Q(i) < Q(i - 1))); }  Output: Q(n) for n = [1..10] is: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] Q(1000) = 502 Q(i) is less than Q(i-1) for i [2..100_000] 49798 times.  ### Faster Version Translation of: Python Same output. import std.stdio, std.algorithm, std.range, std.array; uint Q(in int n) nothrow in { assert(n > 0); } body { __gshared static Appender!(int[]) s = [0, 1, 1]; foreach (immutable i; s.data.length .. n + 1) s ~= s.data[i - s.data[i - 1]] + s.data[i - s.data[i - 2]]; return s.data[n]; } void main() { writeln("Q(n) for n = [1..10] is: ", iota(1, 11).map!Q); writeln("Q(1000) = ", Q(1000)); writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.", iota(2, 100_001).count!(i => Q(i) < Q(i - 1))); }  ### Even Faster Version This code is here to show that you don't have to use all fancy features of D. Straightforward simple code is often clearer, and faster. import std.stdio; int[100_000] Q; void main() { Q[0] = 1; Q[1] = 1; for (int i = 2; i < 100_000; i++) { Q[i] = Q[i - Q[i - 1]] + Q[i - Q[i - 2]]; } write("Q(1..10) : "); for (int i = 0; i < 10; i++) { write(" ", Q[i]); } writeln; write("Q(1000) : "); writeln(Q[999]); int lt = 0; for (int i = 1; i < 100_000; i++) { if( Q[i-1] > Q[i] ) lt++; } writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.", lt); }  ## Dart Naive version using only recursion (Q(1000) fails due to browser script runtime restrictions) int Q(int n) => n>2 ? Q(n-Q(n-1))+Q(n-Q(n-2)) : 1; main() { for(int i=1;i<=10;i++) { print("Q(i)={Q(i)}"); } print("Q(1000)={Q(1000)}"); }  Version featuring caching. class Q { Map<int,int> _table; Q() { _table=new Map<int,int>(); _table[1]=1; _table[2]=1; } int q(int n) { // if the cache is not filled until n-1, fill it starting with the lowest entries first // this avoids doing a recursion from n to 2 (e.g. if you call q(1000000) first) // this doesn't happen in the tasks calls since the cache is filled ascending if(_table[n-1]==null) { for(int i=_table.length;i<n;i++) { q(i); } } if(_table[n]==null) { _table[n]=q(n-q(n-1))+q(n-q(n-2)); } return _table[n]; } } main() { Q q=new Q(); for(int i=1;i<=10;i++) { print("Q(i)={q.q(i)}"); } print("Q(1000)={q.q(1000)}"); int count=0; for(int i=2;i<=100000;i++) { if(q.q(i)<q.q(i-1)) { count++; } } print("value is smaller than previous count times"); }  Output: Q(1)=1 Q(2)=1 Q(3)=2 Q(4)=3 Q(5)=3 Q(6)=4 Q(7)=5 Q(8)=5 Q(9)=6 Q(10)=6 Q(1000)=502 value is smaller than previous 49798 times If the maximum number is known, filling an array is probably the fastest solution. main() { List<int> q=new List<int>(100001); q[1]=q[2]=1; int count=0; for(int i=3;i<q.length;i++) { q[i]=q[i-q[i-1]]+q[i-q[i-2]]; if(q[i]<q[i-1]) { count++; } } for(int i=1;i<=10;i++) { print("Q(i)={q[i]}"); } print("Q(1000)={q[1000]}"); print("value is smaller than previous count times"); }  ## Draco proc nonrec make_Q([*] word q) void: word n; q[1] := 1; q[2] := 1; for n from 3 upto dim(q,1)-1 do q[n] := q[n-q[n-1]] + q[n-q[n-2]] od corp proc nonrec main() void: word MAX = 1000; word i; [MAX+1] word q; make_Q(q); write("The first 10 terms are:"); for i from 1 upto 10 do write(" ", q[i]) od; writeln(); writeln("The 1000th term is: ", q[1000]) corp Output: The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 ## EasyLang Translation of: Lua proc hofstadter limit . q[] . q[] = [ 1 1 ] for n = 3 to limit q[] &= q[n - q[n - 1]] + q[n - q[n - 2]] . . proc count . q[] cnt . for i = 2 to len q[] if q[i] < q[i - 1] cnt += 1 . . . hofstadter 100000 hofq[] for i = 1 to 10 write hofq[i] & " " . print "" print hofq[1000] count hofq[] cnt print cnt ## EchoLisp (define RECURSE_BUMP 500) ;; minimum of chrome:500 safari:1000 firefox:2000 ;; count flips (define (flips N) (for/sum ((n (in-range 2 (1+ N)))) #:when (< (Q n) (Q (1- n))) 1)) (cache-size 120000) (define (Q n) ;; prevent browser stack overflow at low-cost (when (zero? (modulo n RECURSE_BUMP)) (for ((i (in-range 0 n RECURSE_BUMP ))) (Q i))) (+ (Q (- n (Q (1- n)))) (Q (- n (Q (- n 2)))))) (remember 'Q #(1 1 1)) ;; memoize and init ;; first call : check stack OK (Q 100000) → 48157 (for ((i 11)) (write (Q i))) 1 1 1 2 3 3 4 5 5 6 6 (Q 1000) → 502 (flips 100000) → 49798  ## Eiffel class APPLICATION create make feature make -- Test output of the feature hofstadter_q_sequence. local count, i: INTEGER test: ARRAY [INTEGER] do io.put_string ("%NFirst ten numbers: %N") test := hofstadter_q_sequence (10) across test as ar loop io.put_string (ar.item.out + "%T") end test := hofstadter_q_sequence (100000) io.put_string ("1000th:%N") io.put_integer (test [1000]) io.put_string ("%NNumber of Flips:%N") from i := 2 until i > 100000 loop if test [i] < test [i - 1] then count := count + 1 end i := i + 1 end io.put_integer (count) end hofstadter_q_sequence (lim: INTEGER): ARRAY [INTEGER] -- Hofstadter Q Sequence up to 'lim'. require lim_positive: lim > 0 local q: ARRAY [INTEGER] i: INTEGER do create Result.make_filled (1, 1, lim) Result [1] := 1 Result [2] := 1 from i := 3 until i > lim loop Result [i] := Result [i - Result [i - 1]] + Result [i - Result [i - 2]] i := i + 1 end end end  Output: First ten numbers: 1 1 2 3 3 4 5 5 6 6 1000th: 502 Number of Flips: 49798  ## Elixir Translation of: Erlang changed collection (Erlang array => Map) defmodule Hofstadter do defp flip(v2, v1) when v1 > v2, do: 1 defp flip(_v2, _v1), do: 0 defp list_terms(max, n, acc), do: Enum.map_join(n..max, ", ", &acc[&1]) defp hofstadter(n, n, acc, flips) do IO.puts "The first ten terms are: #{list_terms(10, 1, acc)}" IO.puts "The 1000'th term is #{acc[1000]}" IO.puts "Number of flips: #{flips}" end defp hofstadter(max, n, acc, flips) do qn1 = acc[n-1] qn = acc[n - qn1] + acc[n - acc[n-2]] hofstadter(max, n+1, Map.put(acc, n, qn), flips + flip(qn, qn1)) end def main(max \\ 100_000) do acc = %{1 => 1, 2 => 1} hofstadter(max+1, 3, acc, 0) end end Hofstadter.main  Output: The first ten terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 The 1000'th term is 502 Number of flips: 49798  ## Erlang %% @author Jan Willem Luiten <jwl@secondmove.com> %% Hofstadter Q Sequence for Rosetta Code -module(hofstadter). -export([main/0]). -define(MAX, 100000). flip(V2, V1) when V1 > V2 -> 1; flip(_V2, _V1) -> 0. list_terms(N, N, Acc) -> io:format("~w~n", [array:get(N, Acc)]); list_terms(Max, N, Acc) -> io:format("~w, ", [array:get(N, Acc)]), list_terms(Max, N+1, Acc). hofstadter(N, N, Acc, Flips) -> io:format("The first ten terms are: "), list_terms(9, 0, Acc), io:format("The 1000'th term is ~w~n", [array:get(999, Acc)]), io:format("Number of flips: ~w~n", [Flips]); hofstadter(Max, N, Acc, Flips) -> Qn1 = array:get(N-1, Acc), Qn = array:get(N - Qn1, Acc) + array:get(N - array:get(N-2, Acc), Acc), hofstadter(Max, N+1, array:set(N, Qn, Acc), Flips + flip(Qn, Qn1)). main() -> Tmp = array:set(0, 1, array:new(?MAX)), Acc = array:set(1, 1, Tmp), hofstadter(?MAX, 2, Acc, 0).  Output: The first ten terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 The 1000'th term is 502 Number of flips: 49798  ## ERRE ERRE: PROGRAM HOFSTADER_Q ! ! for rosettacode.org ! DIM Q%[10000] PROCEDURE QSEQUENCE(Q,FLAG%->SEQ) ! if FLAG% is true accumulate sequence in SEQ ! (attention to string var lenght=255) ! otherwise calculate values in Q%[] only LOCAL N Q%[1]=1 Q%[2]=1 SEQ="1 1" IF NOT FLAG% THEN Q=NUM END IF FOR N=3 TO Q DO Q%[N]=Q%[N-Q%[N-1]]+Q%[N-Q%[N-2]] IF FLAG% THEN SEQ=SEQ+STR(Q%[N]) END IF END FOR END PROCEDURE BEGIN NUM=10000 QSEQUENCE(10,TRUE->SEQ) PRINT("Q-sequence(1..10) : ";SEQ) QSEQUENCE(1000,FALSE->SEQ) PRINT("1000th number of Q sequence : ";Q%[1000]) FOR N=2 TO NUM DO IF Q%[N]<Q%[N-1] THEN NN+=1 END IF END FOR PRINT("Number of Q(n)<Q(n+1) for n<=10000 : ";NN) END PROGRAM Note: The extra credit was limited to 10000 because memory addressable range is limited to 64K. If you want to implement extra credit for 100,000 you must use external file for array Q%[]. ## Delphi Works with: Delphi version 6.0 type TIntArray = array of integer; procedure FillHofstadterArray(var HA: TIntArray); {Fill array with Hofstader numbers} {Preset array size to the number of terms you want} var I: integer; begin {Starting condition} HA[1]:=1; HA[2]:=1; {Fill array up to last item} for I:=3 to High(HA) do HA[I]:=HA[I-HA[I-1]]+HA[I-HA[I-2]]; end; procedure ShowHofstadterNumbers(Memo: TMemo); {Fill array with a } var I, LessCount: integer; var QArray: TIntArray; begin {Select the number of items we want} SetLength(QArray,100000); {Fill array} FillHofstadterArray(QArray); {Display first 10} for I:=1 to 10 do Memo.Lines.Add(Format('%4d: %4d',[I,QArray[I]])); Memo.Lines.Add(Format('%4d: %4d',[1000,QArray[1000]])); {Count number the number of times Q(n)<Q(n-1)} LessCount:=0; for I:=1 to High(QArray) do if QArray[I]>QArray[I-1] then Inc(LessCount); Memo.Lines.Add('Count of Q(n)<Q(n-1) = '+IntToStr(LessCount)); end;  Output:  1: 1 2: 1 3: 2 4: 3 5: 3 6: 4 7: 5 8: 5 9: 6 10: 6 1000: 502 Count of Q(n)<Q(n-1) = 49997  ## F# ### The function // Populate an array with values of Hofstadter Q sequence. Nigel Galloway: August 26th., 2020 let fQ N=let g=Array.length N in N.