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

1,17

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.

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.

Example

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.

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}]

Crossrefs

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

Sequence in context: A152975 A327439 A244758 * A128262 A140414 A129514

Adjacent sequences: A230490 A230491 A230492 * A230494 A230495 A230496

Keyword

nonn

Author

Zhi-Wei Sun, Oct 20 2013

Status

approved