%I
%S 2,3,2,5,2,7,2,3,2,11,2,13,2,3,2,17,2,19,2,3,2,23,2,5,2,3,2,29,2,31,2,
%T 3,2,5,2,37,2,3,2,41,2,43,2,3,2,47,2,7,2,3,2,53,2,5,2,3,2,59,2,61,2,3,
%U 2,5,2,67,2,3,2,71,2,73,2,3,2,7,2,79,2,3,2,83,2,5,2,3,2,89,2,7,2,3,2,5,2,97
%N a(n) is the smallest number m such that m>1 and m divides n^m+1.
%C a(n)=2 iff n is odd. If n is even then every prime factor of n+1 is a solution of the equation Mod[n^x+1,x]=0 and if n is odd, smallest prime factor of n+1 (2) is a solution of Mod[n^x+1,x]=0, so for each n, a(n) is not greater than the smallest prime factor of n+1. Conjecture 1: All terms of this sequence are primes. We know if n is odd a(n) is the smallest prime factor of n+1.Conjecture 2: For each n, a(n) is the smallest prime factor of n+1 or a(n)=A020639(n+1).
%H Indranil Ghosh, <a href="/A092067/b092067.txt">Table of n, a(n) for n = 1..10000</a>
%F a[n_] := (For[k=2, Mod[n^k+1, k]>0, k++ ];k)
%e a(6)=7 because 7 divides 6^7+1 and there doesn't exist m such that 1<m<7 and m divides 6^m+1.
%t a[n_] := (For[k=2, Mod[n^k+1, k]>0, k++ ];k);Table[a[n], {n, 100}]
%Y Cf. A020639, A092028.
%K nonn
%O 1,1
%A _Farideh Firoozbakht_, Mar 28 2004
