# Fractran

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

FRACTRAN is a Turing-complete esoteric programming language invented by the mathematician John Horton Conway.

A FRACTRAN program is an ordered list of positive fractions ${\displaystyle P=(f_{1},f_{2},\ldots ,f_{m})}$, together with an initial positive integer input ${\displaystyle n}$.

The program is run by updating the integer ${\displaystyle n}$ as follows:

• for the first fraction, ${\displaystyle f_{i}}$, in the list for which ${\displaystyle nf_{i}}$ is an integer, replace ${\displaystyle n}$ with ${\displaystyle nf_{i}}$ ;
• repeat this rule until no fraction in the list produces an integer when multiplied by ${\displaystyle n}$, then halt.

Conway gave a program for primes in FRACTRAN:

${\displaystyle 17/91}$, ${\displaystyle 78/85}$, ${\displaystyle 19/51}$, ${\displaystyle 23/38}$, ${\displaystyle 29/33}$, ${\displaystyle 77/29}$, ${\displaystyle 95/23}$, ${\displaystyle 77/19}$, ${\displaystyle 1/17}$, ${\displaystyle 11/13}$, ${\displaystyle 13/11}$, ${\displaystyle 15/14}$, ${\displaystyle 15/2}$, ${\displaystyle 55/1}$

Starting with ${\displaystyle n=2}$, this FRACTRAN program will change ${\displaystyle n}$ to ${\displaystyle 15=2\times (15/2)}$, then ${\displaystyle 825=15\times (55/1)}$, generating the following sequence of integers:

${\displaystyle 2}$, ${\displaystyle 15}$, ${\displaystyle 825}$, ${\displaystyle 725}$, ${\displaystyle 1925}$, ${\displaystyle 2275}$, ${\displaystyle 425}$, ${\displaystyle 390}$, ${\displaystyle 330}$, ${\displaystyle 290}$, ${\displaystyle 770}$, ${\displaystyle \ldots }$

After 2, this sequence contains the following powers of 2:

${\displaystyle 2^{2}=4}$, ${\displaystyle 2^{3}=8}$, ${\displaystyle 2^{5}=32}$, ${\displaystyle 2^{7}=128}$, ${\displaystyle 2^{11}=2048}$, ${\displaystyle 2^{13}=8192}$, ${\displaystyle 2^{17}=131072}$, ${\displaystyle 2^{19}=524288}$, ${\displaystyle \ldots }$

which are the prime powers of 2.

Write a program that reads a list of fractions in a natural format from the keyboard or from a string, to parse it into a sequence of fractions (i.e. two integers), and runs the FRACTRAN starting from a provided integer, writing the result at each step. It is also required that the number of step is limited (by a parameter easy to find).

Extra credit

Use this program to derive the first 20 or so prime numbers.

For more on how to program FRACTRAN as a universal programming language, see:

• J. H. Conway (1987). Fractran: A Simple Universal Programming Language for Arithmetic. In: Open Problems in Communication and Computation, pages 4–26. Springer.
• J. H. Conway (2010). "FRACTRAN: A simple universal programming language for arithmetic". In Jeffrey C. Lagarias. The Ultimate Challenge: the 3x+1 problem. American Mathematical Society. pp. 249–264. ISBN 978-0-8218-4940-8. Zbl 1216.68068.

with Ada.Text_IO; procedure Fractan is    type Fraction is record Nom: Natural; Denom: Positive; end record;   type Frac_Arr is array(Positive range <>) of Fraction;    function "/" (N: Natural; D: Positive) return Fraction is      Frac: Fraction := (Nom => N, Denom => D);   begin      return Frac;   end "/";    procedure F(List: Frac_Arr; Start: Positive; Max_Steps: Natural) is      N: Positive := Start;      J: Positive;   begin      Ada.Text_IO.Put(" 0:" & Integer'Image(N) & "   ");      for I in 1 .. Max_Steps loop	 J := List'First;	 loop	    if N mod List(J).Denom = 0 then	       N := (N/List(J).Denom) * List(J).Nom;	       exit; -- found fraction	    elsif J >= List'Last then	       return; -- did try out all fractions	    else	       J := J + 1; -- try the next fraction	    end if;	 end loop;	 Ada.Text_IO.Put(Integer'Image(I) & ":" & Integer'Image(N) & "   ");      end loop;   end F; begin   -- F((2/3, 7/2, 1/5, 1/7, 1/9, 1/4, 1/8), 2, 100);    -- output would be "0: 2    1: 7    2: 1" and then terminate    F((17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,       77/19,  1/17, 11/13, 13/11, 15/14,  15/2, 55/1),      2, 15);   -- output is "0: 2    1: 15    2: 825    3: 725   ...   14: 132    15: 116"end Fractan;
Output:
 0: 2    1: 15    2: 825    3: 725    4: 1925    5: 2275    6: 425    7: 390    8: 330    9: 290    10: 770    11: 910    12: 170    13: 156    14: 132    15: 116

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32
# as the numbers required for finding the first 20 primes are quite large, ## we use Algol 68G's LONG LONG INT with a precision of 100 digits          #PR precision 100 PR # mode to hold fractions #MODE FRACTION = STRUCT( INT numerator, INT denominator ); # define / between two INTs to yield a FRACTION #OP / = ( INT a, b )FRACTION: ( a, b ); # mode to define a FRACTRAN progam #MODE FRACTRAN = STRUCT( FLEX[0]FRACTION data                      , LONG LONG INT   n                      , BOOL            halted                      );# prepares a FRACTRAN program for use - sets the initial value of n and halted to FALSE #PRIO STARTAT = 1;OP   STARTAT = ( REF FRACTRAN f, INT start )REF FRACTRAN:BEGIN    halted OF f := FALSE;         n OF f := start;    fEND; # sets n OF f to the next number in the sequence or sets halted OF f to TRUE if the sequence has ended #OP NEXT = ( REF FRACTRAN f )LONG LONG INT:    IF halted OF f    THEN n OF f := 0    ELSE        BOOL          found  := FALSE;        LONG LONG INT result := 0;        FOR pos FROM LWB data OF f TO UPB data OF f WHILE NOT found DO            LONG LONG INT value       = n OF f * numerator OF ( ( data OF f )[ pos ] );            INT           denominator = denominator OF ( ( data OF f )[ pos ] );            IF found := ( value MOD denominator = 0 ) THEN result := value OVER denominator FI        OD;        IF NOT found THEN halted OF f := TRUE FI;        n OF f := result    FI ; # generate and print the sequence of numbers from a FRACTRAN pogram #PROC print fractran sequence = ( REF FRACTRAN f, INT start, INT limit )VOID:BEGIN    VOID( f STARTAT start );    print( ( "0: ", whole( start, 0 ) ) );    FOR i TO limit    WHILE VOID( NEXT f );          NOT halted OF f    DO        print( ( " " + whole( i, 0 ) + ": " + whole( n OF f, 0 ) ) )    OD;    print( ( newline ) )END ; # print the first 16 elements from the primes FRACTRAN program #FRACTRAN pf := ( ( 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14,  15/2, 55/1 ), 0, FALSE );print fractran sequence( pf, 2, 15 ); # find some primes using the pf FRACTRAN progam - n is prime for the members in the sequence that are 2^n #INT primes found := 0;VOID( pf STARTAT 2 );INT pos := 0;print( ( "seq position  prime sequence value", newline ) );WHILE primes found < 20 AND NOT halted OF pf DO    LONG LONG INT value      := NEXT pf;    INT      power of 2 := 0;    pos +:= 1;    WHILE value MOD 2 = 0 AND value > 0 DO power of 2 PLUSAB 1; value OVERAB 2 OD;    IF value = 1 THEN        # found a prime #        primes found +:= 1;        print( ( whole( pos, -12 ) + " " + whole( power of 2, -6 ) + " (" + whole( n OF pf, 0 ) + ")", newline ) )    FIOD
Output:
0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116
seq position  prime sequence value
19      2 (4)
69      3 (8)
280      5 (32)
707      7 (128)
2363     11 (2048)
3876     13 (8192)
8068     17 (131072)
11319     19 (524288)
19201     23 (8388608)
36866     29 (536870912)
45551     31 (2147483648)
75224     37 (137438953472)
101112     41 (2199023255552)
117831     43 (8796093022208)
152025     47 (140737488355328)
215384     53 (9007199254740992)
293375     59 (576460752303423488)
327020     61 (2305843009213693952)
428553     67 (147573952589676412928)
507519     71 (2361183241434822606848)


n := 2, steplimit := 15, numerator := [], denominator := []s := "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1" Loop, Parse, s, % A_Space    if (!RegExMatch(A_LoopField, "^(\d+)/(\d+)$", m)) MsgBox, % "Invalid input string (" A_LoopField ")." else numerator[A_Index] := m1, denominator[A_Index] := m2 SetFormat, FloatFast, 0.0Gui, Add, ListView, R10 W100 -Hdr, |SysGet, VSBW, 2LV_ModifyCol(1, 95 - VSBW), LV_Add( , 0 ": " n)Gui, Show Loop, % steplimit { i := A_Index Loop, % numerator.MaxIndex() if (!Mod(nn := n * numerator[A_Index] / denominator[A_Index], 1)) { LV_Modify(LV_Add( , i ": " (n := nn)), "Vis") continue, 2 } break} Output: 0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 15: 116 ## bash Works with: Bash the "factor" command allows one to decrypt the data. For example, the program below computes the product of a and b, entered as 2a and 3b, the product being 5a×b. Two arrays are computed from the fractions, ns for the numerators and ds for the denominators. Then, every time where the multiplication by a fraction yields an integer, the output of the division is stored into a csv file in factored format. #! /bin/bashprogram="1/1 455/33 11/13 1/11 3/7 11/2 1/3"echo$program | tr " " "\n" | cut -d"/" -f1 | tr "\n" " " > "data"read -a ns < "data"echo $program | tr " " "\n" | cut -d"/" -f2 | tr "\n" " " > "data"read -a ds < "data" t=0n=72echo "steps of computation" > steps.csvwhile [$t -le 6 ]; do	if [ $(($n*${ns[$t]}%${ds[$t]})) -eq 0 ]; then		let "n=$(($n*${ns[$t]}/${ds[$t]}))"		let "t=0"		factor $n >> steps.csv fi let "t=$t+1"done

If at the beginning n=72=23×32 (to compute 3×2), the steps of the computation look like this:

steps of computation
72: 2 2 2 3 3
396: 2 2 3 3 11
5460: 2 2 3 5 7 13
4620: 2 2 3 5 7 11
63700: 2 2 5 5 7 7 13
53900: 2 2 5 5 7 7 11
4900: 2 2 5 5 7 7
2100: 2 2 3 5 5 7
900: 2 2 3 3 5 5
4950: 2 3 3 5 5 11
68250: 2 3 5 5 5 7 13
57750: 2 3 5 5 5 7 11
796250: 2 5 5 5 5 7 7 13
673750: 2 5 5 5 5 7 7 11
61250: 2 5 5 5 5 7 7
26250: 2 3 5 5 5 5 7
11250: 2 3 3 5 5 5 5
61875: 3 3 5 5 5 5 11
853125: 3 5 5 5 5 5 7 13
721875: 3 5 5 5 5 5 7 11
9953125: 5 5 5 5 5 5 7 7 13
8421875: 5 5 5 5 5 5 7 7 11
765625: 5 5 5 5 5 5 7 7
328125: 3 5 5 5 5 5 5 7
140625: 3 3 5 5 5 5 5 5
46875: 3 5 5 5 5 5 5
15625: 5 5 5 5 5 5


This file can be opened with a spreadsheet to draw the successive states of the prime numbers (with countif) and then look how the computation is done in successive steps.

