A230351 Number of ordered ways to write n = p + q (q > 0) with p, 2*p^2 - 1 and 2*q^2 - 1 all prime.

0, 0, 0, 1, 2, 2, 1, 1, 3, 3, 2, 1, 4, 3, 4, 2, 4, 3, 4, 5, 4, 2, 3, 6, 3, 3, 3, 5, 2, 3, 3, 3, 1, 2, 4, 2, 2, 3, 3, 1, 5, 2, 3, 3, 7, 3, 5, 4, 6, 3, 5, 6, 5, 5, 3, 6, 2, 5, 5, 3, 4, 5, 6, 2, 6, 6, 5, 1, 5, 3, 3, 3, 2, 2, 5, 6, 5, 1, 5, 6, 4, 4, 6, 6, 1, 5, 5, 4, 3, 4, 3, 3, 6, 5, 4, 1, 5, 7, 2, 4

Offset: 1

Author

Zhi-Wei Sun, Oct 16 2013

Keywords

• nonn

Comments

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

We have verified this for n up to 2*10^7.

Conjecture verified for n up to 10^9. - Mauro Fiorentini, Aug 07 2023

Examples

			a(7) = 1 since 7 = 3 + 4 with 3, 2*3^2 - 1 = 17, 2*4^2 - 1 = 31 all prime.
a(40) = 1 since 40 = 2 + 38, and 2, 2*2^2 - 1 = 7 , 2*38^2 - 1 = 2887 are all prime.
a(68) = 1 since 68 = 43 + 25, and all the three numbers 43, 2*43^2 - 1 = 3697 and 2*25^2 - 1 = 1249 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, A066049, A106483, A219864, A230252, A230254, A230261.

Programs

Mathematica

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