Hofstadter Figure-Figure sequences: Difference between revisions

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These two sequences of positive integers are defined as:

{\begin{aligned}R(1)&=1\ ;\ S(1)=2\\R(n)&=R(n-1)+S(n-1),\quad n>1.\end{aligned}}

The sequence $S(n)$ is further defined as the sequence of positive integers not present in $R(n)$ .

Sequence $R$ starts:

   1, 3, 7, 12, 18, ...

Sequence $S$ starts:

   2, 4, 5, 6, 8, ...

Task

Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.
(Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).
No maximum value for n should be assumed.
Calculate and show that the first ten values of R are:
1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.

References

Sloane's A005228 and A030124.
Wolfram MathWorld
Wikipedia: Hofstadter Figure-Figure sequences.

11l

Translation of: Python

<lang 11l>V cR = [1] V cS = [2]

F extend_RS()

  V x = :cR[:cR.len-1] + :cS[:cR.len-1]
  :cR [+]= (x)
  :cS [+]= :cS.last+1 .< x
  :cS [+]= (x + 1)

F ff_R(n)

  assert(n > 0)
  L n > :cR.len
     extend_RS()
  R :cR[n - 1]

F ff_S(n)

  assert(n > 0)
  L n > :cS.len
     extend_RS()
  R :cS[n - 1]

print((1..10).map(i -> ff_R(i)))

V arr = [0] * 1001 L(i) (40.<0).step(-1)

  arr[ff_R(i)]++

L(i) (960.<0).step(-1)

  arr[ff_S(i)]++

I all(arr[1..1000].map(a -> a == 1))

  print(‘All Integers 1..1000 found OK’)

E

  print(‘All Integers 1..1000 NOT found only once: ERROR’)</lang>

Output:

[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
All Integers 1..1000 found OK

Ada

Specifying a package providing the functions FFR and FFS: <lang Ada>package Hofstadter_Figure_Figure is

  function FFR(P: Positive) return Positive;

  function FFS(P: Positive) return Positive;

end Hofstadter_Figure_Figure;</lang>

The implementation of the package internally uses functions which generate an array of Figures or Spaces: <lang Ada>package body Hofstadter_Figure_Figure is

  type Positive_Array is array (Positive range <>) of Positive;

  function FFR(P: Positive) return Positive_Array is
     Figures: Positive_Array(1 .. P+1);
     Space: Positive := 2;
     Space_Index: Positive := 2;
  begin
     Figures(1) := 1;
     for I in 2 .. P loop
        Figures(I) := Figures(I-1) + Space;
        Space := Space+1;
        while Space = Figures(Space_Index) loop
           Space := Space + 1;
           Space_Index := Space_Index + 1;
        end loop;
     end loop;
     return Figures(1 .. P);
  end FFR;

  function FFR(P: Positive) return Positive is
     Figures: Positive_Array(1 .. P) := FFR(P);
  begin
     return Figures(P);
  end FFR;

  function FFS(P: Positive) return Positive_Array is
     Spaces:  Positive_Array(1 .. P);
     Figures: Positive_Array := FFR(P+1);
     J: Positive := 1;
     K: Positive := 1;
  begin
     for I in Spaces'Range loop
        while J = Figures(K) loop
           J := J + 1;
           K := K + 1;
        end loop;
        Spaces(I) := J;
        J := J + 1;
     end loop;
     return Spaces;
  end FFS;

  function FFS(P: Positive) return Positive is
     Spaces: Positive_Array := FFS(P);
  begin
     return Spaces(P);
  end FFS;

end Hofstadter_Figure_Figure;</lang>

Finally, a test program for the package, solving the task at hand: <lang Ada>with Ada.Text_IO, Hofstadter_Figure_Figure;

procedure Test_HSS is

  use Hofstadter_Figure_Figure;

  A: array(1 .. 1000) of Boolean := (others => False);
  J: Positive;

begin

  for I in 1 .. 10 loop
     Ada.Text_IO.Put(Integer'Image(FFR(I)));
  end loop;
  Ada.Text_IO.New_Line;

  for I in 1 .. 40 loop
     J := FFR(I);
     if A(J) then
        raise Program_Error with Positive'Image(J) & " used twice";
     end if;
     A(J) := True;
  end loop;

  for I in 1 .. 960 loop
     J := FFS(I);
     if A(J) then
        raise Program_Error with Positive'Image(J) & " used twice";
     end if;
     A(J) := True;
  end loop;

  for I in A'Range loop
     if not A(I) then raise Program_Error with Positive'Image(I) & " unused";
     end if;
  end loop;
  Ada.Text_IO.Put_Line("Test Passed: No overlap between FFR(I) and FFS(J)");

exception

  when Program_Error => Ada.Text_IO.Put_Line("Test Failed"); raise;

end Test_HSS;</lang>

The output of the test program: <lang> 1 3 7 12 18 26 35 45 56 69 Test Passed: No overlap between FFR(I) and FFS(J)</lang>

APL

Works with: Dyalog APL

<lang APL>:Class HFF

   :Field Private Shared RBuf←,1   
   ∇r←ffr n 
       :Access Public Shared
       r←n⊃RBuf←(⊢,⊃∘⌽+≢⊃(⍳1+⌈/)~⊢)⍣(0⌈n-≢RBuf)⊢RBuf
   ∇
   ∇s←ffs n;S 
       :Access Public Shared
       :Repeat
           S←((⍳1+⌈/)~⊢)RBuf
           :If n≤≢S ⋄ :Leave ⋄ :EndIf
           S←ffr 1+≢RBuf
       :EndRepeat
       s←n⊃S
   ∇
   ∇Task;th 
       :Access Public Shared
       ⎕←'R(1 .. 10):', ffr¨⍳10
       :If (⍳1000) ∧.∊ ⊂th←(ffr¨⍳40) ∪ (ffs¨⍳960)
           ⎕←'1..1000 ∊ (ffr 1..40) ∪ (ffs 1..960)'
       :Else
           ⎕←'Missing values: ', (⍳1000)~th
       :EndIf
   ∇

EndClass</lang>

Output:

      HFF.Task
R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
1..1000 ∊ (ffr 1..40) ∪ (ffs 1..960)

AutoHotkey

<lang AutoHotkey>R(n){ if n=1 return 1 return R(n-1) + S(n-1) }

S(n){ static ObjR:=[] if n=1 return 2 ObjS:=[] loop, % n ObjR[R(A_Index)] := true loop, % n-1 ObjS[S(A_Index)] := true Loop if !(ObjR[A_Index]||ObjS[A_Index]) return A_index }</lang> Examples:<lang AutoHotkey>Loop MsgBox, 262144, , % "R(" A_Index ") = " R(A_Index) "`nS(" A_Index ") = " S(A_Index)</lang>

Outputs:

R(1) = 1, 3, 7, 12, 18, 26, 35,...
S(1) = 2, 4, 5,  6,  8,  9, 10,...

AWK

<lang awk># Hofstadter Figure-Figure sequences

R(1) = 1; S(1) = 2;
R(n) = R(n-1) + S(n-1), n > 1
S(n) is the values not in R(n)

BEGIN {

   # start with the first two values of R and S to simplify finding S[n]:
   R[ 1 ] = 1;
   R[ 2 ] = 3;
   S[ 1 ] = 2;
   S[ 2 ] = 4;
   # maximum n we currently have of R and S
   rMax   = 2;
   sMax   = 2;

   # calculate and show the first 10 values of R:
   printf( "R[1..10]:" );
   for( n = 1; n < 11; n ++ )
   {
       printf( " %d", ffr( n ) );
   }
   printf( "\n" );
   # check that R[1..40] and S[1..960] contain the numbers 1..1000 once each
   # add the values of R[ 1..40 ] to the set V
   for( n = 1; n <= 40; n ++ )
   {
       V[ ffr( n ) ] ++;
   }
   # add the values of S[ 1..960 ] to the set V
   for( n = 1; n <= 960; n ++ )
   {
       V[ ffs( n ) ] ++;
   }
   # check all numbers are present and not duplicated
   ok = 1;
   for( n = 1; n <= 1000; n ++ )
   {
       if( ! ( n in V ) )
       {
           printf( "%d not present in R[1..40], S[1..960]\n", n );
           ok = 0;
       }
       else if( V[ n ] != 1 )
       {
           printf( "%d occurs %d times in R[1..40], S[1..960]\n", n, V[ n ] );
           ok = 0;
       }
   }
   if( ok )
   {
       printf( "R[1..40] and S[1..960] uniquely contain all 1..1000\n" );
   }

} # BEGIN

function ffr( n ) {

   # calculate R[n]
   if( ! ( n in R ) )
   {
       # we haven't calculated R[ n ] yet
       R[ n ]  = ffs( n - 1 );
       R[ n ] += ffr( n - 1 );
   }

return R[ n ]; } # ffr

function ffs( n ) {

   # calculate S[n]
   if( ! ( n in S ) )
   {
       # starting at the highest known R, calculate the next one and fill in the S values
       # continuing until we have enough S values
       do
       {
           R[ rMax + 1 ] = R[ rMax ] + S[ rMax ];
           for( sValue = R[ rMax ] + 1; sValue < R[ rMax + 1 ]; sValue ++ )
           {
               S[ sMax ++ ] = sValue;
           }
           rMax ++;
       }
       while( sMax < n );
   }

return S[ n ]; } # ffs</lang>

Output:

R[1..10]: 1 3 7 12 18 26 35 45 56 69
R[1..40] and S[1..960] uniquely contain all 1..1000

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> PRINT "First 10 values of R:"

     FOR i% = 1 TO 10 : PRINT ;FNffr(i%) " "; : NEXT : PRINT
     PRINT "First 10 values of S:"
     FOR i% = 1 TO 10 : PRINT ;FNffs(i%) " "; : NEXT : PRINT
     PRINT "Checking for first 1000 integers:"
     r% = 1 : s% = 1
     ffr% = FNffr(r%)
     ffs% = FNffs(s%)
     FOR wanted% = 1 TO 1000
       CASE TRUE OF
         WHEN wanted% = ffr% : r% += 1 : ffr% = FNffr(r%)
         WHEN wanted% = ffs% : s% += 1 : ffs% = FNffs(s%)
         OTHERWISE: EXIT FOR
       ENDCASE
     NEXT
     IF r% = 41 AND s% = 961 PRINT "Test passed" ELSE PRINT "Test failed"
     END
     
     DEF FNffr(N%)
     LOCAL I%, J%, R%, S%, V%
     DIM V% LOCAL 2*N%+1
     V%?1 = 1
     IF N% = 1 THEN = 1
     R% = 1
     S% = 2
     FOR I% = 2 TO N%
       FOR J% = S% TO 2*N%
         IF V%?J% = 0 EXIT FOR
       NEXT
       V%?J% = 1
       S% = J%
       R% += S%
       IF R% <= 2*N% V%?R% = 1
     NEXT I%
     = R%
     
     DEF FNffs(N%)
     LOCAL I%, J%, R%, S%, V%
     DIM V% LOCAL 2*N%+1
     V%?1 = 1
     IF N% = 1 THEN = 2
     R% = 1
     S% = 2
     FOR I% = 1 TO N%
       FOR J% = S% TO 2*N%
         IF V%?J% = 0 EXIT FOR
       NEXT
       V%?J% = 1
       S% = J%
       R% += S%
       IF R% <= 2*N% V%?R% = 1
     NEXT I%
     = S%</lang>

First 10 values of R:
1 3 7 12 18 26 35 45 56 69
First 10 values of S:
2 4 5 6 8 9 10 11 13 14
Checking for first 1000 integers:
Test passed

