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Least prime p < prime(n) with 2^p + 1 a quadratic nonresidue modulo prime(n), or 0 if such a prime p does not exist.
5

%I #16 Aug 05 2019 02:48:29

%S 0,2,0,2,7,2,2,5,2,11,11,2,7,2,2,2,5,5,2,5,2,5,2,5,2,7,2,2,5,2,2,13,2,

%T 5,13,5,2,2,2,2,5,11,5,2,2,7,5,2,2,23,2,7,5,5,2,2,5,5,2,7,2,2,2,5,2,2,

%U 7,2,2,5,2,7,2,2,11,2,5,2,5,5,5,7,7,2,5,2,5,2,7,2,2,7,2,13,7,2,5,5,2,5

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

%C Conjecture: a(n) > 0 for all n > 3.

%C Note that 2^3 + 1 = 3^2 is a quadratic residue modulo any prime p > 3. Also, there is no prime p < prime(316) = 2089 with 2^p + 1 a primitive root modulo 2089.

%C See also A234972 and A235709 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A235712/b235712.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1405.0290">New observations on primitive roots modulo primes</a>, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

%e a(4) = 2 since 2^2 + 1 = 5 is a quadratic nonresidue modulo prime(4) = 7.

%t Do[Do[If[JacobiSymbol[2^(Prime[k])+1,Prime[n]]==-1,Print[n," ",Prime[k]];Goto[aa]],{k,1,n-1}];

%t Print[n," ",0];Label[aa];Continue,{n,1,100}]

%Y Cf. A000040, A098640, A234972, A235709.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Apr 20 2014