A237284 Number of ordered ways to write 2*n = p + q with p, q and A000720(p) all prime.

0, 0, 1, 2, 2, 1, 2, 3, 2, 2, 4, 3, 1, 3, 2, 1, 5, 3, 1, 3, 3, 3, 4, 5, 2, 3, 4, 1, 4, 3, 3, 6, 2, 1, 6, 6, 3, 4, 7, 1, 4, 6, 3, 5, 6, 2, 4, 4, 2, 6, 5, 3, 5, 4, 3, 7, 8, 2, 4, 8, 1, 4, 5, 3, 6, 5, 4, 2, 7, 5, 6, 6, 3, 4, 6, 2, 5, 7, 2, 4

Offset: 1

Author

Zhi-Wei Sun, Feb 06 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 6, 13, 16, 19, 28, 34, 40, 61, 166, 278.

This is stronger than Goldbach's conjecture.

The conjecture is true for n <= 5*10^8. - Dmitry Kamenetsky, Mar 13 2020

Examples

			a(13) = 1 since 2*13 = 3 + 23 with 3, 23 and A000720(3) = 2 all prime.
a(278) = 1 since 2*278 = 509 + 47 with 509, 47 and A000720(509) = 97 all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Carlos Rivera, Conjecture 85. Conjectures stricter that the Goldbach ones, Prime Puzzles
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.

Crossrefs

Cf. A000040, A000720, A002372, A002375, A006450, A236566.

Programs

Mathematica

a[n_]:=Sum[If[PrimeQ[2n-Prime[Prime[k]]],1,0],{k,1,PrimePi[PrimePi[2n-1]]}]
Table[a[n],{n,1,80}]