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%I #7 Mar 22 2014 13:41:38
%S 0,0,0,0,1,1,2,2,3,2,1,3,3,2,4,1,3,4,2,2,4,3,1,3,3,2,5,2,2,5,2,4,5,2,
%T 5,6,4,4,6,4,4,7,4,1,8,3,3,7,2,4,6,5,4,5,8,5,10,5,3,12,2,4,9,3,4,7,8,
%U 4,9,7,4,9,5,4,10,2,4,8,4,6
%N Number of ways to write 2*n = p + q with p, q and pi(2*p) - pi(p) all prime, where pi(x) denotes the number of primes not exceeding x.
%C Conjecture: (i) a(n) > 0 for all n > 4.
%C (ii) Each integer n > 5 can be written as p + q (q > 0) with p and pi(2*q) - pi(q) both prime.
%C (iii) Any integer n > 2 not equal to 11 can be written as p + q with p prime and pi(2*q) - pi(q) a square.
%C Part (i) is a refinement of Goldbach's conjecture. It implies that there are infinitely many primes p with pi(2*p) - pi(p) prime.
%H Zhi-Wei Sun, <a href="/A239617/b239617.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e a(5) = 1 since 2*5 = 7 + 3 with 7, 3 and pi(2*7) - pi(7) = 6 - 4 = 2 all prime.
%e a(6) = 1 since 2*6 = 7 + 5 with 7, 5 and pi(2*7) - pi(7) = 2 all prime.
%e a(11) = 1 since 2*11 = 11 + 11 with 11 and pi(2*11) - pi(11) = 8 - 5 = 3 both prime.
%e a(16) = 1 since 2*16 = 13 + 19 with 13, 19 and pi(2*13) - pi(13) = 9 - 6 = 3 all prime.
%e a(23) = 1 since 2*23 = 23 + 23 with 23 and pi(2*23) - pi(23) = 14 - 9 = 5 both prime.
%e a(44) = 1 since 2*44 = 59 + 29 with 59, 29 and pi(2*59) - pi(59) = 30 - 17 = 13 all prime.
%e a(166) = 1 since 2*166 = 103 + 229 with 103, 229 and pi(2*103) - pi(103) = 46 - 27 = 19 all prime.
%t p[n_,k_]:=PrimeQ[PrimePi[2*Prime[k]]-k]&&PrimeQ[2n-Prime[k]]
%t a[n_]:=Sum[If[p[n,k],1,0],{k,1,PrimePi[2n-1]}]
%t Table[a[n],{n,1,80}]
%Y Cf. A000040, A000720, A002372, A002375, A237284.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Mar 22 2014