@echo offsetlocal enabledelayedexpansion	::Set the inputsset "code=17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"set "n=2"	::Basic validation of codefor %%. in (!code!) do (	echo.%%.|findstr /r /c:"^[0-9][0-9]*/[1-9][0-9]*$">nul||goto error_code) ::Validate the inputset /a "tst=1*!n!" 2>nulif !tst! lss 0 goto error_inputif !tst! equ 0 (if not "!n!"=="0" (goto error_input)) ::Set the limit outputs set limit=20 ::Execute the codeecho.Input:echo. !n!echo.Output:for /l %%? in (1,1,!limit!) do ( set shouldwehalt=1 for %%A in (!code!) do ( for /f "tokens=1,2 delims=/" %%B in ("%%A") do ( set /a "tst=!n! %% %%C" if !tst! equ 0 ( if !shouldwehalt! equ 1 ( set shouldwehalt=0 set /a "n=n*%%B/%%C" echo. !n! ) ) ) ) if !shouldwehalt! equ 1 goto halt) :haltecho.pauseexit /b 0 :error_codeecho.Syntax error in code.echo.pauseexit /b 1 :error_inputecho.Invalid input.echo.pauseexit /b 1 Output: Input: 2 Output: 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 30 Press any key to continue . . . ## Bracmat This program computes the first twenty primes. It has to do almost 430000 iterations to arrive at the twentieth prime, so instead of immediately writing each number to the terminal, it adds it to a list. After the set number of iterations, the list of numbers is written to a text file numbers.lst (21858548 bytes), so you can inspect it. Because it takes some time to do all iterations, its is advisable to write the source code below in a text file 'fractran' and run it in batch mode in the background, instead of starting Bracmat in interactive mode and typing the program at the prompt. The primes, together with the largest number found, are written to a file FRACTRAN.OUT. (fractran= np n fs A Z fi P p N L M. !arg:(?N,?n,?fs) {Number of iterations, start n, fractions} & :?P:?L {Initialise accumulators.} & whl ' ( -1+!N:>0:?N {Stop when counted down to zero.} & !n !L:?L {Prepend all numbers to result list.} & (2\L!n:#?p&!P !p:?P|) {If log2(n) is rational, append it to list of primes.} & !fs:? (/?fi&!n*!fi:~/:?n) ? {This line does the following (See task description): "for the first fraction, fi, in the list for which nfi is an integer, replace n by nfi ;"} ) & :?M & whl'(!L:%?n ?L&!n !M:?M) {Invert list of numbers. (append to long list is very expensive. Better to prepend and finally invert.} & (!M,!P) {Return the two lists}); ( clk$:?t0& fractran$(430000, 2, 17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1) : (?numbers,?primes)& lst$(numbers,"numbers.lst",NEW)& put$("FRACTRAN found these primes:" !primes "\nThe list of numbers is saved in numbers.txtThe biggest number in the list is" ( 0:?max & !numbers:? (>%@!max:?max&~) ? | !max )str$("\ntime: " flt$(clk$+-1*!t0,4) " sec\n"), "FRACTRAN.OUT",NEW));

In Linux, run the program as follows (assuming bracmat and the file 'fractran' are in the CWD):

./bracmat 'get$fractran' & Output: in FRACTRAN.OUT: FRACTRAN found these primes: 1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 The list of numbers is saved in numbers.txt The biggest number in the list is 1842775069354845065175076326808495219647033145169559640049770986129640260031692378106527030467987060546875 time: 1,8668*10E3 sec  ## C Using GMP. Powers of two are in brackets. For extra credit, pipe the output through | less -S. #include <stdio.h>#include <stdlib.h>#include <gmp.h> typedef struct frac_s *frac;struct frac_s { int n, d; frac next;}; frac parse(char *s){ int offset = 0; struct frac_s h = {0}, *p = &h; while (2 == sscanf(s, "%d/%d%n", &h.n, &h.d, &offset)) { s += offset; p = p->next = malloc(sizeof *p); *p = h; p->next = 0; } return h.next;} int run(int v, char *s){ frac n, p = parse(s); mpz_t val; mpz_init_set_ui(val, v); loop: n = p; if (mpz_popcount(val) == 1) gmp_printf("\n[2^%d = %Zd]", mpz_scan1(val, 0), val); else gmp_printf(" %Zd", val); for (n = p; n; n = n->next) { // assuming the fractions are not reducible if (!mpz_divisible_ui_p(val, n->d)) continue; mpz_divexact_ui(val, val, n->d); mpz_mul_ui(val, val, n->n); goto loop; } gmp_printf("\nhalt: %Zd has no divisors\n", val); mpz_clear(val); while (p) { n = p->next; free(p); p = n; } return 0;} int main(void){ run(2, "17/91 78/85 19/51 23/38 29/33 77/29 95/23 " "77/19 1/17 11/13 13/11 15/14 15/2 55/1"); return 0;} ## C++  #include <iostream>#include <sstream>#include <iterator>#include <vector>#include <cmath> using namespace std; class fractran{public: void run( std::string p, int s, int l ) { start = s; limit = l; istringstream iss( p ); vector<string> tmp; copy( istream_iterator<string>( iss ), istream_iterator<string>(), back_inserter<vector<string> >( tmp ) ); string item; vector< pair<float, float> > v; pair<float, float> a; for( vector<string>::iterator i = tmp.begin(); i != tmp.end(); i++ ) { string::size_type pos = ( *i ).find( '/', 0 ); if( pos != std::string::npos ) { a = make_pair( atof( ( ( *i ).substr( 0, pos ) ).c_str() ), atof( ( ( *i ).substr( pos + 1 ) ).c_str() ) ); v.push_back( a ); } } exec( &v ); } private: void exec( vector< pair<float, float> >* v ) { int cnt = 0; while( cnt < limit ) { cout << cnt << " : " << start << "\n"; cnt++; vector< pair<float, float> >::iterator it = v->begin(); bool found = false; float r; while( it != v->end() ) { r = start * ( ( *it ).first / ( *it ).second ); if( r == floor( r ) ) { found = true; break; } ++it; } if( found ) start = ( int )r; else break; } } int start, limit;};int main( int argc, char* argv[] ){ fractran f; f.run( "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2, 15 ); cin.get(); return 0;}  Output: 0 : 2 1 : 15 2 : 825 3 : 725 4 : 1925 5 : 2275 6 : 425 7 : 390 8 : 330 9 : 290 10 : 770 11 : 910 12 : 170 13 : 156 14 : 132  ## Common Lisp (defun fractran (n frac-list) (lambda () (prog1 n (when n (let ((f (find-if (lambda (frac) (integerp (* n frac))) frac-list))) (when f (setf n (* f n)))))))) ;; test (defvar *primes-ft* '(17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1)) (loop with fractran-instance = (fractran 2 *primes-ft*) repeat 20 for next = (funcall fractran-instance) until (null next) do (print next)) Output: 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 ## D ### Simple Version Translation of: Java import std.stdio, std.algorithm, std.conv, std.array; void fractran(in string prog, int val, in uint limit) { const fracts = prog.split.map!(p => p.split("/").to!(int[])).array; foreach (immutable n; 0 .. limit) { writeln(n, ": ", val); const found = fracts.find!(p => val % p[1] == 0); if (found.empty) break; val = found.front[0] * val / found.front[1]; }} void main() { fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2, 15);} Output: 0: 2 1: 15 2: 825 3: 725 4: 1925 5: 2275 6: 425 7: 390 8: 330 9: 290 10: 770 11: 910 12: 170 13: 156 14: 132 ### Lazy Version import std.stdio, std.algorithm, std.conv, std.array, std.range; struct Fractran { int front; bool empty = false; const int[][] fracts; this(in string prog, in int val) { this.front = val; fracts = prog.split.map!(p => p.split("/").to!(int[])).array; } void popFront() { const found = fracts.find!(p => front % p[1] == 0); if (found.empty) empty = true; else front = found.front[0] * front / found.front[1]; }} void main() { Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2) .take(15).writeln;} Output: [2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132] ## Elixir Translation of: Erlang defmodule Fractran do use Bitwise defp binary_to_ratio(b) do [_, num, den] = Regex.run(~r/(\d+)\/(\d+)/, b) {String.to_integer(num), String.to_integer(den)} end def load(program) do String.split(program) |> Enum.map(&binary_to_ratio(&1)) end defp step(_, []), do: :halt defp step(n, [f|fs]) do {p, q} = mulrat(f, {n, 1}) case q do 1 -> p _ -> step(n, fs) end end def exec(k, n, program) do exec(k-1, n, fn (_) -> true end, program, [n]) |> Enum.reverse end def exec(k, n, pred, program) do exec(k-1, n, pred, program, [n]) |> Enum.reverse end defp exec(0, _, _, _, steps), do: steps defp exec(k, n, pred, program, steps) do case step(n, program) do :halt -> steps m -> if pred.(m), do: exec(k-1, m, pred, program, [m|steps]), else: exec(k, m, pred, program, steps) end end def is_pow2(n), do: band(n, n-1) == 0 def lowbit(n), do: lowbit(n, 0) defp lowbit(n, k) do case band(n, 1) do 0 -> lowbit(bsr(n, 1), k + 1) 1 -> k end end # rational multiplication defp mulrat({a, b}, {c, d}) do {p, q} = {a*c, b*d} g = gcd(p, q) {div(p, g), div(q, g)} end defp gcd(a, 0), do: a defp gcd(a, b), do: gcd(b, rem(a, b))end primegen = Fractran.load("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1")IO.puts "The first few states of the Fractran prime automaton are:\n#{inspect Fractran.exec(20, 2, primegen)}\n"prime = Fractran.exec(26, 2, &Fractran.is_pow2/1, primegen) |> Enum.map(&Fractran.lowbit/1) |> tlIO.puts "The first few primes are:\n#{inspect prime}" Output: The first few states of the Fractran prime automaton are: [2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116, 308, 364, 68, 4] The first few primes are: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]  ## Erlang The exec() function can be passed a predicate which filters steps that satisfy a condition, which for the prime automata is a check to see if the number is a power of 2. #! /usr/bin/escript -mode(native).-import(lists, [map/2, reverse/1]). binary_to_ratio(B) -> {match, [_, Num, Den]} = re:run(B, "([0-9]+)/([0-9]+)"), {binary_to_integer(binary:part(B, Num)), binary_to_integer(binary:part(B, Den))}. load(Program) -> map(fun binary_to_ratio/1, re:split(Program, "[ ]+")). step(_, []) -> halt;step(N, [F|Fs]) -> {P, Q} = mulrat(F, {N, 1}), case Q of 1 -> P; _ -> step(N, Fs) end. exec(K, N, Program) -> reverse(exec(K - 1, N, fun (_) -> true end, Program, [N])).exec(K, N, Pred, Program) -> reverse(exec(K - 1, N, Pred, Program, [N])). exec(0, _, _, _, Steps) -> Steps;exec(K, N, Pred, Program, Steps) -> case step(N, Program) of halt -> Steps; M -> case Pred(M) of true -> exec(K - 1, M, Pred, Program, [M|Steps]); false -> exec(K, M, Pred, Program, Steps) end end. is_pow2(N) -> N band (N - 1) =:= 0. lowbit(N) -> lowbit(N, 0).lowbit(N, K) -> case N band 1 of 0 -> lowbit(N bsr 1, K + 1); 1 -> K end. main(_) -> PrimeGen = load("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"), io:format("The first few states of the Fractran prime automaton are: ~p~n~n", [exec(20, 2, PrimeGen)]), io:format("The first few primes are: ~p~n", [tl(map(fun lowbit/1, exec(26, 2, fun is_pow2/1, PrimeGen)))]). % rational multiplication mulrat({A, B}, {C, D}) -> {P, Q} = {A*C, B*D}, G = gcd(P, Q), {P div G, Q div G}. gcd(A, 0) -> A;gcd(A, B) -> gcd(B, A rem B).  Output: └─$ ▶ ./fractran.erl
The first few states of the Fractran prime automaton are: [2,15,825,725,1925,
2275,425,390,330,
290,770,910,170,
156,132,116,308,
364,68,4]

