A234972 - OEIS

A234972

Least prime p < prime(n) such that 2^p - 1 is a primitive root modulo prime(n), or 0 if such a prime p does not exist.

0, 0, 2, 2, 3, 3, 2, 2, 3, 2, 2, 17, 3, 2, 5, 2, 5, 3, 3, 3, 5, 2, 11, 2, 3, 2, 13, 3, 7, 2, 2, 5, 2, 2, 2, 3, 11, 2, 11, 2, 3, 7, 7, 7, 2, 2, 2, 2, 5, 3, 2, 3, 3, 7, 2, 3, 2, 11, 5, 2, 2, 2, 5, 5, 5, 2, 2, 5, 3, 3, 2, 3, 7, 7, 2, 7, 2, 3, 2, 7, 5, 31, 3, 3, 5, 3, 2, 5, 2, 2, 5, 5, 2, 3, 3, 5, 2, 2, 7, 7

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OFFSET

1,3

COMMENTS

Conjecture: a(n) > 0 for all n > 2.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

EXAMPLE

a(3) = 2 since 2 is a prime smaller than prime(3) = 5 with 2^2 - 1 = 3 a primitive root modulo prime(3) = 5.

MATHEMATICA

gp[g_, p_]:=Mod[g, p]>0&&(Length[Union[Table[Mod[g^k, p], {k, 1, p-1}]]]==p-1)

Do[Do[If[gp[2^(Prime[k])-1, Prime[n]], Print[n, " ", Prime[k]]; Goto[aa]], {k, 1, n-1}]; Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 100}]