C

<lang c>#include <stdio.h>

include <stdlib.h>

// simple extensible array stuff typedef unsigned long long xint;

typedef struct { size_t len, alloc; xint *buf; } xarray;

xarray rs, ss;

void setsize(xarray *a, size_t size) { size_t n = a->alloc; if (!n) n = 1;

while (n < size) n <<= 1; if (a->alloc < n) { a->buf = realloc(a->buf, sizeof(xint) * n); if (!a->buf) abort(); a->alloc = n; } }

void push(xarray *a, xint v) { while (a->alloc <= a->len) setsize(a, a->alloc * 2);

a->buf[a->len++] = v; }

// sequence stuff void RS_append(void);

xint R(int n) { while (n > rs.len) RS_append(); return rs.buf[n - 1]; }

xint S(int n) { while (n > ss.len) RS_append(); return ss.buf[n - 1]; }

void RS_append() { int n = rs.len; xint r = R(n) + S(n); xint s = S(ss.len);

push(&rs, r); while (++s < r) push(&ss, s); push(&ss, r + 1); // pesky 3 }

int main(void) { push(&rs, 1); push(&ss, 2);

int i; printf("R(1 .. 10):"); for (i = 1; i <= 10; i++) printf(" %llu", R(i));

char seen[1001] = { 0 }; for (i = 1; i <= 40; i++) seen[ R(i) ] = 1; for (i = 1; i <= 960; i++) seen[ S(i) ] = 1; for (i = 1; i <= 1000 && seen[i]; i++);

if (i <= 1000) { fprintf(stderr, "%d not seen\n", i); abort(); }

puts("\nfirst 1000 ok"); return 0; }</lang>

C#

Creates an IEnumerable for R and S and uses those to complete the task <lang Csharp>using System; using System.Collections.Generic; using System.Linq;

namespace HofstadterFigureFigure { class HofstadterFigureFigure { readonly List<int> _r = new List<int>() {1}; readonly List<int> _s = new List<int>();

public IEnumerable<int> R() { int iR = 0; while (true) { if (iR >= _r.Count) { Advance(); } yield return _r[iR++]; } }

public IEnumerable<int> S() { int iS = 0; while (true) { if (iS >= _s.Count) { Advance(); } yield return _s[iS++]; } }

private void Advance() { int rCount = _r.Count; int oldR = _r[rCount - 1]; int sVal;

// Take care of first two cases specially since S won't be larger than R at that point switch (rCount) { case 1: sVal = 2; break; case 2: sVal = 4; break; default: sVal = _s[rCount - 1]; break; } _r.Add(_r[rCount - 1] + sVal); int newR = _r[rCount]; for (int iS = oldR + 1; iS < newR; iS++) { _s.Add(iS); } } }

class Program { static void Main() { var hff = new HofstadterFigureFigure(); var rs = hff.R(); var arr = rs.Take(40).ToList();

foreach(var v in arr.Take(10)) { Console.WriteLine("{0}", v); }

var hs = new HashSet<int>(arr); hs.UnionWith(hff.S().Take(960)); Console.WriteLine(hs.Count == 1000 ? "Verified" : "Oops! Something's wrong!"); } } } </lang> Output:

1
3
7
12
18
26
35
45
56
69
Verified

C++

Works with: gcc

Works with: C++ version 11, 14, 17

<lang cpp>#include <iomanip>

include <iostream>
include <set>
include <vector>

using namespace std;

unsigned hofstadter(unsigned rlistSize, unsigned slistSize) {

   auto n = rlistSize > slistSize ? rlistSize : slistSize;
   auto rlist = new vector<unsigned> { 1, 3, 7 };
   auto slist = new vector<unsigned> { 2, 4, 5, 6 };
   auto list = rlistSize > 0 ? rlist : slist;
   auto target_size = rlistSize > 0 ? rlistSize : slistSize;

   while (list->size() > target_size) list->pop_back();

   while (list->size() < target_size)
   {
       auto lastIndex = rlist->size() - 1;
       auto lastr = (*rlist)[lastIndex];
       auto r = lastr + (*slist)[lastIndex];
       rlist->push_back(r);
       for (auto s = lastr + 1; s < r && list->size() < target_size;)
           slist->push_back(s++);
   }

   auto v = (*list)[n - 1];
   delete rlist;
   delete slist;
   return v;

}

ostream& operator<<(ostream& os, const set<unsigned>& s) {

   cout << '(' << s.size() << "):";
   auto i = 0;
   for (auto c = s.begin(); c != s.end();)
   {
       if (i++ % 20 == 0) os << endl;
       os << setw(5) << *c++;
   }
   return os;

}

int main(int argc, const char* argv[]) {

   const auto v1 = atoi(argv[1]);
   const auto v2 = atoi(argv[2]);
   set<unsigned> r, s;
   for (auto n = 1; n <= v2; n++)
   {
       if (n <= v1)
           r.insert(hofstadter(n, 0));
       s.insert(hofstadter(0, n));
   }
   cout << "R" << r << endl;
   cout << "S" << s << endl;

   int m = max(*r.rbegin(), *s.rbegin());
   for (auto n = 1; n <= m; n++)
       if (r.count(n) == s.count(n))
           clog << "integer " << n << " either in both or neither set" << endl;

   return 0;

}</lang>

Output:

<lang sh>% ./hofstadter 40 100 2> /dev/null R(40):

   1    3    7   12   18   26   35   45   56   69   83   98  114  131  150  170  191  213  236  260
 285  312  340  369  399  430  462  495  529  565  602  640  679  719  760  802  845  889  935  982

S(100):

   2    4    5    6    8    9   10   11   13   14   15   16   17   19   20   21   22   23   24   25
  27   28   29   30   31   32   33   34   36   37   38   39   40   41   42   43   44   46   47   48
  49   50   51   52   53   54   55   57   58   59   60   61   62   63   64   65   66   67   68   70
  71   72   73   74   75   76   77   78   79   80   81   82   84   85   86   87   88   89   90   91
  92   93   94   95   96   97   99  100  101  102  103  104  105  106  107  108  109  110  111  112</lang>

CoffeeScript

Translation of: Ruby

<lang coffeescript>R = [ null, 1 ] S = [ null, 2 ]

extend_sequences = (n) ->

 current = Math.max(R[R.length - 1], S[S.length - 1])
 i = undefined
 while R.length <= n or S.length <= n
   i = Math.min(R.length, S.length) - 1
   current += 1
   if current == R[i] + S[i]
     R.push current
   else
     S.push current

ff = (X, n) ->

   extend_sequences n
   X[n]

console.log 'R(' + i + ') = ' + ff(R, i) for i in [1..10] int_array = ([1..40].map (i) -> ff(R, i)).concat [1..960].map (i) -> ff(S, i) int_array.sort (a, b) -> a - b

for i in [1..1000]

 if int_array[i - 1] != i
   throw 'Something\'s wrong!'

console.log '1000 integer check ok.'</lang>

Output:

As JavaScript.

Common Lisp

<lang lisp>;;; equally doable with a list (flet ((seq (i) (make-array 1 :element-type 'integer :initial-element i :fill-pointer 1 :adjustable t)))

 (let ((rr (seq 1)) (ss (seq 2)))
   (labels ((extend-r ()

(let* ((l (1- (length rr))) (r (+ (aref rr l) (aref ss l))) (s (elt ss (1- (length ss))))) (vector-push-extend r rr) (loop while (<= s r) do (if (/= (incf s) r) (vector-push-extend s ss))))))

     (defun seq-r (n)

(loop while (> n (length rr)) do (extend-r)) (elt rr (1- n)))

     (defun seq-s (n)

(loop while (> n (length ss)) do (extend-r)) (elt ss (1- n))))))

(defun take (f n)

 (loop for x from 1 to n collect (funcall f x)))

(format t "First of R: ~a~%" (take #'seq-r 10))

(mapl (lambda (l) (if (and (cdr l) (/= (1+ (car l)) (cadr l))) (error "not in sequence")))

     (sort (append (take #'seq-r 40)

(take #'seq-s 960)) #'<)) (princ "Ok")</lang>

Output:

First of R: (1 3 7 12 18 26 35 45 56 69)
Ok

Cowgol

<lang cowgol>include "cowgol.coh"; include "strings.coh"; include "malloc.coh";

An uint16 is big enough to deal with the figures from the task,
but it is good practice to allow it to be easily redefined.

typedef N is uint16;

There is no extensible vector type included in the standard library,
so it is necessary to define one.

record VecR is

   len: intptr;
   alloc: intptr;
   data: [N];

end record;

typedef Vec is [VecR];

sub NewVec(): (v: Vec) is

   v := Alloc(@bytesof VecR) as Vec;
   MemZero(v as [uint8], @bytesof VecR);
   v.alloc := 256;
   v.data := Alloc(@bytesof N * 256) as [N];
   MemZero(v.data as [uint8], @bytesof N * 256);

end sub;

sub VecGet(v: Vec, i: intptr): (r: N) is

   if i >= v.len then
       print("index error\n");
       ExitWithError();
   end if;
   r := [v.data + i * @bytesof N];

end sub;

sub VecSet(v: Vec, i: intptr, n: N) is

   if i >= v.alloc then
       var newsize := v.alloc;
       while i >= newsize loop
           newsize := newsize + 256;
       end loop;
       var newbytes := newsize * @bytesof N;
       var oldbytes := v.alloc * @bytesof N;
       var newdata := Alloc(newbytes) as [N];
       MemCopy(v.data as [uint8], oldbytes, newdata as [uint8]);
       MemZero(newdata as [uint8] + oldbytes, newbytes - oldbytes);
       Free(v.data as [uint8]);
       v.data := newdata;
       v.alloc := newsize;
   end if;
   [v.data + i * @bytesof N] := n;
   if i >= v.len then
       v.len := i+1;
   end if;

end sub;

sub Last(v: Vec): (r: N) is r := VecGet(v, v.len-1); end sub; sub Append(v: Vec, n: N) is VecSet(v, v.len, n); end sub;

We also need to define a flag array, to avoid taking up 1K of memory
for a thousand bit flags.

sub GetFlag(bitarr: [uint8], n: intptr): (s: uint8) is

   s := ([bitarr + (n >> 3)] >> (n as uint8 & 7)) & 1;

end sub; sub SetFlag(bitarr: [uint8], n: intptr) is

   var p := bitarr + (n >> 3);
   var f: uint8 := 1;
   [p] := [p] | (f << (n as uint8 & 7));

end sub;

Define and initialize vectors holding the R and S sequences

var R := NewVec(); Append(R, 1); var S := NewVec(); Append(S, 2);

Extend the sequences until R(n) is known.

sub Extend(n: intptr) is

   while n > R.len loop
       var newR := Last(R) + VecGet(S, R.len-1);
       Append(R, newR);
       while Last(S) < newR - 1 loop
           Append(S, Last(S) + 1);
       end loop;
       Append(S, newR + 1);
   end loop;

end sub;

Get R

sub ffr(n: intptr): (r: N) is

   Extend(n);
   r := VecGet(R, n-1);

end sub;

Get S

sub ffs(n: intptr): (s: N) is

   while n > S.len loop
       Extend(R.len + 1);
   end loop;
   s := VecGet(S, n-1);

end sub;

Print the first 10 values of R.

print("R(1 .. 10): "); var n: intptr := 1; while n <= 10 loop

   print_i32(ffr(n) as uint32);
   print_char(' ');
   n := n + 1;

end loop; print_nl();

print("Checking that (1 .. 1000) are in R(1 .. 40) U S(1 .. 960)...\n");

Reserve 1000 bits to use as flags, and set them all to zero

var flags: uint8[1000 / 8]; MemZero(&flags[0], @bytesof flags);

Set the flags corresponding to FFR(1 .. 40) and FFS(1 .. 960)

n := 1; while n <= 40 loop

   SetFlag(&flags[0], (ffr(n)-1) as intptr); 
   n := n + 1;

end loop;

n := 1; while n <= 960 loop

   SetFlag(&flags[0], (ffs(n)-1) as intptr);
   n := n + 1;

end loop;

Check all flags

var ok: uint8 := 1; n := 1; while n <= 1000 loop

   if GetFlag(&flags[0], (n-1) as intptr) == 0 then
       print_i32(n as uint32);
       print(" not found!\n");
       ok := 0;
   end if;
   n := n + 1;

end loop;

if ok != 0 then

   print("All numbers 1 .. 1000 found!\n");

end if;</lang>

Output:

R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
Checking that (1 .. 1000) are in R(1 .. 40) U S(1 .. 960)...
All numbers 1 .. 1000 found!

