cp's OEIS Frontend

A236619 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.

%I A236619 #9 Aug 08 2014 17:52:54
%S A236619 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 A236619 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 A236619 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 A236619 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 A236619 Conjecture: a(n) > 0 for every n = 90, 91, ....
%C A236619 We have verified this for n up to 100000.
%C A236619 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 A236619 Zhi-Wei Sun, <a href="/A236619/b236619.txt">Table of n, a(n) for n = 1..10000</a>
%e A236619 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 A236619 p[n_]:=PrimeQ[Prime[n]^3+2*n^3]&&PrimeQ[n^3+2*Prime[n]^3]
%t A236619 f[n_,k_]:=3*EulerPhi[k]+EulerPhi[n-k]-1
%t A236619 a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]
%t A236619 Table[a[n],{n,1,100}]
%Y A236619 Cf. A000010, A000040, A000578, A173587, A220413, A236192, A236574.
%K A236619 nonn
%O A236619 1,3
%A A236619 _Zhi-Wei Sun_, Jan 29 2014