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 A238516 a(n) = |{0 < k < n: (p(k)+1)*p(n) + 1 is prime}|, where p(.) is the partition function (A000041). 6
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified November 29 11:33 EST 2023. Contains 367429 sequences. (Running on oeis4.)