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

These two sequences of positive integers are defined as:

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

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

Sequence ${\displaystyle R}$ starts:

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


Sequence ${\displaystyle S}$ starts:

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


1. 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).
2. No maximum value for n should be assumed.
3. Calculate and show that the first ten values of R are:
1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
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.

References

## 11l

Translation of: Python
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’)
Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
All Integers 1..1000 found OK


## ABC

PUT {[1]: 1} IN r.list
PUT {[1]: 2} IN s.list

HOW TO EXTEND R TO n:
SHARE r.list, s.list
WHILE n > #r.list:
PUT r.list[#r.list] + s.list[#r.list] IN next.r
FOR i IN {s.list[#s.list]+1 .. next.r-1}:
PUT i IN s.list[#s.list+1]
PUT next.r IN r.list[#r.list+1]
PUT next.r + 1 IN s.list[#s.list+1]

HOW TO EXTEND S TO n:
SHARE r.list, s.list
WHILE n > #s.list: EXTEND R TO #r.list + 1

HOW TO RETURN ffr n:
SHARE r.list
IF n > #r.list: EXTEND R TO n
RETURN r.list[n]

HOW TO RETURN ffs n:
SHARE s.list
IF n > #s.list: EXTEND S TO n
RETURN s.list[n]

WRITE "R[1..10]:"
FOR i IN {1..10}: WRITE ffr i
WRITE /

PUT {} IN thousand
FOR i IN {1..40}: INSERT ffr i IN thousand
FOR i IN {1..960}: INSERT ffs i IN thousand
IF thousand = {1..1000}:
WRITE "R[1..40] + S[1..960] = [1..1000]"/
Output:
R[1..10]: 1 3 7 12 18 26 35 45 56 69
R[1..40] + S[1..960] = [1..1000]

Specifying a package providing the functions FFR and FFS:

package Hofstadter_Figure_Figure is

function FFR(P: Positive) return Positive;

function FFS(P: Positive) return Positive;



The implementation of the package internally uses functions which generate an array of Figures or Spaces:

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;



Finally, a test program for the package, solving the task at hand:

with Ada.Text_IO, Hofstadter_Figure_Figure;

procedure Test_HSS is

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

begin
for I in 1 .. 10 loop
end loop;

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;


The output of the test program:

 1 3 7 12 18 26 35 45 56 69
Test Passed: No overlap between FFR(I) and FFS(J)


## APL

Works with: Dyalog 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
∇
: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

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

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
}


Examples:

Loop
MsgBox, 262144, , % "R(" A_Index ") = " R(A_Index) "nS(" A_Index ") = " S(A_Index)


Outputs:

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

## 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

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

      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%

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

#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;
}


## C#

Creates an IEnumerable for R and S and uses those to complete the task

using System;
using System.Collections.Generic;
using System.Linq;

{
{
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)
{
}
yield return _r[iR++];
}
}

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

{
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;
}
int newR = _r[rCount];
for (int iS = oldR + 1; iS < newR; iS++)
{
}
}
}

class Program
{
static void Main()
{
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!");
}
}
}


Output:

1
3
7
12
18
26
35
45
56
69
Verified

## C++

Works with: gcc
Works with: C++ version 11, 14, 17
#include <iomanip>
#include <iostream>
#include <set>
#include <vector>

using namespace std;

{
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)
}
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;
}

Output:
% ./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


## CLU

figfig = cluster is ffr, ffs
rep = null
ai = array[int]
own R: ai := ai$[1] own S: ai := ai$[2]

% Extend R and S until R(n) is known
extend = proc (n: int)
while n > ai$high(R) do next: int := ai$top(R) + S[ai$high(R)] ai$addh(R, next)
while ai$top(S) < next-1 do ai$addh(S, ai$top(S)+1) end ai$addh(S, next+1)
end
end extend

ffr = proc (n: int) returns (int)
extend(n)
return(R[n])
end ffr

ffs = proc (n: int) returns (int)
while n > ai$high(S) do extend(ai$high(R) + 1)
end
return(S[n])
end ffs
end figfig

start_up = proc ()
ai = array[int]
po: stream := stream$primary_output() % Print R[1..10] stream$puts(po, "R[1..10] =")
for i: int in int$from_to(1,10) do stream$puts(po, " " || int$unparse(figfig$ffr(i)))
end
stream$putl(po, "") % Count the occurrences of 1..1000 in R[1..40] and S[1..960] occur: ai := ai$fill(1, 1000, 0)
for i: int in int$from_to(1, 40) do occur[figfig$ffr(i)] := occur[figfig$ffr(i)] + 1 end for i: int in int$from_to(1, 960) do
occur[figfig$ffs(i)] := occur[figfig$ffs(i)] + 1
end