[0]<-1; N.[1]<-1;(for g in 2..g-1 do N.[g]<-N.[g-N.[g-1]]+N.[g-N.[g-2]])  ### The Tasks let Q=Array.zeroCreate<int>10 in fQ Q; printfn "%A" Q let Q=Array.zeroCreate<int>1000 in fQ Q; printfn "%d" (Array.last Q)  Output: [|1; 1; 2; 3; 3; 4; 5; 5; 6; 6|] 502  ### Extra Credit let Q=Array.zeroCreate<int>100000 in fQ Q; printfn "%d" (Q|>Seq.pairwise|>Seq.sumBy(fun(n,g)->if n>g then 1 else 0))  Output: 49798  What is a large number? let Q=Array.zeroCreate<int>2500000000 in fQ Q; printfn "%d" (Array.last Q) let Q=Array.zeroCreate<int>5000000000 in fQ Q; printfn "%d" (Array.last Q)  Output: 121648520 (in 0m14.347s) 247777817 (in 0m37.757s)  ## Factor We define a method next that takes a sequence of the first n Q values and appends the next one to it. Then we perform it 1000 times on { 1 1 } and show the first 10 and 999th (because the list is zero-indexed) elements. ( scratchpad ) : next ( seq -- newseq ) dup 2 tail* over length [ swap - ] curry map [ dupd swap nth ] map 0 [ + ] reduce suffix ; ( scratchpad ) { 1 1 } 1000 [ next ] times dup 10 head . 999 swap nth . { 1 1 2 3 3 4 5 5 6 6 } 502  ## Fermat Func Hq(n) = if n<2 then 1 else Array qq[n+1]; qq[1] := 1; qq[2] := 1; for i = 3, n do qq[i]:=qq[i-qq[i-1]]+qq[i-qq[i-2]] od; Return(qq[n]); fi; . for i=1 to 10 do !Hq(i);!' ' od; Hq(1000)  Output:  1 1 2 3 3 4 5 5 6 6 502 ## Forth Translation of: C 100000 constant N : q ( n -- addr ) cells here + ; : qinit 1 0 q ! 1 1 q ! N 2 do i i 1- q @ - q @ i i 2 - q @ - q @ + i q ! loop ; : flips ." flips: " 0 N 1 do i q @ i 1- q @ < if 1+ then loop . cr ; : qprint ( n -- ) 0 do i q @ . loop cr ; qinit 10 qprint 999 q @ . cr flips bye  Output: 1 1 2 3 3 4 5 5 6 6 502 flips: 49798  ## Fortran The latter-day function COUNT(logical expression) could easily be replaced by a simple test-and-count in the DO-loop preparing the array. One hopes that the compiler produces sensible code rather than creating an auxiliary array of boolean results then counting the true values. Rather more clunky is the need to employ odd structure for the input loop so as to handle possible bad input (text, rather than a valid number, for example) and who knows, end-of-file might happen also. Calculate the Hofstadter Q-sequence, using a big array rather than recursion. INTEGER ENUFF PARAMETER (ENUFF = 100000) INTEGER Q(ENUFF) !Lots of memory these days. Q(1) = 1 !Initial values as per the definition. Q(2) = 1 Q(3:) = -123456789!This will surely cause trouble! DO I = 3,ENUFF !For values beyond the second, Q(I) = Q(I - Q(I - 1)) + Q(I - Q(I - 2)) !Reach back according to the last two values. END DO Cast forth results as per the specification. WRITE (6,1) Q(1:10) !Should be 1 1 2 3 3 4 5 5 6 6... 1 FORMAT ("First ten values:",10I2) !Known to be one-digit numbers. WRITE (6,*) "Q(1000) =",Q(1000) !Should be 502. WRITE (6,3) ENUFF,COUNT(Q(2:ENUFF) < Q(1:ENUFF - 1)) !Please don't create a temporary array! 3 FORMAT ("Count of those elements 2:",I0, 1 " which are less than their predecessor: ",I0) !Should be 49798. Curry favour by allowing enquiries. 10 WRITE (6,11) ENUFF 11 FORMAT ("Nominate an index (in 1:",I0,"): ",) !Obviously, the  says don't start a new line. READ (5,*,END = 999, ERR = 999) I !Ask for a number, with precautions. IF (I.GT.0 .AND. I.LE.ENUFF) THEN !A good number, but, within range? WRITE (6,12) I,Q(I) !Yes. Reveal the requested value. 12 FORMAT ("Q(",I0,") = ",I0) !This should do. GO TO 10 !And ask again. END IF ! WHILE read(5,*) i & i > 0 & i < enuff DO write(6,*) "Q(",i,")=",Q(i); Closedown. 999 WRITE (6,*) "Bye." END  Output: First ten values: 1 1 2 3 3 4 5 5 6 6 Q(1000) = 502 Count of those elements 2:100000 which are less than their predecessor: 49798 Nominate an index (in 1:100000): 100000 Q(100000) = 48157 Nominate an index (in 1:100000): 0 Bye.  ## FreeBASIC Const limite = 100000 Dim As Long Q(limite), i, cont = 0 Q(1) = 1 Q(2) = 1 For i = 3 To limite Q(i) = Q(i-Q(i-1)) + Q(i-Q(i-2)) If Q(i) < Q(i-1) Then cont += 1 Next i Print "Primeros 10 terminos: "; For i = 1 To 10 Print Q(i) &" "; Next i Print Print "Termino numero 1000: "; Q(1000) Print "Terminos menores que los anteriores: " &cont End Output: Primeros 10 terminos: 1 1 2 3 3 4 5 5 6 6 Termino numero 1000: 502 Terminos menores que los anteriores: 49798  ## Fōrmulæ Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition. Programs in Fōrmulæ are created/edited online in its website. In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation. Solution The following function calculate the given number of terms of the Hofstadter Q sequence: Case 1 First 10 terms Case 2 Confirm and display that the 1000th term is 502 Case 3 Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000th term. ## Go Sure there are ways that run faster or handle larger numbers; for the task though, maps and recursion work just fine. package main import "fmt" var m map[int]int func initMap() { m = make(map[int]int) m[1] = 1 m[2] = 1 } func q(n int) (r int) { if r = m[n]; r == 0 { r = q(n-q(n-1)) + q(n-q(n-2)) m[n] = r } return } func main() { initMap() // task for n := 1; n <= 10; n++ { showQ(n) } // task showQ(1000) // extra credit count, p := 0, 1 for n := 2; n <= 1e5; n++ { qn := q(n) if qn < p { count++ } p = qn } fmt.Println("count:", count) // extra credit initMap() showQ(1e6) } func showQ(n int) { fmt.Printf("Q(%d) = %d\n", n, q(n)) }  Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 count: 49798 Q(1000000) = 512066  ## GW-BASIC 10 DIM Q!(1000) 20 Q(1) = 1: Q(2) = 1 30 FOR N = 3 TO 1000 40 Q(N) = Q(N - Q(N - 1)) + Q(N - Q(N - 2)) 50 NEXT N 60 FOR N = 1 TO 10 70 PRINT Q(N) 80 NEXT N 90 PRINT Q(1000) ## Haskell The basic task: qSequence = tail qq where qq = 0 : 1 : 1 : map g [3..] g n = qq !! (n - qq !! (n-1)) + qq !! (n - qq !! (n-2)) -- Output: *Main> (take 10 qSequence, qSequence !! (1000-1)) ([1,1,2,3,3,4,5,5,6,6],502) (0.00 secs, 525044 bytes)  Extra credit task: import Data.Array qSequence n = arr where arr = listArray (1,n)  1:1: map g [3..n] g i = arr!(i - arr!(i-1)) + arr!(i - arr!(i-2)) gradualth m k arr -- gradually precalculate m-th item | m <= v = pre seq arr!m -- in steps of k where -- to prevent STACK OVERFLOW pre = foldl1 (\a b-> a seq arr!b) [u,u+k..m] (u,v) = bounds arr qSeqTest m n = let arr = qSequence  max m n in ( take 10 . elems  arr -- 10 first items , gradualth m 10000  arr -- m-th item , length . filter (> 0) -- reversals in n items . _S (zipWith (-)) tail . take n . elems  arr ) _S f g x = f x (g x)  Output: Prelude Main> qSeqTest 1000 100000 -- reversals in 100,000 ([1,1,2,3,3,4,5,5,6,6],502,49798) (0.09 secs, 18879708 bytes) Prelude Main> qSeqTest 1000000 100000 -- 1,000,000-th item ([1,1,2,3,3,4,5,5,6,6],512066,49798) (2.80 secs, 87559640 bytes)  Using a list (more or less) seemlessly backed up by a double resizing array: q = qq (listArray (1,2) [1,1]) 1 where qq ar n = (arr!n) : qq arr (n+1) where l = snd (bounds ar) step n =arr!(n - (fromIntegral (arr!(n - 1)))) + arr!(n - (fromIntegral (arr!