cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A007542 Successive integers produced by Conway's PRIMEGAME.

Original entry on oeis.org

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
Offset: 1

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Author

Keywords

Comments

Conway's PRIMEGAME produces the terms 2^prime in increasing order.
From Daniel Forgues, Jan 20 2016: (Start)
Pairs (n, a(n)) such that a(n) = 2^k are (1, 2^1), (20, 2^2), (70, 2^3), (282, 2^5), (711, 2^7), (2376, 2^11), (3894, 2^13), (8103, 2^17), ...
Numbers n such that a(n) = 2^k are 1, 20, 70, 282, 711, 2376, 3894, 8103, ... [This is 1 + A007547. - N. J. A. Sloane, Jan 25 2016] (End)

References

  • D. Olivastro, Ancient Puzzles. Bantam Books, NY, 1993, p. 21.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a007542 n = a007542_list !! (n-1)
    a007542_list = iterate a203907 2  -- Reinhard Zumkeller, Jan 24 2012
    
  • Maple
    l:= [17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/2, 1/7, 55]: a:= proc(n) option remember; global l; local p, k; if n=1 then 2 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); # Alois P. Heinz, Aug 12 2009
  • Mathematica
    conwayFracs := {17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/2, 1/7, 55}; a[1] = 2; A007542[n_] := A007542[n] = (p = A007542[n - 1]; k = 1; While[ ! IntegerQ[p * conwayFracs[[k]]], k++]; p * conwayFracs[[k]]); Table[A007542[n], {n, 42}] (* Jean-François Alcover, Jan 23 2012, after Alois P. Heinz *)
  • Python
    from fractions import Fraction
    nums = [17, 78, 19, 23, 29, 77, 95, 77,  1, 11, 13, 15, 1, 55] # A202138
    dens = [91, 85, 51, 38, 33, 29, 23, 19, 17, 13, 11,  2, 7,  1] # A203363
    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(2, PRIMEGAME, steps=42)) # Michael S. Branicky, Feb 15 2021

Formula

a(n+1) = A203907(a(n)), a(1) = 2. [Reinhard Zumkeller, Jan 24 2012]

A007547 Number of steps to compute n-th prime in PRIMEGAME (slow version).

Original entry on oeis.org

19, 69, 281, 710, 2375, 3893, 8102, 11361, 19268, 36981, 45680, 75417, 101354, 118093, 152344, 215797, 293897, 327571, 429229, 508284, 556494, 701008, 809381, 990746, 1274952, 1435957, 1531854, 1712701, 1820085, 2021938, 2835628, 3107393, 3549288, 3723821
Offset: 1

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Author

Keywords

References

  • D. Olivastro, Ancient Puzzles. Bantam Books, NY, 1993, p. 21.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndices)
    a007547 n = a007547_list !! n
    a007547_list = tail $ elemIndices 2 $ map a006530 a007542_list
    -- Reinhard Zumkeller, Jan 24 2012
  • Maple
    a:= proc(n) option remember; local l, p, m, k;
          l:= [17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23,
               77/19, 1/17, 11/13, 13/11, 15/2, 1/7, 55/1]:
          if n=1 then b(0):= 2; a(0):= 0 else a(n-1) fi;
          p:= b(n-1);
          for m do for k while not type(p*l[k], integer) do od;
                   p:= p*l[k];
                   if 2^ilog2(p)=p then break fi
          od:
          b(n):= p;
          m + a(n-1)
        end:
    seq(a(n), n=1..10);  # Alois P. Heinz, May 01 2011
  • Mathematica
    Clear[a]; a[n_] := a[n] = Module[{l, p, m, k}, l = {17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/2, 1/7, 55/1}; If[n == 1, b[0] = 2; a[0] = 0, a[n-1]]; p = b[n-1]; For[m=1, True, m++, For[k=1, !IntegerQ[p*l[[k]]], k++]; p = p*l[[k]]; If[2^(Length[IntegerDigits[p, 2]]-1) == p, Break[]]]; b[n] = p; m + a[n-1]]; Table[Print[a[n]]; a[n], {n, 1, 30}] (* Jean-François Alcover, Nov 25 2014, after Alois P. Heinz *)

Extensions

More terms from Alois P. Heinz, May 01 2011

A202138 Numerators of Conway's PRIMEGAME.

