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

Offset: 1

Author

Zhi-Wei Sun, Apr 20 2014

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A000040, A001348, A001918.

Programs

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}]