A061303 Given a prime p, let s(p,0)=p and let s(p,n+1) be the smallest prime == 1 (mod s(p,n)). Let S(p) be the sequence {s(p,n): n=0,1,...}. Let a(0)=2 and let a(n+1) be the smallest prime not in any of the sequences S(a(0)), ..., S(a(n)).
2, 5, 13, 17, 19, 31, 37, 41, 43, 61, 67, 71, 73, 79, 89, 97, 101, 109, 113, 127, 131, 137, 139, 151, 157, 163, 181, 193, 197, 199, 211, 223, 229, 233, 239, 241, 251, 257, 271, 277, 281, 307, 313, 331, 337, 349, 353, 373, 379, 397, 401, 409, 419, 421, 431, 433
Offset: 0
Keywords
Examples
a(0)=2 so S(a(0))={2,3,7,29,...}, which is A061092. Hence a(1)=5 so S(a(1))={5,11,23,47,...}. Hence a(2)=13 so S(a(2))={13,53,107,643,...}, ...
References
- Amarnath Murthy, On the divisors of Smarandache Unary Sequence. Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000.
- Amarnath Murthy, Smarandache Prime Generator Sequence (to be published in Smarandache Notions Journal).
Links
- Wouter Meeussen, Don Reble's equivalence proof
Programs
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Mathematica
(* start *) s[p_, 0] := s[p, 0]=p; s[p_, n_] := s[p, n]=Module[{q}, For[q=s[p, n-1]+1, !PrimeQ[q], q+=s[p, n-1], Null]; q]; ins[q_, p_] := Module[{k}, For[k=0, s[p, k]<=q, k++, If[s[p, k]==q, Return[True]]]; False]; a[0]=2; a[n_] := a[n]=Module[{i, j, q}, For[i=1, True, i++, q=Prime[i]; For[j=0, j
Wouter Meeussen, Mar 18 2014 *) result = {}; prodEPP = 1; Do[prodEPP *= EulerPhi[Prime[i]]; If[Divisible[EulerPhi[prodEPP], EulerPhi[Prime[i]]], AppendTo[result, Prime[i]]], {i, 1, 1000}]; result (* Richard R. Forberg, Sep 16 2024 *)
Extensions
Edited by Dean Hickerson, Jun 09 2002
Comments