The first few primes are: [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,
67,71,73,79,83,89,97]



## Fortran

### The Plan

As ever, how long is a piece of string: what might be the upper limit for the list of fractions? Similarly, obtaining the source code to be interpreted is a nuisance since as ever, one doesn't know how long it might be and performing a scan to find out will require more messing about than the activity of interest. One could compile in the specified example, but this is not very flexible even though a re-compilation will be swift. So, a pre-emptive requirement: first specify the number of fractions to be read in. Then reading the fractions has its annoyances also (aside from copying from the example text carrying along all the .html formatting) since given 2/3 a slash is not a suitable delimiter for free-format input, indeed it signifies end-of-input. Again, one can devise an input procedure involving a suitable scan, with more code drowning the stuff specific to the task. So, replace the slashes with a space, an acceptable free-format delimiter. As a comma is also acceptable, use those between the fractions, so, 17 91, 78 85, 19 51, 23 38, 29 33, 77 29, 95 23, 77 19, 1 17, 11 13, 13 11, 15 14, 15 2, 55 1 Then, all that remains is to specify the starting number, and as well, a step limit. Naturally, the output can be in a more gracious form. It is presented as the input is read so that should something go awry there would be some indication of what was going on.

### The Code

The source style is F77 except for the use of the I0 format code, though not all F77 compilers will offer INTEGER*8. By not using the MODULE scheme, array parameters can't be declared via P(:) which implies a secret additional parameter giving the size of the array and which can be accessed via the likes of UBOUND(P, DIM = 1) Instead, the old-style specification involves no additional parameters and can be given as P(*) meaning "no statement" as to the upper bound, or P(M) which may be interpreted as the upper bound being the value of M in the compilers that allow this. The actual upper bound of the parameter is unknown and unchecked, so the older style of P(12345) or similar might be used. Rather to my surprise, this compiler (Compaq F90/95) complained if parameter M was declared after the arrays P(M),Q(M) as it is my habit to declare parameters in the order of their appearance.
C:\Nicky\RosettaCode\FRACTRAN\FRACTRAN.for(6) : Warning: This name has not been given an explicit type.   [M]       INTEGER P(M),Q(M)!The terms of the fractions.

So much for multi-pass compilers!

Similarly, without the MODULE protocol, in all calling routines function FRACTRAN would be deemed floating-point so a type declaration is needed in each.
      INTEGER FUNCTION FRACTRAN(N,P,Q,M)	!Notion devised by J. H. Conway.Careful: the rule is N*P/Q being integer. N*6/3 is integer always because this is N*2/1, but 3 may not divide N.Could check GCD(P,Q), dividing out the common denominator so MOD(N,Q) works.       INTEGER*8 N	!The work variable. Modified!       INTEGER M	!The number of fractions supplied.       INTEGER P(M),Q(M)!The terms of the fractions.       INTEGER I	!A stepper.        DO I = 1,M	!Search the supplied fractions, P(i)/Q(i).          IF (MOD(N,Q(I)).EQ.0) THEN	!Does the denominator divide N?            N = N/Q(I)*P(I)	!Yes, compute N*P/Q but trying to dodge overflow.            FRACTRAN = I	!Report the hit.           RETURN		!Done!          END IF	!Otherwise,        END DO		!Try the next fraction in the order supplied.        FRACTRAN = 0	!No hit.      END FUNCTION FRACTRAN	!That's it! Even so, "Turing complete"...       PROGRAM POKE      INTEGER FRACTRAN		!Not the default type of function.      INTEGER P(66),Q(66)	!Holds the fractions as P(i)/Q(i).      INTEGER*8 N		!The working number.      INTEGER I,IT,L,M		!Assistants.       WRITE (6,1)	!Announce.    1 FORMAT ("Interpreter for J.H. Conway's FRACTRAN language.") Chew into an example programme.      OPEN (10,FILE = "Fractran.txt",STATUS="OLD",ACTION="READ")	!Rather than compiled-in stuff.      READ (10,*) L	!I need to know this without having to scan the input.      WRITE (6,2) L	!Reveal in case of trouble.    2 FORMAT (I0," fractions, as follow:")	!Should the input evoke problems.      READ (10,*) (P(I),Q(I),I = 1,L)	!Ask for the specified number of P,Q pairs.      WRITE (6,3) (P(I),Q(I),I = 1,L)	!Show what turned up.    3 FORMAT (24(I0,"/",I0:", "))	!As P(i)/Q(i) pairs. The colon means that there will be no trailing comma.      READ (10,*) N,M			!The start value, and the step limit.      CLOSE (10)			!Finished with input.      WRITE (6,4) N,M			!Hopefully, all went well.    4 FORMAT ("Start with N = ",I0,", step limit ",I0) Commence.      WRITE (6,10) 0,N		!Splat a heading.   10 FORMAT (/,"  Step  #F: N",/,I6,4X,": ",I0)	!Matched FORMAT 11.      DO I = 1,M		!Here we go!        IT = FRACTRAN(N,P,Q,L)		!Do it!        WRITE (6,11) I,IT,N		!Show it!   11   FORMAT (I6,I4,": ",I0)		!N last, as it may be big.        IF (IT.LE.0) EXIT		!No hit, so quit.      END DO			!The next step.      END	!Whee!

### The Results

Output:

Interpreter for J.H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1

Step  #F: N
0    : 2
1  13: 15
2  14: 825
3   5: 725
4   6: 1925
5  11: 2275
6   1: 425
7   2: 390
8  10: 330
9   5: 290
10   6: 770
11  11: 910
12   1: 170
13   2: 156
14  10: 132
15   5: 116
16   6: 308
17  11: 364
18   1: 68
19   9: 4
20  13: 30
21  13: 225
22  14: 12375
23   5: 10875
24   6: 28875
25   5: 25375
26   6: 67375
27  11: 79625
28   1: 14875

Later Fortrans might offer the library function POPCNT(n) which returns the number of on-bits in an integer, most convenient for detecting a straight power of two in a binary computer. Adjusting the interpretation loop to be
      DO I = 1,M		!Here we go!        IT = FRACTRAN(N,P,Q,L)		!Do it!        IF (POPCNT(N).EQ.1) WRITE (6,11) I,IT,N		!Show it!   11   FORMAT (I6,I4,": ",I0)		!N last, as it may be big.        IF (IT.LE.0) EXIT		!No hit, so quit.        IF (N.LE.0) THEN		!Otherwise, worry about overflow.          WRITE (6,*) "Integer overflow!"	!Justified. The test is not certain.          WRITE (6,11) I,IT,N			!Alas, the step failed.          EXIT					!Give in.        END IF				!So much for overflow.      END DO			!The next step.

Interpreter for J.H. Conway's FRACTRAN language.
14 fractions, as follow:
17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1

Step  #F: N
0    : 2
19   9: 4
69   9: 8
280   9: 32
707   9: 128
2363   9: 2048
Integer overflow!
2863   6: -3816591145242754741


And no, fraction #9 is not always the one with the power of two only, for instance step 130 with N = 24x7. So, even a 64-bit integer is not enough, and with a 32-bit integer,

 Integer overflow!
296   6: -1406264171


In the absence of a proper test such as the IF OVERFLOW ...  of First Fortran (1957), the check for overflow relies on the improper appearance of negative numbers during two's-complement binary arithmetic that should only produce positive numbers. This will catch about half the events: N*M might or might not leave the sign bit on when the result overflows.

### Revised Plan

One could introduce a scheme for multi-precision (or "bignum") arithmetic, but there is no need to maintain a unified number N because there are no additions or subtractions, only multiplications and division without remainders. In other words, why not mess about with collections of prime factors? That is, represent N, P, and Q as the list of prime numbers with their powers for each case. To calculate N*P, just add the list of prime number powers for P to that for N, and similarly, for N/Q subtract. At no stage (except perhaps to decorate output or for debugging) need the list be multiplied out to give the unified number and so there is no need for multi-precision multiplies and divides. To determine if N is divisible by Q (that is, if N*fraction = N*P/Q is integral), check only that the primes listed by Q have powers no greater than those of the corresponding primes for N.

To facilitate access without searching the list of primes for N, instead its list is represented by an array of powers, NPPOW, with most entries zero. Thus, NPPOW(i) has the power for PRIME(i) in the factorisation of N, and for N = 2, NPPOW(1) = 1 with all other elements zero. But for FP (the factorisation of P) and FQ (for Q) there is a proper list provided via type FACTORED, whereby FP.PLIST(0) is the count of prime factors and FP.PLIST(1:FP.PLIST(0)) fingers the prime numbers that are factors, with FP.PPOW(i) having their corresponding powers. Thus, FP.PLIST(2) has the index of the second prime factor of P (should it have so many), which is PRIME(FP.PLIST(2)), and its power is FP.PPOW(2). Accordingly, to determine if N (as NPPOW) is divisible by one of the fractions FQ, the appropriate elements of NPPOW (that give its powers) must be compared to the corresponding powers in FQ.PPOW, and if ALL of the powers in NPPOW fingered by FQ.PLIST are greater than or equal to those in FQ.PPOW, then a hit! For this purpose, the factorisation of a prime number includes that prime number. Note that the prime number factorisation of one has no elements in its list, and happily, the ALL operation applied for no tests yields true as desired, because one divides any number. Thus, in the FRACTRAN op-code system, a fraction P/1 is always a match, and no fractions beyond it will ever be considered.

As a part of the preparation, for each fraction the greatest common divisor is divided out to remove the possibility that converting the test N times fraction being integral via fraction = P/Q to Q divides N will behave differently. For example, N*6/3 will always be integral, but N may not be divisible by three. Reducing 6/3 to 2/1 however will work as will reducing 55/25 to 11/5. The example contains no such occasions, but the possibility nags.

For output, the value of N will not be shown multiplied out but via a schedule showing the powers of the first few prime numbers that form its factorisation. Rather than staring in puzzlement at opaque monster strings of digits, one can view each separate prime factor's power counting up and down as the calculation proceeds. A simple scan of all the factorisations soon determines the highest prime employed, and this never changes. An extension of this checks for which primes are omitted, and in this example, none are. However, a further extension modifies the output of the schedule of powers so as to blank out those that are zero. This could be achieved by replacing the likes of WRITE (...) NPPOW by WRITE (...) I6FMT(NPPOW) where function I6FMT writes out its integer with I6 format if the value is positive, and supplies six spaces if not, were it not that few Fortran systems allow such re-entrant usage of the system for formatted output. So, prepare the output with a straightforward WRITE to a CHARACTER variable, blank out the portions where unwanted values appear, and write the result.