D

Translation of: Go

<lang d>int delegate(in int) nothrow ffr, ffs;

nothrow static this() {

   auto r = [0, 1], s = [0, 2];

   ffr = (in int n) nothrow {
       while (r.length <= n) {
           immutable int nrk = r.length - 1;
           immutable int rNext = r[nrk] + s[nrk];
           r ~= rNext;
           foreach (immutable sn; r[nrk] + 2 .. rNext)
               s ~= sn;
           s ~= rNext + 1;
       }
       return r[n];
   };

   ffs = (in int n) nothrow {
       while (s.length <= n)
           ffr(r.length);
       return s[n];
   };

}

void main() {

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

   iota(1, 11).map!ffr.writeln;
   auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs);
   t.array.sort().equal(iota(1, 1001)).writeln;

}</lang>

Output:

[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
true

Alternative version

Translation of: Python

(Same output) <lang d>import std.stdio, std.array, std.range, std.algorithm;

struct ffr {

   static r = [int.min, 1];

   static int opCall(in int n) nothrow {
       assert(n > 0);
       if (n < r.length) {
           return r[n];
       } else {
           immutable int ffr_n_1 = ffr(n - 1);
           immutable int lastr = r[$ - 1];
           // Extend s up to, and one past, last r.
           ffs.s ~= iota(ffs.s[$ - 1] + 1, lastr).array;
           if (ffs.s[$ - 1] < lastr)
               ffs.s ~= lastr + 1;
           // Access s[n - 1] temporarily extending s if necessary.
           immutable size_t len_s = ffs.s.length;
           immutable int ffs_n_1 = (len_s > n) ?
                                   ffs.s[n - 1] :
                                   (n - len_s) + ffs.s[$ - 1];
           immutable int ans = ffr_n_1 + ffs_n_1;
           r ~= ans;
           return ans;
       }
   }

}

struct ffs {

   static s = [int.min, 2];

   static int opCall(in int n) nothrow {
       assert(n > 0);
       if (n < s.length) {
           return s[n];
       } else {
           foreach (immutable i; ffr.r.length .. n + 2) {
               ffr(i);
               if (s.length > n)
                   return s[n];
           }
           assert(false, "Whoops!");
       }
   }

}

void main() {

   iota(1, 11).map!ffr.writeln;
   auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs);
   t.array.sort().equal(iota(1, 1001)).writeln;

}</lang>

EchoLisp

<lang scheme>(define (FFR n) (+ (FFR (1- n)) (FFS (1- n))))

(define (FFS n) (define next (1+ (FFS (1- n)))) (for ((k (in-naturals next))) #:break (not (vector-search* k (cache 'FFR))) => k ))

(remember 'FFR #(0 1)) ;; init cache (remember 'FFS #(0 2)) </lang>

Output:

<lang scheme> (define-macro m-range [a .. b] (range a (1+ b)))

(map FFR [1 .. 10])

   → (1 3 7 12 18 26 35 45 56 69)

checking

(equal? [1 .. 1000] (list-sort < (append (map FFR [1 .. 40]) (map FFS [1 .. 960]))))

   → #t</lang>

Euler Math Toolbox

<lang Euler Math Toolbox> >function RSstep (r,s) ... $ n=cols(r); $ r=r|(r[n]+s[n]); $ s=s|(max(s[n]+1,r[n]+1):r[n+1]-1); $ return {r,s}; $ endfunction >function RS (n) ... $ if n==1 then return {[1],[2]}; endif; $ if n==2 then return {[1,3],[2]}; endif; $ r=[1,3]; s=[2,4]; $ loop 3 to n; {r,s}=RSstep(r,s); end; $ return {r,s}; $ endfunction >{r,s}=RS(10); >r

[ 1  3  7  12  18  26  35  45  56  69 ]

>{r,s}=RS(50); >all(sort(r[1:40]|s[1:960])==(1:1000))

</lang>

F#

The function

<lang fsharp> // Populate R and S with values of Hofstadter Figure Figure sequence. Nigel Galloway: August 28th., 2020 let fF q=let R,S=Array.zeroCreate<int>q,Array.zeroCreate<int>q

        R.[0]<-1;S.[0]<-2
        let rec fN n g=match n=q with true->(R,S)
                                     |_->R.[n]<-R.[n-1]+S.[n-1]
                                         match S.[n-1]+1 with i when i<>R.[g]->S.[n]<-i; fN (n+1) g
                                                             |i->S.[n]<-i+1; fN (n+1) (g+1)
        fN 1 1

</lang>

The Tasks

<lang fsharp> let ffr,ffs=fF 960 ffr|>Seq.take 10|>Seq.iter(printf "%d "); printfn ""

let N=Array.concat [|ffs;(Array.take 40 ffr)|] in printfn "Unique values=%d Minimum value=%d Maximum Value=%d" ((Array.distinct N).Length)(Array.min N)(Array.max N) </lang>

Output:

1 3 7 12 18 26 35 45 56 69
Unique values=1000 Minimum value=1 Maximum Value=1000

Unbounded n?

n is bounded in this implementation because it is an signed 32 integer. Within such limit the 10 millionth value will have to be sufficiently unbounded. It can be found in 43 thousandths of sec. <lang fsharp> let ffr,ffs=fF 10000000 printfn "%d\n%d (Array.last ffr) (Array.last ffs) </lang> 1584961838 10004416

Output:

Factor

We keep lists S and R, and increment them when necessary. <lang factor>SYMBOL: S V{ 2 } S set SYMBOL: R V{ 1 } R set

next ( s r -- news newr )

2dup [ last ] bi@ + suffix dup [

 [ dup last 1 + dup ] dip member? [ 1 + ] when suffix

] dip ;

inc-SR ( n -- )

dup 0 <= [ drop ] [ [ S get R get ] dip [ next ] times R set S set ] if ;

ffs ( n -- S(n) )

dup S get length - inc-SR 1 - S get nth ;

ffr ( n -- R(n) )

dup R get length - inc-SR 1 - R get nth ;</lang>

<lang factor>( scratchpad ) 10 iota [ 1 + ffr ] map . { 1 3 7 12 18 26 35 45 56 69 } ( scratchpad ) 40 iota [ 1 + ffr ] map 960 iota [ 1 + ffs ] map append 1000 iota 1 v+n set= . t</lang>

Go

<lang go>package main

import "fmt"

var ffr, ffs func(int) int

// The point of the init function is to encapsulate r and s. If you are // not concerned about that or do not want that, r and s can be variables at // package level and ffr and ffs can be ordinary functions at package level. func init() {

   // task 1, 2
   r := []int{0, 1}
   s := []int{0, 2}

   ffr = func(n int) int {
       for len(r) <= n {
           nrk := len(r) - 1       // last n for which r(n) is known
           rNxt := r[nrk] + s[nrk] // next value of r:  r(nrk+1)
           r = append(r, rNxt)     // extend sequence r by one element
           for sn := r[nrk] + 2; sn < rNxt; sn++ {
               s = append(s, sn)   // extend sequence s up to rNext
           }
           s = append(s, rNxt+1)   // extend sequence s one past rNext
       }
       return r[n]
   }

   ffs = func(n int) int {
       for len(s) <= n {
           ffr(len(r))
       }
       return s[n]
   }

}

func main() {

   // task 3
   for n := 1; n <= 10; n++ {
       fmt.Printf("r(%d): %d\n", n, ffr(n))
   }
   // task 4
   var found [1001]int
   for n := 1; n <= 40; n++ {
       found[ffr(n)]++
   }
   for n := 1; n <= 960; n++ {
       found[ffs(n)]++
   }
   for i := 1; i <= 1000; i++ {
       if found[i] != 1 {
           fmt.Println("task 4: FAIL")
           return
       }
   }
   fmt.Println("task 4: PASS")

}</lang>

Output:

r(1): 1
r(2): 3
r(3): 7
r(4): 12
r(5): 18
r(6): 26
r(7): 35
r(8): 45
r(9): 56
r(10): 69
task 4: PASS

The following defines two mutually recursive generators without caching results. Each generator will end up dragging a tree of closures behind it, but due to the odd nature of the two series' growth pattern, it's still a heck of a lot faster than the above method when producing either series in sequence. <lang go>package main import "fmt"

type xint int64 func R() (func() (xint)) { r, s := xint(0), func() (xint) (nil) return func() (xint) { switch { case r < 1: r = 1 case r < 3: r = 3 default: if s == nil { s = S() s() } r += s() } if r < 0 { panic("r overflow") } return r } }

func S() (func() (xint)) { s, r1, r := xint(0), xint(0), func() (xint) (nil) return func() (xint) { if s < 2 { s = 2 } else { if r == nil { r = R() r() r1 = r() } s++ if s > r1 { r1 = r() } if s == r1 { s++ } } if s < 0 { panic("s overflow") } return s } }

func main() { r, sum := R(), xint(0) for i := 0; i < 10000000; i++ { sum += r() } fmt.Println(sum) }</lang>

Haskell

<lang haskell>import Data.List (delete, sort)

-- Functions by Reinhard Zumkeller ffr :: Int -> Int ffr n = rl !! (n - 1)

 where
   rl = 1 : fig 1 [2 ..]
   fig n (x:xs) = n_ : fig n_ (delete n_ xs)
     where
       n_ = n + x

ffs :: Int -> Int ffs n = rl !! n

 where
   rl = 2 : figDiff 1 [2 ..]
   figDiff n (x:xs) = x : figDiff n_ (delete n_ xs)
     where
       n_ = n + x

main :: IO () main = do

 print $ ffr <$> [1 .. 10]
 let i1000 = sort (fmap ffr [1 .. 40] ++ fmap ffs [1 .. 960])
 print (i1000 == [1 .. 1000])</lang>

Output:

[1,3,7,12,18,26,35,45,56,69]
True

Defining R and S literally: <lang haskell>import Data.List (sort)

r :: [Int] r = scanl (+) 1 s

s :: [Int] s = 2 : 4 : tail (complement (tail r))

 where
   complement = concat . interval
   interval x = zipWith (\x y -> [succ x .. pred y]) x (tail x)

main :: IO () main = do

 putStr "R: "
 print (take 10 r)
 putStr "S: "
 print (take 10 s)
 putStr "test 1000: "
 print $ [1 .. 1000] == sort (take 40 r ++ take 960 s)</lang>

output:

R: [1,3,7,12,18,26,35,45,56,69]
S: [2,4,5,6,8,9,10,11,13,14]
test 1000: True

Icon and Unicon

<lang Icon>link printf,ximage

procedure main()

  printf("Hofstader ff sequences R(n:= 1 to %d)\n",N := 10)
  every printf("R(%d)=%d\n",n := 1 to N,ffr(n))

  L := list(N := 1000,0)
  zero := dup := oob := 0
  every n := 1 to (RN := 40) do 
     if not L[ffr(n)] +:= 1 then    # count R occurrence
        oob +:= 1                   # count out of bounds

  every n := 1 to (N-RN) do 
     if not L[ffs(n)] +:= 1 then    # count S occurrence 
        oob +:= 1                   # count out of bounds  
  
  every zero +:= (!L = 0)           # count zeros / misses
  every dup  +:= (!L > 1)           # count > 1's / duplicates
     
  printf("Results of R(1 to %d) and S(1 to %d) coverage is ",RN,(N-RN))
  if oob+zero+dup=0 then 
     printf("complete.\n")
  else 
     printf("flawed\noob=%i,zero=%i,dup=%i\nL:\n%s\nR:\n%s\nS:\n%s\n",
            oob,zero,dup,ximage(L),ximage(ffr(ffr)),ximage(ffs(ffs)))

end

procedure ffr(n) static R,S initial {

  R := [1]
  S := ffs(ffs)               # get access to S in ffs
  }
  
  if n === ffr then return R  # secret handshake to avoid globals :)
  
  if integer(n) > 0 then 
     return R[n] | put(R,ffr(n-1) + ffs(n-1))[n]

end

procedure ffs(n) static R,S initial {

  S := [2] 
  R := ffr(ffr)               # get access to R in ffr
  }
  
  if n === ffs then return S  # secret handshake to avoid globals :)
  
  if integer(n) > 0 then {
     if S[n] then return S[n]
     else {
        t := S[*S]  
        until *S = n do 
           if (t +:= 1) = !R then next # could be optimized with more code
           else return put(S,t)[*S]    # extend S
        }
  }

end</lang>

Library: Icon Programming Library

printf.icn provides formatting ximage.icn allows formatting entire structures

Output:

Hofstader ff sequences R(n:= 1 to 10)
R(1)=1
R(2)=3
R(3)=7
R(4)=12
R(5)=18
R(6)=26
R(7)=35
R(8)=45
R(9)=56
R(10)=69
Results of R(1 to 40) and S(1 to 960) coverage is complete.

