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A238393
a(n) = |{0 < k <= n: 2*p(k)*p(n) + 1 is prime}|, where p(.) is the partition function (A000041).
6
1, 1, 3, 2, 3, 2, 5, 2, 6, 3, 3, 5, 1, 9, 3, 4, 5, 5, 6, 2, 7, 3, 5, 8, 3, 4, 8, 10, 7, 10, 6, 7, 9, 8, 8, 6, 6, 4, 12, 10, 10, 8, 6, 6, 5, 7, 8, 6, 10, 5, 9, 9, 11, 7, 7, 6, 9, 11, 8, 7, 11, 6, 9, 8, 4, 8, 5, 18, 14, 10, 9, 7, 8, 6, 13, 9, 4, 7, 7, 15
OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 13.
(ii) If n > 1, then 2*p(k)*p(n) - 1 is prime for some 0 < k < n.
(iii) For any integer n > 0, p(k)*(p(n)+1) + 1 is prime for some k = 1, ..., n.
LINKS
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.
EXAMPLE
a(2) = 1 since 2*p(1)*p(2) + 1 = 2*1*2 + 1 = 5 is prime.
a(13) = 1 since 2*p(3)*p(13) + 1 = 2*3*101 + 1 = 607 is prime.
MATHEMATICA
p[n_, k_]:=PrimeQ[2*PartitionsP[n]*PartitionsP[k]+1]
a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n}]
Table[a[n], {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 01 2014
STATUS
approved