OFFSET
1,1
COMMENTS
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.
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).
From Charlie Neder, Jun 16 2019: (Start)
Theorem: a(n) = A020639(n+1).
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)
LINKS
Indranil Ghosh, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
MATHEMATICA
a[n_] := (For[k=2, Mod[n^k+1, k]>0, k++ ]; k); Table[a[n], {n, 100}]
snm[n_]:=Module[{m=2}, While[PowerMod[n, m, m]!=m-1, m++]; m]; Array[snm, 100] (* Harvey P. Dale, Jul 31 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Mar 28 2004
STATUS
approved