A344173 Number of primes p < prime(n) of the form x^3 + 2*y^3 with x nonnegative and y + 1 prime such that p is a quadratic nonresidue modulo prime(n).

0, 1, 2, 1, 1, 1, 1, 2, 1, 3, 3, 3, 3, 3, 2, 2, 2, 4, 3, 1, 3, 4, 2, 3, 2, 3, 2, 3, 2, 4, 3, 4, 4, 4, 3, 1, 3, 5, 2, 4, 2, 3, 2, 3, 3, 3, 5, 1, 3, 3, 4, 1, 3, 4, 3, 2, 4, 4, 4, 2, 4, 4, 4, 3, 3, 5, 3, 3, 2, 3, 1, 2, 6, 4, 6, 2, 4, 3, 4, 3, 4, 5, 4, 2, 4, 5, 4, 1, 5, 3, 3, 6, 4, 4, 3, 4, 3, 3, 5, 4

Offset: 1

Author

Zhi-Wei Sun, May 10 2021

Keywords

• nonn

Comments

In 2001 Heath-Brown proved that there are infinitely many primes of the form x^3 + 2*y^3 with x and y nonnegative integers.

Conjecture: (i) a(n) > 0 for all n > 1. In other words, for each odd prime p, there is a prime q < p of the form x^3 + 2*y^3 with x nonnegative and y + 1 prime such that q is a quadratic nonresidue modulo p.

(ii) For any prime p > 5 not equal to 29, there is a prime q < p of the form x^3 + 2*y^3 with x nonnegative and y + 1 prime such that q is a quadratic residue modulo p.

Part (i) of the conjecture verified for all odd primes p < 2*10^9.

We even conjecture further that for any prime p > 5 there is a prime q < p of the form x^3 + 2*y^3 with x nonnegative and y + 1 prime such that q is a primitive root modulo p.

See also A344174 for a similar conjecture.

Examples

			a(2) = 1, and the prime 0^3 + 2*(2-1)^3 = 2 is a quadratic nonresidue modulo prime(2) = 3.
a(6) = 1, and the prime 0^3 + 2*(2-1)^3 = 2 is a quadratic nonresidue modulo prime(6) = 13.
a(20) = 1, and the prime 1^3 + 2*(2-1)^3 = 17 is a quadratic nonresidue modulo prime(20) = 71.
a(48) = 1, and the prime 1^3 + 2*(2-1)^3 = 3 is a quadratic nonresidue modulo prime(48) = 223.
a(88) = 1, and the prime 3^3 + 2*(3-1)^3 = 43 is a quadratic nonresidue modulo prime(88) = 457.
		

Links

• D. R. Heath-Brown, Primes represented by x^3 + 2y^3, Acta Mathematica 186 (2001), 1-84.

Crossrefs

Cf. A000040, A000578, A173587, A220413, A344174.

Programs

Mathematica

tab={0};Do[p:=p=Prime[n];tt={};Do[If[PrimeQ[b+1]&&PrimeQ[a^3+2b^3]&&JacobiSymbol[a^3+2b^3,p]==-1,tt=Append[tt,a^3+2b^3]],{a,0,(p-1)^(1/3)},{b,1,((p-1-a^3)/2)^(1/3)}];tab=Append[tab,Length[Union[tt]]],{n,2,100}];Print[tab]