Hofstadter Q sequence

From Rosetta Code
Task
Hofstadter Q sequence
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:

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.


Task
  • 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).

360 Assembly[edit]

Translation of: PL/I
*        Hofstrader q sequence for any n -   18/10/2015
HOFSTRAD CSECT
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
BR R14 return to caller
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

Ada[edit]

with Ada.Text_IO;
 
procedure Hofstadter_Q_Sequence is
 
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
Ada.Text_IO.Put(Positive'Image(P));
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
Ada.Text_IO.New_Line;
 
Q(1000, 1000, Print'Access);
-- the 1000'th term is: 502
Ada.Text_IO.New_Line;
 
Q(2, 100_000, Decrease_Count'Access);
Ada.Text_IO.Put_Line(Integer'Image(Decrease_Counter));
-- 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
end Hofstadter_Q_Sequence;
Output:
 1 1 2 3 3 4 5 5 6 6
 502
 49798

ALGOL 68[edit]

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.


File: Hofstadter_Q_sequence.a68
#!/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

AutoHotkey[edit]

SetBatchLines, -1
Q := HofsQSeq(100000)
 
Loop, 10
Out .= Q[A_Index] ", "
 
MsgBox, % "First ten:`t" Out "`n"
. "1000th:`t`t" Q[1000] "`n"
. "Flips:`t`t" 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[edit]

#!/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

BBC BASIC[edit]

      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

Bracmat[edit]

( 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. }
)
& ( !(!arg$memo):>0 { Set index to !arg. Return value at index if > 0 }
| Q$(!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$!i)
& out$(Q$1000)
& 0:?previous:?lessThan:?i
& whl
' ( 1+!i:~>100000:?i
& Q$!i
 : ( <!previous&1+!lessThan:?lessThan
| ?
)
 : ?previous
)
& out$!lessThan
);

Output:

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

C[edit]

#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++[edit]

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!

C#[edit]

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.

Clojure[edit]

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

CoffeeScript[edit]

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

Common Lisp[edit]

(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)))

D[edit]

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[edit]

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[edit]

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[edit]

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");
}

EchoLisp[edit]

 
(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[edit]

 
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[edit]

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[edit]

%% @author Jan Willem Luiten <[email protected]>
%% 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[edit]

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%[].

F#[edit]

let memoize f =
let cache = System.Collections.Generic.Dictionary<_,_>()
fun x ->
match cache.TryGetValue(x) with
| (true, v) -> v
| (_, _) ->
let v = f x
cache.[x] <- v
v
 
let rec q = memoize (fun i ->
if i < 3I then 1I
else q (i - q (i - 1I)) + q (i - q(i - 2I)))
 
printf "q(1 .. 10) ="; List.iter (q >> (printf " %A")) [1I .. 10I]
printfn ""
printfn "q(1000) = %A" (q 1000I)
printfn "descents(100000) = %A" (Seq.sum (Seq.init 100000 (fun i -> if q(bigint(i)) > q(bigint(i+1)) then 1 else 0)))
Output:
q(1 .. 10) = 1 1 2 3 3 4 5 5 6 6
q(1000) = 502
descents(100000) = 49798

Factor[edit]

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

Fortran[edit]

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.

Go[edit]

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

Haskell[edit]

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[edit]

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

printf.icn provides formatting

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

J[edit]

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[edit]

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[edit]

ES5[edit]

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[edit]

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[edit]

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[edit]

 
$ 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[edit]

The following implementation accepts an argument that is a single integer, an array of integers, or a range:

function Q(n)
N = maximum(n)
q = Array(Int, N)
q[1], q[2] = 1, 1
for i = 3:N
q[i] = q[i - q[i-1]] + q[i - q[i-2]]
end
return q[n]
end
Output:
julia> Q(1:10)
10-element Array{Int64,1}:
1
1
2
3
3
4
5
5
6
6
 
julia> Q(1000)
502

And we can also count the number of times a value is less than its predecessor by, for example:

julia> sum(diff(Q(1:10^5)) .< 0)
49798

Since the implementation is non-recursive, there is no issue with recursion limits.

Kotlin[edit]

// 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

Maple[edit]

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[edit]

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[edit]

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

Nim[edit]

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[edit]

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

Oforth[edit]

: 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[edit]

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[edit]

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[edit]

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

Perl 6[edit]

OO solution[edit]

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[edit]

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.)

PicoLisp[edit]

(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[edit]

 
/* 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;

Python[edit]

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[edit]

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)]))

Racket[edit]

 
#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

REXX[edit]

non-recursive[edit]

