A238516 a(n) = |{0 < k < n: (p(k)+1)*p(n) + 1 is prime}|, where p(.) is the partition function (A000041).

0, 1, 1, 1, 2, 3, 3, 3, 4, 5, 5, 2, 4, 6, 5, 3, 3, 3, 4, 1, 7, 7, 2, 6, 3, 8, 7, 4, 1, 6, 3, 4, 5, 8, 4, 4, 2, 2, 4, 9, 7, 6, 3, 6, 4, 2, 6, 6, 3, 8, 5, 6, 4, 7, 7, 4, 8, 7, 9, 1, 6, 7, 7, 3, 3, 7, 2, 5, 4, 10, 8, 5, 1, 8, 9, 1, 4, 6, 7, 12, 3, 2, 4, 10, 4, 4

Offset

1,5

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Also, for any integer n > 4 there is a positive integer k < n with (p(k)-1)*p(n) - 1 prime.

(ii) Let q(.) be the strict partition function (A000009). If n > 5, then p(n)*q(k) + 1 is prime for some 3 < k < n. If n > 6, then p(n)*q(k) - 1 is prime for some 0 < k < n. If n > 1, then q(n)*q(k) + 1 is prime for some 0 < k < n. If n > 3, then q(n)*q(k) - 1 is prime for some 0 < k < n.

We have verified that a(n) > 0 for all n = 2, 3, ..., 60000.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Example

a(4) = 1 since (p(1)+1)*p(4) + 1 = 2*5 + 1 = 11 is prime.
a(20) = 1 since (p(12)+1)*p(20) + 1 = 78*627 + 1 = 48907 is prime.
a(246) = 1 since (p(45)+1)*p(246) + 1 = 89135*169296722391554 + 1 = 15090263350371165791 is prime.

Mathematica

p[n_, k_]:=PrimeQ[PartitionsP[n]*(PartitionsP[k]+1)+1]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000009, A000040, A000041, A233417, A238457, A238509.

Sequence in context: A097087 A153161 A346373 * A282692 A269371 A287355

Adjacent sequences: A238513 A238514 A238515 * A238517 A238518 A238519

Keyword

nonn

Author

Zhi-Wei Sun, Feb 28 2014

Status

approved