% See if they all occur exactly once
begin
for i: int in int$from_to(1, 1000) do if occur[i] ~= 1 then exit wrong(i) end end stream$putl(po,
"All numbers 1..1000 occur exactly once in R[1..40] U S[1..960].")
end except when wrong(i: int):
stream$putl(po, "Error: " || int$unparse(i) || " occurs " || int$unparse(occur[i]) || " times.") end end start_up Output: R[1..10] = 1 3 7 12 18 26 35 45 56 69 All numbers 1..1000 occur exactly once in R[1..40] U S[1..960]. ## CoffeeScript Translation of: Ruby 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.'  Output: As JavaScript. ## Common 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")  Output: First of R: (1 3 7 12 18 26 35 45 56 69) Ok ## 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; 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 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; }  Output: [1, 3, 7, 12, 18, 26, 35, 45, 56, 69] true ### Alternative version Translation of: Python (Same output) 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[$- 1]; immutable int ans = ffr_n_1 + ffs_n_1; r ~= ans; return ans; } } } struct ffs { static s = [int.min, 2]; static int opCall(in int n) nothrow { assert(n > 0); if (n < s.length) { return s[n]; } else { foreach (immutable i; ffr.r.length .. n + 2) { ffr(i); if (s.length > n) return s[n]; } assert(false, "Whoops!"); } } } void main() { iota(1, 11).map!ffr.writeln; auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs); t.array.sort().equal(iota(1, 1001)).writeln; }  ## EasyLang Translation of: C global rs[] ss[] . procdecl RS_append . . func R n . while n > len rs[] RS_append . return rs[n] . func S n . while n > len ss[] RS_append . return ss[n] . proc RS_append . . n = len rs[] r = R n + S n s = S len ss[] rs[] &= r repeat s += 1 until s = r ss[] &= s . ss[] &= r + 1 . rs[] = [ 1 ] ss[] = [ 2 ] write "R(1 .. 10): " for i to 10 write R i & " " . print "" len seen[] 1000 for i to 40 seen[R i] = 1 . for i to 960 seen[S i] = 1 . for i to 1000 if seen[i] = 0 print i & " not seen" return . . print "first 1000 ok" Output: R(1 .. 10): 1 3 7 12 18 26 35 45 56 69 first 1000 ok  ## EchoLisp (define (FFR n) (+ (FFR (1- n)) (FFS (1- n)))) (define (FFS n) (define next (1+ (FFS (1- n)))) (for ((k (in-naturals next))) #:break (not (vector-search* k (cache 'FFR))) => k )) (remember 'FFR #(0 1)) ;; init cache (remember 'FFS #(0 2))  Output: (define-macro m-range [a .. b] (range a (1+ b))) (map FFR [1 .. 10]) → (1 3 7 12 18 26 35 45 56 69) ;; checking (equal? [1 .. 1000] (list-sort < (append (map FFR [1 .. 40]) (map FFS [1 .. 960])))) → #t  ## Euler Math Toolbox >function RSstep (r,s) ...$  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))
1

## F#

### The function

// 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


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)

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.

let ffr,ffs=fF 10000000
printfn "%d\n%d (Array.last ffr) (Array.last ffs)


1584961838 10004416

Output:

## Factor

We keep lists S and R, and increment them when necessary.

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 ;

( 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


## FreeBASIC

Translation of: BBC BASIC
function ffr( n as integer ) as integer
if n = 1 then return 1
dim as integer i, j, r=1, s=1, v(1 to 2*n+1)
v(1) = 1
for i = 2 to n
for j = s to 2*n
if v(j) = 0 then exit for
next j
v(j) = 1
s = j
r += s
if r <= 2*n then v(r) = 1
next i
return r
end function

function ffs( n as integer ) as integer
if n = 1 then return 2
dim as integer i, j, r=1, s=2, v(1 to 2*n+1)
for i = 1 to n
for j = s to 2*n
if v(j) = 0 then exit for
next j
v(j) = 1
s = j
r += s
if r <= 2*n then v(r) = 1
next i
return s
end function

dim as integer i
print " R"," S"
print
for i = 1 to 10
print ffr(i), ffs(i)
next i

dim as boolean found(1 to 1000), failed
for i = 1 to 40
found(ffr(i)) = true
next i
for i = 1 to 960
found(ffs(i)) = true
next i
for i = 1 to 1000
if found(i) = false then failed = true
next i

if failed then print "Oh no!" else print "All integers from 1 to 1000 accounted for"

Output:

R             S

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

All integers from 1 to 1000 accounted for



## 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() {
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() {
for n := 1; n <= 10; n++ {
fmt.Printf("r(%d): %d\n", n, ffr(n))
}
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 {
return
}
}
}

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.

