A180871 Index of term in Sylvester's sequence A000058 divisible by prime A007996(n).

0, 1, 2, 4, 3, 11, 4, 9, 6, 6, 6, 29, 64, 42, 9, 59, 10, 80, 39, 103, 140, 41, 137, 53, 69, 146, 104, 14, 92, 15, 117, 199, 75, 98, 316, 233, 28, 92, 281, 44, 136, 26, 258, 7, 38, 6, 176, 126, 74, 59, 89, 61, 45, 79, 13, 448, 119, 180, 290, 184, 348, 502, 508, 161, 7, 265, 229

Offset: 1

Author

T. D. Noe, Sep 25 2010

Keywords

• nonn

Comments

Because all terms of Sylvester's sequence are coprime to each other, each prime in A007996 divides only one term of A000058. The Mathematica program computes both the primes in A007996 and the terms in this sequence. Using modular arithmetic, it is easy to see that if prime p divides A000058(k) for some k, then we must have k < p. In practice, k < 5*sqrt(p).

An open problem is to prove that all terms of Sylvester's sequence are squarefree or to find a counterexample. Using the p from A007996 and k found here, it is simple to determine whether A000058(k) = 0 (mod p^2). No p < 10^10 was found to have this property.

Examples

			A000058(4) = 1807 = 43 * 181 = A007996(4) * A007996(7), so a(4) = a(7) = 4. - _Jonathan Sondow_, Jan 26 2014
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..12046 (first 8181 terms are also given at the Andersen link)
• Jens Kruse Andersen, Factorization of Sylvester's sequence

Crossrefs

Cf. A000058, A007996, A126263.

Programs

Mathematica

t={}; p=1; While[Length[t]<100, p=NextPrime[p]; s=Mod[2,p]; k=0; modSet={}; While[s>0 && !MemberQ[modSet,s], AppendTo[modSet,s]; k++; s=Mod[s^2-s+1,p]]; If[s==0, AppendTo[t,{p,k}]]]; Transpose[t][[2]]

Formula

A000058(a(n)) == 0 (mod A007996(n)) implies a(n) < A007996(n). - Jonathan Sondow, Jan 26 2014

Extensions

Definition clarified by Jonathan Sondow, Jan 26 2014