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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(p-1) 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
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(p-1) - 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(p-1) 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[n-k]/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, n-1}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 28 2013
STATUS
approved