(n - 2)))) arr :: Array Int Integer arr | n <= l = ar | otherwise = listArray (1, l*2) ([ar!i | i <- [1..l]] ++ [step i | i <- [l+1..l*2]]) main = do putStr("first 10: "); print (take 10 q) putStr("1000-th: "); print (q !! 999) putStr("flips: ") print  length  filter id  take 100000 (zipWith (>) q (tail q))  Output: first 10: [1,1,2,3,3,4,5,5,6,6] 1000-th: 502 flips: 49798  List backed up by a list of arrays, with nominal constant lookup time. Somehow faster than the previous method. import Data.Array import Data.Int (Int64) q = qq [listArray (1,2) [1,1]] 1 where qq a n = seek aa n : qq aa (1 + n) where aa | n <= l = a | otherwise = listArray (l+1,l*2) (take l  drop 2 lst):a where l = snd (bounds  head a) lst = seek a (l-1):seek a l:(ext lst (l+1)) ext (q1:q2:qs) i = (g (i-q2) + g (i-q1)):ext (q2:qs) (1+i) g = seek aa seek (ar:ars) n | n >= fst (bounds ar) = ar ! n | otherwise = seek ars n -- Only a perf test. Task can be done exactly the same as above main = print  sum qqq where qqq :: [Int64] qqq = map fromIntegral  take 3000000 q  ## Icon and Unicon link printf procedure main() V := [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] every i := 1 to *V do if Q(i) ~= V[i] then stop("Assertion failure for position ",i) printf("Q(1 to %d) - verified.\n",*V) q := Q(n := 1000) v := 502 printf("Q[%d]=%d - %s.\n",n,v,if q = v then "verified" else "failed") invcount := 0 every i := 2 to (n := 100000) do if Q(i) < Q(i-1) then { printf("Q(%d)=%d < Q(%d)=%d\n",i,Q(i),i-1,Q(i-1)) invcount +:= 1 } printf("There were %d inversions in Q up to %d\n",invcount,n) end procedure Q(n) #: Hofstader Q sequence static S initial S := [1,1] if q := S[n] then return q else { q := Q(n - Q(n - 1)) + Q(n - Q(n - 2)) if *S = n - 1 then { put(S,q) return q } else runerr(500,n) } end  Output: Q(1 to 10) - verified. Q[1000]=502 - verified. Q(16)=9 < Q(15)=10 Q(25)=14 < Q(24)=16 Q(32)=17 < Q(31)=20 Q(36)=19 < Q(35)=21 ... Q(99996)=48252 < Q(99995)=50276 Q(99999)=48456 < Q(99998)=50901 Q(100000)=48157 < Q(99999)=48456 There were 49798 inversions in Q up to 100000 ## IS-BASIC 100 PROGRAM "QSequen.bas" 110 LET LIMIT=1000 120 NUMERIC Q(1 TO LIMIT) 130 LET Q(1),Q(2)=1 140 FOR I=3 TO LIMIT 150 LET Q(I)=Q(I-Q(I-1))+Q(I-Q(I-2)) 160 NEXT 170 PRINT "First 10 terms:" 180 FOR I=1 TO 10 190 PRINT Q(I); 200 NEXT 210 PRINT :PRINT "Term 1000:";Q(1000) ## J Solution (bottom-up):  Qs=:0 1 1 Q=: verb define n=. >./,y while. n>:#Qs do. Qs=: Qs,+/(-_2{.Qs){Qs end. y{Qs )  Solution (top-down):  Q=: 1:(+&:/@:- :@-& 1 2)@.(>&2)"0 M.  Example:  Q 1+i.10 1 1 2 3 3 4 5 5 6 6 Q 1000 502 +/2>/\ Q 1+i.100000 49798  Note: The bottom-up solution uses iteration and doesn't risk failure due to recursion limits or cache overflows. The top-down solution uses recursion, and likely hews closer to the spirit of the task. While this latter uses memoization/caching, at some point it will still hit a recursion limit (depends on the environment; in mine, it barfs at N=4402). We use the bottom up version for the extra credit part of this task (the expression which compares adjacent numbers and gave us the result 49798). It happens to be that the bottom-up version is written in the "explicit" style of code and the top-down version is written in the "tacit" (aka "point-free") style. This is incidental and it's possible to write bottom-up tacitly and/or top-down explicitly. The top-down version may be interesting as an example of algebraic factorization of code: taking advantage of some unique function composition operations in J, it manages to only mention : (aka recursion aka "Q") twice. ## Java Works with: Java version 1.5+ This example also counts the number of times each n is used as an argument up to 100000 and reports the one that was used the most. import java.util.HashMap; import java.util.Map; public class HofQ { private static Map<Integer, Integer> q = new HashMap<Integer, Integer>(){{ put(1, 1); put(2, 1); }}; private static int[] nUses = new int[100001];//not part of the task public static int Q(int n){ nUses[n]++;//not part of the task if(q.containsKey(n)){ return q.get(n); } int ans = Q(n - Q(n - 1)) + Q(n - Q(n - 2)); q.put(n, ans); return ans; } public static void main(String[] args){ for(int i = 1; i <= 10; i++){ System.out.println("Q(" + i + ") = " + Q(i)); } int last = 6;//value for Q(10) int count = 0; for(int i = 11; i <= 100000; i++){ int curr = Q(i); if(curr < last) count++; last = curr; if(i == 1000) System.out.println("Q(1000) = " + curr); } System.out.println("Q(i) is less than Q(i-1) for i <= 100000 " + count + " times"); //Optional stuff below here int maxUses = 0, maxN = 0; for(int i = 1; i<nUses.length;i++){ if(nUses[i] > maxUses){ maxUses = nUses[i]; maxN = i; } } System.out.println("Q(" + maxN + ") was called the most with " + maxUses + " calls"); } } Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 Q(i) is less than Q(i-1) for i <= 100000 49798 times Q(44710) was called the most with 19 calls ## JavaScript ### ES5 Based on memoization example from 'JavaScript: The Good Parts'. var hofstadterQ = function() { var memo = [1,1,1]; var Q = function (n) { var result = memo[n]; if (typeof result !== 'number') { result = Q(n - Q(n-1)) + Q(n - Q(n-2)); memo[n] = result; } return result; }; return Q; }(); for (var i = 1; i <=10; i += 1) { console.log('Q('+ i +') = ' + hofstadterQ(i)); } console.log('Q(1000) = ' + hofstadterQ(1000));  Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 ### ES6 Memoising with the accumulator of a fold (() => { 'use strict'; // hofQSeq :: Int -> [Int] const hofQSeq = x => x > 2 ? tail(foldl((Q, n) => n < 3 ? Q : Q.concat( Q[n - Q[n - 1]] + Q[n - Q[n - 2]] ), [0, 1, 1], range(1, x))) : (x > 0 ? take(x, [1, 1]) : undefined); // GENERIC FUNCTIONS ------------------------------------------- // foldl :: (b -> a -> b) -> b -> [a] -> b const foldl = (f, a, xs) => xs.reduce(f, a), // range :: Int -> Int -> [Int] range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i), // tail :: [a] -> [a] tail = xs => xs.length ? xs.slice(1) : undefined, // last :: [a] -> a last = xs => xs.length ? xs.slice(-1)[0] : undefined, // Int -> [a] -> [a] take = (n, xs) => xs.slice(0, n); // TEST -------------------------------------------------------- return { firstTen: hofQSeq(10), thousandth: last(hofQSeq(1000)), 'Q<Q-1UpTo10E5': hofQSeq(100000) .reduce((a, x, i, xs) => x < xs[i - 1] ? a + 1 : a, 0) }; })();  Output: {"firstTen":[1, 1, 2, 3, 3, 4, 5, 5, 6, 6], "thousandth":502, "Q<Q-1UpTo10E5":49798}  ## jq For the tasks related to evaluating Q(n) directy, a recursive implementation is used, firstly because the task requirements refer to "recursion limits", and secondly to demonstrate one way to handle a cache in a functional language. To count the number of inversions, a non-recursive approach is used as it is faster and scales linearly. For simplicity, we also define Q(0) = 1, so that the defining formula also holds for n == 2, and so that we can cache Q(n) at the n-th position of an array with index origin 0. # For n>=2, Q(n) = Q(n - Q(n-1)) + Q(n - Q(n-2)) def Q: def Q(n): n as n | (if . == null then [1,1,1] else . end) as q | if q[n] != null then q else q | Q(n-1) as q1 | q1 | Q(n-2) as q2 | q2 | Q(n - q2[n - 1]) as q3 # Q(n - Q(n-1)) | q3 | Q(n - q3[n - 2]) as q4 # Q(n - Q(n-2)) | (q4[n - q4[n-1]] + q4[n - q4[n -2]]) as ans | q4 | setpath( [n]; ans) end ; . as n | null | Q(n) | .[n]; # count the number of times Q(i) > Q(i+1) for 0 < i < n def flips(n): (reduce range(3; n) as n ([1,1,1]; . + [ .[n - .[n-1]] + .[n - .[n - 2 ]] ] )) as q | reduce range(0; n) as i (0; . + (if q[i] > q[i + 1] then 1 else 0 end)) ; # The three tasks: ((range(0;11), 1000) | "Q(\(.)) = \( . | Q)"), (100000 | "flips(\(.)) = \(flips(.))") ### Transcript  uname -a Darwin Mac-mini 13.3.0 Darwin Kernel Version 13.3.0: Tue Jun 3 21:27:35 PDT 2014; root:xnu-2422.110.17~1/RELEASE_X86_64 x86_64  time jq -r -n -f hofstadter.jq Q(0) = 1 Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 flips(100000) = 49798 real 0m0.562s user 0m0.541s sys 0m0.011s  ## Julia The following implementation accepts an argument that is a single integer, an array of integers, or a range: function hofstQseq(n, typerst::Type=Int) nmax = maximum(n) r = Vector{typerst}(nmax) r[1] = 1 if nmax ≥ 2 r[2] = 1 end for i in 3:nmax r[i] = r[i - r[i - 1]] + r[i - r[i - 2]] end return r[n] end println("First ten elements of sequence: ", join(hofstQseq(1:10), ", ")) println("1000-th element: ", hofstQseq(1000))  Output: First ten elements of sequence: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 1000-th element: 502 And we can also count the number of times a value is less than its predecessor by, for example: seq = hofstQseq(1:100_000) cnt = count(diff(seq) .