%I #28 Mar 21 2023 06:09:21
%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 (n^x + 1) mod x = 0, and if n is odd, the smallest prime factor of n+1 (2) is a solution of (n^x + 1) mod x = 0, so for each n, a(n) is not greater than the smallest prime factor of n+1.
%C 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.
%C Conjecture 2: For each n, a(n) is the smallest prime factor of n+1 or a(n)=A020639(n+1).
%C From _Charlie Neder_, Jun 16 2019: (Start)
%C Theorem: a(n) = A020639(n+1).
%C Proof: If a(n) is composite (kp, say) then n^(kp) == -1 (mod p), but then n^k is also congruent to -1 (mod p) by Fermat's little theorem, contradicting the assumption that a(n) was minimal. Thus, a(n) must be prime, and using Fermat's little theorem again shows that n^p == -1 (mod p) iff n == -1 (mod p), and A020639(n+1) gives the least p such that this is the case. (End)
%C The theorem plus the conjecture 2 in A092028 imply a(n) = A092028(n+2). - _R. J. Mathar_, Mar 21 2023
%H Indranil Ghosh, <a href="/A092067/b092067.txt">Table of n, a(n) for n = 1..10000</a>
%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}]
%t snm[n_]:=Module[{m=2},While[PowerMod[n,m,m]!=m-1,m++];m]; Array[snm,100] (* _Harvey P. Dale_, Jul 31 2021 *)
%Y Cf. A020639, A092028.
%Y Row n=2 of A333429.
%K nonn
%O 1,1
%A _Farideh Firoozbakht_, Mar 28 2004