cp's OEIS Frontend

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

%I A180871 #20 Mar 27 2023 17:47:07
%S A180871 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,
%T A180871 146,104,14,92,15,117,199,75,98,316,233,28,92,281,44,136,26,258,7,38,
%U A180871 6,176,126,74,59,89,61,45,79,13,448,119,180,290,184,348,502,508,161,7,265,229
%N A180871 Index of term in Sylvester's sequence A000058 divisible by prime A007996(n).
%C A180871 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).
%C A180871 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.
%H A180871 Max Alekseyev, <a href="/A180871/b180871.txt">Table of n, a(n) for n = 1..12046</a> (first 8181 terms are also given at the Andersen link)
%H A180871 Jens Kruse Andersen, <a href="http://primerecords.dk/sylvester-factors.htm">Factorization of Sylvester's sequence</a>
%F A180871 A000058(a(n)) == 0 (mod A007996(n)) implies a(n) < A007996(n). - _Jonathan Sondow_, Jan 26 2014
%e A180871 A000058(4) = 1807 = 43 * 181 = A007996(4) * A007996(7), so a(4) = a(7) = 4. - _Jonathan Sondow_, Jan 26 2014
%t A180871 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]]
%Y A180871 Cf. A000058, A007996, A126263.
%K A180871 nonn
%O A180871 1,3
%A A180871 _T. D. Noe_, Sep 25 2010
%E A180871 Definition clarified by _Jonathan Sondow_, Jan 26 2014