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)
}


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])


Output:

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

Defining R and S literally:

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)  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 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  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 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,>./@,)  Required examples:  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  ## Java Code: 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"); } }  Output: R(): 1 3 7 12 18 26 35 45 56 69 Done ## JavaScript Translation of: Ruby var R = [null, 1]; var S = [null, 2]; var extend_sequences = function (n) { var current = Math.max(R[R.length-1],S[S.length-1]); var i; while (R.length <= n || 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); } } } 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."); } }  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. ## jq Works with: jq Works with gojq, the Go implementation of jq In this entry, the functions ffr and ffs are defined as per the task requirements, but neither is used. Instead, for efficiency, a function for extending the two sequences is defined. This function is then used to create generators, which are then harnessed to accomplish the specific tasks. def init: {r: [0, 1], s: [0, 2] }; # input: {r,s} # output: {r,s,emit} where .emit is either null or the next R and where either .r or .s on output has been extended. # .emit is provided in case an unbounded stream of R values is desired. def extend_ff: (.r|length) as$rn
| if .s[$rn - 1] then .emit = .r[$rn - 1] + .s[$rn - 1] | .r[$rn] = .emit
| reduce range( [.r[$rn-1], .s[-1]] | max + 1; .r[$rn] ) as $i (.; .s += [$i] )
else .emit = null
| .s += [.r[$rn - 1] + 1] end; def ffr($n):
first(init | while(true; extend_ff) | select(.r[$n])).r[$n] ;

def ffs($n): first(init | while(true; extend_ff) | select(.s[$n])).s[$n] ; def task1($n):
"The first \($n) values of R are:", (init | until( .r | length >$n; extend_ff) | .r[1:]) ;

"The result of checking that the first 40 values of R and the first 960 of S together cover the interval [1,1000] is:",
( init | until( (.r|length) > 40 and (.s|length) > 960; extend_ff)
| (.r[1:41] + .s[1:961] | sort) == [range(1;1001)] ) ;

task1(10), task2
Output:
The first 10 values of R are:
[1,3,7,12,18,26,35,45,56,69]
The result of checking that the first 40 values of R and the first 960 of S together cover the interval [1,1000] is:
true


## 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

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


Main

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.")

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.

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") }  ## 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]]  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}  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  ## 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.  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;  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 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"  Output: /home/deen/git/nim-unsorted/hofstadter 1 3 7 12 18 26 35 45 56 69 All Integers 1..1000 found OK ## 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 ; 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. #!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";
}


## Phix

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

with javascript_semantics
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)
printf(1,"The first ten values of R: %v\n",{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

(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 ) ) ) )

Test:

: (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

## PL/I

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;

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

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;

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:

   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');

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

:- use_module(library(chr)).

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


 ?- 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)
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)].


hofstadter :-
% 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.


 ?- hofstadter.
ok
false.


## 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:

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

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")


output

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


### Using cyclic iterators

Defining R and S as mutually recursive generators. Follows directly from the definition of the R and S sequences.

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))

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


## Quackery

from, index, and end are defined at Loops/Increment loop index within loop body#Quackery.

As with the Phix solution, initialising to the first few elements simplified things significantly.

  [ ' [ 1 3 7 ]
' [ 2 4 5 6 ] ]      is initialise (       --> r s   )

[ over size 1 -
over swap peek
dip [ over -1 peek ]
+ swap dip join
over -2 split nip do
temp put
1 + from
[ temp share
index = iff
end done
index join ]
temp release ]       is extend     (   r s --> r s   )

[ temp put
[ over size
temp share < while
extend again ]
over
temp take 1 - peek ] is ffr        ( r s n --> r s n )

[ temp put
[ dup size
temp share < while
extend again ]
dup
temp take 1 - peek ] is ffs        ( r s n --> r s n )

initialise
say "R(1)..R(10): "
10 times
[ i^ 1+ ffr echo sp ]
cr cr
960 ffs drop
960 split drop
dip [ 40 split drop ]
join sort
[] 1000 times
[ i^ 1+ join ]
!=
say "All integers from 1 to 1000"
if say " not"
say " found once and only once."
Output:
R(1)..R(10): 1 3 7 12 18 26 35 45 56 69

All integers from 1 to 1000 found once and only once.

## 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.

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])
}

invisible(ffr(10))
print(rValues)

#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])))

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


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.

