Hofstadter Figure-Figure sequences: Difference between revisions

Content added Content deleted

Inline

Revision as of 12:57, 21 November 2016

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.

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>

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,...

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++

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>

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

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

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>

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 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 n = rl !! n where

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

main = do

   print $ map ffr [1 .. 10]
   let i1000 = sort (map ffr [1 .. 40] ++ map 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 = scanl (+) 1 s s = 2:4:tail (compliment (tail r)) where compliment = concat.interval interval x = zipWith (\x y -> [x+1..y-1]) x (tail x)

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>package hofstadter

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 =

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

proc ffs(n): 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>

Perl 6

<lang perl6>my @ffr; my @ffs;

@ffr.plan: 0, 1, gather take @ffr[$_] + @ffs[$_] for 1..*; @ffs.plan: 0, 2, 4..6, gather take @ffr[$_] ^..^ @ffr[$_+1] for 3..*;

say @ffr[1..10];

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

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

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

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

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

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

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"); 10.of { |i|

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

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

var seen = Hash()

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

if (seen.count {|k,v| (k.to_i >= 1) && (k.to_i <= 1000) && (v == 1) } == 1000) {

   say "All occured exactly once.";

} else {

   var missed = (1..1000 -> grep { !seen.has_key(_) })
   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

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!!!

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

Output:

True<-- should be True

@@ Line 2: / Line 2: @@
 These two sequences of positive integers are defined as:
-:::: <big><big><math> \begin{align}
+:::: <big><math>\begin{align}
-  R(1)&=1\ \\
+R(1)&=1\ ;\ S(1)=2 \\
-  S(1)&=2\ \\
+R(n)&=R(n-1)+S(n-1), \quad n>1.
+\end{align}</math></big>
-  R(n)&=R(n-1)+S(n-1), \quad n>1.
-\end{align} </math></big></big>
 <br>
-The sequence &nbsp; <big><big><math> S(n) </math></big></big> &nbsp; is further defined as the sequence of positive integers &nbsp; ''not'' &nbsp; present in &nbsp; <big><big><math> R(n).</math></big></big>
+The sequence <big><math>S(n)</math></big> is further defined as the sequence of positive integers '''''not''''' present in <big><math>R(n)</math></big>.
-Sequence &nbsp; <big><big><math> R </math></big></big> &nbsp; starts:
+Sequence <big><math>R</math></big> starts:
 , 3, 7, 12, 18, ...
-Sequence &nbsp; <big><big><math> S </math></big></big> &nbsp; starts:
+Sequence <big><math>S</math></big> starts:
 , 4, 5, 6, 8, ...
 ;Task:
-# Create two functions named &nbsp; <big> '''ffr''' </big> &nbsp;  and &nbsp; <big> '''ffs''' </big> &nbsp; that when given &nbsp; <big><big><math> n </math></big></big> &nbsp; return &nbsp; <big><big><math> R(n) </math></big></big> &nbsp; or &nbsp; <big><big><math> S(n) </math></big></big> &nbsp; respectively. <br>(Note that &nbsp; R(1) = 1 &nbsp; and &nbsp; S(1) = 2 &nbsp; to avoid off-by-one errors).
+# Create two functions named '''ffr''' and '''ffs''' that when given '''n''' return '''R(n)''' or '''S(n)''' respectively.<br>(Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).
-# No maximum value for &nbsp; <big><big><math> n </math></big></big> &nbsp; should be assumed.
+# No maximum value for '''n''' should be assumed.
-# Calculate and show that the first ten values of &nbsp; <big><big><math> R </math></big></big> &nbsp; are: <br>1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
+# Calculate and show that the first ten values of '''R''' are:<br> 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
-# Calculate and show that the first 40 values of &nbsp; <big> '''ffr''' </big> &nbsp; plus the first 960 values of &nbsp; <big> '''ffs''' </big> &nbsp; include all the integers from '''1''' to '''1000''' exactly once.
+# 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.