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A238393 a(n) = |{0 < k <= n: 2*p(k)*p(n) + 1 is prime}|, where p(.) is the partition function (A000041).

%I A238393 #18 Mar 01 2014 09:52:02
%S A238393 1,1,3,2,3,2,5,2,6,3,3,5,1,9,3,4,5,5,6,2,7,3,5,8,3,4,8,10,7,10,6,7,9,
%T A238393 8,8,6,6,4,12,10,10,8,6,6,5,7,8,6,10,5,9,9,11,7,7,6,9,11,8,7,11,6,9,8,
%U A238393 4,8,5,18,14,10,9,7,8,6,13,9,4,7,7,15
%N A238393 a(n) = |{0 < k <= n: 2*p(k)*p(n) + 1 is prime}|, where p(.) is the partition function (A000041).
%C A238393 Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 13.
%C A238393 (ii) If n > 1, then 2*p(k)*p(n) - 1 is prime for some 0 < k < n.
%C A238393 (iii) For any integer n > 0, p(k)*(p(n)+1) + 1 is prime for some k = 1, ..., n.
%H A238393 Zhi-Wei Sun, <a href="/A238393/b238393.txt">Table of n, a(n) for n = 1..10000</a>
%H A238393 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e A238393 a(2) = 1 since 2*p(1)*p(2) + 1 = 2*1*2 + 1 = 5 is prime.
%e A238393 a(13) = 1 since 2*p(3)*p(13) + 1 = 2*3*101 + 1 = 607 is prime.
%t A238393 p[n_,k_]:=PrimeQ[2*PartitionsP[n]*PartitionsP[k]+1]
%t A238393 a[n_]:=Sum[If[p[n,k],1,0],{k,1,n}]
%t A238393 Table[a[n],{n,1,80}]
%Y A238393 Cf. A000040, A000041, A238457, A238509, A238516.
%K A238393 nonn
%O A238393 1,3
%A A238393 _Zhi-Wei Sun_, Mar 01 2014