⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

{{task}}The [[wp:Hofstadter_sequence#Hofstadter_Q_sequence|Hofstadter Q sequence]] is defined as: :: \begin\left\{align\right\} Q\left(1\right)&=Q\left(2\right)=1, \ Q\left(n\right)&=Q\big\left(n-Q\left(n-1\right)\big\right)+Q\big\left(n-Q\left(n-2\right)\big\right), \quad n>2. \end\left\{align\right\}

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 '''n'''th 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

{{trans|PL/I}}

*        Hofstrader q sequence for any n -   18/10/2015
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
N        DC     F'1000'            n=1000
Q        DS     1000F              array q(1000)
YREGS


{{out}}

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

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


{{out}}

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


## ALGOL 68

{{trans|C}} Note: This specimen retains the original [[Hofstadter_Q_sequence#C|C]] coding style.
{{works with|ALGOL 68|Revision 1 - no extension to language used.}}
{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-2.3.5 algol68g-2.3.5].}}
{{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of '''format'''[ted] ''transput''.}}

algol68
#!/usr/local/bin/a68g --script #

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

q := q := 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));

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

printf(($"flips: "g(0)l$, flip))
)


{{out}}

txt

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



## APL

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
⎕←(+/ 2>/seq),'terms were preceded by a larger term.'
∇


{{out}}

txt

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.



## AutoHotkey

AutoHotkey
SetBatchLines, -1
Q := HofsQSeq(100000)

Loop, 10
Out .= Q[A_Index] ", "

MsgBox, % "First ten:t" Out "n"
. "1000th:tt" Q "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
}


{{out}}

txt
First ten:	1, 1, 2, 3, 3, 4, 5, 5, 6, 6,
1000th:		502
Flips:		49798


## AWK

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
for (n=2; n<=N; n++) {
if (Q[n]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 : ( #include #define N 100000 int main() { int i, flip, *q = (int*)malloc(sizeof(int) * N) - 1; q = q = 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); for (flip = 0, i = 1; i < N; i++) flip += q[i] > q[i + 1]; printf("flips: %d\n", flip); return 0; }  {{out}} txt 1 1 2 3 3 4 5 5 6 6 502 flips: 49798  ## C++ solution modeled after Perl solution cpp #include 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; }  {{out}} txt 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# C sharp using System; using System.Collections.Generic; namespace HofstadterQSequence { class Program { // Initialize the dictionary with the first two indices filled. private static readonly Dictionary QList = new Dictionary { {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; } } }  {{out}} txt 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 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]. clojure (defn qs [q] (let [n (count q)] (condp = n 0  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))  {{out}} first 10: [1 1 2 3 3 4 5 5 6 6] 1000th: 502 extra credit: 49798  ## CoffeeScript {{trans|JavaScript}} coffeescript 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)  {{out}} txt 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 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)))  {{out}} txt 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: lisp (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 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))); }  {{out}} txt 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 {{trans|Python}} Same output. d 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. d import std.stdio; int[100_000] Q; void main() { Q = 1; Q = 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); 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) dart 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.

dart
class Q {
Map _table;

Q() {
_table=new Map();
_table=1;
_table=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 q=new List(100001);
q=q=1;

int count=0;
for(int i=3;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 up to 'lim'.
require
lim_positive: lim > 0
local
q: ARRAY [INTEGER]
i: INTEGER
do
create Result.make_filled (1, 1, lim)
Result  := 1
Result  := 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



{{out}}

txt

First ten numbers:
1 1 2 3 3 4 5 5 6 6
1000th:
502
Number of Flips:
49798



## Elixir

{{trans|Erlang}}
changed collection (Erlang array => Map)

elixir
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}"
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}
end
end



{{out}}

txt

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

erlang
%% @author Jan Willem Luiten
%% Hofstadter Q Sequence for Rosetta Code

-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).

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]);
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),



{{out}}

txt

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

{{output|ERRE}}

ERRE

!
! for rosettacode.org
!

