

A234567


Number of ways to write n = k + m with k > 0 and m > 0 such that p = phi(k) + phi(m)/2 + 1 and P(p1) are both prime, where phi(.) is Euler's totient function and P(.) is the partition function (A000041).


14



0, 0, 0, 1, 2, 1, 1, 3, 2, 2, 3, 2, 4, 2, 4, 4, 2, 4, 3, 5, 1, 3, 2, 3, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 1, 2, 1, 2, 3, 2, 8, 2, 1, 2, 2, 3, 3, 1, 2, 7, 0, 2, 3, 3, 4, 5, 7, 3, 4, 1, 9, 1, 4, 3, 1, 2
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OFFSET

1,5


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 727.
(ii) For the strict partition function q(.) (cf. A000009), any n > 93 can be written as k + m with k > 0 and m > 0 such that p = phi(k) + phi(m)/2 + 1 and q(p1)  1 are both prime.
(iii) If n > 75 is not equal to 391, then n can be written as k + m with k > 0 and m > 0 such that f(k,m)  1, f(k,m) + 1 and q(f(k,m)) + 1 are all prime, where f(k,m) = phi(k) + phi(m)/2.
Part (i) of the conjecture implies that there are infinitely many primes p with P(p1) prime.


LINKS



EXAMPLE

a(21) = 1 since 21 = 6 + 15 with phi(6) + phi(15)/2 + 1 = 7 and P(6) = 11 both prime.
a(700) = 1 since 700 = 247 + 453 with phi(247) + phi(453)/2 + 1 = 367 and P(366) = 790738119649411319 both prime.
a(945) = 1 since 945 = 687 + 258 with phi(687) + phi(258)/2 + 1 = 499 and P(498) = 2058791472042884901563 both prime.


MATHEMATICA

f[n_, k_]:=EulerPhi[k]+EulerPhi[nk]/2
q[n_, k_]:=PrimeQ[f[n, k]+1]&&PrimeQ[PartitionsP[f[n, k]]]
a[n_]:=Sum[If[q[n, k], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



