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Number of ordered ways to write 2*n = p + q with p, q and A000720(p) all prime.
10

%I #14 Mar 13 2020 13:28:38

%S 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,

%T 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,

%U 7,5,6,6,3,4,6,2,5,7,2,4

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

%C 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.

%C This is stronger than Goldbach's conjecture.

%C The conjecture is true for n <= 5*10^8. - _Dmitry Kamenetsky_, Mar 13 2020

%H Zhi-Wei Sun, <a href="/A237284/b237284.txt">Table of n, a(n) for n = 1..10000</a>

%H Carlos Rivera, <a href="https://www.primepuzzles.net/conjectures/conj_085.htm">Conjecture 85. Conjectures stricter that the Goldbach ones</a>, Prime Puzzles

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014-2016.

%e a(13) = 1 since 2*13 = 3 + 23 with 3, 23 and A000720(3) = 2 all prime.

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

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

%t Table[a[n],{n,1,80}]

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

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Feb 06 2014