### Revised Code

Because this scheme requires a supply of prime numbers, it is convenient to employ the routines prepared for the extensible prime generator via module PRIMEBAG. So, this means escalating to the F90 style, and given that, some compound data structures can be used (for better mnemonics) in place of collections of arrays.
       MODULE CONWAYSIDEA	!Notion devised by J. H. Conway.       USE PRIMEBAG		!This is a common need.       INTEGER LASTP,ENUFF	!Some size allowances.       PARAMETER (LASTP = 66, ENUFF = 66)	!Should suffice for the example in mind.       INTEGER NPPOW(1:LASTP)	!Represent N as a collection of powers of prime numbers.       TYPE FACTORED		!But represent P and Q of freaction = P/Q        INTEGER PNUM(0:LASTP)	!As a list of prime number indices with PNUM(0) the count.        INTEGER PPOW(LASTP)	!And the powers. for the fingered primes.       END TYPE FACTORED	!Rather than as a simple number multiplied out.       TYPE(FACTORED) FP(ENUFF),FQ(ENUFF)	!Thus represent a factored fraction, P(i)/Q(i).       INTEGER PLIVE(ENUFF),NL	!Helps subroutine SHOWN display NPPOW.       CONTAINS		!Now for the details.        SUBROUTINE SHOWFACTORS(N)	!First, to show an internal data structure.         TYPE(FACTORED) N	!It is supplied as a list of prime factors.         INTEGER I		!A stepper.          DO I = 1,N.PNUM(0)	!Step along the list.            IF (I.GT.1) WRITE (MSG,"('x',$)") !Append a glyph for "multiply". WRITE (MSG,"(I0,$)") PRIME(N.PNUM(I))	!The prime fingered in the list.            IF (N.PPOW(I).GT.1) WRITE (MSG,"('^',I0,$)") N.PPOW(I) !With an interesting power? END DO !On to the next element in the list. WRITE (MSG,1) N.PNUM(0) !End the line 1 FORMAT (": Factor count ",I0) !With a count of prime factors. END SUBROUTINE SHOWFACTORS !Hopefully, this will not be needed often. TYPE(FACTORED) FUNCTION FACTOR(IT) !Into a list of primes and their powers. INTEGER IT,N !The number and a copy to damage. INTEGER P,POW !A stepper and a power. INTEGER F,NF !A factor and a counter. IF (IT.LE.0) STOP "Factor only positive numbers!" !Or else... N = IT !A copy I can damage. NF = 0 !No factors found. P = 0 !Because no primes have been tried. PP:DO WHILE (N.GT.1) !Step through the possibilities. P = P + 1 !Another prime impends. F = PRIME(P) !Grab a possible factor. POW = 0 !It has no power yet. FP:DO WHILE(MOD(N,F).EQ.0) !Well? POW = POW + 1 !Count a factor.. N = N/F !Reduce the number. END DO FP !The P'th prime's power's produced. IF (POW.GT.0) THEN !So, was it a factor? IF (NF.GE.LASTP) THEN !Yes. Have I room in the list? WRITE (MSG,1) IT,LASTP !Alas. 1 FORMAT ("Factoring ",I0," but with provision for only ", 1 I0," prime factors!") FACTOR.PNUM(0) = NF !Place the count so far, CALL SHOWFACTORS(FACTOR)!So this can be invoked. STOP "Not enough storage!" !Quite. END IF !But normally, NF = NF + 1 !Admit another factor. FACTOR.PNUM(NF) = P !Identify the prime. NOT the prime itself. FACTOR.PPOW(NF) = POW !Place its power. END IF !So much for that factor. END DO PP !Try another prime, if N > 1 still. FACTOR.PNUM(0) = NF !Place the count. END FUNCTION FACTOR !Thus, a list of primes and their powers. INTEGER FUNCTION GCD(I,J) !Greatest common divisor. INTEGER I,J !Of these two integers. INTEGER N,M,R !Workers. N = MAX(I,J) !Since I don't want to damage I or J, M = MIN(I,J) !These copies might as well be the right way around. 1 R = MOD(N,M) !Divide N by M to get the remainder R. IF (R.GT.0) THEN !Remainder zero? N = M !No. Descend a level. M = R !M-multiplicity has been removed from N. IF (R .GT. 1) GO TO 1 !No point dividing by one. END IF !If R = 0, M divides N. GCD = M !There we are. END FUNCTION GCD !Euclid lives on! INTEGER FUNCTION FRACTRAN(L) !Applies Conway's idea to a list of fractions.Could abandon all parameters since global variables have the details... INTEGER L !The last fraction to consider. INTEGER I,NF !Assistants. DO I = 1,L !Step through the fractions in the order they were given. NF = FQ(I).PNUM(0) !How many factors are listed in FQ(I)? IF (ALL(NPPOW(FQ(I).PNUM(1:NF)) !Can N (as NPPOW) be divided by Q (as FQ)? 1 .GE. FQ(I).PPOW(1:NF))) THEN !By comparing the supplies of prime factors. FRACTRAN = I !Yes! NPPOW(FQ(I).PNUM(1:NF)) = NPPOW(FQ(I).PNUM(1:NF)) !Remove prime powers from N 1 - FQ(I).PPOW(1:NF) !Corresponding to Q. NF = FP(I).PNUM(0) !Add powers to N NPPOW(FP(I).PNUM(1:NF)) = NPPOW(FP(I).PNUM(1:NF)) !Corresponding to P. 1 + FP(I).PPOW(1:NF) !Thus, N = N/Q*P. RETURN !That's all it takes! No multiplies nor divides! END IF !So much for that fraction. END DO !This relies on ALL(zero tests) yielding true, as when Q = 1. FRACTRAN = 0 !No hit. END FUNCTION FRACTRAN !No massive multi-precision arithmetic! SUBROUTINE SHOWN(S,F) !Service routine to show the state after a step is calculated.Could imaging a function I6FMT(23) that returns " 23" and " " for non-positive numbers.Can't do it, as if this were invoked via a WRITE statement, re-entrant use of WRITE usually fails. INTEGER S,F !Step number, Fraction number. INTEGER I !A stepper. CHARACTER*(9+4+1 + NL*6) ALINE !A scratchpad matching FORMAT 103. WRITE (ALINE,103) S,F,NPPOW(PLIVE(1:NL)) !Show it! 103 FORMAT (I9,I4,":",<NL>I6) !As a sequence of powers of primes. IF (F.LE.0) ALINE(10:13) = "" !Scrub when no fraction is fingered. DO I = 1,NL !Step along the live primes. IF (NPPOW(PLIVE(I)).GT.0) CYCLE !Ignoring the empowered ones. ALINE(15 + (I - 1)*6:14 + I*6) = "" !Blank out zero powers. END DO !On to the next. WRITE (MSG,"(A)") ALINE !Reveal at last. END SUBROUTINE SHOWN !A struggle. END MODULE CONWAYSIDEA !Simple... PROGRAM POKE USE CONWAYSIDEA !But, where does he get his ideas from? INTEGER P(ENUFF),Q(ENUFF) !Holds the fractions as P(i)/Q(i). INTEGER N !The working number. INTEGER LF !Last fraction given. INTEGER LP !Last prime needed. INTEGER MS !Maximum number of steps. INTEGER I,IT !Assistants. LOGICAL*1 PUSED(ENUFF) !Track the usage of prime numbers, MSG = 6 !Standard output. WRITE (6,1) !Announce. 1 FORMAT ("Interpreter for J. H. Conway's FRACTRAN language.") Chew into an example programme. 10 OPEN (10,FILE = "Fractran.txt",STATUS="OLD",ACTION="READ") !Rather than compiled-in stuff. READ (10,*) LF !I need to know this without having to scan the input. WRITE (MSG,11) LF !Reveal in case of trouble. 11 FORMAT (I0," fractions, as follow:") !Should the input evoke problems. READ (10,*) (P(I),Q(I),I = 1,LF) !Ask for the specified number of P,Q pairs. WRITE (MSG,12) (P(I),Q(I),I = 1,LF) !Show what turned up. 12 FORMAT (24(I0,"/",I0:", ")) !As P(i)/Q(i) pairs. The colon means that there will be no trailing comma. READ (10,*) N,MS !The start value, and the step limit. CLOSE (10) !Finished with input. WRITE (MSG,13) N,MS !Hopefully, all went well. 13 FORMAT ("Start with N = ",I0,", step limit ",I0) IF (.NOT.GRASPPRIMEBAG(66)) STOP "Gan't grab my file of primes!" !Attempt in hope. Convert the starting number to a more convenient form, an array of powers of successive prime numbers. 20 FP(1) = FACTOR(N) !Borrow one of the factor list variables. NPPOW = 0 !Clear all prime factor counts. DO I = 1,FP(1).PNUM(0) !Now find what they are. NPPOW(FP(1).PNUM(I)) = FP(1).PPOW(I) !Convert from a variable-length list END DO !To a fixed-length random-access array. PUSED = NPPOW.GT.0 !Note which primes have been used. LP = FP(1).PNUM(FP(1).PNUM(0)) !Recall the last prime required. More later.Convert the supplied P(i)/Q(i) fractions to lists of prime number factors and powers in FP(i) and FQ(i). DO I = 1,LF !Step through the fractions. IT = GCD(P(I),Q(I)) !Suspicion. IF (IT.GT.1) THEN !Justified? WRITE (MSG,21) I,P(I),Q(I),IT !Alas. Complain. The rule is N*(P/Q) being integer. 21 FORMAT ("Fraction ",I3,", ",I0,"/",I0,!N*6/3 is integer always because this is N*2/1, but 3 may not divide N. 1 " has common factor ",I0,"!") !By removing IT, P(I) = P(I)/IT !The test need merely check if N is divisible by Q. Q(I) = Q(I)/IT !And, as N is factorised in NPPOW END IF !And Q in FQ, subtractions of powers only is needed. FP(I) = FACTOR(P(I)) !Righto, form the factor list for P. PUSED(FP(I).PNUM(1:FP(I).PNUM(0))) = .TRUE. !Mark which primes it fingers. LP = MAX(LP,FP(I).PNUM(FP(I).PNUM(0))) !One has no prime factors: PNUM(0) = 0. FQ(I) = FACTOR(Q(I)) !And likewise for Q. PUSED(FQ(I).PNUM(1:FQ(I).PNUM(0))) = .TRUE. !Some primes may be omitted. LP = MAX(LP,FQ(I).PNUM(FQ(I).PNUM(0))) !If no prime factors, PNUM(0) fingers element zero, which is zero. END DO !All this messing about saves on multiplication and division.Check which primes are in use, preparing an index of live primes.. NL = 0 !No live primes. DO I = 1,LP !Check up to the last prime. IF (PUSED(I)) THEN !This one used? NL = NL + 1 !Yes. Another. PLIVE(NL) = I !Fingered. END IF !So much for that prime. END DO !On to the next. WRITE (MSG,22) NL,LP,PRIME(LP) !Remark on usage. 22 FORMAT ("Require ",I0," primes only, up to Prime(",I0,") = ",I0) !Presume always more than one prime. IF (LP.GT.LASTP) STOP "But, that's too many for array NPPOW!" Cast forth a heading. 100 WRITE (MSG,101) (PRIME(PLIVE(I)), I = 1,NL) !Splat a heading. 