DIM Q%

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
Q%=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%) FOR N=2 TO NUM DO IF Q%[N]() 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)))  {{out}} txt q(1 .. 10) = 1 1 2 3 3 4 5 5 6 6 q(1000) = 502 descents(100000) = 49798  ## 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. factor ( 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  =={{header|Fōrmulæ}}== In [https://wiki.formulae.org/Hofstadter_Q_sequence this] page you can see the solution of this task. Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). 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 transportation effects more than visualization and edition. The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code. ## 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. Fortran 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: txt 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 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  {{out}} txt Primeros 10 terminos: 1 1 2 3 3 4 5 5 6 6 Termino numero 1000: 502 Terminos menores que los anteriores: 49798  ## Go Sure there are ways that run faster or handle larger numbers; for the task though, maps and recursion work just fine. go package main import "fmt" var m map[int]int func initMap() { m = make(map[int]int) m = 1 m = 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)) }  {{out}} txt 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 The basic task: Haskell 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: Haskell 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))

| 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)


{{out}}

Haskell>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:

haskell
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))


{{out}}

txt

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.

haskell
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

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


[http://www.cs.arizona.edu/icon/library/src/procs/printf.icn printf.icn provides formatting]

{{out}}

txt
Q(1 to 10) - verified.
Q=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


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''):
j
Qs=:0 1 1
Q=: verb define
n=. >./,y
while. n>:#Qs do.
Qs=: Qs,+/(-_2{.Qs){Qs
end.
y{Qs
)


'''Solution''' (''top-down''):
j
Q=: 1:(+&$:/@:-$:@-& 1 2)@.(>&2)"0 M.


'''Example''':
j
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 [[Category:Memoization]]{{works with|Java|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. java5 import java.util.HashMap; import java.util.Map; public class HofQ { private static Map q = new HashMap(){{ put(1, 1); put(2, 1); }}; private static int[] nUses = new int;//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 maxUses){ maxUses = nUses[i]; maxN = i; } } System.out.println("Q(" + maxN + ") was called the most with " + maxUses + " calls"); } }  {{out}} txt 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'. JavaScript 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));  {{out}} txt 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 JavaScript (() => { '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) : undefined, // Int -> [a] -> [a] take = (n, xs) => xs.slice(0, n); // TEST -------------------------------------------------------- return { firstTen: hofQSeq(10), thousandth: last(hofQSeq(1000)), 'Q x < xs[i - 1] ? a + 1 : a, 0) }; })();  {{Out}} JavaScript {"firstTen":[1, 1, 2, 3, 3, 4, 5, 5, 6, 6], "thousandth":502, "Q=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)) ;

((range(0;11), 1000) | "Q(\(.)) = \( . | Q)"),

(100000 | "flips(\(.)) = \(flips(.))")


### Transcript

bash

$uname -a Darwin Mac-mini 13.3.0 Darwin Kernel Version 13.3.0: Tue Jun 3 21:27:35 PDT 2014; root:xnu-2422.110.17~1/RELEASE_X86_64 x86_64$ 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:

julia
function hofstQseq(n, typerst::Type=Int)
nmax = maximum(n)
r = Vector{typerst}(nmax)
r = 1
if nmax ≥ 2 r = 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))



{{out}}

txt
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:

julia
seq = hofstQseq(1:100_000)
cnt = count(diff(seq) .< 0)
println("$cnt elements are less than the preceding one.")  {{out}} txt 49798 elements are less than the preceding one.  Since the implementation is non-recursive, there is no issue with recursion limits. ## Kotlin scala // version 1.1.4 fun main(args: Array) { val q = IntArray(100_001) q = 1 q = 1 for (n in 3..100_000) q[n] = q[n - q[n - 1]] + q[n - q[n - 2]] print("The first 10 terms are : ") for (i in 1..10) print("${q[i]}  ")
println("\n\nThe 1000th term is : ${q}") val flips = (2..100_000).count { q[it] < q[it - 1] } println("\nThe number of flips for the first 100,000 terms is :$flips")
}


{{out}}

txt

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

We use automatic memoisation ("option remember") in the following.  The use of "option system" assures that memoised values can be garbage collected.

Maple
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:

Maple>
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.

Maple>
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.

Maple
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.

