cp's OEIS Frontend

A234503 Number of ways to write n = k + m with k > 0 and m > 0 such that 3^(phi(k)/2 + phi(m)/12) + 2 is prime, where phi(.) is Euler's totient function.

%I A234503 #22 Apr 16 2018 03:05:16
%S A234503 0,0,0,0,0,0,0,1,1,1,1,0,0,0,1,2,1,1,2,3,2,1,1,1,2,1,2,2,3,2,4,4,4,2,
%T A234503 3,2,1,3,4,8,3,4,4,4,6,3,4,6,3,5,5,3,2,2,6,5,3,2,3,7,4,3,4,4,3,4,4,4,
%U A234503 5,2,5,2,6,5,7,3,5,7,6,13,5,7,7,10,6,8,8,9,6,7,8,6,6,5,7,9,6,7,8,10
%N A234503 Number of ways to write n = k + m with k > 0 and m > 0 such that 3^(phi(k)/2 + phi(m)/12) + 2 is prime, where phi(.) is Euler's totient function.
%C A234503 It might seem that a(n) > 0 for all n > 14, but a(43905) = 0. If a(n) > 0 infinitely often, then there are infinitely many primes of the form 3^m + 2.
%C A234503 Similarly, it might seem that for n > 26 there is a positive integer k < n such that m = phi(k)/2 + phi(n-k)/12 is an integer with 3^m - 2 prime, but n = 41213 is a counterexample.
%C A234503 See also A234451 and A236358 for similar sequences.
%H A234503 Zhi-Wei Sun, <a href="/A234503/b234503.txt">Table of n, a(n) for n = 1..4000</a>
%e A234503 a(15) = 1 since 15 = 1 + 14 with 3^(phi(1)/2 + phi(14)/12) + 2 = 3 + 2 = 5 prime.
%e A234503 a(23) = 1 since 23 = 10 + 13 with 3^(phi(10)/2 + phi(13)/12) + 2 = 3^3 + 2 = 29 prime.
%e A234503 a(24) = 1 since 24 = 3 + 21 with 3^(phi(3)/2 + phi(21)/12) + 2 = 3^2 + 2 = 11 prime.
%e A234503 a(37) = 1 since 37 = 9 + 28 with 3^(phi(9)/2 + phi(28)/12) + 2 = 3^4 + 2 = 83 prime.
%t A234503 f[n_,k_]:=3^(EulerPhi[k]/2+EulerPhi[n-k]/12)+2
%t A234503 a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
%t A234503 Table[a[n],{n,1,100}]
%Y A234503 Cf. A000010, A000040, A000244, A014232, A057735, A079363, A111974, A234344, A234346, A234347, A234361, A234451, A234470, A234475, A234504, A236358.
%K A234503 nonn
%O A234503 1,16
%A A234503 _Zhi-Wei Sun_, Dec 26 2013