A235709 Least prime p < prime(n) with 2^p - 1 a quadratic residue modulo prime(n), or 0 if such a number does not exist.

0, 0, 0, 0, 2, 2, 7, 3, 2, 3, 3, 2, 5, 5, 2, 3, 2, 2, 7, 2, 2, 5, 2, 23, 2, 5, 3, 2, 2, 3, 5, 2, 3, 3, 3, 5, 2, 11, 2, 5, 2, 2, 2, 2, 3, 3, 11, 3, 2, 2, 3, 2, 2, 2, 5, 2, 7, 3, 2, 3, 3, 5, 3, 2, 2, 3, 5, 2, 2, 2, 7, 2, 3, 2, 7, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 11, 5, 2, 2, 5, 2, 5, 2, 7, 5, 3, 2

Offset: 1

Author

Zhi-Wei Sun, Apr 20 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4.

We have verified this for all n = 5, ..., 10^8.

Note that the conjecture in A234972 implies that for any prime p > 3 there is a prime q < p with 2^q - 1 a quadratic nonresidue modulo p.

Examples

			a(8) = 3 since 2^3 - 1 = 7 is a quadratic residue modulo prime(8) = 19, but 2^2 - 1 = 3 is not.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A001348, A234972.

Programs

Mathematica

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