

A092067


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


2



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
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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}]


CROSSREFS

Cf. A020639, A092028.
Sequence in context: A135679 A092028 A020639 * A214606 A325643 A079879
Adjacent sequences: A092064 A092065 A092066 * A092068 A092069 A092070


KEYWORD

nonn


AUTHOR

Farideh Firoozbakht, Mar 28 2004


STATUS

approved