J

<lang j>R=: 1 1 3 S=: 0 2 4 FF=: 3 :0

 while. +./y>:R,&#S do.
   R=: R,({:R)+(<:#R){S
   S=: (i.<:+/_2{.R)-.R
 end.
 R;S

) ffr=: { 0 {:: FF@(>./@,) ffs=: { 1 {:: FF@(0,>./@,)</lang>

Required examples:

<lang j> ffr 1+i.10 1 3 7 12 18 26 35 45 56 69

  (1+i.1000) -: /:~ (ffr 1+i.40), ffs 1+i.960

1</lang>

Java

Code:

<lang java>import java.util.*;

class Hofstadter {

 private static List<Integer> getSequence(int rlistSize, int slistSize)
 {
   List<Integer> rlist = new ArrayList<Integer>();
   List<Integer> slist = new ArrayList<Integer>();
   Collections.addAll(rlist, 1, 3, 7);
   Collections.addAll(slist, 2, 4, 5, 6);
   List<Integer> list = (rlistSize > 0) ? rlist : slist;
   int targetSize = (rlistSize > 0) ? rlistSize : slistSize;
   while (list.size() > targetSize)
     list.remove(list.size() - 1);
   while (list.size() < targetSize)
   {
     int lastIndex = rlist.size() - 1;
     int lastr = rlist.get(lastIndex).intValue();
     int r = lastr + slist.get(lastIndex).intValue();
     rlist.add(Integer.valueOf(r));
     for (int s = lastr + 1; (s < r) && (list.size() < targetSize); s++)
       slist.add(Integer.valueOf(s));
   }
   return list;
 }
 
 public static int ffr(int n)
 {  return getSequence(n, 0).get(n - 1).intValue();  }
 
 public static int ffs(int n)
 {  return getSequence(0, n).get(n - 1).intValue();  }
 
 public static void main(String[] args)
 {
   System.out.print("R():");
   for (int n = 1; n <= 10; n++)
     System.out.print(" " + ffr(n));
   System.out.println();
   
   Set<Integer> first40R = new HashSet<Integer>();
   for (int n = 1; n <= 40; n++)
     first40R.add(Integer.valueOf(ffr(n)));
     
   Set<Integer> first960S = new HashSet<Integer>();
   for (int n = 1; n <= 960; n++)
     first960S.add(Integer.valueOf(ffs(n)));
   
   for (int i = 1; i <= 1000; i++)
   {
     Integer n = Integer.valueOf(i);
     if (first40R.contains(n) == first960S.contains(n))
       System.out.println("Integer " + i + " either in both or neither set");
   }
   System.out.println("Done");
 }

}</lang>

Output:

R(): 1 3 7 12 18 26 35 45 56 69
Done

JavaScript

Translation of: Ruby

<lang JavaScript>var R = [null, 1]; var S = [null, 2];

var ffr = function(n) { extend_sequences(n); return R[n]; };

var ffs = function(n) { extend_sequences(n); return S[n]; };

for (var i = 1; i <=10; i += 1) {

  console.log('R('+ i +') = ' + ffr(i));

}

var int_array = [];

for (var i = 1; i <= 40; i += 1) { int_array.push(ffr(i)); } for (var i = 1; i <= 960; i += 1) { int_array.push(ffs(i)); }

int_array.sort(function(a,b){return a-b;});

for (var i = 1; i <= 1000; i += 1) { if (int_array[i-1] !== i) { throw "Something's wrong!" } else { console.log("1000 integer check ok."); } }</lang> Output:

R(1) = 1
R(2) = 3
R(3) = 7
R(4) = 12
R(5) = 18
R(6) = 26
R(7) = 35
R(8) = 45
R(9) = 56
R(10) = 69
1000 integer check ok.

Julia

Much of this task would seem to lend itself to an iterator based solution. However, the first step calls for ffr(n) and ffs(n), which imply that the series values are to be "randomly" rather than "sequentially" accessed. Given this implied requirement, I chose to implement ffr and ffs as closures containing the type (data structure) FigureFigure, which are used to calculate their values as required. I address task requirement 2 (no maximum n) by having these functions extend this data structure as needed to accommodate values of n larger than those used for their creation.

Functions <lang Julia> type FigureFigure{T<:Integer}

   r::Array{T,1}
   rnmax::T
   snmax::T
   snext::T

end

function grow!{T<:Integer}(ff::FigureFigure{T}, rnmax::T=100)

   ff.rnmax < rnmax || return nothing
   append!(ff.r, zeros(T, (rnmax-ff.rnmax)))
   snext = ff.snext
   for i in (ff.rnmax+1):rnmax
       ff.r[i] = ff.r[i-1] + snext
       snext += 1
       while snext in ff.r
           snext += 1
       end
   end
   ff.rnmax = rnmax
   ff.snmax = ff.r[end] - rnmax
   ff.snext = snext
   return nothing

end

function FigureFigure{T<:Integer}(rnmax::T=10)

   ff = FigureFigure([1], 1, 0, 2)
   grow!(ff, rnmax)
   return ff

end

function FigureFigure{T<:Integer}(rnmax::T, snmax::T)

   ff = FigureFigure(rnmax)
   while ff.snmax < snmax
       grow!(ff, 2ff.rnmax)
   end
   return ff

end

function make_ffr{T<:Integer}(nmax::T=10)

   ff = FigureFigure(nmax)
   function ffr{T<:Integer}(n::T)
       if n > ff.rnmax
           grow!(ff, 2n)
       end
       ff.r[n]
   end

end

function make_ffs{T<:Integer}(nmax::T=100)

   ff = FigureFigure(13, nmax)
   function ffs{T<:Integer}(n::T)
       while ff.snmax < n
           grow!(ff, 2ff.rnmax)
       end
       s = n
       for r in ff.r
           r <= s || return s
           s += 1
       end
   end

end </lang>

Main <lang Julia> NR = 40 NS = 960 ffr = make_ffr(NR) ffs = make_ffs(NS)

hi = 10 print("The first ", hi, " values of R are:\n ") for i in 1:hi

   print(ffr(i), "  ")

end println()

tally = falses(NR+NS) iscontained = true for i in 1:NR

   try
       tally[ffr(i)] = true
   catch
       iscontained = false
   end

end for i in 1:NS

   try
       tally[ffs(i)] = true
   catch
       iscontained = false
   end

end

println() print("The first ", NR, " values of R and ", NS, " of S are ") if !iscontained

   print("not ")

end println("contained in the interval 1:", NR+NS, ".") print("These values ") if !all(tally)

   print("do not ")

end println("cover the entire interval.") </lang>

Output:

The first 10 values of R are:
    1  3  7  12  18  26  35  45  56  69  

The first 40 values of R and 960 of S are contained in the interval 1:1000.
These values cover the entire interval.

Kotlin

Translated from Java. <lang scala>fun ffr(n: Int) = get(n, 0)[n - 1]

fun ffs(n: Int) = get(0, n)[n - 1]

internal fun get(rSize: Int, sSize: Int): List<Int> {

   val rlist = arrayListOf(1, 3, 7)
   val slist = arrayListOf(2, 4, 5, 6)
   val list = if (rSize > 0) rlist else slist
   val targetSize = if (rSize > 0) rSize else sSize

   while (list.size > targetSize)
       list.removeAt(list.size - 1)
   while (list.size < targetSize) {
       val lastIndex = rlist.lastIndex
       val lastr = rlist[lastIndex]
       val r = lastr + slist[lastIndex]
       rlist += r
       var s = lastr + 1
       while (s < r && list.size < targetSize)
           slist += s++
   }
   return list

}

fun main(args: Array<String>) {

   print("R():")
   (1..10).forEach { print(" " + ffr(it)) }
   println()

   val first40R = (1..40).map { ffr(it) }
   val first960S = (1..960).map { ffs(it) }
   val indices = (1..1000).filter  { it in first40R == it in first960S }
   indices.forEach { println("Integer $it either in both or neither set") }
   println("Done")

}</lang>

Mathematica / Wolfram Language

1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.

   The instructions call for two functions.
   Because S[n] is generated while computing R[n], one would normally avoid redundancy by combining 
   R and S into a single function that returns both sequences.

2. No maximum value for n should be assumed.

ffr[j_] := Module[{R = {1}, S = 2, k = 1},
   Do[While[Position[R, S] != {}, S++]; k = k + S; S++;
   R = Append[R, k], {n, 1, j - 1}]; R]

ffs[j_] := Differences[ffr[j + 1]]

</lang>

3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

ffr[10]

(* out *)
{1, 3, 7, 12, 18, 26, 35, 45, 56, 69}

</lang>

4. Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.

t = Sort[Join[ffr[40], ffs[960]]];

t == Range[1000]

(* out *)
True

</lang>

MATLAB / Octave

1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively. 2. No maximum value for n should be assumed.

<lang MATLAB> function [R,S] = ffr_ffs(N)

   t = [1,0]; 
   T = 1;
   n = 1; 
   %while T<=1000,
   while n<=N,
       R = find(t,n);
       S = find(~t,n);
       T = R(n)+S(n);

       % pre-allocate memory, this improves performance

if T > length(t), t = [t,zeros(size(t))]; end;

       t(T) = 1; 
       n = n + 1;
   end; 
   if nargout>0, 
     r = max(R); 
     s = max(S);
   else 
     printf('Sequence R:\n'); disp(R);
     printf('Sequence S:\n'); disp(S);
   end; 
 end; </lang>

3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

>>ffr_ffs(10)
Sequence R:
    1    3    7   12   18   26   35   45   56   69
Sequence S:
    2    4    5    6    8    9   10   11   13   14

4. This is self-evident from the function definition, but also because R and S are complementary in t and ~t. However, one can also Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once. Modify the function above in such a way that, instead of r and s, R and S are returned, and run

  [R1,S1] = ffr_ffs(40);	
  [R2,S2] = ffr_ffs(960);	
  all(sort([R1,S2])==1:1000) 
ans =  1

Nim

<lang nim>var cr = @[1] var cs = @[2]

proc extendRS =

 let x = cr[cr.high] + cs[cr.high]
 cr.add x
 for y in cs[cs.high] + 1 ..< x: cs.add y
 cs.add x + 1

proc ffr(n: int): int =

 assert n > 0
 while n > cr.len: extendRS()
 cr[n - 1]

proc ffs(n: int): int =

 assert n > 0
 while n > cs.len: extendRS()
 cs[n - 1]

for i in 1..10: stdout.write ffr i," " echo ""

var bin: array[1..1000, int] for i in 1..40: inc bin[ffr i] for i in 1..960: inc bin[ffs i] var all = true for x in bin:

 if x != 1:
   all = false
   break

if all: echo "All Integers 1..1000 found OK" else: echo "All Integers 1..1000 NOT found only once: ERROR"</lang> Output:

/home/deen/git/nim-unsorted/hofstadter 
1 3 7 12 18 26 35 45 56 69 
All Integers 1..1000 found OK

Oforth

<lang oforth>tvar: R ListBuffer new 1 over add R put

tvar: S ListBuffer new 2 over add S put

buildnext

| r s current i |

  R at ->r 
  S at ->s
  r last  r size s at  + dup ->current  r add 
  s last 1+  current 1-  for: i [ i s add ]
  current 1+ s add ;

ffr(n)

  while ( R at size n < ) [ buildnext ]
  n R at at ;

ffs(n)

  while ( S at size n < ) [ buildnext ]
  n S at at ;</lang>

Output :

>#[ ffr . ] 10 seqEach
1 3 7 12 18 26 35 45 56 69
ok
>#ffr 40 seq map  #ffs 960 seq map  + sort 1000 seq == .
1 ok

Perl

The program produces a table with the first 10 values of R and S. It also calculates R(40) which is 982, S(960) which is 1000, and R(41) which is 1030.

Then we go through the first 1000 outputs, mark those which are seen, then check if all values in the range one through one thousand were seen.

