%I #10 Apr 06 2014 10:46:24
%S 0,0,0,0,0,1,1,0,1,1,1,1,1,2,1,0,0,1,2,3,3,4,5,3,4,4,7,4,5,5,3,3,4,5,
%T 4,3,6,8,3,3,3,7,3,7,4,5,3,6,3,2,3,6,3,3,2,5,1,4,6,4,3,3,7,5,3,3,3,4,
%U 1,5,4,3,2,4,3,6,2,5,6,4,5,2,1,6,4,4,2,11,1,6,3,5,6,7,2,4,4,2,3,2
%N a(n) = |{0 < k < n: m = phi(k)/2 + phi(n-k)/8 is an integer with p(m)^2 + q(m)^2 prime}|, where phi(.) is Euler's totient, p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).
%C Conjecture: a(n) > 0 for all n > 17.
%C We have verified this for n up to 65000.
%C The conjecture implies that there are infinitely positive integers m with p(m)^2 + q(m)^2 prime. See A236413 for a list of such numbers m. See also A236414 for primes of the form p(m)^2 + q(m)^2.
%H Zhi-Wei Sun, <a href="/A236412/b236412.txt">Table of n, a(n) for n = 1..10000</a>
%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014
%e a(15) = 1 since phi(2)/2 + phi(13)/8 = 1/2 + 12/8 = 2 with p(2)^2 + q(2)^2 = 2^2 + 1^2 = 5 prime.
%e a(69) = 1 since phi(5)/2 + phi(64)/8 = 2 + 4 = 6 with p(6)^2 + q(6)^2 = 11^2 + 4^2 = 137 prime.
%e a(89) = 1 since phi(73)/2 + phi(16)/8 = 36 + 1 = 37 with p(37)^2 + q(37)^2 = 21637^2 + 760^2 = 468737369 prime.
%t p[n_]:=IntegerQ[n]&&PrimeQ[PartitionsP[n]^2+PartitionsQ[n]^2]
%t f[n_,k_]:=EulerPhi[k]/2+EulerPhi[n-k]/8
%t a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A000041, A233307, A236374, A236413, A236414.
%K nonn
%O 1,14
%A _Zhi-Wei Sun_, Jan 24 2014