(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))))


Tests:

(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"))


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
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..$n), max(%s.values)+1..*)[0] } my @ffr = map &ffr, 1..*; my @ffs = map &ffs, 1..*; say @ffr[^10]; say "Rawks!" if 1...1000 eqv sort |@ffr[^40], |@ffs[^960];  Output: 1 3 7 12 18 26 35 45 56 69 Rawks! ## REXX ### version 1 This REXX example makes use of sparse arrays. Over a third of the program was for verification of the first thousand numbers in the Hofstadter Figure-Figure sequences. This REXX version is over 15,000% faster than REXX version 2. /*REXX program calculates and verifies the Hofstadter Figure─Figure sequences. */ parse arg x top bot . /*obtain optional arguments from the CL*/ if x=='' | x=="," then x= 10 /*Not specified? Then use the default.*/ if top=='' | top=="," then top=1000 /* " " " " " " */ if bot=='' | bot=="," then bot= 40 /* " " " " " " */ low=1; if x<0 then low=abs(x) /*only display a single │X│ value? */ r.=0; r.1=1; rr.=r.; rr.1=1; s.=r.; s.1=2 /*initialize the R, RR, and S arrays.*/ errs=0 /*the number of errors found (so far).*/ do i=low to abs(x) /*display the 1st X values of R & S.*/ say right('R('i") =",20) right(FFR(i),7) right('S('i") =",20) right(FFS(i),7) end /*i*/ /* [↑] list the 1st X Fig─Fig numbers.*/ if x<1 then exit /*if X isn't positive, then we're done.*/$.=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


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

/* 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

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 = []
for n = 1 to 10
hofr[n+1] = hofr[n] + hofs[n]
if n = 1
else
for p = hofr[n] + 1 to hofr[n+1] - 1
if p != hofs[n]
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

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


## RPL

Works with: Halcyon Calc version 4.2.8
RPL code Comment
 ≪ { 1 3 } 'R' STO { 2 } 'S' STO
≫ 'INITFF' STO

≪
S DUP SIZE GET
DO 1 + UNTIL R OVER POS NOT END
S OVER + 'S' STO
R DUP SIZE GET + R SWAP + 'R' STO
≫ 'NXTFF' STO

≪
WHILE R SIZE OVER < REPEAT NXTFF END
R SWAP GET
≫ 'FFR' STO

≪
WHILE S SIZE OVER < REPEAT NXTFF END
S SWAP GET
≫ 'FFS' STO

≪ INITFF
40 FFR DROP R
960 FFS DROP S +
1 SF 1 1000 FOR j
IF DUP j POS NOT THEN 1 CF END NEXT DROP
1 FS? "Passed" "Failed" IFTE

INITFF ( -- )
Initialize R(1..2) and S(1)

NXTFF ( -- )
n = last stored item of S()
n += 1 until n not in R()
append n to S()
append (n + last item of R()) to R()

FFR ( n -- R(n) )
if R(n) not stored, develop R()
Get R(n)

FFS ( n -- S(n) )
if S(n) not stored, develop S()
Get S(n)

Get R(40) and put R(1..40) in stack
Get S(960), append S(1..960) to R(1..40)
set flag ; for j=1 to 1000
if j not in the merged list then clear flag
Flag is still set iff all 1..1000 were in list once


Input:
10 FFR DROP R

Output:
2: { 1 3 7 12 18 26 35 45 56 69 }
1: "Passed"


## Ruby

Translation of: Tcl
$r = [nil, 1]$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'

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


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
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

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


## 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");
}

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
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())
}


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

## SETL

program figure_figure;
init R := [1], S := [2];

print("R(1..10):", [ffr(n) : n in [1..10]]);
print("R(1..40) + S(1..960) = {1..1000}:",
{ffr(n) : n in {1..40}} + {ffs(n) : n in {1..960}} = {1..1000});

proc ffr(n);
loop while n > #R do
nextR := R(#R) + S(#R);
S +:= [S(#S)+1 .. nextR-1];
R with:= nextR;
S with:= nextR + 1;
end loop;
return R(n);
end proc;

proc ffs(n);
loop while n > #S do
ffr(#R + 1);
end loop;
return S(n);
end proc;
end program;
Output:
R(1..10): [1 3 7 12 18 26 35 45 56 69]
R(1..40) + S(1..960) = {1..1000}: #T

## Sidef

Translation of: Perl
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()

{|i| seen{ffr(i)} := 0 ++ } << 1..40
{|i| seen{ffs(i)} := 0 ++ } << 1..960

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}"
}

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)
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}}]"


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.

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));
Next

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

For x = 1 To 10
Print " ";FUNC(_Sx(x));
Next

Print : Print                          ' Terminate and skip line

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
Next

For x = 1 to 40                        ' Prove there are only 40 R(x) values
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)

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

Private Function ffr(n As Long) As Long
Dim R As New Collection
Dim S As New Collection
'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
Next j
For j = R(i) + 1 To R(i) + S(i - 1)
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
'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
Next j
For j = R(i) + 1 To R(i) + S(i - 1)
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
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

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

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

'Set initial values of r.

'Set initial values of s.

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

'r function
Function ffr(n)
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
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
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

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
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).")

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

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

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");
Output:
True<-- should be True
`