101 FORMAT (/,14X,"N as powers of prime factors",/, !The prime heading, 1 5X,"Step F#:",<LP>I6) !With primes beneath. CALL SHOWN(0,0) !Initial state of N as NPPOW. Step zero, no fraction. Commence! DO I = 1,MS !Here we go! IT = FRACTRAN(LF) !Do it! CALL SHOWN(I,IT) !Show it! IF (IT.LE.0) EXIT !Quit it? END DO !The next step.Complete! END !Whee!  ### Revised Results Edited in are >>> markers for the prime powers of two. Interpreter for J. H. Conway's FRACTRAN language. 14 fractions, as follow: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1 Start with N = 2, step limit 100 Require 10 primes only, up to Prime(10) = 29 N as powers of prime factors Step F#: 2 3 5 7 11 13 17 19 23 29 0 : 1 1 13: 1 1 2 14: 1 2 1 3 5: 2 1 4 6: 2 1 1 5 11: 2 1 1 6 1: 2 1 7 2: 1 1 1 1 8 10: 1 1 1 1 9 5: 1 1 1 10 6: 1 1 1 1 11 11: 1 1 1 1 12 1: 1 1 1 13 2: 2 1 1 14 10: 2 1 1 15 5: 2 1 16 6: 2 1 1 17 11: 2 1 1 18 1: 2 1 >>> 19 9: 2 20 13: 1 1 1 21 13: 2 2 22 14: 2 3 1 23 5: 1 3 1 24 6: 1 3 1 1 25 5: 3 1 1 26 6: 3 2 1 27 11: 3 2 1 28 1: 3 1 1 29 2: 1 1 2 1 1 30 1: 1 1 2 1 31 2: 2 2 1 1 32 10: 2 2 1 1 33 5: 2 1 1 1 34 6: 2 1 1 1 1 35 5: 2 1 1 1 36 6: 2 1 2 1 37 11: 2 1 2 1 38 1: 2 1 1 1 39 2: 3 1 1 1 40 1: 3 1 1 41 3: 3 1 42 4: 2 1 43 7: 2 1 1 44 4: 1 1 1 45 7: 1 2 1 46 4: 2 1 47 7: 3 1 48 8: 3 1 1 49 11: 3 1 1 50 1: 3 1 51 2: 1 1 2 1 52 10: 1 1 2 1 53 5: 1 2 1 54 6: 1 2 1 1 55 11: 1 2 1 1 56 1: 1 2 1 57 2: 2 1 1 1 58 10: 2 1 1 1 59 5: 2 1 1 60 6: 2 1 1 1 61 11: 2 1 1 1 62 1: 2 1 1 63 2: 3 1 1 64 10: 3 1 1 65 5: 3 1 66 6: 3 1 1 67 11: 3 1 1 68 1: 3 1 >>> 69 9: 3 70 13: 2 1 1 71 13: 1 2 2 72 13: 3 3 73 14: 3 4 1 74 5: 2 4 1 75 6: 2 4 1 1 76 5: 1 4 1 1 77 6: 1 4 2 1 78 5: 4 2 1 79 6: 4 3 1 80 11: 4 3 1 81 1: 4 2 1 82 2: 1 1 3 2 1 83 1: 1 1 3 1 1 84 2: 2 2 2 1 1 85 1: 2 2 2 1 86 2: 3 3 1 1 87 10: 3 3 1 1 88 5: 3 2 1 1 89 6: 3 2 1 1 1 90 5: 3 1 1 1 1 91 6: 3 1 1 2 1 92 5: 3 1 2 1 93 6: 3 1 3 1 94 11: 3 1 3 1 95 1: 3 1 2 1 96 2: 4 1 2 1 97 1: 4 1 1 1 98 3: 4 1 1 99 4: 3 1 1 100 7: 3 1 1 1  This time, restricting output to only occasions when N is a power of two requires no peculiar bit-counting function. Just change the interpretation loop to  DO I = 1,MS !Here we go! IT = FRACTRAN(LF) !Do it! IF (ALL(NPPOW(2:LP).EQ.0)) CALL SHOWN(I,IT) !Show it! IF (IT.LE.0) EXIT !Quit it? END DO !The next step. Output: Interpreter for J. H. Conway's FRACTRAN language. 14 fractions, as follow: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1 Start with N = 2, step limit 6666666 Require 10 primes only, up to Prime(10) = 29 N as powers of prime factors Step F#: 2 3 5 7 11 13 17 19 23 29 0 : 1 19 9: 2 69 9: 3 280 9: 5 707 9: 7 2363 9: 11 3876 9: 13 8068 9: 17 11319 9: 19 19201 9: 23 36866 9: 29 45551 9: 31 75224 9: 37 101112 9: 41 117831 9: 43 152025 9: 47 215384 9: 53 293375 9: 59 327020 9: 61 428553 9: 67 507519 9: 71 555694 9: 73 700063 9: 79 808331 9: 83 989526 9: 89 1273490 9: 97 1434366 9: 101 1530213 9: 103 1710923 9: 107 1818254 9: 109 2019962 9: 113 2833089 9: 127 3104685 9: 131 3546320 9: 137 3720785 9: 139 4549718 9: 149 4755581 9: 151 5329874 9: 157 5958403 9: 163 6400897 9: 167 Execution took about two seconds. ### Add and Multiply Examples taken from Wikipaedia... Interpreter for J. H. Conway's FRACTRAN language. 1 fraction, as follows: 3/2 Start with N = 72, step limit 66 Require 2 primes only, up to Prime(2) = 3 N as powers of prime factors Step F#: 2 3 0 : 3 2 1 1: 2 3 2 1: 1 4 3 1: 5 4 : 5  The initial value is 72 = 23x32 so that "register" two holds 3 and register three holds 2. On completion, register three holds 5, the sum of 2 and 3. Interpreter for J. H. Conway's FRACTRAN language. 6 fractions, as follow: 455/33, 11/13, 1/11, 3/7, 11/2, 1/3 Start with N = 72, step limit 66 Require 6 primes only, up to Prime(6) = 13 N as powers of prime factors Step F#: 2 3 5 7 11 13 0 : 3 2 1 5: 2 2 1 2 1: 2 1 1 1 1 3 2: 2 1 1 1 1 4 1: 2 2 2 1 5 2: 2 2 2 1 6 3: 2 2 2 7 4: 2 1 2 1 8 4: 2 2 2 9 5: 1 2 2 1 10 1: 1 1 3 1 1 11 2: 1 1 3 1 1 12 1: 1 4 2 1 13 2: 1 4 2 1 14 3: 1 4 2 15 4: 1 1 4 1 16 4: 1 2 4 17 5: 2 4 1 18 1: 1 5 1 1 19 2: 1 5 1 1 20 1: 6 2 1 21 2: 6 2 1 22 3: 6 2 23 4: 1 6 1 24 4: 2 6 25 6: 1 6 26 6: 6 27 : 6  Here, register two holds 3 and register three holds 2. Their product appears in register five. ## FreeBASIC Added a compiler condition to make the program work with the old GMP.bi header file ' version 06-07-2015' compile with: fbc -s console' uses gmp #Include Once "gmp.bi" ' in case the two #define's are missing from 'gmp.bi' define them now#Ifndef mpq_numref #Define mpq_numref(Q) (@(Q)->_mp_num) #Define mpq_denref(Q) (@(Q)->_mp_den)#EndIf Dim As String prog(0 To ...) = {"17/91", "78/85", "19/51", "23/38", "29/33",_"77/29", "95/23", "77/19", "1/17", "11/13", "13/11", "15/14", "15/2", "55/1"} Dim As UInteger i, j, c, max = UBound(prog)Dim As Integer scanbit Dim As ZString Ptr gmp_str : gmp_str = Allocate(10000)Dim As Mpq_ptr in_, out_in_ = Allocate(Len(__mpq_struct)) : Mpq_init(in_)out_ = Allocate(Len(__mpq_struct)) : Mpq_init(out_)Dim As mpz_ptr num, dennum = Allocate(Len(__mpz_struct)) : Mpz_init(num)den = Allocate(Len(__mpz_struct)) : Mpz_init(den) Dim As mpq_ptr instruction(max)For i = 0 To max instruction(i) = Allocate(Len(__mpq_struct)) mpq_init(instruction(i)) mpq_set_str(instruction(i), prog(i), 10 )Next mpq_set_str(in_ ,"2",10)i = 0 : j = 0Print "2";Do mpq_mul(out_, instruction(i), in_) i = i + 1 den = mpq_denref(out_) If mpz_cmp_ui(den, 1) = 0 Then Mpq_get_str(gmp_str, 10, out_) Print ", ";*gmp_str; mpq_swap(in_, out_) i = 0 j = j + 1 End IfLoop Until j > 14 ' this one only display if the integer is 2^p, p being primempq_set_str(in_ ,"2",10)i = 0 : j = 0 : c = 0Print : Print : PrintPrint "count iterations prime 2^prime" Do mpq_mul(out_, instruction(i), in_) i = i + 1 j = j + 1 den = mpq_denref(out_) If mpz_cmp_ui(den, 1) = 0 Then num = mpq_numref(out_) scanbit = mpz_scan1(num, 0) ' if scanbit = 0 then number is odd If scanbit > 0 Then ' return from mpz_scan1(num, scanbit+1) is -1 for power of 2 If mpz_scan1(num, scanbit +1) = -1 Then If c <= 20 Then Mpq_get_str(gmp_str, 10, out_) Else *gmp_str = "" c = c + 1 Print Using "##### ################### ######## "; c; j; scanbit; Print *gmp_str If InKey <> "" Then Exit Do End If End If mpq_swap(in_, out_) i = 0 End IfLoop ' Loop Until scanbit > 300' Loop Until InKey <> ""' Loop Until scanbit > 300 Or InKey <> ""' stopping conditions will slow down the hole loop' loop will check for key if it's printing a result ' empty keyboard bufferWhile InKey <> "" : WendPrint : Print "hit any key to end program"SleepEnd Output: 2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116 count iterations prime 2^prime 1 129 2 4 2 425 3 8 3 1563 5 32 4 3735 7 128 5 11674 11 2048 6 18811 13 8192 7 38010 17 131072 8 52854 19 524288 9 88134 23 8388608 10 166070 29 536870912 11 204575 31 2147483648 12 333931 37 137438953472 13 446506 41 2199023255552 14 519556 43 8796093022208 15 667496 47 140737488355328 16 940183 53 9007199254740992 17 1274660 59 576460752303423488 18 1419935 61 2305843009213693952 19 1853979 67 147573952589676412928 20 2191673 71 2361183241434822606848 shorten output file 42 34533967 181 43 40326168 191 ## Go Basic task: This compiles to produce a program that reads the limit, starting number n, and list of fractions as command line arguments, with the list of fractions as a single argument. package main import ( "fmt" "log" "math/big" "os" "strconv" "strings") func compile(src string) ([]big.Rat, bool) { s := strings.Fields(src) r := make([]big.Rat, len(s)) for i, s1 := range s { if _, ok := r[i].SetString(s1); !ok { return nil, false } } return r, true} func exec(p []big.Rat, n *big.Int, limit int) { var q, r big.Intrule: for i := 0; i < limit; i++ { fmt.Printf("%d ", n) for j := range p { q.QuoRem(n, p[j].Denom(), &r) if r.BitLen() == 0 { n.Mul(&q, p[j].Num()) continue rule } } break } fmt.Println()} func usage() { log.Fatal("usage: ft <limit> <n> <prog>")} func main() { if len(os.Args) != 4 { usage() } limit, err := strconv.Atoi(os.Args[1]) if err != nil { usage() } var n big.Int _, ok := n.SetString(os.Args[2], 10) if !ok { usage() } p, ok := compile(os.Args[3]) if !ok { usage() } exec(p, &n, limit)} Command line usage, with program compiled as "ft": > ft 15 2 "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1" 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132  Extra credit: This invokes above program with appropriate arguments, and processes the output to obtain the 20 primes. package main import ( "fmt" "log" "math/big" "os" "os/exec") func main() { c := exec.Command("ft", "1000000", "2", 17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1) c.Stderr = os.Stderr r, err := c.StdoutPipe() if err != nil { log.Fatal(err) } if err = c.Start(); err != nil { log.Fatal(err) } var n big.Int for primes := 0; primes < 20; { if _, err = fmt.Fscan(r, &n); err != nil { log.Fatal(err) } l := n.BitLen() - 1 n.SetBit(&n, l, 0) if n.BitLen() == 0 && l > 1 { fmt.Printf("%d ", l) primes++ } } fmt.Println()} Output: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71  ## Haskell ### Running the program import Data.List (find)import Data.Ratio (Ratio, (%), denominator) fractran :: (Integral a) => [Ratio a] -> a -> [a]fractran fracts n = n : case find (\f -> n mod denominator f == 0) fracts of Nothing -> [] Just f -> fractran fracts$ truncate (fromIntegral n * f)