< 0) println("cnt elements are less than the preceding one.")  Output: 49798 elements are less than the preceding one. Since the implementation is non-recursive, there is no issue with recursion limits. ## Kotlin // version 1.1.4 fun main(args: Array<String>) { val q = IntArray(100_001) q[1] = 1 q[2] = 1 for (n in 3..100_000) q[n] = q[n - q[n - 1]] + q[n - q[n - 2]] print("The first 10 terms are : ") for (i in 1..10) print("{q[i]} ") println("\n\nThe 1000th term is : {q[1000]}") val flips = (2..100_000).count { q[it] < q[it - 1] } println("\nThe number of flips for the first 100,000 terms is : flips") }  Output: The first 10 terms are : 1 1 2 3 3 4 5 5 6 6 The 1000th term is : 502 The number of flips for the first 100,000 terms is : 49798  ## Lua Here, the whole sequence up to the 100,000th term is generated for the first task, so this is where we risk hitting the recursion limit. As it happens, we do not. The function is called using 'pcall' so that any error would be caught. By increasing the argument on line 19 from 1e5 to 1e8, we can cause LuaJIT to run out of memory, but that is not necessary for this task. function hofstadter (limit) local Q = {1, 1} for n = 3, limit do Q[n] = Q[n - Q[n - 1]] + Q[n - Q[n - 2]] end return Q end function countDescents (t) local count = 0 for i = 2, #t do if t[i] < t[i - 1] then count = count + 1 end end return count end local noError, hofSeq = pcall(hofstadter, 1e5) if noError == false then print("The sequence could not be calculated up to the specified limit.") os.exit() end for i = 1, 10 do io.write(hofSeq[i] .. " ") end print("\n" .. hofSeq[1000]) print(countDescents(hofSeq))  Output: 1 1 2 3 3 4 5 5 6 6 502 49798 ## MAD  NORMAL MODE IS INTEGER VECTOR VALUES FMT = 2HQ(,I4,3H) =,I4* DIMENSION Q(1000) Q(1) = 1 Q(2) = 1 THROUGH FILL, FOR N=3, 1, N.G.1000 FILL Q(N) = Q(N-Q(N-1)) + Q(N-Q(N-2)) THROUGH SHOW, FOR N=1, 1, N.G.10 SHOW PRINT FORMAT FMT, N, Q(N) PRINT FORMAT FMT, 1000, Q(1000) END OF PROGRAM Output: Q( 1) = 1 Q( 2) = 1 Q( 3) = 2 Q( 4) = 3 Q( 5) = 3 Q( 6) = 4 Q( 7) = 5 Q( 8) = 5 Q( 9) = 6 Q( 10) = 6 Q(1000) = 502 ## Maple We use automatic memoisation ("option remember") in the following. The use of "option system" assures that memoised values can be garbage collected. Q := proc( n ) option remember, system; if n = 1 or n = 2 then 1 else thisproc( n - thisproc( n - 1 ) ) + thisproc( n - thisproc( n - 2 ) ) end if end proc: From this we get: > seq( Q( i ), i = 1 .. 10 ); 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 > Q( 1000 ); 502 To determine the number of "flips", we proceed as follows. > flips := 0: > for i from 2 to 100000 do > if L[ i ] < L[ i - 1 ] then > flips := 1 + flips > end if > end do: > flips; 49798 Alternatively, we can build the sequence in an array. Qflips := proc( n ) local a := Array( 1 .. n ); a[ 1 ] := 1; a[ 2 ] := 1; for local i from 3 to n do a[ i ] := a[ i - a[ i - 1 ] ] + a[ i - a[ i - 2 ] ] end do; local flips := 0; for i from 2 to n do if a[ i ] < a[ i - 1 ] then flips := 1 + flips end if end do; flips end proc: This gives the same result. > Qflips( 10^5 ); 49798 ## Mathematica / Wolfram Language Hofstadter[1] = Hofstadter[2] = 1; Hofstadter[n_Integer?Positive] := Hofstadter[n] = Block[{RecursionLimit = Infinity}, Hofstadter[n - Hofstadter[n - 1]] + Hofstadter[n - Hofstadter[n - 2]] ]  Output: Hofstadter /@ Range[10] {1,1,2,3,3,4,5,5,6,6} Hofstadter[1000] 502 Count[Differences[Hofstadter /@ Range[100000]], _?Negative] 49798  ## MATLAB / Octave This solution pre-allocates memory and is an iterative solution, so caching or recursion limits do not apply. function Q = Qsequence(N) %% zeros are used to pre-allocate memory, this is not strictly necessary but can significantly improve performance for large N Q = [1,1,zeros(1,N-2)]; for n=3:N Q(n) = Q(n-Q(n-1))+Q(n-Q(n-2)); end; end;  Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6 >> Qsequence(10) ans = 1 1 2 3 3 4 5 5 6 6  Confirm and display that the 1000'th term is: 502 >> Q=Qsequence(1000); Q(end) ans = 502  Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000'th term. >> sum(diff(Qsequence(100000))<0) ans = 49798  ## Maxima /* Function that return the terms of the Hofstadter Q sequence */ hofstadter(n):=block( if member(n,[1,2]) then L[n]:1 else L[n]:L[n-L[n-1]]+L[n-L[n-2]], L[n]) /* Test cases */ /* First ten terms */ makelist(hofstadter(i),i,1,10); /* 1000th term */ last(makelist(hofstadter(i),i,1,1000));  Output: [1,1,2,3,3,4,5,5,6,6] 502  ## Modula-2 MODULE QSequence; FROM InOut IMPORT WriteString, WriteCard, WriteLn; VAR n: CARDINAL; Q: ARRAY [1..1000] OF CARDINAL; BEGIN Q[1] := 1; Q[2] := 1; FOR n := 3 TO 1000 DO Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]]; END; WriteString("The first 10 terms are:"); FOR n := 1 TO 10 DO WriteCard(Q[n],2); END; WriteLn(); WriteString("The 1000th term is:"); WriteCard(Q[1000],4); WriteLn(); END QSequence.  Output: The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 ## MiniScript cache = {1:1, 2:1} Q = function(n) if not cache.hasIndex(n) then q = Q(n - Q(n-1)) + Q(n - Q(n-2)) cache[n] = q end if return cache[n] end function for i in range(1,10) print "Q(" + i + ") = " + Q(i) end for print "Q(1000) = " + Q(1000)  Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 ## Miranda main :: [sys_message] main = [Stdout (lay (map showq ([1..10] ++ [1000])))] where showq n = "q!" ++ show n ++ " = " ++ show (q!n) q :: [num] q = 0 : 1 : 1 : map f [3..] where f n = q!(n - q!(n-1)) + q!(n - q!(n-2)) Output: q!1 = 1 q!2 = 1 q!3 = 2 q!4 = 3 q!5 = 3 q!6 = 4 q!7 = 5 q!8 = 5 q!9 = 6 q!10 = 6 q!1000 = 502 ## Nim var q = @[1, 1] for n in 2 ..< 100_000: q.add q[n-q[n-1]] + q[n-q[n-2]] echo q[0..9] assert q[0..9] == @[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] echo q[999] assert q[999] == 502 var lessCount = 0 for n in 1 ..< 100_000: if q[n] < q[n-1]: inc lessCount echo lessCount  Output: @[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798 ## Oberon-2 Works with oo2c version 2 MODULE Hofstadter; IMPORT Out; VAR i,count,q,prev: LONGINT; founds: ARRAY 100001 OF LONGINT; PROCEDURE Q(n: LONGINT): LONGINT; BEGIN IF founds[n] = 0 THEN CASE n OF 1 .. 2: founds[n] := 1 ELSE founds[n] := Q(n - Q(n - 1)) + Q(n - Q(n - 2)) END END; RETURN founds[n] END Q; BEGIN (* first ten numbers in the sequence *) FOR i := 1 TO 10 DO Out.String("At ");Out.LongInt(i,0);Out.String(":> ");Out.LongInt(Q(i),4);Out.Ln END; Out.String("1000th value: ");Out.LongInt(Q(1000),4);Out.Ln; prev := 1; FOR i := 2 TO 100000 DO q := Q(i); IF q < prev THEN INC(count) END; prev := q END; Out.String("terms less than the previous: ");Out.LongInt(count,4);Out.Ln END Hofstadter. Output: At 1:> 1 At 2:> 1 At 3:> 2 At 4:> 3 At 5:> 3 At 6:> 4 At 7:> 5 At 8:> 5 At 9:> 6 At 10:> 6 1000th value: 502 terms less than the previous: 49798  ## OCaml (* valid results for n in 0..119628 *) let seq_hofstadter_q n = let a = Bigarray.(Array1.create int16_unsigned c_layout n) in let () = for i = 0 to pred n do a.{i} <- if i < 2 then 1 else a.{i - a.{pred i}} + a.{i - a.{i - 2}} done in Seq.init n (Bigarray.Array1.get a) let () = let count_backflip (a, c) b = b, if b < a then succ c else c and hq = seq_hofstadter_q 100_000 in let () = Seq.(hq |> take 10 |> iter (Printf.printf " %u")) in let () = Seq.