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

Author

Zhi-Wei Sun, Feb 28 2014

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

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

Programs

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}]