<lang perl>#!perl use strict; use warnings;

my @r = ( undef, 1 ); my @s = ( undef, 2 );

sub ffsr {

 my $n = shift;
 while( $#r < $n ) {
   push @r, $s[$#r]+$r[-1];
   push @s, grep { $s[-1]<$_ } $s[-1]+1..$r[-1]-1, $r[-1]+1;
 }
 return $n;

}

sub ffr { $r[ffsr shift] } sub ffs { $s[ffsr shift] }

printf " i: R(i) S(i)\n"; printf "==============\n"; printf "%3d: %3d %3d\n", $_, ffr($_), ffs($_) for 1..10; printf "\nR(40)=%3d S(960)=%3d R(41)=%3d\n", ffr(40), ffs(960), ffr(41);

my %seen; $seen{ffr($_)}++ for 1 .. 40; $seen{ffs($_)}++ for 1 .. 960; if( 1000 == keys %seen and grep $seen{$_}, 1 .. 1000 ) { print "All occured exactly once.\n"; } else { my @missed = grep !$seen{$_}, 1 .. 1000; my @dupped = sort { $a <=> $b} grep $seen{$_}>1, keys %seen; print "These were missed: @missed\n"; print "These were duplicated: @dupped\n"; } </lang>

Phix

Initialising such that length(S)>length(F) simplified things significantly.

sequence F = {1,3,7},
         S = {2,4,5,6}
integer fmax = 3 -- (ie F[3], ==7, already in S)
 
forward function ffs(integer n)
 
function ffr(integer n)
    integer l = length(F)
    while n>l do
        F &= F[l]+ffs(l)
        l += 1
    end while
    return F[n]
end function
 
function ffs(integer n)
    while n>length(S) do
        fmax += 1
        if fmax>length(F) then {} = ffr(fmax) end if
        S &= tagset(lim:=F[fmax]-1,start:=F[fmax-1]+1)
        -- ie/eg if fmax was 3, then F[2..3] being {3,7} 
        --       ==> tagset(lim:=6,start:=4), ie {4,5,6}.
    end while
    return S[n]
end function
 
{} = ffr(10)    -- (or collect one by one)
?{"The first ten values of R",F[1..10]}
{} = ffr(40)    -- (not actually needed)
{} = ffs(960)
if sort(F[1..40]&S[1..960])=tagset(1000) then
    puts(1,"test passed\n")
else
    puts(1,"some error!\n")
end if

Output:

{"The first ten values of R",{1,3,7,12,18,26,35,45,56,69}}
test passed

PicoLisp

<lang PicoLisp>(setq *RNext 2)

(de ffr (N)

  (cache '(NIL) N
     (if (= 1 N)
        1
        (+ (ffr (dec N)) (ffs (dec N))) ) ) )

(de ffs (N)

  (cache '(NIL) N
     (if (= 1 N)
        2
        (let S (inc (ffs (dec N)))
           (when (= S (ffr *RNext))
              (inc 'S)
              (inc '*RNext) )
           S ) ) ) )</lang>

Test: <lang PicoLisp>: (mapcar ffr (range 1 10)) -> (1 3 7 12 18 26 35 45 56 69)

(=

  (range 1 1000)
  (sort (conc (mapcar ffr (range 1 40)) (mapcar ffs (range 1 960)))) )

-> T</lang>

PL/I

<lang pli>ffr: procedure (n) returns (fixed binary(31));

  declare n fixed binary (31);
  declare v(2*n+1) bit(1);
  declare (i, j) fixed binary (31);
  declare (r, s) fixed binary (31);

  v = '0'b;
  v(1) = '1'b;

  if n = 1 then return (1);

  r = 1;
  do i = 2 to n;
     do j = 2 to 2*n;
        if v(j) = '0'b then leave;
     end;
     v(j) = '1'b;
     s = j;
     r = r + s;
     if r <= 2*n then v(r) = '1'b;
  end;
  return (r);

end ffr;</lang> Output:

Please type a value for n: 
    1    3    7   12   18   26   35   45   56   69   83   98  114  131  150
  170  191  213  236  260  285  312  340  369  399  430  462  495  529  565
  602  640  679  719  760  802  845  889  935  982

<lang pli>ffs: procedure (n) returns (fixed binary (31));

  declare n fixed binary (31);
  declare v(2*n+1) bit(1);
  declare (i, j) fixed binary (31);
  declare (r, s) fixed binary (31);

  v = '0'b;
  v(1) = '1'b;

  if n = 1 then return (2);

  r = 1;
  do i = 1 to n;
     do j = 2 to 2*n;
        if v(j) = '0'b then leave;
     end;
     v(j) = '1'b;
     s = j;
     r = r + s;
     if r <= 2*n then v(r) = '1'b;
  end;
  return (s);

end ffs;</lang> Output of first 960 values:

Please type a value for n: 
    2    4    5    6    8    9   10   11   13   14   15   16   17   19   20
   21   22   23   24   25   27   28   29   30   31   32   33   34   36   37
  ...
  986  987  988  989  990  991  992  993  994  995  996  997  998  999 1000

Verification using the above procedures: <lang pli>

  Dcl t(1000) Bit(1) Init((1000)(1)'0'b);
  put skip list ('Verification that the first 40 FFR numbers and the first');
  put skip list ('960 FFS numbers result in the integers 1 to 1000 only.');
  do i = 1 to 40;
     j = ffr(i);
     if t(j) then put skip list ('error, duplicate value at ' || i);
     else t(j) = '1'b;
  end;
  do i = 1 to 960;
     j = ffs(i);
     if t(j) then put skip list ('error, duplicate value at ' || i);
     else t(j) = '1'b;
  end;
  if all(t = '1'b) then put skip list ('passed test');

</lang> Output:

Verification that the first 40 FFR numbers and the first 
960 FFS numbers result in the integers 1 to 1000 only. 
passed test

Prolog

Constraint Handling Rules

CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker <lang Prolog>:- use_module(library(chr)).

- chr_constraint ffr/2, ffs/2, hofstadter/1,hofstadter/2.

- chr_option(debug, off).

- chr_option(optimize, full).

% to remove duplicates ffr(N, R1) \ ffr(N, R2) <=> R1 = R2 | true. ffs(N, R1) \ ffs(N, R2) <=> R1 = R2 | true.

% compute ffr ffr(N, R), ffr(N1, R1), ffs(N1,S1) ==>

        N > 1, N1 is N - 1 |

R is R1 + S1.

% compute ffs ffs(N, S), ffs(N1,S1) ==>

        N > 1, N1 is N - 1 |

V is S1 + 1, ( find_chr_constraint(ffr(_, V)) -> S is V+1; S = V).

% init hofstadter(N) ==> ffr(1,1), ffs(1,2). % loop hofstadter(N), ffr(N1, _R), ffs(N1, _S) ==> N1 < N, N2 is N1 +1 | ffr(N2,_), ffs(N2,_).

</lang> Output for first task :

 ?- hofstadter(10), bagof(ffr(X,Y), find_chr_constraint(ffr(X,Y)), L).
ffr(10,69)
ffr(9,56)
ffr(8,45)
ffr(7,35)
ffr(6,26)
ffr(5,18)
ffr(4,12)
ffr(3,7)
ffr(2,3)
ffr(1,1)
ffs(10,14)
ffs(9,13)
ffs(8,11)
ffs(7,10)
ffs(6,9)
ffs(5,8)
ffs(4,6)
ffs(3,5)
ffs(2,4)
ffs(1,2)
hofstadter(10)
L = [ffr(10,69),ffr(9,56),ffr(8,45),ffr(7,35),ffr(6,26),ffr(5,18),ffr(4,12),ffr(3,7),ffr(2,3),ffr(1,1)].

Code for the second task <lang Prolog>hofstadter :- hofstadter(960), % fetch the values of ffr bagof(Y, X^find_chr_constraint(ffs(X,Y)), L1), % fetch the values of ffs bagof(Y, X^(find_chr_constraint(ffr(X,Y)), X < 41), L2), % concatenate then append(L1, L2, L3), % sort removing duplicates sort(L3, L4), % check the correctness of the list ( (L4 = [1|_], last(L4, 1000), length(L4, 1000)) -> writeln(ok); writeln(ko)), % to remove all pending constraints fail. </lang> Output for second task

 ?- hofstadter.
ok
false.

Python

<lang python>def ffr(n):

   if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
   try:
       return ffr.r[n]
   except IndexError:
       r, s = ffr.r, ffs.s
       ffr_n_1 = ffr(n-1)
       lastr = r[-1]
       # extend s up to, and one past, last r 
       s += list(range(s[-1] + 1, lastr))
       if s[-1] < lastr: s += [lastr + 1]
       # access s[n-1] temporarily extending s if necessary
       len_s = len(s)
       ffs_n_1 = s[n-1] if len_s > n else (n - len_s) + s[-1]
       ans = ffr_n_1 + ffs_n_1
       r.append(ans)
       return ans

ffr.r = [None, 1]

def ffs(n):

   if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
   try:
       return ffs.s[n]
   except IndexError:
       r, s = ffr.r, ffs.s
       for i in range(len(r), n+2):
           ffr(i)
           if len(s) > n:
               return s[n]
       raise Exception("Whoops!")

ffs.s = [None, 2]

if __name__ == '__main__':

   first10 = [ffr(i) for i in range(1,11)]
   assert first10 == [1, 3, 7, 12, 18, 26, 35, 45, 56, 69], "ffr() value error(s)"
   print("ffr(n) for n = [1..10] is", first10)
   #
   bin = [None] + [0]*1000
   for i in range(40, 0, -1):
       bin[ffr(i)] += 1
   for i in range(960, 0, -1):
       bin[ffs(i)] += 1
   if all(b == 1 for b in bin[1:1000]):
       print("All Integers 1..1000 found OK")
   else:
       print("All Integers 1..1000 NOT found only once: ERROR")</lang>

Output

ffr(n) for n = [1..10] is [1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
All Integers 1..1000 found OK

Alternative

<lang python>cR = [1] cS = [2]

def extend_RS(): x = cR[len(cR) - 1] + cS[len(cR) - 1] cR.append(x) cS += range(cS[-1] + 1, x) cS.append(x + 1)

def ff_R(n): assert(n > 0) while n > len(cR): extend_RS() return cR[n - 1]

def ff_S(n): assert(n > 0) while n > len(cS): extend_RS() return cS[n - 1]

tests

print([ ff_R(i) for i in range(1, 11) ])

s = {} for i in range(1, 1001): s[i] = 0 for i in range(1, 41): del s[ff_R(i)] for i in range(1, 961): del s[ff_S(i)]

the fact that we got here without a key error

print("Ok")</lang>output<lang>[1, 3, 7, 12, 18, 26, 35, 45, 56, 69] Ok</lang>

Using cyclic iterators

Translation of: Haskell

Defining R and S as mutually recursive generators. Follows directly from the definition of the R and S sequences. <lang python>from itertools import islice

def R(): n = 1 yield n for s in S(): n += s yield n;

def S(): yield 2 yield 4 u = 5 for r in R(): if r <= u: continue; for x in range(u, r): yield x u = r + 1

def lst(s, n): return list(islice(s(), n))

print "R:", lst(R, 10) print "S:", lst(S, 10) print sorted(lst(R, 40) + lst(S, 960)) == list(range(1,1001))

perf test case
print sum(lst(R, 10000000))</lang>

Output:

R: [1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
S: [2, 4, 5, 6, 8, 9, 10, 11, 13, 14]
True

R

Global variables aren't idiomatic R, but this is otherwise an ideal task for the language. Comments aside, this is easily one of the shortest solutions on this page. This is mostly due to how R treats most things as a vector. For example, rValues starts out as the number 1, but repeatedly has new values appended to it without much ceremony. <lang R>rValues <- 1 sValues <- 2 ffr <- function(n) {

 if(!is.na(rValues[n])) rValues[n] else (rValues[n] <<- ffr(n-1) + ffs(n-1))

}

In theory, generating S requires computing ALL values not in R.
That would be infinitely many values.
However, to generate S(n) we only need to observe that its value cannot exceed R(n)+1.

ffs <- function(n) {

 if(!is.na(sValues[n])) sValues[n] else (sValues[n] <<- setdiff(seq_len(1 + ffr(n)), rValues)[n])

}

Task 1

invisible(ffr(10)) print(rValues)

Task 2
If we try to call ffs(960) directly, R will complain about the stack being too big.
Calling ffs(500) first solves this problem.

invisible(ffs(500)) invisible(ffs(960))

In R, "the first 40 values of ffr plus the first 960 values of ffs" can easily be misread.
rValues[1:40]+sValues[1:960] is valid R code. It will duplicate the first 40 rValues 23
times, append them to the original, and add that vector to the first 960 sValues.