(hq |> drop 999 |> take 1 |> iter (Printf.printf "\n%u\n")) in hq |> Seq.fold_left count_backflip (0, 0) |> snd |> Printf.printf "%u\n"  Output:  1 1 2 3 3 4 5 5 6 6 502 49798  ## Oforth : QSeqTask | q i | ListBuffer newSize(100000) dup add(1) dup add(1) ->q 0 3 100000 for: i [ q add(q at(i q at(i 1-) -) q at(i q at(i 2 -) -) +) q at(i) q at(i 1-) < ifTrue: [ 1+ ] ] q left(10) println q at(1000) println println ; Output: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798  ## PARI/GP Straightforward, unoptimized version; about 1 ms. Q=vector(1000);Q[1]=Q[2]=1;for(n=3,#Q,Q[n]=Q[n-Q[n-1]]+Q[n-Q[n-2]]); Q1=vecextract(Q,"1..10"); print("First 10 terms: "Q1,if(Q1==[1, 1, 2, 3, 3, 4, 5, 5, 6, 6]," (as expected)"," (in error)")); print("1000-th term: "Q[1000],if(Q[1000]==502," (as expected)"," (in error)")); Output: First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] (as expected) 1000-th term: 502 (as expected) ## Pascal Program HofstadterQSequence (output); const limit = 100000; var q: array [1..limit] of longint; i, flips: longint; begin q[1] := 1; q[2] := 1; for i := 3 to limit do q[i] := q[i - q[i - 1]] + q[i - q[i - 2]]; for i := 1 to 10 do write(q[i], ' '); writeln; writeln(q[1000]); flips := 0; for i := 1 to limit - 1 do if q[i] > q[i+1] then inc(flips); writeln('Flips: ', flips); end.  Output: :> ./HofstadterQSequence 1 1 2 3 3 4 5 5 6 6 502 Flips: 49798  ## Perl my @Q = (0,1,1); push @Q, Q[-Q[-1]] + Q[-Q[-2]] for 1..100_000; say "First 10 terms: [@Q[1..10]]"; say "Term 1000: Q[1000]"; say "Terms less than preceding in first 100k: ",scalar(grep { Q[_] < Q[_-1] } 2..100000);  Output: First 10 terms: [1 1 2 3 3 4 5 5 6 6] Term 1000: 502 Terms less than preceding in first 100k: 49798 A more verbose and less idiomatic solution: #!/usr/bin/perl use warnings; use strict; my @hofstadters = ( 1 , 1 ); while ( @hofstadters < 100000 ) { my nextn = @hofstadters + 1; # array index counting starts at 0 , so we have to subtract 1 from the numbers! push @hofstadters , hofstadters [ nextn - 1 - hofstadters[ nextn - 1 - 1 ] ] + hofstadters[ nextn - 1 - hofstadters[ nextn - 2 - 1 ]]; } for my i ( 0..9 ) { print "hofstadters[ i ]\n"; } print "The 1000'th term is hofstadters[ 999 ]!\n"; my less_than_preceding = 0; for my i ( 0..99998 ) { less_than_preceding++ if hofstadters[ i + 1 ] < hofstadters[ i ]; } print "Up to and including the 100000'th term, less_than_preceding terms are less " . "than their preceding terms!\n";  Output: 1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502! Up to and including the 100000'th term, 49798 terms are less than their preceding terms!  This different solution uses tie to make the Q sequence look like a regular array, and only fills the cache on demand. Some pre-allocation is done which provides a minor speed increase for the extra credit. I could have chosen to do recursion instead of iteration, as perl has no limit on how deeply one may recurse, but did not see the benefit of doing so. #!perl use strict; use warnings; package Hofstadter; sub TIEARRAY { bless [undef, 1, 1], shift; } sub FETCH { my (self, n) = @_; die if n < 1; if( n > #self ) { my start = #self + 1; #self = n; # pre-allocate for efficiency for my nn ( start .. n ) { my (a, b) = (1, 2); _ = self->[ nn - _ ] for a, b; _ = self->[ nn - _ ] for a, b; self->[nn] = a + b; } } self->[n]; } package main; tie my (@q), "Hofstadter"; print "@q[1..10]\n"; print q[1000], "\n"; my count = 0; for my n ( 2 .. 100_000 ) { count++ if q[n] < q[n - 1]; } print "Extra credit: count\n";  Output: 1 1 2 3 3 4 5 5 6 6 502 Extra credit: 49798  ## Phix Just to be flash, I also (on the desktop only) calculated the 100 millionth term - the only limiting factor here is the length of Q (theoretically 402,653,177 on 32 bit). with javascript_semantics sequence Q = {1,1} function q(integer n) integer l = length(Q) while n>l do l += 1 Q &= Q[l-Q[l-1]]+Q[l-Q[l-2]] end while return Q[n] end function {} = q(10) -- (or collect one by one) printf(1,"First ten terms: %v\n",{Q[1..10]}) printf(1,"1000th: %d\n",q(1000)) printf(1,"100,000th: %,d\n",q(100_000)) integer n = 0 for i=2 to 100_000 do n += Q[i]<Q[i-1] end for printf(1,"Flips up to 100,000: %,d\n",{n}) if platform()!=JS then atom t0 = time() printf(1,"100,000,000th: %,d (%3.2fs)\n",{q(100_000_000),time()-t0}) end if  Output: First ten terms: {1,1,2,3,3,4,5,5,6,6} 1000th: 502 100,000th: 48,157 Flips up to 100,000: 49,798 100,000,000th: 50,166,508 (8.52s)  The last line shows fine under pwa/p2js, but would take about 20s. ## Picat go => println([q(I) : I in 1..10]), println(q1000=q(1000)), Q = {q(I) : I in 1..100_000}, println(flips=sum({1 : I in 2..100_000, Q[I-1] > Q[I]})), nl. table q(1) = 1. q(2) = 1. q(N) = q(N-q(N-1)) + q(N-q(N-2)). Output: [1,1,2,3,3,4,5,5,6,6] q1000 = 502 flips = 49798 ## PicoLisp (de q (N) (cache '(NIL) N (if (>= 2 N) 1 (+ (q (- N (q (dec N)))) (q (- N (q (- N 2)))) ) ) ) ) Test: : (mapcar q (range 1 10)) -> (1 1 2 3 3 4 5 5 6 6) : (q 1000) -> 502 : (let L (mapcar q (range 1 100000)) (cnt < (cdr L) L) ) -> 49798 ## PL/I /* Hofstrader Q sequence for any "n". */ H: procedure options (main); /* 28 January 2012 */ declare n fixed binary(31); put ('How many values do you want? :'); get (n); begin; declare Q(n) fixed binary (31); declare i fixed binary (31); Q(1), Q(2) = 1; do i = 1 upthru n; if i >= 3 then Q(i) = ( Q(i - Q(i-1)) + Q(i - Q(i-2)) ); if i <= 20 then put skip list ('n=' || trim(i), Q(i)); end; put skip list ('n=' || trim(i), Q(i)); end; end H; Output: How many values do you want? : n=1 1 n=2 1 n=3 2 n=4 3 n=5 3 n=6 4 n=7 5 n=8 5 n=9 6 n=10 6 n=11 6 n=12 8 n=13 8 n=14 8 n=15 10 n=16 9 n=17 10 n=18 11 n=19 11 n=20 12 n=1000 502  Output: for n=100,000 n=100000 48157  Bonus to produce the count of unordered values:  declare tally fixed binary (31) initial (0); do i = 1 to n-1; if Q(i) > Q(i+1) then tally = tally + 1; end; put skip data (tally);  Output: n=100000 48157 TALLY= 49798;  ## PL/M 100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT; PRINTNUMBER: PROCEDURE (N); DECLARE S (7) BYTE INITIAL ('..... '); DECLARE (N, P) ADDRESS, C BASED P BYTE; P = .S(5); DIGIT: P = P - 1; C = N MOD 10 + '0'; N = N / 10; IF N > 0 THEN GO TO DIGIT; CALL PRINT(P); END PRINTNUMBER; DECLARE Q (1001) ADDRESS; DECLARE N ADDRESS; Q(1)=1; Q(2)=1; DO N=3 TO LAST(Q); Q(N) = Q(N-Q(N-1)) + Q(N-Q(N-2)); END; CALL PRINT(.'THE FIRST 10 TERMS ARE: '); DO N=1 TO 10; CALL PRINTNUMBER(Q(N)); END; CALL PRINT(.(13,10,'THE 1000TH TERM IS: ')); CALL PRINTNUMBER(Q(1000)); CALL EXIT; EOF Output: THE FIRST 10 TERMS ARE: 1 1 2 3 3 4 5 5 6 6 THE 1000TH TERM IS: 502 ## PureBasic If Not OpenConsole("Hofstadter Q sequence") End 1 EndIf #N = 100000 Define i.i, flip.i = 0 Dim q.i(#N) q(1) = 1 q(2) = 1 For i = 3 To #N q(i) = q(i - q(i - 1)) + q(i - q(i - 2)) Next For i = 1 To #N - 1 flip + Bool(q(i) > q(i + 1)) Next Print(~"First ten:\t") For i = 1 To 10 : Print(LSet(Str(q(i)), 3)) : Next PrintN(~"\n1000th:\t\t" + Str(q(1000))) PrintN(~"Flips:\t\t" + Str(flip)) Input() End Output: First ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 Flips: 49798 ## Python def q(n): if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1") try: return q.seq[n] except IndexError: ans = q(n - q(n - 1)) + q(n - q(n - 2)) q.seq.append(ans) return ans q.seq = [None, 1, 1] if __name__ == '__main__': first10 = [q(i) for i in range(1,11)] assert first10 == [1, 1, 2, 3, 3, 4, 5, 5, 6, 6], "Q() value error(s)" print("Q(n) for n = [1..10] is:", ', '.join(str(i) for i in first10)) assert q(1000) == 502, "Q(1000) value error" print("Q(1000) =", q(1000))  Extra credit If you try and initially compute larger values of n then you tend to hit the Python recursion limit. The function q1 gets around this by calling function q to extend the Q series in increments below the recursion limit. The following code is to be concatenated to the code above: from sys import getrecursionlimit def q1(n): if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1") try: return q.seq[n] except IndexError: len_q, rlimit = len(q.seq), getrecursionlimit() if (n - len_q) > (rlimit // 5): for i in range(len_q, n, rlimit // 5): q(i) ans = q(n - q(n - 1)) + q(n - q(n - 2)) q.seq.append(ans) return ans if __name__ == '__main__': tmp = q1(100000) print("Q(i+1) < Q(i) for i [1..100000] is true %i times." % sum(k1 < k0 for k0, k1 in zip(q.seq[1:], q.seq[2:])))  Combined output: Q(n) for n = [1..10] is: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 Q(1000) = 502 Q(i+1) < Q(i) for i [1..10000] is true 49798 times. ### Alternative def q(n): l = len(q.seq) while l <= n: q.seq.append(q.seq[l - q.seq[l - 1]] + q.seq[l - q.seq[l - 2]]) l += 1 return q.seq[n] q.seq = [None, 1, 1] print("Q(n) for n = [1..10] is:", [q(i) for i in range(1, 11)]) print("Q(1000) =", q(1000)) q(100000) print("Q(i+1) < Q(i) for i [1..100000] is true %i times." % sum([q.seq[i] > q.seq[i + 1] for i in range(1, 100000)]))  ## Quackery [ 2dup swap size dup negate swap within not if [ drop size 1+ number  "Term " swap join  " of the Q sequence is not defined." join message put bail ] peek ] is qpeek ( [ n --> x ) [ dup dup -1 qpeek negate qpeek dip [ dup dup -2 qpeek negate qpeek ] + join ] is next-q ( [ --> [ ) [ dup size 2 < iff [ drop 0 ] done 0 swap behead swap witheach [ tuck > if [ dip 1+ ] ] drop ] is drops ( [ --> n ) 0 backup [ ' [ 1 1 ] 998 times next-q dup -1 split swap 10 split drop witheach [ echo sp ] say "... " 0 peek echo cr 99000 times next-q drops echo say " decreasing terms" ] bailed if [ message take cr echo cr ] Output: 1 1 2 3 3 4 5 5 6 6 ... 502 49798 decreasing terms ## R cache <- vector("integer", 0) cache[1] <- 1 cache[2] <- 1 Q <- function(n) { if (is.na(cache[n])) { value <- Q(n-Q(n-1)) + Q(n-Q(n-2)) cache[n] <<- value } cache[n] } for (i in 1:1e5) { Q(i) } for (i in 1:10) { cat(Q(i)," ",sep = "") } cat("\n") cat(Q(1000),"\n") count <- 0 for (i in 2:1e5) { if (Q(i) < Q(i-1)) count <- count + 1 } cat(count,"terms is less than its preceding term\n") Output: 1 1 2 3 3 4 5 5 6 6 502 49798 terms is less than its preceding term  ## Racket #lang racket (define t (make-hash)) (hash-set! t 0 0) (hash-set! t 1 1) (hash-set! t 2 1) (define (Q n) (hash-ref! t n (λ() (+ (Q (- n (Q (- n 1)))) (Q (- n (Q (- n 2)))))))) (for/list ([i (in-range 1 11)]) (Q i)) (Q 1000) ;; extra credit (for/sum ([i 100000]) (if (< (Q (add1 i)) (Q i)) 1 0))  Output: '(1 1 2 3 3 4 5 5 6 6) 502 49798  ## Raku (formerly Perl 6) ### OO solution Works with: rakudo version 2016.03 Similar concept as the perl5 solution, except that the cache is only filled on demand. class Hofstadter { has @!c = 1,1; method AT-POS (me: Int i) { @!c.push(me[@!c.elems-me[@!c.elems-1]] + me[@!c.elems-me[@!c.elems-2]]) until @!c[i]:exists; return @!c[i]; } } # Testing: my Hofstadter Q .= new(); say "first ten: Q[^10]"; say "1000th: Q[999]"; my count = 0; count++ if Q[_ +1 ] < Q[_] for ^99_999; say "In the first 100_000 terms, count terms are less than their preceding terms";  Output: first ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 In the first 100_000 terms, 49798 terms are less than their preceding terms ### Idiomatic solution Works with: rakudo version 2015-11-22 With a lazily generated array, we automatically get caching. my @Q = 1, 1, -> a, b { (state n = 1)++; @Q[n - a] + @Q[n - b] } ... *; # Testing: say "first ten: ", @Q[^10]; say "1000th: ", @Q[999]; say "In the first 100_000 terms, ", [+](@Q[1..100000] Z< @Q[0..99999]), " terms are less than their preceding terms";  (Same output.) ## REXX ### non-recursive The REXX language doesn't allow expressions for stemmed array indices, so a temporary variable must be used. /*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a=='' | a=="," then a= 10 /*Not specified? Then use the default.*/ if b=='' | b=="," then b= -1000 /* " " " " " " */ if c=='' | c=="," then c= -100000 /* " " " " " " */ if d=='' | d=="," then d= -1000000 /* " " " " " " */ @.= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1 downs= downs + (@.j<@.jm) end /*j*/ say commas(downs) ' HofstatdterQ terms are less then the previous term,' , ' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac) call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/ /* [↑] OX is the same as X. */ x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/ do j=1 for x /* [↓] use short─circuit IF test*/ if j>2 then if @.j==1 then do; jm1= j - 1; jm2= j - 2 one= j - @.jm1; two= j - @.jm2 @.j= @.one + @.two end if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j))) end /*j*/ return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4))  output when using the internal default inputs: HofstadterQ( 1): 1 HofstadterQ( 2): 1 HofstadterQ( 3): 2 HofstadterQ( 4): 3 HofstadterQ( 5): 3 HofstadterQ( 6): 4 HofstadterQ( 7): 5 HofstadterQ( 8): 5 HofstadterQ( 9): 6 HofstadterQ(10): 6 HofstadterQ 1,000th term is: 502 49,798 HofstatdterQ terms are less then the previous term, HofstatdterQ(100,000th) term is: 48,157 The 1,000,000th HofstatdterQ term is: 512,066  ### non-recursive, simpler This REXX example is identical to the first version except that it uses a function to retrieve array elements which may have index expressions. /*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a=='' | a=="," then a= 10 /*Not specified? Then use the default.*/ if b=='' | b=="," then b= -1000 /* " " " " " " */ if c=='' | c=="," then c= -100000 /* " " " " " " */ if d=='' | d=="," then d= -1000000 /* " " " " " " */ @.= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1 downs= downs + (@.j<@.jm) end /*j*/ say commas(downs) ' HofstatdterQ terms are less then the previous term,' , ' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac) call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/ /* [↑] OX is the same as X. */ x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/ do j=1 for x if j>2 then if @.j==1 then @.j= @(j - @(j-1)) + @(j - @(j-2)) if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j))) end /*j*/ return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ @: parse arg ?; return @.? /*return value of @.? to invoker.*/ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4)) commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _  output is identical to the 1st REXX version. Because of the additional subroutine (function) invokes, this REXX version is about half as fast as the 1st REXX version. ### recursive /*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a=='' | a=="," then a= 10 /*Not specified? Then use the default.*/ if b=='' | b=="," then b= -1000 /* " " " " " " */ if c=='' | c=="," then c= -100000 /* " " " " " " */ if d=='' | d=="," then d= -1000000 /* " " " " " " */ @.= 0; @.1= 1; @.2= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1 downs= downs + (@.j<@.jm) end /*j*/ say commas(downs) ' HofstatdterQ terms are less then the previous term,' , ' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac) call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/ /* [↑] OX is the same as X. */ x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/ do j=1 for x if @.j==0 then @.j= QR(j) /*Not defined? Then define it.*/ if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j))) end /*j*/ return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ QR: procedure expose @.; parse arg n /*this QR function is recursive.*/ if @.n==0 then @.n= QR(n-QR(n-1)) + QR(n-QR(n-2)) /*Not defined? Then define it.