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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. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,16

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.

LINKS

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

EXAMPLE

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.

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}]

CROSSREFS

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

Sequence in context: A072782 A122563 A204030 * A236325 A080345 A285202

Adjacent sequences:  A234500 A234501 A234502 * A234504 A234505 A234506

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 26 2013

STATUS

approved

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Last modified September 18 07:50 EDT 2019. Contains 327168 sequences. (Running on oeis4.)