OFFSET

1,5

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.

LINKS

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

EXAMPLE

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

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}]

CROSSREFS

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Apr 20 2014

STATUS

approved