*/ return @.n /*return the value to the invoker*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4)) commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _  output is identical to the 1st REXX version. The recursive version is almost ten times slower than the (1st) non-recursive version. ## Ring n = 20 aList = list(n) aList[1] = 1 aList[2] = 1 for i = 1 to n if i >= 3 aList[i] = ( aList[i - aList[i-1]] + aList[i - aList[i-2]] ) ok if i <= 20 see "n = " + string(i) + " : "+ aList[i] + nl ok next ## RPL Works with: Halcyon Calc version 4.2.7 RPL code Comment ≪ { 1 1 } 3 WHILE DUP 4 PICK ≤ REPEAT DUP2 2 - GETI ROT ROT GET → n q2 q1 ≪ DUP n q1 - GET OVER n q2 - GET + + n 1 + SWAP ≫ END DROP ≫ 'HOFST' STO  HOFST ( m -- { Q(1)..Q(m) } ) initialize stack with Q1, Q2 and loop index n loop store n, Q(n-2) and Q(n-1) get Q(n-Q(n-1)) add Q(n-Q(n-2)) and add result to list put back n+1 in stack  Input: 10 HOFST 1000 HOSFT DUP SIZE GET  Output: 2: { 1 1 2 3 3 4 5 5 6 6 } 1: 502  ## Ruby @cache = [] def Q(n) if @cache[n].nil? case n when 1, 2 then @cache[n] = 1 else @cache[n] = Q(n - Q(n-1)) + Q(n - Q(n-2)) end end @cache[n] end puts "first 10 numbers in the sequence: #{(1..10).map {|n| Q(n)}}" puts "1000'th term: #{Q(1000)}" prev = Q(1) count = 0 2.upto(100_000) do |n| q = Q(n) count += 1 if q < prev prev = q end puts "number of times in the first 100,000 terms where Q(i)<Q(i-1): #{count}"  Output: first 10 numbers in the sequence: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 1000'th term: 502 number of times in the first 100,000 terms where Q(i)<Q(i-1): 49798 ## Run BASIC input "How many values do you want? :";n dim Q(n) Q(1) = 1 Q(2) = 1 for i = 1 to n if i >= 3 then Q(i) = ( Q(i - Q(i-1)) + Q(i - Q(i-2)) ) if i <= 20 then print "n=";using("####",i);" ";using("###",Q(i)) next i if i > 20 then print "n=";using("####",i);using("####",Q(i)) end Output: How many values do you want? :?1000 n= 1 1 n= 2 1 n= 3 2 n= 4 3 n= 5 3 n= 6 4 n= 7 5 n= 8 5 n= 9 6 n= 10 6 n= 11 6 n= 12 8 n= 13 8 n= 14 8 n= 15 10 n= 16 9 n= 17 10 n= 18 11 n= 19 11 n= 20 12 n=1000 502  ## Rust Rust doesn't allow static Vec's (but there's lazy_static crate), thus memoization storage is allocated in main. fn hofq(q: &mut Vec<u32>, x : u32) -> u32 { let cur_len=q.len()-1; let i=x as usize; if i>cur_len { // extend storage q.reserve(i+1); for j in (cur_len+1)..(i+1) { let qj=(q[j-q[j-1] as usize]+q[j-q[j-2] as usize]) as u32; q.push(qj); } } q[i] } fn main() { let mut q_memo: Vec<u32>=vec![0,1,1]; let mut q=|i| {hofq(&mut q_memo, i)}; for i in 1..11 { println!("Q({})={}", i, q(i)); } println!("Q(1000)={}", q(1000)); let q100001=q(100_000); // precompute all println!("Q(100000)={}", q100000); let nless=(1..100_000).fold(0,|s,i|{if q(i+1)<q(i) {s+1} else {s}}); println!("Term is less than preceding term {} times", nless); }  Output: Q(1)=1 Q(2)=1 Q(3)=2 Q(4)=3 Q(5)=3 Q(6)=4 Q(7)=5 Q(8)=5 Q(9)=6 Q(10)=6 Q(1000)=502 Q(100001)=53471 Term is less than preceding term 49798 times  ## Scala Works with: Scala version 2.9.1 Naive but elegant version using only recursion doesn't work because runtime is excessive increasing ... object HofstadterQseq extends App { val Q: Int => Int = n => { if (n <= 2) 1 else Q(n-Q(n-1))+Q(n-Q(n-2)) } (1 to 10).map(i=>(i,Q(i))).foreach(t=>println("Q("+t._1+") = "+t._2)) println("Q("+1000+") = "+Q(1000)) }  Unfortunately the function Q isn't tail recursiv, therefore the compiler can't optimize it. Thus we are forced to use a caching featured version. object HofstadterQseq extends App { val HofQ = scala.collection.mutable.Map((1->1),(2->1)) val Q: Int => Int = n => { if (n < 1) 0 else { val res = HofQ.keys.filter(_==n).toList match { case Nil => {val v = Q(n-Q(n-1))+Q(n-Q(n-2)); HofQ += (n->v); v} case xs => HofQ(n) } res } } (1 to 10).map(i=>(i,Q(i))).foreach(t=>println("Q("+t._1+") = "+t._2)) println("Q("+1000+") = "+Q(1000)) println((3 to 100000).filter(i=>Q(i)<Q(i-1)).size) }  Output: Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 49798 ## Scheme I wish there were a portable way to define-syntax, or to resize arrays, or to do formated output--anything to make the code less silly looking while still run under more than one interpreter. (define qc '#(0 1 1)) (define filled 3) (define len 3) ;; chicken scheme: vector-resize! ;; gambit: vector-append (define (extend-qc) (let* ((new-len (* 2 len)) (new-qc (make-vector new-len))) (let copy ((n 0)) (if (< n len) (begin (vector-set! new-qc n (vector-ref qc n)) (copy (+ 1 n))))) (set! len new-len) (set! qc new-qc))) (define (q n) (let loop () (if (>= filled len) (extend-qc)) (if (>= n filled) (begin (vector-set! qc filled (+ (q (- filled (q (- filled 1)))) (q (- filled (q (- filled 2)))))) (set! filled (+ 1 filled)) (loop)) (vector-ref qc n)))) (display "Q(1 .. 10): ") (let loop ((i 1)) ;; (print) behave differently regarding newline across compilers (display (q i)) (display " ") (if (< i 10) (loop (+ 1 i)) (newline))) (display "Q(1000): ") (display (q 1000)) (newline) (display "bumps up to 100000: ") (display (let loop ((s 0) (i 1)) (if (>= i 100000) s (loop (+ s (if (> (q i) (q (+ 1 i))) 1 0)) (+ 1 i))))) (newline)  Output: Q(1 .. 10): 1 1 2 3 3 4 5 5 6 6 Q(1000): 502 bumps up to 100000: 49798 ## Seed7  include "seed7_05.s7i"; const type: intHash is hash [integer] integer; var intHash: qHash is intHash.value; const func integer: q (in integer: n) is func result var integer: q is 1; begin if n in qHash then q := qHash[n]; else if n > 2 then q := q(n - q(pred(n))) + q(n - q(n - 2)); end if; qHash @:= [n] q; end if; end func; const proc: main is func local var integer: n is 0; var integer: less_than_preceding is 0; begin writeln("q(n) for n = 1 .. 10:"); for n range 1 to 10 do write(q(n) <& " "); end for; writeln; writeln("q(1000)=" <& q(1000)); for n range 2 to 100000 do if q(n) < q(pred(n)) then incr(less_than_preceding); end if; end for; writeln("q(n) < q(n-1) for n = 2 .. 100000: " <& less_than_preceding); end func; Output: q(n) for n = 1 .. 10: 1 1 2 3 3 4 5 5 6 6 q(1000)=502 q(n) < q(n-1) for n = 2 .. 100000: 49798  ## SETL program hofstadter_q; q := [1,1]; loop for n in [3..100000] do q(n) := q(n-q(n-1)) + q(n-q(n-2)); end loop; print("First 10 terms: " + q(1..10)); print("1000th term: " + q(1000)); print("q(x) < q(x-1): " + #[x : x in [2..#q] | q(x) < q(x-1)]); end program; Output: First 10 terms: [1 1 2 3 3 4 5 5 6 6] 1000th term: 502 q(x) < q(x-1): 49798 ## Sidef Using a memoized function: func Q(n) is cached { n <= 2 ? 1 : Q(n - Q(n-1))+Q(n-Q(n-2)) } say "First 10 terms: #{ {|n| Q(n) }.map(1..10) }" say "Term 1000: #{Q(1000)}" say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q(i)<Q(i-1)}}"  Using an array: var Q = [0, 1, 1] 100_000.times { Q << (Q[-Q[-1]] + Q[-Q[-2]]) } say "First 10 terms: #{Q.slice(1).first(10)}" say "Term 1000: #{Q[1000]}" say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q[i]<Q[i-1]}}"  Output: First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] Term 1000: 502 Terms less than preceding in first 100k: 49798  ## Swift Translation of: C let n = 100000 var q = Array(repeating: 0, count: n) q[0] = 1 q[1] = 1 for i in 2..<n { q[i] = q[i - q[i - 1]] + q[i - q[i - 2]] } print("First 10 elements of the sequence: \(q[0..<10])") print("1000th element of the sequence: \(q[999])") var count = 0 for i in 1..<n { if q[i] < q[i - 1] { count += 1 } } print("Number of times a member of the sequence is less than the preceding term for terms up to and including the 100,000th term: \(count)")  Output: First 10 elements of the sequence: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 1000th element of the sequence: 502 Number of times a member of the sequence is less than the preceding term for terms up to and including the 100,000th term: 49798  ## Tailspin templates q def outputFrom: (1); def until: (2); @: [1,1]; 1..until -> # when <@::length~..> do ..|@: @( - @( - 1)) + @( - @( - 2));  -> # when <outputFrom..> do @() ! end q [1,10] -> q -> '; ' -> !OUT::write ' ' -> !OUT::write [1000,1000] -> q -> '; ' -> !OUT::write templates countDownSteps @: 0; def qs: ; 2..qs::length -> # @ ! when <?(qs() <..~qs(-1)>)> do @: @ + 1; end countDownSteps [[1, 100000] -> q] -> countDownSteps -> 'Less than previous ; times' -> !OUT::write Output: 1 1 2 3 3 4 5 5 6 6 502 Less than previous 49798 times  ## Tcl package require Tcl 8.5 # Index 0 is not used, but putting it in makes the code a bit shorter set tcl::mathfunc::Qcache {Q:-> 1 1} proc tcl::mathfunc::Q {n} { variable Qcache if {n >= [llength Qcache]} { lappend Qcache [expr {Q(n - Q(n-1)) + Q(n - Q(n-2))}] } return [lindex Qcache n] } # Demonstration code for {set i 1} {i <= 10} {incr i} { puts "Q(i) == [expr {Q(i)}]" } # This runs very close to recursion limit... puts "Q(1000) == [expr Q(1000)]" # This code is OK, because the calculations are done step by step set q [expr Q(1)] for {set i 2} {i <= 100000} {incr i} { incr count [expr {q > [set q [expr {Q(i)}]]}] } puts "Q(i)<Q(i-1) for i \[2..100000$ is true \$count times"