Example:

λ> let prog = [17 % 91,78 % 85,19 % 51,23 % 38,29 % 33,77 % 29,95 % 23,77 % 19,1 % 17,11 % 13,13 % 11,15 % 14,15 % 2,55 % 1]

λ> take 15 $fractran prog 2 [2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132] ### Reading the program Additional import import Data.List.Split (splitOn) readProgram :: String -> [Ratio a]readProgram = map (toFrac . splitOn "/") . splitOn "," where toFrac [n,d] = read n % read d Example of running the program: λ> let prog = readProgram "17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23 , 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1" λ> take 15$ fractran prog 2
[2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132]

### Generation of primes

import Data.Maybe (mapMaybe)
primes = mapMaybe log2 $fractran prog 2 where prog = [17 % 91, 78 % 85, 19 % 51, 23 % 38, 29 % 33 ,77 % 29, 95 % 23, 77 % 19, 1 % 17, 11 % 13 ,13 % 11, 15 % 14, 15 % 2, 55 % 1] log2 = fmap (+ 1) . findIndex (== 2) . takeWhile even . iterate (div 2) λ> take 20 primes [1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67] ## Icon and Unicon Works in both languages: record fract(n,d) procedure main(A) fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2)end procedure fractran(s, n, limit) execute(parse(s),n, limit)end procedure parse(s) f := [] s ? while not pos(0) do { tab(upto(' ')|0) ? put(f,fract(tab(upto('/')), (move(1),tab(0)))) move(1) } return fend procedure execute(f,d,limit) /limit := 15 every !limit do { if d := (d%f[i := !*f].d == 0, (writes(" ",d)/f[i].d)*f[i].n) then {} else break write() } write()end Output: ->fractan 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 ->  ## J Solution: toFrac=: '/r' 0&"[email protected] ] NB. read fractions from stringfractran15=: ({~ (= <.) i. 1:)@([email protected][ * ]) ^:(<15) NB. return first 15 Fractran results Example:  taskstr=: '17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1' taskstr fractran15 22 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 ## Java import java.util.Vector;import java.util.regex.Matcher;import java.util.regex.Pattern; public class Fractran{ public static void main(String []args){ new Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1", 2); } final int limit = 15; Vector<Integer> num = new Vector<>(); Vector<Integer> den = new Vector<>(); public Fractran(String prog, Integer val){ compile(prog); dump(); exec(2); } void compile(String prog){ Pattern regexp = Pattern.compile("\\s*(\\d*)\\s*\\/\\s*(\\d*)\\s*(.*)"); Matcher matcher = regexp.matcher(prog); while(matcher.find()){ num.add(Integer.parseInt(matcher.group(1))); den.add(Integer.parseInt(matcher.group(2))); matcher = regexp.matcher(matcher.group(3)); } } void exec(Integer val){ int n = 0; while(val != null && n<limit){ System.out.println(n+": "+val); val = step(val); n++; } } Integer step(int val){ int i=0; while(i<den.size() && val%den.get(i) != 0) i++; if(i<den.size()) return num.get(i)*val/den.get(i); return null; } void dump(){ for(int i=0; i<den.size(); i++) System.out.print(num.get(i)+"/"+den.get(i)+" "); System.out.println(); }} ## JavaScript  var num = new Array();var den = new Array();var val ; function compile(prog){ var regex = /\s*(\d*)\s*\/\s*(\d*)\s*(.*)/m; while(regex.test(prog)){ num.push(regex.exec(prog)[1]); den.push(regex.exec(prog)[2]); prog = regex.exec(prog)[3]; }} function dump(prog){ for(var i=0; i<num.length; i++) document.body.innerHTML += num[i]+"/"+den[i]+" "; document.body.innerHTML += "<br>";} function step(val){ var i=0; while(i<den.length && val%den[i] != 0) i++; return num[i]*val/den[i];} function exec(val){ var i = 0; while(val && i<limit){ document.body.innerHTML += i+": "+val+"<br>"; val = step(val); i ++; }} // Maincompile("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1");dump(); var limit = 15;exec(2);  ## Julia Works with: Julia version 0.6 function fractran(n::Integer, ratios::Vector{<:Rational}, steplim::Integer) rst = zeros(BigInt, steplim) for i in 1:steplim rst[i] = n if (pos = findfirst(x -> isinteger(n * x), ratios)) > 0 n *= ratios[pos] else break end end return rstend using IterToolsmacro ratio_str(s) a = split(s, r"[\s,/]+") return collect(parse(BigInt, n) // parse(BigInt, d) for (n, d) in partition(a, 2))end fracs = ratio"""17 / 91, 78 / 85, 19 / 51, 23 / 38, 29 / 33, 77 / 29, 95 / 23, 77 / 19, 1 / 17, 11 / 13, 13 / 11, 15 / 14, 15 / 2, 55 / 1"""println("The first 20 in the series are ", fractran(2, fracs, 20)) prmfound = 0n = big(2)while prmfound < 20 if isinteger(log2(n)) prmfound += 1 println("Prime$prmfound found: $n is 2 ^$(Int(log2(n)))")    end    n = fractran(n, fracs, 2)[2]end
Output:
The first 20 in the series are BigInt[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116, 308, 364, 68, 4]
Prime 1 found: 2 is 2 ^ 1
Prime 2 found: 4 is 2 ^ 2
Prime 3 found: 8 is 2 ^ 3
Prime 4 found: 32 is 2 ^ 5
Prime 5 found: 128 is 2 ^ 7
Prime 6 found: 2048 is 2 ^ 11
Prime 7 found: 8192 is 2 ^ 13
Prime 8 found: 131072 is 2 ^ 17
Prime 9 found: 524288 is 2 ^ 19
Prime 10 found: 8388608 is 2 ^ 23
Prime 11 found: 536870912 is 2 ^ 29
Prime 12 found: 2147483648 is 2 ^ 31
Prime 13 found: 137438953472 is 2 ^ 37
Prime 14 found: 2199023255552 is 2 ^ 41
Prime 15 found: 8796093022208 is 2 ^ 43
Prime 16 found: 140737488355328 is 2 ^ 47
Prime 17 found: 9007199254740992 is 2 ^ 53
Prime 18 found: 576460752303423488 is 2 ^ 59
Prime 19 found: 2305843009213693952 is 2 ^ 61
Prime 20 found: 147573952589676412928 is 2 ^ 67

## Kotlin

// version 1.1.3 import java.math.BigInteger class Fraction(val num: BigInteger, val denom: BigInteger) {    operator fun times(n: BigInteger) = Fraction (n * num, denom)     fun isIntegral() = num % denom == BigInteger.ZERO} fun String.toFraction(): Fraction {    val split = this.split('/')    return Fraction(BigInteger(split[0]), BigInteger(split[1]))} val BigInteger.isPowerOfTwo get() = this.and(this - BigInteger.ONE) == BigInteger.ZERO val log2 = Math.log(2.0) fun fractran(program: String, n: Int, limit: Int, primesOnly: Boolean): List<Int> {    val fractions = program.split(' ').map { it.toFraction() }    val results = mutableListOf<Int>()    if (!primesOnly) results.add(n)    var nn = BigInteger.valueOf(n.toLong())    while (results.size < limit) {        val frac = fractions.find { (it * nn).isIntegral() } ?: break        nn = nn * frac.num / frac.denom        if (!primesOnly) {           results.add(nn.toInt())        }        else if (primesOnly && nn.isPowerOfTwo) {           val prime = (Math.log(nn.toDouble()) / log2).toInt()           results.add(prime)        }    }    return results} fun main(args: Array<String>) {    val program = "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1"    println("First twenty numbers:")    println(fractran(program, 2, 20, false))    println("\nFirst twenty primes:")    println(fractran(program, 2, 20, true))}
Output:
First twenty numbers:
[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116, 308, 364, 68, 4]

First twenty primes:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]


## Mathematica / Wolfram Language

This isn't as efficient as possible for long lists of fractions, since it doesn't stop doing n*listelements once it finds an integer. Instead, it computes "is integer?" for n*{all list elements}. For short lists that's probably not a big deal.

fractionlist = {17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1};n = 2;steplimit = 20;j = 0;break = False;While[break == False && j <= steplimit, newlist = n fractionlist; isintegerlist = IntegerQ[#] & /@ newlist;  truepositions = Position[isintegerlist, True]; If[Length[truepositions] == 0,  break = True,  Print[ToString[j] <> ": " <> ToString[n]];   n = newlist[[truepositions[[1, 1]]]]; j++;  ] ]
Output:
0: 2
1: 15
2: 825
3: 725
4: 1925
5: 2275
6: 425
7: 390
8: 330
9: 290
10: 770
11: 910
12: 170
13: 156
14: 132
15: 116
16: 308
17: 364
18: 68
19: 4
20: 30

## OCaml

This reads a Fractran program from standard input (keyboard or file) and runs it with the input given by the command line arguments, using arbitrary-precision numbers and fractions.

open Num let get_input () =   num_of_int (     try int_of_string Sys.argv.(1)     with _ -> 10) let get_max_steps () =   try int_of_string Sys.argv.(2)   with _ -> 50 let read_program () =   let line = read_line () in   let words = Str.split (Str.regexp " +") line in   List.map num_of_string words let is_int n = n =/ (integer_num n) let run_program num prog =    let replace n =      let rec step = function      | [] -> None      | h :: t ->            let n' = h */ n in            if is_int n' then Some n' else step t in      step prog in    let rec repeat m lim =      Printf.printf "  %s\n" (string_of_num m);      if lim = 0 then print_endline "Reached max step limit" else         match replace m with         | None -> print_endline "Finished"         | Some x -> repeat x (lim-1)   in    let max_steps = get_max_steps () in   repeat num max_steps let () =   let num = get_input () in   let prog = read_program () in   run_program num prog

The program

17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1

saved in a file can can be run with the command:

ocaml nums.cma str.cma fractran.ml 2 30 < primes.fct
Output:
  2
15
825
725
1925
2275
425
390
330
290
770
910
170
156
132
116
308
364
68
4
30
225
12375
10875
28875
25375
67375
79625
14875
13650
2550
Reached max step limit


## PARI/GP

In this version ideas were borrowed from C++, Java and JavaScript.