This gives an output of length 960, which clearly cannot contain 1000 different values.

Presumably, the task wants us to append rValues[1:40] and sValues[1:960].

print(table(c(rValues[1:40], sValues[1:960])))</lang>

Output:

> print(rValues)
 [1]  1  3  7 12 18 26 35 45 56 69
> print(table(c(rValues[1:40], sValues[1:960])))

   1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16   17   18   19 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  20   21   22   23   24   25   26   27   28   29   30   31   32   33   34   35   36   37   38 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  39   40   41   42   43   44   45   46   47   48   49   50   51   52   53   54   55   56   57 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  58   59   60   61   62   63   64   65   66   67   68   69   70   71   72   73   74   75   76 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  77   78   79   80   81   82   83   84   85   86   87   88   89   90   91   92   93   94   95 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  96   97   98   99  100  101  102  103  104  105  106  107  108  109  110  111  112  113  114 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 115  116  117  118  119  120  121  122  123  124  125  126  127  128  129  130  131  132  133 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 134  135  136  137  138  139  140  141  142  143  144  145  146  147  148  149  150  151  152 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 153  154  155  156  157  158  159  160  161  162  163  164  165  166  167  168  169  170  171 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 172  173  174  175  176  177  178  179  180  181  182  183  184  185  186  187  188  189  190 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 191  192  193  194  195  196  197  198  199  200  201  202  203  204  205  206  207  208  209 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 210  211  212  213  214  215  216  217  218  219  220  221  222  223  224  225  226  227  228 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 229  230  231  232  233  234  235  236  237  238  239  240  241  242  243  244  245  246  247 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 248  249  250  251  252  253  254  255  256  257  258  259  260  261  262  263  264  265  266 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 267  268  269  270  271  272  273  274  275  276  277  278  279  280  281  282  283  284  285 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 286  287  288  289  290  291  292  293  294  295  296  297  298  299  300  301  302  303  304 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 305  306  307  308  309  310  311  312  313  314  315  316  317  318  319  320  321  322  323 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 324  325  326  327  328  329  330  331  332  333  334  335  336  337  338  339  340  341  342 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 343  344  345  346  347  348  349  350  351  352  353  354  355  356  357  358  359  360  361 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 362  363  364  365  366  367  368  369  370  371  372  373  374  375  376  377  378  379  380 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 381  382  383  384  385  386  387  388  389  390  391  392  393  394  395  396  397  398  399 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 400  401  402  403  404  405  406  407  408  409  410  411  412  413  414  415  416  417  418 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 419  420  421  422  423  424  425  426  427  428  429  430  431  432  433  434  435  436  437 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 438  439  440  441  442  443  444  445  446  447  448  449  450  451  452  453  454  455  456 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 457  458  459  460  461  462  463  464  465  466  467  468  469  470  471  472  473  474  475 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 476  477  478  479  480  481  482  483  484  485  486  487  488  489  490  491  492  493  494 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 495  496  497  498  499  500  501  502  503  504  505  506  507  508  509  510  511  512  513 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 514  515  516  517  518  519  520  521  522  523  524  525  526  527  528  529  530  531  532 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 533  534  535  536  537  538  539  540  541  542  543  544  545  546  547  548  549  550  551 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 552  553  554  555  556  557  558  559  560  561  562  563  564  565  566  567  568  569  570 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 571  572  573  574  575  576  577  578  579  580  581  582  583  584  585  586  587  588  589 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 590  591  592  593  594  595  596  597  598  599  600  601  602  603  604  605  606  607  608 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 609  610  611  612  613  614  615  616  617  618  619  620  621  622  623  624  625  626  627 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 628  629  630  631  632  633  634  635  636  637  638  639  640  641  642  643  644  645  646 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 647  648  649  650  651  652  653  654  655  656  657  658  659  660  661  662  663  664  665 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 666  667  668  669  670  671  672  673  674  675  676  677  678  679  680  681  682  683  684 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 685  686  687  688  689  690  691  692  693  694  695  696  697  698  699  700  701  702  703 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 704  705  706  707  708  709  710  711  712  713  714  715  716  717  718  719  720  721  722 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 723  724  725  726  727  728  729  730  731  732  733  734  735  736  737  738  739  740  741 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 742  743  744  745  746  747  748  749  750  751  752  753  754  755  756  757  758  759  760 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 761  762  763  764  765  766  767  768  769  770  771  772  773  774  775  776  777  778  779 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 780  781  782  783  784  785  786  787  788  789  790  791  792  793  794  795  796  797  798 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 799  800  801  802  803  804  805  806  807  808  809  810  811  812  813  814  815  816  817 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 818  819  820  821  822  823  824  825  826  827  828  829  830  831  832  833  834  835  836 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 837  838  839  840  841  842  843  844  845  846  847  848  849  850  851  852  853  854  855 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 856  857  858  859  860  861  862  863  864  865  866  867  868  869  870  871  872  873  874 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 875  876  877  878  879  880  881  882  883  884  885  886  887  888  889  890  891  892  893 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 894  895  896  897  898  899  900  901  902  903  904  905  906  907  908  909  910  911  912 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 913  914  915  916  917  918  919  920  921  922  923  924  925  926  927  928  929  930  931 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 932  933  934  935  936  937  938  939  940  941  942  943  944  945  946  947  948  949  950 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 951  952  953  954  955  956  957  958  959  960  961  962  963  964  965  966  967  968  969 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 970  971  972  973  974  975  976  977  978  979  980  981  982  983  984  985  986  987  988 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 989  990  991  992  993  994  995  996  997  998  999 1000 
   1    1    1    1    1    1    1    1    1    1    1    1

Racket

Translation of: Java

We store the values of r and s in hash-tables. The first values are added by hand. The procedure extend-r-s! adds more values.

<lang Racket>#lang racket/base

(define r-cache (make-hash '((1 . 1) (2 . 3) (3 . 7)))) (define s-cache (make-hash '((1 . 2) (2 . 4) (3 . 5) (4 . 6))))

(define (extend-r-s!)

 (define r-count (hash-count r-cache))
 (define s-count (hash-count s-cache))
 (define last-r (ffr r-count))
 (define new-r (+ (ffr r-count) (ffs r-count)))
 (hash-set! r-cache (add1 r-count) new-r)
 (define offset (- s-count last-r))
 (for ([val (in-range (add1 last-r) new-r)])
   (hash-set! s-cache (+ val offset) val)))</lang>

The functions ffr and ffs simply retrieve the value from the hash table if it exist, or call extend-r-s until they are long enought.

<lang Racket>(define (ffr n)

 (hash-ref r-cache n (lambda () (extend-r-s!) (ffr n))))

(define (ffs n)

 (hash-ref s-cache n (lambda () (extend-r-s!) (ffs n))))</lang>

Tests: <lang Racket>(displayln (map ffr (list 1 2 3 4 5 6 7 8 9 10))) (displayln (map ffs (list 1 2 3 4 5 6 7 8 9 10)))

(displayln "Checking for first 1000 integers:") (displayln (if (equal? (sort (append (for/list ([i (in-range 1 41)])

                                      (ffr i))
                                    (for/list ([i (in-range 1 961)])
                                      (ffs i)))
                            <)
                      (for/list ([i (in-range 1 1001)])
                        i))
              "Test passed"
              "Test failed"))</lang>

Sample Output:

(1 3 7 12 18 26 35 45 56 69)
(2 4 5 6 8 9 10 11 13 14)
Checking for first 1000 integers: Test passed

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.03

<lang perl6>my %r = 1 => 1; my %s = 1 => 2;

sub ffr ($n) { %r{$n} //= ffr($n - 1) + ffs($n - 1) } sub ffs ($n) { %s{$n} //= (grep none(map &ffr, 1..$n), max(%s.values)+1..*)[0] }

my @ffr = map &ffr, 1..*; my @ffs = map &ffs, 1..*;

say @ffr[^10]; say "Rawks!" if 1...1000 eqv sort |@ffr[^40], |@ffs[^960];</lang> Output:

1 3 7 12 18 26 35 45 56 69
Rawks!

REXX

version 1

This REXX example makes use of sparse arrays.

Over a third of the program was for verification of the first thousand numbers in the Hofstadter Figure-Figure sequences. <lang rexx>/*REXX program calculates and verifies the Hofstadter Figure─Figure sequences. */ parse arg x top bot . /*obtain optional arguments from the CL*/ if x== | x=="," then x= 10 /*Not specified? Then use the default.*/ if top== | top=="," then top=1000 /* " " " " " " */ if bot== | bot=="," then bot= 40 /* " " " " " " */ low=1; if x<0 then low=abs(x) /*only display a single │X│ value? */ r.=0; r.1=1; rr.=r.; rr.1=1; s.=r.; s.1=2 /*initialize the R, RR, and S arrays.*/ errs=0 /*the number of errors found (so far).*/

            do i=low  to abs(x)                 /*display the 1st  X  values of  R & S.*/
            say right('R('i") =",20) right(FFR(i),7) right('S('i") =",20) right(FFS(i),7)
            end   /*i*/
                                                /* [↑]  list the 1st X Fig─Fig numbers.*/

if x<1 then exit /*if X isn't positive, then we're done.*/ $.=0 /*initialize the memoization ($) array.*/

            do m=1  for  bot;  r=FFR(m);  $.r=1 /*calculate the first forty  R  values.*/
            end   /*m*/                         /* [↑]  ($.)  is used for memoization. */
                                                /* [↓]  check for duplicate #s in R & S*/
            do n=1  for top-bot;     s=FFS(n)   /*calculate the value of  FFS(n).      */
            if $.s  then call ser 'duplicate number in R and S lists:' s;   $.s=1
            end   /*n*/                         /* [↑]  calculate the 1st 960 S values.*/
                                                /* [↓]  check for missing values in R│S*/
            do v=1  for top;  if \$.v  then  call ser     'missing R │ S:'    v
            end   /*v*/                         /* [↑]  are all 1≤ numbers ≤1k present?*/

say if errs==0 then say 'verification completed for all numbers from 1 ──►' top " [inclusive]."

           else say 'verification failed with'      errs      "errors."

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ FFR: procedure expose r. rr. s.; parse arg n /*obtain the number from the arguments.*/

    if r.n\==0  then return r.n                 /*R.n  defined?  Then return the value.*/
    _=FFR(n-1) + FFS(n-1)                       /*calculate the  FFR  and  FFS  values.*/
    r.n=_;       rr._=1;        return _        /*assign the value to R & RR;   return.*/

/*──────────────────────────────────────────────────────────────────────────────────────*/ FFS: procedure expose r. s. rr.; parse arg n /*search for not null R or S number. */

    if s.n==0  then do k=1  for n               /* [↓]  1st  IF  is a  SHORT CIRCUIT.  */
                    if s.k\==0  then if r.k\==0  then iterate       /*are both defined?*/
                    call FFR k                  /*define  R.k  via the  FFR  subroutine*/
                    km=k-1;     _=s.km+1        /*calc. the next  S  number,  possibly.*/
                    _=_+rr._;   s.k=_           /*define an element of  the  S  array. */
                    end   /*k*/
    return s.n                                  /*return   S.n   value to the invoker. */

/*──────────────────────────────────────────────────────────────────────────────────────*/ ser: errs=errs+1; say '***error***' arg(1); return</lang> To see the talk section about this REXX program's timings, click here: timings for the REXX solutions.

output when using the default inputs:

              R(1) =       1               S(1) =       2
              R(2) =       3               S(2) =       4
              R(3) =       7               S(3) =       5
              R(4) =      12               S(4) =       6
              R(5) =      18               S(5) =       8
              R(6) =      26               S(6) =       9
              R(7) =      35               S(7) =      10
              R(8) =      45               S(8) =      11
              R(9) =      56               S(9) =      13
             R(10) =      69              S(10) =      14

verification completed for all numbers from  1 ──► 1000   [inclusive].