Maple>
Qflips( 10^5 );
49798


Mathematica
Hofstadter[n_Integer?Positive] := Hofstadter[n] = Block[{$RecursionLimit = Infinity}, Hofstadter[n - Hofstadter[n - 1]] + Hofstadter[n - Hofstadter[n - 2]] ]  {{out}} Mathematica Hofstadter /@ Range {1,1,2,3,3,4,5,5,6,6} Hofstadter 502 Count[Differences[Hofstadter /@ Range], _?Negative] 49798  =={{header|MATLAB}} / {{header|Octave}}== This solution pre-allocates memory and is an iterative solution, so caching or recursion limits do not apply. MATLAB 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 txt >> Qsequence(10) ans = 1 1 2 3 3 4 5 5 6 6  Confirm and display that the 1000'th term is: 502 txt >> 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. txt >> sum(diff(Qsequence(100000))<0) ans = 49798  ## MiniScript 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)  {{out}} txt 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 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 assert q == 502 var lessCount = 0 for n in 1 .. <100_000: if q[n] < q[n-1]: inc lessCount echo lessCount  {{out}} txt @[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798  =={{header|Oberon-2}}== Works with oo2c version 2 oberon2 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: txt 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 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 ;  {{out}} txt [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798  ## PARI/GP Straightforward, unoptimized version; about 1 ms. parigp Q=vector(1000);Q=Q=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,if(Q==502," (as expected)"," (in error)"));  {{out}} txt First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] (as expected) 1000-th term: 502 (as expected)  ## Pascal pascal Program HofstadterQSequence (output); const limit = 100000; var q: array [1..limit] of longint; i, flips: longint; begin q := 1; q := 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); flips := 0; for i := 1 to limit - 1 do if q[i] > q[i+1] then inc(flips); writeln('Flips: ', flips); end.  {{out}} txt :> ./HofstadterQSequence 1 1 2 3 3 4 5 5 6 6 502 Flips: 49798  ## Perl 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";
say "Terms less than preceding in first 100k: ",scalar(grep { $Q[$_] < $Q[$_-1] } 2..100000);


{{out}}

txt
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:

Perl
#!/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, $less_than_preceding terms are less " . "than their preceding terms!\n";  {{out}} txt 1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502! Up to and including the 100000'th term, 49798 terms are less than their preceding terms!  This different solution uses tie to make the Q sequence look like a regular array, and only fills the cache on demand. Some pre-allocation is done which provides a minor speed increase for the extra credit. I could have chosen to do recursion instead of iteration, as perl has no limit on how deeply one may recurse, but did not see the benefit of doing so. Perl #!perl use strict; use warnings; package Hofstadter; sub TIEARRAY { bless [undef, 1, 1], shift; } sub FETCH { my ($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;

print "@q[1..10]\n";
print $q, "\n"; my$count = 0;
for my $n ( 2 .. 100_000 ) {$count++ if $q[$n] < $q[$n - 1];
}
print "Extra credit: $count\n";  {{out}} txt 1 1 2 3 3 4 5 5 6 6 502 Extra credit: 49798  ## Perl 6 ### OO solution {{Works with|rakudo|2016.03}} Similar concept as the perl5 solution, except that the cache is only filled on demand. perl6 class Hofstadter { has @!c = 1,1; method AT-POS ($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";

my $count = 0;$count++ if $Q[$_ +1 ] < $Q[$_] for  ^99_999;
say "In the first 100_000 terms, $count terms are less than their preceding terms";  {{out}} txt first ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 In the first 100_000 terms, 49798 terms are less than their preceding terms  ### Idiomatic solution {{Works with|rakudo|2015-11-22}} With a lazily generated array, we automatically get caching. perl6 my @Q = 1, 1, ->$a, $b { (state$n = 1)++;
@Q[$n -$a] + @Q[$n -$b]
} ... *;

# Testing:

say "first ten: ", @Q[^10];
say "1000th: ", @Q;
say "In the first 100_000 terms, ",
[+](@Q[1..100000] Z< @Q[0..99999]),
" terms are less than their preceding terms";


(Same output.)

## Phix

Just to be flash, I also calculated the 100 millionth term - the only limiting factor here would be the length of Q
(on 32 bit, theoretical max sequence length is 402,653,177).

Phix
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: %s\n",{sprint(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]= 2 N)
1
(+
(q (- N (q (dec N))))
(q (- N (q (- N 2)))) ) ) ) )


Test:

PicoLisp
: (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

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;



{{out}}

txt

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



{{out}} for n=100,000

txt

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);



{{out}}

txt

n=100000                         48157
TALLY=         49798;



## PureBasic

PureBasic
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


{{out}}

txt
First ten:      1  1  2  3  3  4  5  5  6  6
1000th:         502
Flips:          49798


## Python

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:

python
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:])))


{{out|Combined output}}

txt
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

python
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)]))


## R

rsplus

cache <- vector("integer", 0)
cache <- 1
cache <- 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")



{{out}}

txt

1 1 2 3 3 4 5 5 6 6
502
49798 terms is less than its preceding term



## Racket

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))



{{out}}

txt

'(1 1 2 3 3 4 5 5 6 6)
502
49798



## REXX

===non-recursive===
The REXX language doesn't allow expressions for stemmed array indices, so a temporary variable must be used.

rexx
/*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              /* "      "         "   "   "      "   */
q.= 1;                 ac=   abs(c)              /* [↑]  negative #'s don't show values.*/
call HofstadterQ  b;   say;    say  abs(b)th(b)      'value is:'      result;          say
downs= 0;              do j=2  for ac-1;        jm= j - 1
downs= downs + (q.j2   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))


