A230230 Number of ways to write 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime.

0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 3, 4, 3, 3, 5, 1, 5, 5, 3, 4, 5, 3, 2, 6, 4, 3, 6, 3, 3, 6, 3, 5, 6, 3, 6, 5, 4, 4, 9, 5, 4, 9, 5, 3, 9, 4, 4, 6, 4, 5, 6, 5, 5, 10, 4, 8, 10, 3, 7, 12, 3, 6, 11, 5, 7, 8, 3, 4, 6, 6, 4, 7, 2, 7, 9, 2, 10, 9, 3, 9, 8, 3, 5, 14, 8, 4, 12, 5, 5, 11, 5, 6, 8, 3, 8, 7, 4, 9, 11, 3

Offset: 1

Author

Zhi-Wei Sun, Oct 12 2013

Keywords

• nonn

Comments

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

This is stronger than Goldbach's conjecture for even numbers. If 2*n = p + q with p, q, 3*p - 10, 3*q + 10 all prime, then 6*n is the sum of the two primes 3*p - 10 and 3*q + 10.

Conjecture verified for 2*n up to 10^9. - Mauro Fiorentini, Jul 08 2023

Examples

			a(5) = 1 since 2*5 = 7 + 3 with 3*7 - 10 = 11 and 3*3 + 10 = 19 both prime.
a(16) = 1 since 2*16 = 13 + 19 with 3*13 - 10 = 29 and 3*19 + 10 = 67 both 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. A002375, A187757, A199920, A227923, A230227, A230140, A230141, A230217, A230219, A230223, A230224.

Programs

Mathematica

PQ[n_]:=n>3&&PrimeQ[3n-10]
SQ[n_]:=PrimeQ[n]&&PrimeQ[3n+10]
a[n_]:=Sum[If[PQ[Prime[i]]&&SQ[2n-Prime[i]],1,0],{i,1,PrimePi[2n-2]}]
Table[a[n],{n,1,100}]