Version 2 from PL/I

<lang rexx>/* REXX **************************************************************

21.11.2012 Walter Pachl transcribed from PL/I
- - - - /

 Call time 'R'
 Say 'Verification that the first 40 FFR numbers and the first'
 Say '960 FFS numbers result in the integers 1 to 1000 only.'
 t.=0
 num.=
 do i = 1 to 40
   j = ffr(i)
   if t.j then Say 'error, duplicate value at ' || i
   else t.j = 1
   num.i=j
   end
 nn=0
 Say time('E') 'seconds elapsed'
 Do i=1 To 3
   ol=
   Do j=1 To 15
     nn=nn+1
     ol=ol right(num.nn,3)
     End
   Say ol
   End
 do i = 1 to 960
   j = ffs(i)
   if t.j then
     Say 'error, duplicate value at ' || i
   else t.j = 1
   end
 Do i=1 To 1000
   if t.i=0 Then
     Say i 'was not set'
   End
 If i>1000 Then
   Say 'passed test'
 Say time('E') 'seconds elapsed'
 Exit

ffr: procedure Expose v.
  Parse Arg n
  v.= 0
  v.1 = 1
  if n = 1 then return 1
  r = 1
  do i = 2 to n
    do j = 2 to 2*n
      if v.j = 0 then leave
      end
    v.j = 1
    s = j
    r = r + s
    if r <= 2*n then v.r = 1
    end
  return r

ffs: procedure Expose v.
  Parse Arg n
  v.= 0
  v.1 = 1
  if n = 1 then return 2
  r = 1
  do i = 1 to n
    do j = 2 to 2*n
      if v.j = 0 then leave
      end
    v.j = 1
    s = j
    r = r + s
    if r <= 2*n then v.r = 1
    end
  return s</lang>

Output:

Verification that the first 40 FFR numbers and the first
960 FFS numbers result in the integers 1 to 1000 only.
0.011000 seconds elapsed
   1   3   7  12  18  26  35  45  56  69  83  98 114 131 150
 170 191 213 236 260 285 312 340 369 399 430 462 495 529 565
 602 640 679 719 760 802 845 889 935 982                    
passed test
Windows (ooRexx)  33.183000 seconds elapsed
Windows (Regina)  22.627000 seconds elapsed
TSO interpreted: 139.699246 seconds elapsed
TSO compiled:      9.749457 seconds elapsed

Ring

Project : Hofstadter Figure-Figure sequences

hofr = list(20) hofr[1] = 1 hofs = [] add(hofs,2) for n = 1 to 10

     hofr[n+1] = hofr[n] + hofs[n]
     if n = 1
        add(hofs,4)
     else
        for p = hofr[n] + 1 to hofr[n+1] - 1
              if p != hofs[n]
                 add(hofs,p)
              ok
        next
     ok

next see "First 10 values of R:" + nl showarray(hofr) see "First 10 values of S:" + nl showarray(hofs)

func showarray(vect)

        svect = ""
       for n = 1 to 10
             svect = svect + vect[n] + " "
       next
       svect = left(svect, len(svect) - 1)
       see svect + nl

</lang> Output:

First 10 values of R:
1 3 7 12 18 26 35 45 56 69
First 10 values of S:
2 4 5 6 8 9 10 11 13 14

Ruby

Translation of: Tcl

<lang ruby>$r = [nil, 1] $s = [nil, 2]

def buildSeq(n)

 current = [ $r[-1], $s[-1] ].max
 while $r.length <= n || $s.length <= n
   idx = [ $r.length, $s.length ].min - 1
   current += 1
   if current == $r[idx] + $s[idx]
     $r << current
   else
     $s << current
   end
 end

end

def ffr(n)

 buildSeq(n)
 $r[n]

end

def ffs(n)

 buildSeq(n)
 $s[n]

end

require 'set' require 'test/unit'

class TestHofstadterFigureFigure < Test::Unit::TestCase

 def test_first_ten_R_values
   r10 = 1.upto(10).map {|n| ffr(n)}
   assert_equal(r10, [1, 3, 7, 12, 18, 26, 35, 45, 56, 69])
 end

 def test_40_R_and_960_S_are_1_to_1000
   rs_values = Set.new
   rs_values.merge( 1.upto(40).collect  {|n| ffr(n)} )
   rs_values.merge( 1.upto(960).collect {|n| ffs(n)} )
   assert_equal(rs_values, Set.new( 1..1000 ))
 end

end</lang>

outputs

Loaded suite hofstadter.figurefigure
Started
..
Finished in 0.511000 seconds.

2 tests, 2 assertions, 0 failures, 0 errors, 0 skips

Using cyclic iterators

Translation of: Python

<lang ruby>R = Enumerator.new do |y|

 y << n = 1
 S.each{|s_val| y << n += s_val}

end

S = Enumerator.new do |y|

 y << 2
 y << 4
 u = 5
 R.each do |r_val|
   next if u > r_val
   (u...r_val).each{|r| y << r}
   u = r_val+1
 end

end

p R.take(10) p S.take(10) p (R.take(40)+ S.take(960)).sort == (1..1000).to_a </lang>

Output:

[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
[2, 4, 5, 6, 8, 9, 10, 11, 13, 14]
true

Rust

<lang rust> use std::collections::HashMap;

struct Hffs {

   sequence_r: HashMap<usize, usize>,
   sequence_s: HashMap<usize, usize>,

}

impl Hffs {

   fn new() -> Hffs {
       Hffs {
           sequence_r: HashMap::new(),
           sequence_s: HashMap::new(),
       }
   }
   fn ffr(&mut self, n: usize) -> usize {
       // first try the cache
       let new_r = if let Some(result) = self.sequence_r.get(&n) {
           *result
       } else if n == 0 {
           1
       } else {
           // call recursively
           self.ffr(n - 1) + self.ffs(n - 1)
       };

       // insert into the cache and return value
       *self.sequence_r.entry(n).or_insert(new_r)
   }

   fn ffs(&mut self, n: usize) -> usize {
       // first try the cache
       let new_s = if let Some(result) = self.sequence_s.get(&n) {
           *result
       } else if n == 0 {
           2
       } else {
           let lower = self.ffs(n - 1) + 1_usize;
           let upper = self.ffr(n) + 1_usize;
           let mut min_s: usize = 0;
           // find next available S
           for i in lower..=upper {
               if !self.sequence_r.values().any(|&val| val == i) {
                   min_s = i;
                   break;
               }
           }
           min_s
       };

       // insert into the cache and return value
       *self.sequence_s.entry(n).or_insert(new_s)
   }

}

impl Default for Hffs {

   fn default() -> Self {
       Self::new()
   }

} fn main() {

   let mut hof = Hffs::new();

   for i in 0..10 {
       println!("H:{} -> R: {}, S: {}", i, hof.ffr(i), hof.ffs(i));
   }

   let r40 = (0..40).map(|i| hof.ffr(i)).collect::<Vec<_>>();
   let mut s960 = (0..960).map(|i| hof.ffs(i)).collect::<Vec<_>>();

   s960.extend(&r40);
   s960.sort_unstable();
   let f1000 = (1_usize..=1000).collect::<Vec<_>>();

   assert_eq!(f1000, s960, "Does NOT match");

} </lang>

Output:

H:0 -> R: 1, S: 2
H:1 -> R: 3, S: 4
H:2 -> R: 7, S: 5
H:3 -> R: 12, S: 6
H:4 -> R: 18, S: 8
H:5 -> R: 26, S: 9
H:6 -> R: 35, S: 10
H:7 -> R: 45, S: 11
H:8 -> R: 56, S: 13
H:9 -> R: 69, S: 14

Scala

Translation of: Go

<lang Scala>object HofstadterFigFigSeq extends App {

 import scala.collection.mutable.ListBuffer

 val r = ListBuffer(0, 1)
 val s = ListBuffer(0, 2)

 def ffr(n: Int): Int = {
   val ffri: Int => Unit = i => {
     val nrk = r.size - 1
     val rNext = r(nrk)+s(nrk)
     r += rNext
     (r(nrk)+2 to rNext-1).foreach{s += _}
     s += rNext+1
   }

   (r.size to n).foreach(ffri(_))
   r(n)
 }

 def ffs(n:Int): Int = {
   while (s.size <= n) ffr(r.size)
   s(n)
 }

 (1 to 10).map(i=>(i,ffr(i))).foreach(t=>println("r("+t._1+"): "+t._2))
 println((1 to 1000).toList.filterNot(((1 to 40).map(ffr(_))++(1 to 960).map(ffs(_))).contains)==List())

}</lang> Output:

r(1): 1
r(2): 3
r(3): 7
r(4): 12
r(5): 18
r(6): 26
r(7): 35
r(8): 45
r(9): 56
r(10): 69
true

Sidef

Translation of: Perl

<lang ruby>var r = [nil, 1] var s = [nil, 2]

func ffsr(n) {

 while(r.end < n) {
   r << s[r.end]+r[-1]
   s << [(s[-1]+1 .. r[-1]-1)..., r[-1]+1].grep{ s[-1] < _ }...
 }
 return n

}

func ffr(n) { r[ffsr(n)] } func ffs(n) { s[ffsr(n)] }

printf(" i: R(i) S(i)\n") printf("==============\n") { |i|

   printf("%3d:  %3d  %3d\n", i, ffr(i), ffs(i))

} << 1..10 printf("\nR(40)=%3d S(960)=%3d R(41)=%3d\n", ffr(40), ffs(960), ffr(41))

var seen = Hash()

if (seen.count {|k,v| (k.to_i >= 1) && (k.to_i <= 1000) && (v == 1) } == 1000) { say "All occured exactly once." } else { var missed = { !seen.has_key(_) }.grep(1..1000) var dupped = seen.grep { |_, v| v > 1 }.keys.sort say "These were missed: #{missed}" say "These were duplicated: #{dupped}" }</lang>

Output:

  i: R(i) S(i)
==============
  1:    1    2
  2:    3    4
  3:    7    5
  4:   12    6
  5:   18    8
  6:   26    9
  7:   35   10
  8:   45   11
  9:   56   13
 10:   69   14

R(40)=982 S(960)=1000 R(41)=1030
All occured exactly once.

Tcl

Library: Tcllib (Package: struct::set)

<lang tcl>package require Tcl 8.5 package require struct::set

Core sequence generator engine; stores in $R and $S globals

set R {R:-> 1} set S {S:-> 2} proc buildSeq {n} {

   global R S
   set ctr [expr {max([lindex $R end],[lindex $S end])}]
   while {[llength $R] <= $n || [llength $S] <= $n} {

set idx [expr {min([llength $R],[llength $S]) - 1}] if {[incr ctr] == [lindex $R $idx]+[lindex $S $idx]} { lappend R $ctr } else { lappend S $ctr }

}

Accessor procedures

proc ffr {n} {

   buildSeq $n
   lindex $::R $n

} proc ffs {n} {

   buildSeq $n
   lindex $::S $n

}

Show some things about the sequence

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

   puts "R($i) = [ffr $i]"

} puts "Considering {1..1000} vs {R(i)|i\u2208\[1,40\]}\u222a{S(i)|i\u2208\[1,960\]}" for {set i 1} {$i <= 1000} {incr i} {lappend numsInSeq $i} for {set i 1} {$i <= 40} {incr i} {

   lappend numsRS [ffr $i]

} for {set i 1} {$i <= 960} {incr i} {

   lappend numsRS [ffs $i]

} puts "set sizes: [struct::set size $numsInSeq] vs [struct::set size $numsRS]" puts "set equality: [expr {[struct::set equal $numsInSeq $numsRS]?{yes}:{no}}]"</lang> Output:

R(1) = 1
R(2) = 3
R(3) = 7
R(4) = 12
R(5) = 18
R(6) = 26
R(7) = 35
R(8) = 45
R(9) = 56
R(10) = 69
Considering {1..1000} vs {R(i)|i∈[1,40]}∪{S(i)|i∈[1,960]}
set sizes: 1000 vs 1000
set equality: yes

uBasic/4tH

Note that uBasic/4tH has no dynamic memory facilities and only one single array of 256 elements. So the only way to cram over a 1000 values there is to use a bitmap. This bitmap consists of an R range and an S range. In each range, a bit represents a positional value (bit 0 = "1", bit 1 = "2", etc.). The R(x) and S(x) functions simply count the number of bits set they encountered. To determine whether all integers between 1 and 1000 are complementary, both ranges are XORed, which would result in a value other than 2³¹-1 if there were any discrepancies present. An extra check determines if there are exactly 40 R values. <lang>Proc _SetBitR(1) ' Set the first R value Proc _SetBitS(2) ' Set the first S value