Original entry on oeis.org

17, 78, 19, 23, 29, 77, 95, 77, 1, 11, 13, 15, 1, 55
Offset: 1

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Author

Alonso del Arte, Dec 31 2011

Keywords

Comments

Denominators are in A203363.
Conway's PRIMEGAME (also called "Conway's prime producing machine") is a fascinating (and very inefficient) method for obtaining the prime numbers.
The "machine" consists of 14 rational numbers. Starting with 2, one finds the first number in the machine that multiplied by 2 gives an integer; then for that integer we find the first number in the machine that generates another integer. This process is repeated for each new integer obtained. Thus A007542 is generated. Except for the initial 2, each number in A007542 having an integer for a binary logarithm is a prime number.
Note that in R. K. Guy's 1983 paper, the last four numbers of the machine are 13/11, 15/14, 15/2 and 55 rather than 13/11, 15/2, 1/7 and 55.

Crossrefs

Programs

  • Haskell
    a202138_list = [17, 78, 19, 23, 29, 77, 95, 77, 1, 11, 13, 15, 1, 55]
    -- Reinhard Zumkeller, Jan 24 2012

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

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Dec 26 2010

Keywords

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.

Crossrefs

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

A183133 Number of steps to compute the n-th prime in PRIMEGAME using Kilminster's Fractran program with only nine fractions.

Original entry on oeis.org

10, 46, 196, 500, 1428, 2488, 4588, 6840, 10546, 17118, 23064, 33332, 44472, 55848, 70330, 90836, 115136, 137912, 168802, 201000, 233542, 276680, 320332, 373198, 439722, 503810, 568334, 640092, 712314, 792186, 917090, 1023878, 1146632, 1263818, 1419298
Offset: 1

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Author

Alois P. Heinz, Dec 26 2010

Keywords

References

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

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
          local l,p,m,k;
          l:= [3/11, 847/45, 143/6, 7/3, 10/91, 3/7, 36/325, 1/2, 36/5]:
          if n=1 then b(0):= 10; a(0):= 0
                 else a(n-1)
          fi;
          p:= b(n-1);
          for m do
             for k while not type(p*l[k], integer)
             do od; p:= p*l[k];
             if 10^ilog10(p)=p then break fi
          od:
          b(n):= p;
          m + a(n-1)
        end:
    seq(a(n), n=1..20);
  • Mathematica
    a[n_] := a[n] = Module[{l, p, m, k},
         l = {3/11, 847/45, 143/6, 7/3, 10/91, 3/7, 36/325, 1/2, 36/5};
         If[n == 1, b[0] = 10; a[0] = 0, a[n - 1]]; p = b[n - 1];
         For[m = 1, True, m++,
              For[k = 1, !IntegerQ[p*l[[k]]], k++];
              p = p*l[[k]];If[10^(Length@IntegerDigits[p]-1) == p, Break[]]];
         b[n] = p; m + a[n - 1]];
    Array[a, 20] (* Jean-François Alcover, Apr 02 2021, after Alois P. Heinz *)

A267572 Number of steps J. H. Conway's Fractran program needs to calculate the n-th prime.

Original entry on oeis.org

19, 50, 211, 577, 2083, 3469, 7361, 10395, 17915, 35249, 43188, 72392, 97236, 113324, 146556, 209098, 285307, 317925, 417234, 494939, 541264, 684114, 789130, 968524, 1249354, 1408123, 1500944, 1679217, 1781388, 1980305
Offset: 1

Views

Author

Ivan N. Ianakiev, Jan 17 2016

Keywords

Comments

The sieve consists of the 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}.

Examples

			For n = 1, start with 2^n and find the first fraction (fraction1 = 15/2) where the product (2^n)*fraction1 is an integer (integer1 = 15). With integer1 repeat the above, i.e., find the first fraction (fraction2 = 55/1) where integer1*fraction2 is an integer (integer2 = 825). Repeat until a power of 2 is reached (2^2 in this case). The exponent of 2 is prime(1) and a(1) = 19 is the number of steps to reach it.
		

References

  • Dominic Olivastro, Ancient Puzzles, Bantam Books, 1993, pp. 20-21.

Crossrefs

Programs

  • Mathematica
    fracList = {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};
    stepCount[n_] := n * fracList[[First[Flatten[Position[n * fracList, First[Select[n * fracList, IntegerQ]]]]]]];
    A267572[n_] := Length[NestWhileList[stepCount[#] &, 2^n, stepCount[#] != 2^Prime[n] &]];
    Table[tempVar = A267572[n]; Print["a(", n,") = ", tempVar]; tempVar, {n, 30}]
Showing 1-6 of 6 results.