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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Dec 26 2013

Keywords

• nonn

Comments

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.

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.

See also A234451 and A236358 for similar sequences.

Examples

			a(15) = 1 since 15 = 1 + 14 with 3^(phi(1)/2 + phi(14)/12) + 2 = 3 + 2 = 5 prime.
a(23) = 1 since 23 = 10 + 13 with 3^(phi(10)/2 + phi(13)/12) + 2 = 3^3 + 2 = 29 prime.
a(24) = 1 since 24 = 3 + 21 with 3^(phi(3)/2 + phi(21)/12) + 2 = 3^2 + 2 = 11 prime.
a(37) = 1 since 37 = 9 + 28 with 3^(phi(9)/2 + phi(28)/12) + 2 = 3^4 + 2 = 83 prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..4000

Crossrefs

Cf. A000010, A000040, A000244, A014232, A057735, A079363, A111974, A234344, A234346, A234347, A234361, A234451, A234470, A234475, A234504, A236358.

Programs

Mathematica

f[n_,k_]:=3^(EulerPhi[k]/2+EulerPhi[n-k]/12)+2
a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]