cp's OEIS Frontend

A238766 Number of ordered ways to write n = k + m (k > 0 and m > 0) such that prime(prime(k)) - prime(k) + 1, prime(prime(2*k+1)) - prime(2*k+1) + 1 and prime(prime(m)) - prime(m) + 1 are all prime.

%I A238766 #9 Mar 05 2014 06:57:54
%S A238766 0,1,1,2,3,2,4,3,2,4,1,4,3,4,6,3,6,3,3,4,3,3,2,6,4,4,5,3,3,5,4,4,4,3,
%T A238766 4,3,6,5,2,6,3,4,6,1,3,3,6,4,6,6,4,4,5,5,1,5,3,3,6,5,6,4,7,6,8,6,8,3,
%U A238766 9,8,9,10,8,11,6,10,10,4,5,4
%N A238766 Number of ordered ways to write n = k + m (k > 0 and m > 0) such that prime(prime(k)) - prime(k) + 1, prime(prime(2*k+1)) - prime(2*k+1) + 1 and prime(prime(m)) - prime(m) + 1 are all prime.
%C A238766 Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 11, 44, 55, 149, 371.
%C A238766 This suggests that there are infinitely many prime pairs {p, q} with 2*pi(p) + 1 = pi(q) such that prime(p) - p + 1 and prime(q) - q + 1 are both prime.
%H A238766 Zhi-Wei Sun, <a href="/A238766/b238766.txt">Table of n, a(n) for n = 1..10000</a>
%H A238766 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e A238766 a(3) = 1 since 3 = 1 + 2 with prime(prime(1)) - prime(1) + 1 = prime(2) - 2 + 1 = 2, prime(prime(2*1+1)) - prime(2*1+1) + 1 = prime(5) - 5 + 1 = 7 and prime(prime(2)) - prime(2) + 1 = prime(3) - 3 + 1 = 3 all prime.
%e A238766 a(371) = 1 since 371 = 66 + 305 with prime(prime(66)) - prime(66) + 1 = prime(317) - 317 + 1 = 2099 - 316 = 1783, prime(prime(2*66+1)) - prime(2*66+1) + 1 = prime(751) - 751 + 1 = 5701 - 750 = 4951 and prime(prime(305)) - prime(305) + 1 = prime(2011) - 2011 + 1 = 17483 - 2010 = 15473 all prime.
%t A238766 pq[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]
%t A238766 a[n_]:=Sum[If[pq[k]&&pq[2k+1]&&pq[n-k],1,0],{k,1,n-1}]
%t A238766 Table[a[n],{n,1,80}]
%Y A238766 Cf. A000040, A234694, A234695, A235189, A236832, A238134, A238756, A238776.
%K A238766 nonn
%O A238766 1,4
%A A238766 _Zhi-Wei Sun_, Mar 05 2014