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 \datatype(a, 'W') then a= 10 /*Not specified? Then use the default.*/
if \datatype(b, 'W') then b= -1000 /* " " " " " " */
if \datatype(c, 'W') then c= -100000 /* " " " " " " */
if \datatype(d, 'W') then d= -1000000 /* " " " " " " */
q.=1; ac= abs(c) /* [↑] negative #'s don't show values.*/
call HofstadterQ a
call HofstadterQ b; say; say abs(b)th(b) 'value is:' result; say
call HofstadterQ c
downs=0; do j=2 for ac-1; jm=j-1
downs=downs + (q.j<q.jm)
end /*j*/
 
say downs 'terms are less then the previous term,' ac || th(ac) 'term is:' q.ac
call HofstadterQ d; ad=abs(d); say
say 'The' ad || th(ad) 'term is' q.ad
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
HofstadterQ: procedure expose q.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x=abs(x) /*use the absolute value for X. */
w=length(x) /*use for right justified output.*/
do j=1 for x /* [↓] use short─circuit IF test*/
if j>2 then if q.j==1 then do; jm1=j-1; jm2=j-2
_1=j - q.jm1; _2=j - q.jm2
q.j=q._1 + q._2
end
if ox>0 then say right(j,w) right(q.j,w) /*display the number if OX > 0. */
end /*j*/
return q.x /*return the │X│th term to caller*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
th: procedure; x=abs(arg(1)); return word('th st nd rd',1+x//10*(x//100%10\==1)*(x//10<4))
Output:
 1  1
 2  1
 3  2
 4  3
 5  3
 6  4
 7  5
 8  5
 9  6
10  6

1000th value is: 502

49798 terms are less then the previous term, 100000th term is: 48157

The 1000000th term is 512066

non-recursive, simpler[edit]

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 \datatype(a, 'W') then a= 10 /*Not specified? Then use the default.*/
if \datatype(b, 'W') then b= -1000 /* " " " " " " */
if \datatype(c, 'W') then c= -100000 /* " " " " " " */
if \datatype(d, 'W') then d= -1000000 /* " " " " " " */
q.=1; ac= abs(c) /* [↑] negative #'s don't show values.*/
call HofstadterQ a
call HofstadterQ b; say; say abs(b)th(b) 'value is:' result; say
call HofstadterQ c
downs=0; do j=2 for ac-1; jm=j-1
downs=downs + (q.j<q.jm)
end /*j*/
 
say downs 'terms are less then the previous term,' ac || th(ac) 'term is:' q.ac
call HofstadterQ d; ad=abs(d); say
say 'The' ad || th(ad) 'term is' q.ad
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
HofstadterQ: procedure expose q.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x=abs(x) /*use the absolute value for X. */
w=length(x) /*use for right justified output.*/
do j=1 for x
if j>2 then if q.j==1 then q.j=q(j-q(j-1)) + q(j-q(j-2))
if ox>0 then say right(j,w) right(q.j,w) /*if X>0, tell*/
end /*j*/
return q.x /*return the │X│th term to caller*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
q: parse arg ?; return q.? /*return value of Q.? to invoker.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
th: procedure; x=abs(arg(1)); return word('th st nd rd',1+x//10*(x//100%10\==1)*(x//10<4))

output is identical to the first version.

Because of the additional subroutine (function) invokes, this REXX version is about half as fast as the 1st REXX version.

recursive[edit]

/*REXX program generates the    Hofstadter  Q     sequence for any specified   N.       */
parse arg a b c d . /*obtain optional arguments from the CL*/
if \datatype(a, 'W') then a= 10 /*Not specified? Then use the default.*/
if \datatype(b, 'W') then b= -1000 /* " " " " " " */
if \datatype(c, 'W') then c= -100000 /* " " " " " " */
if \datatype(d, 'W') then d= -1000000 /* " " " " " " */
q.=0; q.1=1; q.2=1; ac= abs(c) /* [↑] negative #'s don't show values.*/
call HofstadterQ a
call HofstadterQ b; say; say abs(b)th(b) 'value is:' result; say
call HofstadterQ c
downs=0; do j=2 for ac-1; jm=j-1
downs=downs + (q.j<q.jm)
end /*j*/
 
say downs 'terms are less then the previous term,' ac || th(ac) "term is:" q.ac
call HofstadterQ d; ad=abs(d); say
say 'The' ad || th(ad) "term is" q.ad
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
HofstadterQ: procedure expose q.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x=abs(x) /*use the absolute value for X. */
w=length(x) /*use for right justified output.*/
do j=1 for x
if q.j==0 then q.j=QR(j) /*Not defined? Then define it.*/
if ox>0 then say right(j,w) right(q.j,w) /*show if OX>0*/
end /*j*/
return q.x /*return the │X│th term to caller*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
QR: procedure expose q.; parse arg n /*this QR function is recursive.*/
if q.n==0 then q.n=QR(n-QR(n-1)) + QR(n-QR(n-2)) /*Not defined? Then define it.*/
return q.n /*return the value to the invoker*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
th: procedure; x=abs(arg(1)); return word('th st nd rd',1+x//10*(x//100%10\==1)*(x//10<4))

output is identical to the first version.

The recursive version is almost ten times slower than the (1st) non-recursive version.

Ring[edit]

 
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
 

Ruby[edit]

@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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

$ 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

Sidef[edit]

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.ft(1, 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

Tcl[edit]

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

VBScript[edit]

 
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

uBasic/4tH[edit]

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 [email protected] < 3 Then Return (1)
If [email protected] = 3 Then Return (2)
 
@(0) = 1 : @(1) = 1 : @(2) = 2
[email protected] = 0
 
For [email protected] = 3 To [email protected]
@([email protected]) = @([email protected] - @([email protected])) + @([email protected] - @([email protected]))
If @([email protected]) < @([email protected]) Then [email protected] = [email protected] + 1
Next
 
Return (@([email protected]))
Output:
First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6
256th term = 123

0 OK, 0:320

Visual FoxPro[edit]

 
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

XPL0[edit]

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[edit]

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[edit]

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