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a(n) = |{0 < k < n-2: A000009(m)^2 + A047967(m)^2 is prime with m = k + phi(n-k)/2}|, where phi(.) is Euler's totient function.
2

%I #11 Apr 06 2014 10:47:57

%S 0,0,0,1,2,3,3,2,2,2,1,1,1,1,2,2,1,1,4,2,3,2,3,5,4,3,2,6,6,4,2,1,8,4,

%T 4,3,1,6,4,3,3,3,3,3,4,4,5,3,4,5,3,3,7,4,5,5,5,11,7,6,3,7,8,6,5,5,8,6,

%U 7,11,7,5,7,8,7,7,5,10,10,5,6,8,6,10,8,6,8,11,10,6,10,7,7,9,4,9,11,8,13,7

%N a(n) = |{0 < k < n-2: A000009(m)^2 + A047967(m)^2 is prime with m = k + phi(n-k)/2}|, where phi(.) is Euler's totient function.

%C Conjecture: a(n) > 0 for all n > 3.

%C We have verified this for n up to 50000.

%C The conjecture implies that there are infinitely many positive integers m with A000009(m)^2 + A047967(m)^2 prime. See A236440 for such numbers m.

%H Zhi-Wei Sun, <a href="/A236439/b236439.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(14) = 1 since 2 + phi(12)/2 = 4 with A000009(4)^2 + A047967(4)^2 = 2^2 + 3^2 = 13 prime.

%e a(17) = 1 since 10 + phi(7)/2 = 13 with A000009(13)^2 + A047967(13)^2 = 18^2 + 83^2 = 7213 prime.

%t p[n_]:=PrimeQ[PartitionsQ[n]^2+(PartitionsP[n]-PartitionsQ[n])^2]

%t a[n_]:=Sum[If[p[k+EulerPhi[n-k]/2],1,0],{k,1,n-3}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000009, A000010, A000040, A047967, A236412, A236417, A236419, A236440.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Jan 25 2014