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Least prime p < prime(n) with 2^p - 1 a quadratic residue modulo prime(n), or 0 if such a number does not exist.
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%I #8 Apr 20 2014 19:57:58

%S 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,

%T 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,

%U 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

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

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

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

%C 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.

%H Zhi-Wei Sun, <a href="/A235709/b235709.txt">Table of n, a(n) for n = 1..10000</a>

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

%t 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}]

%Y Cf. A000040, A001348, A234972.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Apr 20 2014