A092067 a(n) is the smallest number m such that m > 1 and m divides n^m + 1.

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, 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, 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

Offset: 1

Author

Farideh Firoozbakht, Mar 28 2004

Keywords

• nonn

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)

The theorem plus the conjecture 2 in A092028 imply a(n) = A092028(n+2). - R. J. Mathar, Mar 21 2023

Examples

			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.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A020639, A092028.

Row n=2 of A333429.

Programs

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 *)