Works with: PARI/GP version 2.7.4 and above
 \\ FRACTRAN \\ 4/27/16 aevfractran(val,ft,lim)={my(ftn=#ft,fti,di,L=List(),j=0);while(val&&j<lim, listput(L,val);  for(i=1,ftn, fti=ft[i]; di=denominator(fti);      if(val%di==0, break));\\fend i  val= numerator(fti)*val/di; j++);\\wend jreturn(Vec(L));} {\\ Executing:my(v=[17/91,78/85,19/51,23/38,29/33,77/29,95/23,77/19,1/17,11/13,13/11,15/14,15/2,55/1]);print(fractran(2,v,15));}
Output:
[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132]


## Perl

Instead of printing all steps, I chose to only print those steps which were a power of two. This makes the fact that it's a prime-number-generating program much clearer.

use strict;use warnings;use Math::BigRat; my ($n, @P) = map Math::BigRat->new($_), qw{2 17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1}; $|=1;MAIN: for( 1 .. 5000 ) { print " " if$_ > 1;	my ($pow,$rest) = (0, $n->copy); until($rest->is_odd ) {		++$pow;$rest->bdiv(2);	}	if( $rest->is_one ) { print "2**$pow";	} else {		#print $n; } for my$f_i (@P) {		my $nf_i =$n * $f_i; next unless$nf_i->is_int;		$n =$nf_i;		next MAIN;	}	last;} print "\n";
If you uncomment the
#print $n , it will print all the steps. ## Perl 6 Works with: rakudo version 2015-11-03 A Fractran program potentially returns an infinite list, and infinite lists are a common data structure in Perl 6. The limit is therefore enforced only by slicing the infinite list. sub fractran(@program) { 2, { first Int, map (* *$_).narrow, @program } ... 0}say fractran(<17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11        15/14 15/2 55/1>)[^100];
Output:
(2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 30 225 12375 10875 28875 25375 67375 79625 14875 13650 2550 2340 1980 1740 4620 4060 10780 12740 2380 2184 408 152 92 380 230 950 575 2375 9625 11375 2125 1950 1650 1450 3850 4550 850 780 660 580 1540 1820 340 312 264 232 616 728 136 8 60 450 3375 185625 163125 433125 380625 1010625 888125 2358125 2786875 520625 477750 89250 81900 15300 14040 11880 10440 27720 24360 64680 56840 150920 178360 33320 30576 5712 2128 1288)

Extra credit: We can weed out all the powers of two into another infinite constant list based on the first list. In this case the sequence is limited only by our patience, and a ^C from the terminal. The .msb method finds the most significant bit of an integer, which conveniently is the base-2 log of the power-of-two in question.

for fractran <17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11        15/14 15/2 55/1> {        say $++, "\t", .msb, "\t",$_ if 1 +< .msb == $_;} Output: 0 1 2 1 2 4 2 3 8 3 5 32 4 7 128 5 11 2048 6 13 8192 7 17 131072 8 19 524288 9 23 8388608 ^C ## Python ### Python: Generate series from a fractran program from fractions import Fraction def fractran(n, fstring='17 / 91, 78 / 85, 19 / 51, 23 / 38, 29 / 33,' '77 / 29, 95 / 23, 77 / 19, 1 / 17, 11 / 13,' '13 / 11, 15 / 14, 15 / 2, 55 / 1'): flist = [Fraction(f) for f in fstring.replace(' ', '').split(',')] n = Fraction(n) while True: yield n.numerator for f in flist: if (n * f).denominator == 1: break else: break n *= f if __name__ == '__main__': n, m = 2, 15 print('First %i members of fractran(%i):\n ' % (m, n) + ', '.join(str(f) for f,i in zip(fractran(n), range(m)))) Output: First 15 members of fractran(2): 2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132 #### Python: Generate primes Use fractran above as a module imported into the following program.  from fractran import fractran def fractran_primes(): for i, fr in enumerate(fractran(2), 1): binstr = bin(fr)[2:] if binstr.count('1') == 1: prime = binstr.count('0') if prime > 1: # Skip 2**0 and 2**1 yield prime, i if __name__ == '__main__': for (prime, i), j in zip(fractran_primes(), range(15)): print("Generated prime %2i from the %6i'th member of the fractran series" % (prime, i)) Output: Generated prime 2 from the 20'th member of the fractran series Generated prime 3 from the 70'th member of the fractran series Generated prime 5 from the 281'th member of the fractran series Generated prime 7 from the 708'th member of the fractran series Generated prime 11 from the 2364'th member of the fractran series Generated prime 13 from the 3877'th member of the fractran series Generated prime 17 from the 8069'th member of the fractran series Generated prime 19 from the 11320'th member of the fractran series Generated prime 23 from the 19202'th member of the fractran series Generated prime 29 from the 36867'th member of the fractran series Generated prime 31 from the 45552'th member of the fractran series Generated prime 37 from the 75225'th member of the fractran series Generated prime 41 from the 101113'th member of the fractran series Generated prime 43 from the 117832'th member of the fractran series Generated prime 47 from the 152026'th member of the fractran series ## Racket Translation of: D Simple version, without sequences. #lang racket (define (displaysp x) (display x) (display " ")) (define (read-string-list str) (map string->number (string-split (string-replace str " " "") ","))) (define (eval-fractran n list) (for/or ([e (in-list list)]) (let ([en (* e n)]) (and (integer? en) en)))) (define (show-fractran fr n s) (printf "First ~a members of fractran(~a):\n" s n) (displaysp n) (for/fold ([n n]) ([i (in-range (- s 1))]) (let ([new-n (eval-fractran n fr)]) (displaysp new-n) new-n)) (void)) (define fractran (read-string-list (string-append "17 / 91, 78 / 85, 19 / 51, 23 / 38, 29 / 33," "77 / 29, 95 / 23, 77 / 19, 1 / 17, 11 / 13," "13 / 11, 15 / 14, 15 / 2, 55 / 1"))) (show-fractran fractran 2 15) Output: First 15 members of fractran(2): 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 ## REXX Programming note: extra blanks can be inserted in the fractions before and/or after the solidus [/]. ### showing all terms /*REXX program runs FRACTRAN for a given set of fractions and from a specified N. */numeric digits 2000 /*be able to handle larger numbers. */parse arg N terms fracs /*obtain optional arguments from the CL*/if N=='' | N=="," then N=2 /*Not specified? Then use the default.*/if terms=='' | terms=="," then terms=100 /* " " " " " " */if fracs='' then fracs= '17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,', '77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1' /* [↑] The default for the fractions. */f=space(fracs,0) /*remove all blanks from the FRACS list*/ do #=1 while f\==''; parse var f n.# '/' d.# "," f end /*#*/ /* [↑] parse all the fractions in list*/#=#-1 /*the number of fractions just found. */say # 'fractions:' fracs /*display number and actual fractions. */say 'N is starting at ' N /*display the starting number N. */say terms ' terms are being shown:' /*display a kind of header/title. */ do j=1 for terms /*perform the DO loop for each term. */ do k=1 for # /* " " " " " " fraction*/ if N//d.k\==0 then iterate /*Not an integer? Then ignore it. */ say right('term' j, 35) "──► " N /*display the Nth term with the N. */ N=N % d.k * n.k /*calculate next term (use %≡integer ÷)*/ iterate j /*go start calculating the next term. */ end /*k*/ /* [↑] if an integer, we found a new N*/ end /*j*/ /*stick a fork in it, we're all done. */ output using the default input: 14 fractions: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1 N is starting at 2 100 terms are being shown: term 1 ──► 2 term 2 ──► 15 term 3 ──► 825 term 4 ──► 725 term 5 ──► 1925 term 6 ──► 2275 term 7 ──► 425 term 8 ──► 390 term 9 ──► 330 term 10 ──► 290 term 11 ──► 770 term 12 ──► 910 term 13 ──► 170 term 14 ──► 156 term 15 ──► 132 term 16 ──► 116 term 17 ──► 308 term 18 ──► 364 term 19 ──► 68 term 20 ──► 4 term 21 ──► 30 term 22 ──► 225 term 23 ──► 12375 term 24 ──► 10875 term 25 ──► 28875 term 26 ──► 25375 term 27 ──► 67375 term 28 ──► 79625 term 29 ──► 14875 term 30 ──► 13650 term 31 ──► 2550 term 32 ──► 2340 term 33 ──► 1980 term 34 ──► 1740 term 35 ──► 4620 term 36 ──► 4060 term 37 ──► 10780 term 38 ──► 12740 term 39 ──► 2380 term 40 ──► 2184 term 41 ──► 408 term 42 ──► 152 term 43 ──► 92 term 44 ──► 380 term 45 ──► 230 term 46 ──► 950 term 47 ──► 575 term 48 ──► 2375 term 49 ──► 9625 term 50 ──► 11375 term 51 ──► 2125 term 52 ──► 1950 term 53 ──► 1650 term 54 ──► 1450 term 55 ──► 3850 term 56 ──► 4550 term 57 ──► 850 term 58 ──► 780 term 59 ──► 660 term 60 ──► 580 term 61 ──► 1540 term 62 ──► 1820 term 63 ──► 340 term 64 ──► 312 term 65 ──► 264 term 66 ──► 232 term 67 ──► 616 term 68 ──► 728 term 69 ──► 136 term 70 ──► 8 term 71 ──► 60 term 72 ──► 450 term 73 ──► 3375 term 74 ──► 185625 term 75 ──► 163125 term 76 ──► 433125 term 77 ──► 380625 term 78 ──► 1010625 term 79 ──► 888125 term 80 ──► 2358125 term 81 ──► 2786875 term 82 ──► 520625 term 83 ──► 477750 term 84 ──► 89250 term 85 ──► 81900 term 86 ──► 15300 term 87 ──► 14040 term 88 ──► 11880 term 89 ──► 10440 term 90 ──► 27720 term 91 ──► 24360 term 92 ──► 64680 term 93 ──► 56840 term 94 ──► 150920 term 95 ──► 178360 term 96 ──► 33320 term 97 ──► 30576 term 98 ──► 5712 term 99 ──► 2128 term 100 ──► 1288  ### showing prime numbers Programming note: if the number of terms specified (the 2nd argument) is negative, then only powers of two are displayed. /*REXX program runs FRACTRAN for a given set of fractions and from a specified N. */numeric digits 999; w=length(digits()) /*be able to handle gihugeic numbers. */parse arg N terms fracs /*obtain optional arguments from the CL*/if N=='' | N=="," then N=2 /*Not specified? Then use the default.*/if terms=='' | terms=="," then terms=100 /* " " " " " " */if fracs='' then fracs= '17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,', '77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1' /* [↑] The default for the fractions. */f=space(fracs, 0) /*remove all blanks from the FRACS list*/ do #=1 while f\==''; parse var f n.# '/' d.# "," f end /*#*/ /* [↑] parse all the fractions in list*/#=#-1 /*adjust the number of fractions found.*/tell= terms>0 /*flag: show number or a power of 2.*/!.=0; _=1 /*the default value for powers of 2. */if \tell then do p=1 until length(_)>digits(); _=_+_; !._=1 if p==1 then @._=left('',w+9) "2**"left(p,w) ' ' else @._='(prime' right(p,w)") 2**"left(p,w) ' ' end /*p*/ /* [↑] build powers of 2 tables. */L=length(N) /*length in decimal digits of integer N*/say # 'fractions:' fracs /*display number and actual fractions. */say 'N is starting at ' N /*display the starting number N. */if tell then say terms ' terms are being shown:' /*display hdr.*/ else say 'only powers of two are being shown:' /* " " */q='(max digits used:' /*a literal used in the SAY below. */ do j=1 for abs(terms) /*perform DO loop once for each term. */ do k=1 for # /* " " " " " " fraction*/ if N//d.k\==0 then iterate /*Not an integer? Then ignore it. */ if tell then say right('term' j, 35) "──► " N /*display Nth term and N.*/ else if !.N then say right('term' j,15) "──►" @.N q right(L,w)") " N N=N % d.k * n.k /*calculate next term (use %≡integer ÷)*/ L=max(L, length(N)) /*the maximum number of decimal digits.*/ iterate j /*go start calculating the next term. */ end /*k*/ /* [↑] if an integer, we found a new N*/ end /*j*/ /*stick a fork in it, we're done. */ output using the input of: , -50000000 (negative fifty million) 14 fractions: 17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1 N is starting at 2 only powers of two are being shown: term 1 2**1 (max digits used: 1) 2 term 20 ──► (prime 2) 2**2 (max digits used: 4) 4 term 70 ──► (prime 3) 2**3 (max digits used: 5) 8 term 281 ──► (prime 5) 2**5 (max digits used: 8) 32 term 708 ──► (prime 7) 2**7 (max digits used: 12) 128 term 2364 ──► (prime 11) 2**11 (max digits used: 18) 2048 term 3877 ──► (prime 13) 2**13 (max digits used: 21) 8192 term 8069 ──► (prime 17) 2**17 (max digits used: 27) 131072 term 11320 ──► (prime 19) 2**19 (max digits used: 30) 524288 term 19202 ──► (prime 23) 2**23 (max digits used: 36) 8388608 term 36867 ──► (prime 29) 2**29 (max digits used: 46) 536870912 term 45552 ──► (prime 31) 2**31 (max digits used: 49) 2147483648 term 75225 ──► (prime 37) 2**37 (max digits used: 58) 137438953472 ... (some output elided.) ... term 193455490 ──► (prime 523) 2**523 (max digits used: 808) 27459190640522438859927603196325572869077741200573221637577853836742172733590624208490238562645818219909185245565923432148487951998866575250296113164460228608  Output note: There are intermediary numbers (that aren't powers of two) that are hundreds of digits long. ## Ruby ar = %w[17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1]FractalProgram = ar.map(&:to_r) #=> array of rationals Runner = Enumerator.new do |y| num = 2 loop{ y << num *= FractalProgram.detect{|f| (num*f).denominator == 1} }end prime_generator = Enumerator.new do |y| Runner.each do |num| l = Math.log2(num) y << l.to_i if l.floor == l endend # demop Runner.take(20).map(&:numerator)p prime_generator.take(20) Output: [15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132, 116, 308, 364, 68, 4, 30] [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]  ## Scala class TestFractran extends FunSuite { val program = Fractran("17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1") val expect = List(2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290, 770, 910, 170, 156, 132) test("find first fifteen fractran figures") { assert((program .execute(2) take 15 toList) === expect) }} object Fractran { val pattern = """\s*(\d+)\s*/\s*(\d+)\s*""".r def parse(m: Match) = ((m group 1).toInt, (m group 2).toInt) def apply(program: String) = new Fractran( pattern.findAllMatchIn(program).map(parse).toList)} class Fractran(val numDem: List[(Int,Int)]) { def execute(value: Int) = unfold(value) { v => numDem indexWhere(v % _._2 == 0) match { case i if i > -1 => Some(v, numDem(i)._1 * v / numDem(i)._2) case _ => None } }} ## Scheme Scheme naturally handles fractions, translating to integers as required. The first part of the code translates from a string representation, as required, but equally the user could type the list of fractions in directly as a list.  (import (scheme base) (scheme inexact) (scheme read) (scheme write)) (define *string-fractions* ; string input of fractions "17/91 78/85 19/51 23/38 29/33 77/29 95/23 77/19 1/17 11/13 13/11 15/14 15/2 55/1") (define *fractions* ; create vector of fractions from string input (list->vector ; convert result to a vector, for constant access times (read (open-input-string ; read from the string of fractions, as a list (string-append "(" *string-fractions* ")"))))) ;; run a fractran interpreter, returning the next number for n;; or #f if no next number available;; assume fractions: ordered vector of positive fractions;; n: a positive integer(define (fractran fractions n) (let ((max-n (vector-length fractions))) (let loop ((i 0)) (cond ((= i max-n) #f) ((integer? (* n (vector-ref fractions i))) (* n (vector-ref fractions i))) (else (loop (+ 1 i))))))) ;; Task(define (display-result max-n) (do ((i 0 (+ 1 i)) (n 2 (fractran *fractions* n))) ((= i max-n) (newline)) (display n) (display " "))) (display "Task: ")(display-result 20) ; show first 20 numbers ;; Extra Credit: derive first 20 prime numbers(define (generate-primes target-number initial-n) (define (is-power-of-two? n) (and (> n 2) (integer? (log n 2)))) (define (extract-prime n) (exact (log n 2))) ; (let loop ((count 0) (n initial-n)) (when (< count target-number) (cond ((eq? n #f) (display "-- FAILED TO COMPUTE N --\n")) ((is-power-of-two? n) (display (extract-prime n)) (display " ") (loop (+ 1 count) (fractran *fractions* n))) (else (loop count (fractran *fractions* n)))))) (newline)) (display "Primes:\n")(generate-primes 20 2) ; create first 20 primes  Output: Task: 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 Primes: 2 3 5 7 11 13 17 19 23 37 41 43 53 61 67 71 73 79 83 89  ## Seed7 $ include "seed7_05.s7i";  include "rational.s7i"; const func array integer: fractran (in integer: limit, in var integer: number, in array rational: program) is func  result    var array integer: output is 0 times 0;  local    var integer: index is 1;    var rational: newNumber is 0/1;  begin    output := [] (number);    while index <= length(program) and length(output) <= limit do      newNumber := rat(number) * program[index];      if newNumber = rat(trunc(newNumber)) then        number := trunc(newNumber);        output &:= number;        index := 1;      else        incr(index);      end if;    end while;  end func; const proc: main is func  local    const array rational: program is []        (17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1);    var array integer: output is 0 times 0;    var integer: number is 0;  begin    output := fractran(15, 2, program);    for number range output do      write(number <& " ");    end for;    writeln;  end func;
Output:
2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116


