A230493 Number of ways to write n = (2-(n mod 2))*p + q + r with p <= q <= r such that p, q, r, 2*p^2 - 1, 2*q^2 - 1, 2*r^2 - 1 are all prime.

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

Offset: 1

Author

Zhi-Wei Sun, Oct 20 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 6.

This is stronger than Goldbach's weak conjecture which was finally proved by H. Helfgott in 2013. It also implies that there are infinitely many primes p with 2*p^2 - 1 also prime.

We have verified the conjecture for n up to 10^6.

Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 22 2023

See also A230351, A230494 and A230502 for similar conjectures.

Examples

			a(14) = 1 since 14 = 2*2 + 3 + 7 with 2, 3, 7, 2*2^2 - 1 = 7, 2*3^2 - 1 = 17, 2*7^2 - 1 = 97 all prime.
a(19) = 1 since 19 = 3 + 3 + 13, and 3, 13, 2*3^2 - 1 = 17 and 2*13^2 - 1 = 337 are all prime.
a(53) = 1 since 53 = 3 + 7 + 43, and all the six numbers 3, 7, 43, 2*3^2 - 1 = 17, 2*7^2 - 1 = 97, 2*43^2 - 1 = 3697 are prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588 [math.NT], 2012-2017.

Crossrefs

Cf. A000040, A068307, A106483, A230219, A230351, A230494, A230502.

Programs

Mathematica

pp[n_]:=PrimeQ[2n^2-1]
pq[n_]:=PrimeQ[n]&&pp[n]
a[n_]:=Sum[If[pp[Prime[i]]&&pp[Prime[j]]&&pq[n-(2-Mod[n,2])Prime[i]-Prime[j]],1,0],{i,1,PrimePi[n/(4-Mod[n,2])]},{j,i,PrimePi[(n-(2-Mod[n,2])Prime[i])/2]}]
Table[a[n],{n,1,100}]