A235712 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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Apr 20 2014

Keywords

• nonn

Comments

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

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.

See also A234972 and A235709 for similar conjectures.

Examples

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Crossrefs

Cf. A000040, A098640, A234972, A235709.

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}]