Program to compute prime numbers with fractran (The program has no limit, use CTRL-C to terminate it):

$include "seed7_05.s7i"; include "bigrat.s7i"; const proc: fractran (in var bigInteger: number, in array bigRational: program) is func local var integer: index is 1; var bigRational: newNumber is 0_/1_; begin while index <= length(program) do newNumber := rat(number) * program[index]; if newNumber = rat(trunc(newNumber)) then number := trunc(newNumber); if 2_ ** ord(log2(number)) = number then writeln(log2(number)); end if; index := 1; else incr(index); end if; end while; end func; const proc: main is func local const array bigRational: program is [] (17_/91_, 78_/85_, 19_/51_, 23_/38_, 29_/33_, 77_/29_, 95_/23_, 77_/19_, 1_/17_, 11_/13_, 13_/11_, 15_/14_, 15_/2_, 55_/1_); begin fractran(2_, program); end func; Output: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101  ## Sidef Translation of: Ruby var str ="17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1"const FractalProgram = str.split(',').map{.num} #=> array of rationals func runner(n, callback) { var num = 2 n.times { callback(num *= FractalProgram.find { |f| f * num -> is_int }) }} func prime_generator(n, callback) { var x = 0; runner(Inf, { |num| var l = num.log2 if (l.floor == l) { callback(l.int) ++x == n && return nil } })} STDOUT.autoflush(true) runner(20, {|n| print (n, ' ') })print "\n" prime_generator(20, {|n| print (n, ' ') })print "\n" Output: 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132 116 308 364 68 4 30 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71  ## Tcl Works with: Tcl version 8.6 package require Tcl 8.6 oo::class create Fractran { variable fracs nco constructor {fractions} { set fracs {} foreach frac$fractions {	    if {[regexp {^(\d+)/(\d+),?$}$frac -> num denom]} {		lappend fracs $num$denom	    } else {		return -code error "$frac is not a supported fraction" } } if {![llength$fracs]} {	    return -code error "need at least one fraction"	}    }     method execute {n {steps 15}} {	set co [coroutine [incr nco] my Generate $n] for {set i 0} {$i < $steps} {incr i} { lappend result [$co]	}	catch {rename $co ""} return$result    }     method Step {n} {	foreach {num den} $fracs { if {$n % $den} continue return [expr {$n * $num /$den}]	}	return -code break    }    method Generate {n} {	yield [info coroutine]	while 1 {	    yield $n set n [my Step$n]	}	return -code break    }} set ft [Fractran new {    17/91 78/85 19/51 23/38 29/33 77/29 95/23    77/19 1/17 11/13 13/11 15/14 15/2 55/1}]puts [$ft execute 2] Output: 2 15 825 725 1925 2275 425 390 330 290 770 910 170 156 132  You can just collect powers of 2 by monkey-patching in something like this: oo::objdefine$ft method pow2 {n} {    set co [coroutine [incr nco] my Generate 2]    set pows {}    while {[llength $pows] <$n} {	set item [$co] if {($item & ($item-1)) == 0} { lappend pows$item	}    }    return $pows}puts [$ft pow2 10]

Which will then produce this additional output:

2 4 8 32 128 2048 8192 131072 524288 8388608


## zkl

var fracs="17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17,"   "11/13,  13/11, 15/14, 15/2, 55/1";fcn fractranW(n,fracsAsOneBigString){ //-->Walker (iterator)  fracs:=(fracsAsOneBigString-" ").split(",").apply(	 fcn(frac){ frac.split("/").apply("toInt") }); //( (n,d), (n,d), ...)   Walker(fcn(rn,fracs){      n:=rn.value;      foreach a,b in (fracs){	 if(n*a%b == 0){ 	    rn.set(n*a/b);	    return(n);	 }      }   }.fp(Ref(n),fracs))}
fractranW(2,fracs).walk(20).println();
Output:
L(2,15,825,725,1925,2275,425,390,330,290,770,910,170,156,132,116,308,364,68,4)

Translation of: Python
var [const] BN=Import("zklBigNum");  // libGMPfcn fractranPrimes{   foreach n,fr in ([1..].zip(fractranW(BN(2),fracs))){      if(fr.num1s==1){	 p:=(fr.toString(2) - "1").len();  // count zeros	 if(p>1)	    println("Prime %3d from the nth Fractran(%8d): %d".fmt(p,n,fr));      }   }} fractranPrimes();
Output:
Prime   2 from the nth Fractran(      20): 4
Prime   3 from the nth Fractran(      70): 8
Prime   5 from the nth Fractran(     281): 32
...
Prime 227 from the nth Fractran(15956646): 215679573337205118357336120696157045389097155380324579848828881993728
Prime 229 from the nth Fractran(16429799): 862718293348820473429344482784628181556388621521298319395315527974912
Prime 233 from the nth Fractran(17293373): 13803492693581127574869511724554050904902217944340773110325048447598592
Prime 239 from the nth Fractran(18633402): 883423532389192164791648750371459257913741948437809479060803100646309888
...