A230194 Number of ways to write n = x + y + z (x, y, z > 0) such that all the 11 integers 6*x-1, 6*x+1, 6*x-5, 6*x+5, 6*y-1, 6*y-5, 6*y+5, 6*(x+y)+5, 6*z-1, 6*z-5 and 6*z+5 are prime.

0, 0, 0, 0, 0, 1, 1, 3, 3, 3, 4, 3, 6, 5, 3, 3, 3, 5, 4, 4, 4, 2, 9, 10, 9, 7, 5, 12, 8, 2, 8, 6, 6, 7, 9, 4, 3, 10, 11, 2, 4, 6, 10, 9, 11, 9, 4, 10, 17, 9, 1, 4, 7, 6, 6, 6, 2, 5, 14, 13, 7, 5, 14, 6, 3, 5, 4, 12, 11, 14, 5, 2, 16, 11, 5, 9, 6, 8, 11, 23, 15, 3, 23, 18, 17, 9, 8, 20, 5, 10, 14, 3, 14, 15, 16, 9, 8, 24, 10, 7

Offset: 1

Author

Zhi-Wei Sun, Oct 11 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 5.

Let S be the set of those primes p with p-4 and p+6 also prime. Since each element of S has the form 6*k-1 with k > 0, the conjecture implies that 6*n-3 with n > 5 can be expressed as a sum of three primes in the set S. If n = x + y + z, then 6*n = (6*(x+y)+5) + (6*z-5). So a(n) > 0 implies Goldbach's conjecture for the even number 6*n.

Let T be the set of those primes p with p+4 and p-6 also prime. Clearly each element of T has the form 6*k+1 with k > 0. We conjecture that 6*n+3 with n > 5 can be expressed as a sum of three primes in the set T.

Examples

			a(30) = 2 since 30 = 2 + 14 + 14 = 18 + 4 + 8, and 6*2-1 = 11, 6*2+1 = 13, 6*2-5 = 7, 6*2+5 = 17, 6*14-1 = 83, 6*14-5 = 79, 6*14+5 = 89, 6*(2+14)+5 = 101, 6*18-1 = 107, 6*18+1 = 109, 6*18-5 = 103, 6*18+5 = 113, 6*4-1 = 23, 6*4-5 = 19, 6*4+5 = 29, 6*(18+4)+5 = 137, 6*8-1 = 47, 6*8-5 = 43 and 6*8+5 = 53 are all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588 [math.NT], 2012-2017.

Crossrefs

Cf. A230140, A230141.

Programs

Mathematica

SQ[n_]:=PrimeQ[6n-1]&&PrimeQ[6n-5]&&PrimeQ[6n+5]
a[n_]:=Sum[If[SQ[i]&&PrimeQ[6i+1]&&SQ[j]&&PrimeQ[6(i+j)+5]&&SQ[n-i-j],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]