cp's OEIS Frontend

A237715 Number of ordered ways to write n = p + q (q > 0) with p, prime(p) - p + 1 and prime(prime(q)) - prime(q) + 1 all prime.

%I A237715 #40 Mar 06 2014 09:05:11
%S A237715 0,0,1,2,2,3,2,3,4,2,3,2,2,4,2,5,2,3,3,4,3,2,3,3,4,5,4,2,3,4,4,4,2,4,
%T A237715 2,4,5,2,2,3,4,4,4,5,5,3,6,2,6,5,4,4,4,4,5,2,3,2,4,4,5,3,6,5,9,5,6,4,
%U A237715 4,7,6,5,7,3,8,5,7,4,5,3
%N A237715 Number of ordered ways to write n = p + q (q > 0) with p, prime(p) - p + 1 and prime(prime(q)) - prime(q) + 1 all prime.
%C A237715 Conjecture: a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3.
%H A237715 Zhi-Wei Sun, <a href="/A237715/b237715.txt">Table of n, a(n) for n = 1..10000</a>
%H A237715 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e A237715 a(3) = 1 since 3 = 2 + 1 with 2, prime(2) - 2 + 1 = 3 - 1 = 2 and prime(prime(1)) - prime(1) + 1 = prime(2) - 2 + 1  = 2 all prime.
%e A237715 a(7) = 2 since 7 = 3 + 4 with 3, prime(3) - 3 + 1 = 5 - 2 = 3 and prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 are all prime, and 7 = 5 + 2 with 5, prime(5) - 5 + 1 = 11 - 4 = 7 and prime(prime(2)) - prime(2) + 1 = prime(3) - 3 + 1 = 5 - 2 = 3 all prime.
%t A237715 pq[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]
%t A237715 a[n_]:=Sum[If[pq[k]&&pq[n-Prime[k]],1,0],{k,1,PrimePi[n-1]}]
%t A237715 Table[a[n],{n,1,80}]
%Y A237715 Cf. A000040, A234694, A234695, A238766, A238776, A238814.
%K A237715 nonn
%O A237715 1,4
%A A237715 _Zhi-Wei Sun_, Mar 06 2014