A183132 Successive integers produced by Conway's PRIMEGAME using Kilminster's Fractran program with only nine fractions.

10, 5, 36, 858, 234, 5577, 1521, 3549, 8281, 910, 100, 50, 25, 180, 3388, 924, 252, 6006, 1638, 39039, 10647, 24843, 57967, 6370, 700, 300, 7150, 1950, 46475, 12675, 29575, 3250, 360, 6776, 1848, 504, 12012, 3276, 78078, 21294, 507507, 138411, 322959, 753571

Offset: 1

Author

Alois P. Heinz, Dec 26 2010

Keywords

• easy
• look
• nonn

Comments

The exponents of exact powers of 10 in this sequence are 1, followed by the successive primes (A008578).

References

• D. Olivastro, Ancient Puzzles. Bantam Books, NY, 1993, p. 21.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10547
• J. H. Conway, FRACTRAN: a simple universal programming language for arithmetic, in T. M. Cover and Gopinath, eds., Open Problems in Communication and Computation, Springer, NY 1987, pp. 4-26.
• Esolang wiki, Fractran
• Chaim Goodman-Strauss, Can’t Decide? Undecide!, Notices of the AMS, Volume 57, Number 3, pp. 343-356, March 2010.
• R. K. Guy, Conway's prime producing machine, Math. Mag. 56 (1983), no. 1, 26-33.
• Eric Weisstein's World of Mathematics, FRACTRAN.
• Wikipedia, FRACTRAN.

Crossrefs

Cf. A183133, A008578, A007542, A007546, A007547.

Programs

Maple

l:= [3/11, 847/45, 143/6, 7/3, 10/91, 3/7, 36/325, 1/2, 36/5]:
a:= proc(n) option remember;
      global l;
      local p, k;
      if n=1 then 10
             else p:= a(n-1);
                  for k while not type(p*l[k], integer)
                  do od; p*l[k]
      fi
    end:
seq(a(n), n=1..50);

Mathematica

l = {3/11, 847/45, 143/6, 7/3, 10/91, 3/7, 36/325, 1/2, 36/5};
a[n_] := a[n] = Module[{p, k}, If[n == 1, 10, p = a[n - 1]; For[k = 1, !IntegerQ[p*l[[k]]], k++]; p*l[[k]]]];
Array[a, 50] (* Jean-François Alcover, May 28 2018, from Maple *)

Python

from fractions import Fraction
nums = [ 3, 847, 143, 7, 10, 3,  36, 1, 36]
dens = [11,  45,   6, 3, 91, 7, 325, 2,  5]
PRIMEGAME = [Fraction(num, den) for num, den in zip(nums, dens)]
def succ(n, program):
    for i in range(len(program)):
      if (n*program[i]).denominator == 1: return (n*program[i]).numerator
def orbit(start, program, steps):
    orb = [start]
    for s in range(1, steps): orb.append(succ(orb[-1], program))
    return orb
print(orbit(10, PRIMEGAME, steps=44)) # Michael S. Branicky, Oct 05 2021