{{out|output|text=  when using the internal default inputs:}}

txt

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===
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
/*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              /* "      "         "   "   "      "   */
q.= 1;                 ac=   abs(c)              /* [↑]  negative #'s don't show values.*/
call HofstadterQ  b;   say;    say  abs(b)th(b)       'value is:'       result;        say
downs=0;                       do j=2  for ac-1;    jm=j-1
downs= downs + (q.j2   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))


{{out|output|text=  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
/*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              /* "      "         "   "   "      "   */
q.=0;  q.1=1;  q.2=1;  ac=   abs(c)              /* [↑]  negative #'s don't show values.*/
call HofstadterQ  b;   say;    say  abs(b)th(b)      'value is:'      result;          say
downs=0;                       do j=2  for ac-1;    jm= j-1
downs= downs + (q.j0   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))


{{out|output|text=  is identical to the 1st REXX version.}}

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

## Ring

ring

n = 20
aList = list(n)
aList = 1
aList = 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

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)= 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



{{out}}

txt

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.

rust
fn hofq(q: &mut Vec, 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=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) 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.

scala

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)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.

lisp
(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)


{{out}}

txt
Q(1 .. 10): 1 1 2 3 3 4 5 5 6 6
Q(1000): 502
bumps up to 100000: 49798


## Seed7

seed7
$include "seed7_05.s7i"; const type: intHash is hash [integer] integer; var intHash: qHash is intHash.value; const func integer: q (in integer: n) is func result var integer: q is 1; begin if n in qHash then q := qHash[n]; else if n > 2 then q := q(n - q(pred(n))) + q(n - q(n - 2)); end if; qHash @:= [n] q; end if; end func; const proc: main is func local var integer: n is 0; var integer: less_than_preceding is 0; begin writeln("q(n) for n = 1 .. 10:"); for n range 1 to 10 do write(q(n) <& " "); end for; writeln; writeln("q(1000)=" <& q(1000)); for n range 2 to 100000 do if q(n) < q(pred(n)) then incr(less_than_preceding); end if; end for; writeln("q(n) < q(n-1) for n = 2 .. 100000: " <& less_than_preceding); end func;  {{out}} txt q(n) for n = 1 .. 10: 1 1 2 3 3 4 5 5 6 6 q(1000)=502 q(n) < q(n-1) for n = 2 .. 100000: 49798  ## Sidef Using a memoized function: ruby func Q(n) is cached { n <= 2 ? 1 : Q(n - Q(n-1))+Q(n-Q(n-2)) } say "First 10 terms: #{ {|n| Q(n) }.map(1..10) }" say "Term 1000: #{Q(1000)}" say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q(i)count{|i|Q[i] # <$@::length~..>
..|@: $@($ - $@($ - 1)) + $@($ - $@($ - 2));
$-> # <$outputFrom..>
$@($) !
end q

[1,10] -> q -> '$; ' -> !OUT::write ' ' -> !OUT::write [1000,1000] -> q -> '$;
' -> !OUT::write

templates countDownSteps
@: 0;
def qs: $; 2..$qs::length -> #
$@ ! )> @:$@ + 1;
end countDownSteps

[[1, 100000] -> q] -> countDownSteps -> 'Less than previous $; times' -> !OUT::write  {{out}} txt 1 1 2 3 3 4 5 5 6 6 502 Less than previous 49798 times  ## Tcl tcl package require Tcl 8.5 # Index 0 is not used, but putting it in makes the code a bit shorter set tcl::mathfunc::Qcache {Q:-> 1 1} proc tcl::mathfunc::Q {n} { variable Qcache if {$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)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))


{{out}}

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

0 OK, 0:320


## VBA

vb
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

{{out}}

txt
First ten terms:
1  1  2  3  3  4  5  5  6  6
The 1000th term is: 502
Number of times smaller: 49798


## Visual FoxPro

vfp

LOCAL p As Integer, i As Integer
CLEAR
p = 0
? "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
IF n > 1
aq = 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



{{out}}

txt

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

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);
]


{{out}}

txt

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



## zkl

{{trans|ALGOL 68}}

zkl
const n = 0d100_000;
q:=(n+1).pump(List.createLong(n+1).write); // (0,1,2,...,n) base 1
q = q = 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);

flip := 0;
foreach i in (n){ flip += (q[i] > q[i + 1]) }
println("flips: ",flip);


{{out}}

txt

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



## ZX Spectrum Basic

{{trans|BBC_BASIC}}
Extra credit 100000 is not implemented because of memory limitations.

zxbasic
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)