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a(n) = |{0 < k < n: prime(m)^3 + 2*m^3 and m^3 + 2*prime(m)^3 are both prime with m = 3*phi(k) + phi(n-k) - 1}|, where phi(.) is Euler's totient function.
1

%I #9 Aug 08 2014 17:52:54

%S 0,1,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,1,0,0,1,2,0,0,2,2,1,1,0,1,1,2,2,1,2,

%U 4,0,1,3,0,2,3,3,2,3,1,3,2,3,3,2,4,3,4,2,0,2,5,4,2,4,2,2,3,5,5,6

%N a(n) = |{0 < k < n: prime(m)^3 + 2*m^3 and m^3 + 2*prime(m)^3 are both prime with m = 3*phi(k) + phi(n-k) - 1}|, where phi(.) is Euler's totient function.

%C Conjecture: a(n) > 0 for every n = 90, 91, ....

%C We have verified this for n up to 100000.

%C The conjecture implies that there are infinitely many positive integers m with prime(m)^3 + 2*m^3 and m^3 + 2*prime(m)^3 both prime.

%H Zhi-Wei Sun, <a href="/A236619/b236619.txt">Table of n, a(n) for n = 1..10000</a>

%e a(51) = 1 since 3*phi(35) + phi(51-35) - 1 = 3*24 + 8 - 1 = 79 with prime(79)^3 + 2*79^3 = 401^3 + 2*79^3 = 65467279 and 79^3 + 2*prime(79)^3 = 79^3 + 2*401^3 = 129455441 both prime.

%t p[n_]:=PrimeQ[Prime[n]^3+2*n^3]&&PrimeQ[n^3+2*Prime[n]^3]

%t f[n_,k_]:=3*EulerPhi[k]+EulerPhi[n-k]-1

%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, A000578, A173587, A220413, A236192, A236574.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Jan 29 2014