Output:
Q(1) == 1
Q(2) == 1
Q(3) == 2
Q(4) == 3
Q(5) == 3
Q(6) == 4
Q(7) == 5
Q(8) == 5
Q(9) == 6
Q(10) == 6
Q(1000) == 502
Q(i)<Q(i-1) for i [2..100000] is true 49798 times


## uBasic/4tH

Translation of: BBC BASIC

uBasic/4tH simply lacks the memory to make it through to the 1000th term. 256 is the best it can do.

Print "First 10 terms of Q = " ;
For i = 1 To 10 : Print FUNC(_q(i));" "; : Next : Print
Print "256th term = ";FUNC(_q(256))

End

_q Param(1)
Local(2)

If a@ < 3 Then Return (1)
If a@ = 3 Then Return (2)

@(0) = 1 : @(1) = 1 : @(2) = 2
c@ = 0

For b@ = 3 To a@-1
@(b@) = @(b@ - @(b@-1)) + @(b@ - @(b@-2))
If @(b@) < @(b@-1) Then c@ = c@ + 1
Next

Return (@(a@-1))

Output:
First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6
256th term = 123

0 OK, 0:320

## VBA

Public Q(100000) As Long
Dim n As Long, smaller As Long
Q(1) = 1
Q(2) = 1
For n = 3 To 100000
Q(n) = Q(n - Q(n - 1)) + Q(n - Q(n - 2))
If Q(n) < Q(n - 1) Then smaller = smaller + 1
Next n
Debug.Print "First ten terms:"
For i = 1 To 10
Debug.Print Q(i);
Next i
Debug.print
Debug.Print "The 1000th term is:"; Q(1000)
Debug.Print "Number of times smaller:"; smaller
End Sub
Output:
First ten terms:
1  1  2  3  3  4  5  5  6  6
The 1000th term is: 502
Number of times smaller: 49798 

## VBScript

Sub q_sequence(n)
Dim Q()
ReDim Q(n)
Q(1)=1 : Q(2)=1 : Q(3)=2
less_precede = 0
For i = 4 To n
Q(i)=Q(i-Q(i-1))+Q(i-Q(i-2))
If Q(i) < Q(i-1) Then
less_precede = less_precede + 1
End If
Next
WScript.StdOut.Write "First 10 terms of the sequence: "
For j = 1 To 10
If j < 10 Then
WScript.StdOut.Write Q(j) & ", "
Else
WScript.StdOut.Write "and " & Q(j)
End If
Next
WScript.StdOut.WriteLine
WScript.StdOut.Write "1000th term of the sequence: " & Q(1000)
WScript.StdOut.WriteLine
WScript.StdOut.Write "Number of times the member of the sequence is less than its preceding term: " &_
less_precede
End Sub

q_sequence(100000)
Output:
First 10 terms of the sequence: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6
1000th term of the sequence: 502
Number of times the member of the sequence is less than its preceding term: 49798


## Visual FoxPro

LOCAL p As Integer, i As Integer
CLEAR
p = 0
? "Hofstadter Q Sequence"
? "First 10 terms:"
FOR i = 1 TO 10
?? Q(i, @p)
ENDFOR
? "1000th term:", Q(1000, @p)
? "100000th term:", q(100000, @p)
? "Number of terms less than the preceding term:", p

FUNCTION Q(n As Integer, k As Integer) As Integer
LOCAL i As Integer
LOCAL ARRAY aq[n]
aq[1] = 1
IF n > 1
aq[2] = 1
ENDIF
k = 0
FOR i = 3 TO n
aq[i] = aq[i - aq[i-1]] + aq[i-aq[i-2]]
IF aq(i) < aq(i-1)
k = k + 1
ENDIF
ENDFOR
RETURN aq[n]
ENDFUNC

Output:
Hofstadter Q Sequence
First 10 terms:  1    1    2    3    3    4    5    5    6   6
1000th term:     502
100000th term:   48157
Number of terms less than the preceding term:  49798


## Wren

var N = 1e5
var q = List.filled(N + 1, 0)
q[1] = 1
q[2] = 1
for (n in 3..N) q[n] = q[n - q[n-1]] + q[n - q[n-2]]

System.print("The first ten terms of the Hofstadter Q sequence are:")
System.print(q[1..10])
System.print("\nThe thousandth term is %(q[1000]).")
var flips = 0
for (n in 2..N) {
if (q[n] < q[n-1]) flips = flips + 1
}
System.print("\nThere are %(flips) flips in the first %(N) terms.")

Output:
The first ten terms of the Hofstadter Q sequence are:
[1, 1, 2, 3, 3, 4, 5, 5, 6, 6]

The thousandth term is 502.

There are 49798 flips in the first 100000 terms.


## XPL0

code ChOut=8, CrLf=9, IntOut=11;
int  N, C, Q(100_001);
[Q(1):= 1;  Q(2):= 1;  C:= 0;
for N:= 3 to 100_000 do
[Q(N):= Q(N-Q(N-1)) + Q(N-Q(N-2));
if Q(N) < Q(N-1) then C:= C+1;
];
for N:= 1 to 10 do
[IntOut(0, Q(N));  ChOut(0, ^ )];
CrLf(0);
IntOut(0, Q(1000));  CrLf(0);
IntOut(0, C);  CrLf(0);
]
Output:
1 1 2 3 3 4 5 5 6 6
502
49798


## zkl

Translation of: ALGOL 68
const n = 0d100_000;
q:=(n+1).pump(List.createLong(n+1).write); // (0,1,2,...,n) base 1
q[1] = q[2] = 1;

foreach i in ([3..n]) { q[i] = q[i - q[i - 1]] + q[i - q[i - 2]] }

q[1,10].concat(" ").println();
println(q[1000]);

flip := 0;
foreach i in (n){ flip += (q[i] > q[i + 1]) }
println("flips: ",flip);
Output:
1 1 2 3 3 4 5 5 6 6
502
flips: 49798


## ZX Spectrum Basic

Translation of: BBC_BASIC

Extra credit 100000 is not implemented because of memory limitations.

10 PRINT "First 10 terms of Q = "
20 FOR i=1 TO 10: GO SUB 1000: PRINT s;" ";: NEXT i: PRINT
30 LET i=1000
40 PRINT "1000th term = ";: GO SUB 1000: PRINT s
50 PRINT "Term is less than preceding term ";c;" times"
100 STOP
1000 REM Qsequence subroutine
1010 IF i<3 THEN LET s=1: RETURN
1020 IF i=3 THEN LET s=2: RETURN
1030 DIM q(i)
1040 LET q(1)=1: LET q(2)=1: LET q(3)=2
1050 LET c=0
1060 FOR j=3 TO i
1070 LET q(j)=q(j-q(j-1))+q(j-q(j-2))
1080 IF q(j)<q(j-1) THEN LET c=c+1
1090 NEXT j
1100 LET s=q(i)
1110 RETURN