Print "Creating bitmap, wait.." ' Create the bitmap Proc _MakeBitMap Print

Print "R(1 .. 10):"; ' Print first 10 R-values

For x = 1 To 10

 Print " ";FUNC(_Rx(x));

For x = 1 To 10

 Print " ";FUNC(_Sx(x));

For x = 0 To (1000/31) ' Check the first 1000 values

 Print "Checking ";(x*31)+1;" to ";(x*31)+31;":\t";
 If XOR(@(x), @(x+64)) = 2147483647 Then
    Print "OK"                        ' XOR R() and S() ranges
 Else                                 ' should deliver MAX-N
    Print "Fail!"                     ' or we did have an error
 EndIf

 If FUNC(_Rx(x)) > 1000 Then Print "R(";x;") value greater than 1000"

Next ' below 1000

If FUNC(_Rx(x)) < 1001 Then Print "R(";x;") value also below 1000" End

_MakeBitMap ' Create the bitmap

 Local (4)

 a@ = 1                               ' Previous R(x) level
 b@ = 1                               ' Previous R(x) value

 Do Until b@ > (1000/31)*32           ' Fill up an entire array element
                                      ' calculate R(x+1) level
   c@ = FUNC(_Rx(a@)) + FUNC(_Sx(a@))
   Proc _SetBitR (c@)                 ' Set R(x+1) in the bitmap

   For d@ = b@ + 1 To c@ - 1          ' Set all intermediate S() values
     Proc _SetBitS (d@)               ' between R(x) and R(x+1)
   Next

   Proc _SetBitS (c@+1)               ' Number after R(x) is always S()
   b@ = c@                            ' R(x+1) now becomes R(x)
   a@ = a@ + 1                        ' Increment level
 Loop                                 ' Now do it again

Return

_Rx Param(1) ' Return value R(x)

 Local(2)

 b@ = 0                               ' No value found so far

 For c@ = 1 To (64*31)-1              ' Check the entire bitmap
   If (FUNC(_GetBitR(c@))) Then b@ = b@ + 1
   Until b@ = a@                      ' If a value found, increment counter
 Next                                 ' Until the required level is reached

Return (c@) ' Return position in bitmap

_Sx Param(1) ' Return value S(x)

 Local(2)

 b@ = 0                               ' No value found so far

 For c@ = 1 To (64*31)-1              ' Check the entire bitmap
   If (FUNC(_GetBitS(c@))) Then b@ = b@ + 1
   Until b@ = a@                      ' If a value found, increment counter
 Next                                 ' Until the required level is reached

Return (c@) ' Return position in bitmap

_SetBitR Param(1) ' Set bit n-1 in R-bitmap

 a@ = a@ - 1
 @(a@/31) = OR(@(a@/31), SHL(1,a@%31))

Return

_GetBitR Param(1) ' Return bit n-1 in R-bitmap

 a@ = a@ - 1

Return (AND(@(a@/31), SHL(1,a@%31))#0)

_SetBitS Param(1) ' Set bit n-1 in S-bitmap

 a@ = a@ - 1
 @(64+a@/31) = OR(@(64+a@/31), SHL(1,a@%31))

Return

_GetBitS Param(1) ' Return bit n-1 in S-bitmap

 a@ = a@ - 1

Return (AND(@(64+a@/31), SHL(1,a@%31))#0)</lang>

Output:

Creating bitmap, wait..

R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
S(1 .. 10): 2 4 5 6 8 9 10 11 13 14

Checking 1 to 31:       OK
Checking 32 to 62:      OK
Checking 63 to 93:      OK
Checking 94 to 124:     OK
Checking 125 to 155:    OK
Checking 156 to 186:    OK
Checking 187 to 217:    OK
Checking 218 to 248:    OK
Checking 249 to 279:    OK
Checking 280 to 310:    OK
Checking 311 to 341:    OK
Checking 342 to 372:    OK
Checking 373 to 403:    OK
Checking 404 to 434:    OK
Checking 435 to 465:    OK
Checking 466 to 496:    OK
Checking 497 to 527:    OK
Checking 528 to 558:    OK
Checking 559 to 589:    OK
Checking 590 to 620:    OK
Checking 621 to 651:    OK
Checking 652 to 682:    OK
Checking 683 to 713:    OK
Checking 714 to 744:    OK
Checking 745 to 775:    OK
Checking 776 to 806:    OK
Checking 807 to 837:    OK
Checking 838 to 868:    OK
Checking 869 to 899:    OK
Checking 900 to 930:    OK
Checking 931 to 961:    OK
Checking 962 to 992:    OK
Checking 993 to 1023:   OK

0 OK, 0:875

VBA

<lang vb>Private Function ffr(n As Long) As Long

   Dim R As New Collection
   Dim S As New Collection
   R.Add 1
   S.Add 2
   'return R(n)
   For i = 2 To n
       R.Add R(i - 1) + S(i - 1)
       For j = S(S.Count) + 1 To R(i) - 1
           S.Add j
       Next j
       For j = R(i) + 1 To R(i) + S(i - 1)
           S.Add j
       Next j
   Next i
   ffr = R(n)
   Set R = Nothing
   Set S = Nothing

End Function Private Function ffs(n As Long) As Long

   Dim R As New Collection
   Dim S As New Collection
   R.Add 1
   S.Add 2
   'return S(n)
   For i = 2 To n
       R.Add R(i - 1) + S(i - 1)
       For j = S(S.Count) + 1 To R(i) - 1
           S.Add j
       Next j
       For j = R(i) + 1 To R(i) + S(i - 1)
           S.Add j
       Next j
       If S.Count >= n Then Exit For
   Next i
   ffs = S(n)
   Set R = Nothing
   Set S = Nothing

End Function Public Sub main()

   Dim i As Long
   Debug.Print "The first ten values of R are:"
   For i = 1 To 10
       Debug.Print ffr(i);
   Next i
   Debug.Print
   Dim x As New Collection
   For i = 1 To 1000
       x.Add i, CStr(i)
   Next i
   For i = 1 To 40
       x.Remove CStr(ffr(i))
   Next i
   For i = 1 To 960
       x.Remove CStr(ffs(i))
   Next i
   Debug.Print "The first 40 values of ffr plus the first 960 values of ffs "
   Debug.Print "include all the integers from 1 to 1000 exactly once is "; Format(x.Count = 0)

End Sub</lang>

Output:

The first ten values of R are:
 1  3  7  12  18  26  35  45  56  69 
The first 40 values of ffr plus the first 960 values of ffs 
include all the integers from 1 to 1000 exactly once is True

VBScript

<lang vb> 'Initialize the r and the s arrays. Set r = CreateObject("System.Collections.ArrayList") Set s = CreateObject("System.Collections.ArrayList")

'Set initial values of r. r.Add "" : r.Add 1

'Set initial values of s. s.Add "" : s.Add 2

'Populate the r and the s arrays. For i = 2 To 1000 ffr(i) ffs(i) Next

'r function Function ffr(n) r.Add r(n-1)+s(n-1) End Function

's function Function ffs(n) 'index is the value of the last element of the s array. index = s(n-1)+1 Do

               'Add to s if the current index is not in the r array.

If r.IndexOf(index,0) = -1 Then s.Add index Exit Do Else index = index + 1 End If Loop End Function

'Display the first 10 values of r. WScript.StdOut.Write "First 10 Values of R:" WScript.StdOut.WriteLine For j = 1 To 10 If j = 10 Then WScript.StdOut.Write "and " & r(j) Else WScript.StdOut.Write r(j) & ", " End If Next WScript.StdOut.WriteBlankLines(2)

'Show that the first 40 values of r plus the first 960 values of s include all the integers from 1 to 1000 exactly once. 'The idea here is to create another array(integer) with 1000 elements valuing from 1 to 1000. Go through the first 40 values 'of the r array and remove the corresponding element in the integer array. Do the same thing with the first 960 values of 'the s array. If the resultant count of the integer array is 0 then it is a pass. Set integers = CreateObject("System.Collections.ArrayList") For k = 1 To 1000 integers.Add k Next For l = 1 To 960 If l <= 40 Then integers.Remove(r(l)) End If integers.Remove(s(l)) Next WScript.StdOut.Write "Test for the first 1000 integers: " If integers.Count = 0 Then WScript.StdOut.Write "Passed!!!" WScript.StdOut.WriteLine Else WScript.StdOut.Write "Miserably Failed!!!" WScript.StdOut.WriteLine End If </lang>

Output:

First 10 Values of R:
1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

Test for the first 1000 integers: Passed!!!

Wren

Translation of: Go

<lang ecmascript>var r = [0, 1] var s = [0, 2]

var ffr = Fn.new { |n|

   while (r.count <= n) {
       var nrk = r.count - 1         // last n for which r[n] is known
       var rNxt = r[nrk] + s[nrk]    // r[nrk+1]
       r.add(rNxt)                   // extend r by one element
       for (sn in r[nrk]+2...rNxt) {
           s.add(sn)                 // extend sequence s up to rNxt
       }
       s.add(rNxt + 1)               // extend sequence s one past rNxt
   }
   return r[n]

}

var ffs = Fn.new { |n|

   while (s.count <= n) ffr.call(r.count)
   return s[n]

}

System.print("The first 10 values of R are:") for (i in 1..10) System.write(" %(ffr.call(i))") System.print() var present = List.filled(1001, false) for (i in 1..40) present[ffr.call(i)] = true for (i in 1..960) present[ffs.call(i)] = true var allPresent = present.skip(1).all { |i| i == true } System.print("\nThe first 40 values of ffr plus the first 960 values of ffs") System.print("includes all integers from 1 to 1000 exactly once is %(allPresent).")</lang>

Output:

The first 10 values of R are:
 1 3 7 12 18 26 35 45 56 69

The first 40 values of ffr plus the first 960 values of ffs
includes all integers from 1 to 1000 exactly once is true.

zkl

<lang zkl>fcn genRS(reset=False){ //-->(n,R,S)

 var n=0, Rs=L(0,1), S=2;
 if(True==reset){ n=0; Rs=L(0,1); S=2; return(); }

 if (n==0) return(n=1,1,2);
 R:=Rs[-1] + S; Rs.append(R);
 foreach s in ([S+1..]){
    if(not Rs.holds(s)) { S=s; break; } // trimming Rs doesn't save space
 }
 return(n+=1,R,S);

} fcn ffrs(n) { genRS(True); do(n){ n=genRS() } n[1,2] } //-->( R(n),S(n) )</lang>

Output:

(0).pump(10,List,genRS).apply("get",1).println();
L(1,3,7,12,18,26,35,45,56,69)

<lang zkl>genRS(True); // reset sink:=(0).pump(40,List, 'wrap(ns){ T(Void.Write,Void.Write,genRS()[1,*]) }); sink= (0).pump(960-40,sink,'wrap(ns){ T(Void.Write,genRS()[2]) }); (sink.sort()==[1..1000].pump(List)) // [1..n].pump(List)-->(1,2,3...)

  .println("<-- should be True");</lang>

Output:

True<-- should be True

@@ Line 2,236: / Line 2,236: @@
 invisible(ffs(960))
 #In R, "the first 40 values of ffr plus the first 960 values of ffs" can easily be misread.
-#rValues[1:40]+sValues[1:960] is valid R code. It will duplicate the first 40 rValues 24
+#rValues[1:40]+sValues[1:960] is valid R code. It will duplicate the first 40 rValues 23
-#times, append them, and add that vector to the first 960 sValues. This gives an output of
+#times, append them to the original, and add that vector to the first 960 sValues.
-#length 960, which clearly cannot contain 1000 different values.
+This gives an output of length 960, which clearly cannot contain 1000 different values.
 #Presumably, the task wants us to append rValues[1:40] and sValues[1:960].
 print(table(c(rValues[1:40], sValues[1:960])))</lang>