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 A233417 a(n) = |{0 < k <= n/2: q(k)*q(n-k) + 1 is prime}|, where q(.) is the strict partition function (A000009). 6
 0, 1, 1, 2, 2, 2, 2, 3, 2, 2, 4, 5, 3, 1, 5, 7, 1, 3, 4, 4, 3, 2, 5, 3, 6, 6, 1, 6, 8, 6, 6, 4, 7, 7, 3, 5, 5, 6, 6, 5, 5, 3, 7, 8, 7, 7, 8, 8, 6, 4, 8, 8, 5, 3, 8, 8, 5, 15, 6, 8, 3, 9, 5, 6, 7, 9, 4, 6, 8, 9, 5, 4, 7, 8, 7, 6, 10, 9, 9, 8, 6, 6, 9, 9, 7, 12, 5, 10, 7, 7, 5, 3, 8, 10, 7, 5, 9, 7, 4, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 5, there is a positive integer k < n with q(k)*q(n-k) - 1 prime. (ii) Let n > 1 be an integer. Then p(k) + q(n-k)^2 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n is not equal to 8, then k^3 + q(n-k)^2 is prime for some 0 < k < n. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014. EXAMPLE a(14) = 1 since q(1)*q(13) + 1 = 1*18 + 1 = 19 is prime. a(17) = 1 since q(4)*q(13) + 1 = 2*18 + 1 = 37 is prime. a(27) = 1 since q(13)*q(14) + 1 = 18*22 + 1 = 397 is prime. MATHEMATICA a[n_]:=Sum[If[PrimeQ[PartitionsQ[k]*PartitionsQ[n-k]+1], 1, 0], {k, 1, n/2}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000009, A000040, A232504, A233307, A233346, A233359, A233390, A233393. Sequence in context: A309865 A128764 A324818 * A299741 A074589 A199800 Adjacent sequences: A233414 A233415 A233416 * A233418 A233419 A233420 KEYWORD nonn AUTHOR Zhi-Wei Sun, Dec 09 2013 STATUS approved

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