

A239214


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


4



0, 1, 2, 3, 1, 3, 3, 2, 3, 3, 5, 4, 4, 3, 3, 6, 2, 4, 5, 4, 1, 2, 3, 6, 6, 6, 2, 4, 6, 9, 2, 7, 8, 6, 6, 2, 2, 2, 10, 4, 4, 7, 5, 7, 1, 4, 9, 9, 9, 4, 6, 8, 7, 8, 6, 4, 13, 10, 3, 6, 10, 7, 13, 12, 12, 8, 6, 8, 5, 11, 5, 3, 4, 5, 11, 7, 6, 12, 16, 4
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OFFSET

1,3


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For each n = 2, 3, ... there is a positive integer k < n with p(k)*p(n)*(p(n)1) + 1 prime. If n > 2, then p(k)*p(n)*(p(n)1)1 is prime for some 0 < k < n.
(iii) For any n > 1, there is a positive integer k < n with 2*p(k)*p(n)*A000009(n)*A047967(n) + 1 prime.
We have verified that a(n) > 0 for all n = 2, ..., 10^5.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.


EXAMPLE

a(2) = 1 since p(1)*p(2)*(p(2)+1)  1 = 1*2*3  1 = 5 is prime.
a(5) = 1 since p(3)*p(5)*(p(5)+1)  1 = 3*7*8  1 = 167 is prime.
a(21) = 1 since p(10)*p(21)*(p(21)+1)  1 = 42*792*793  1 = 26378351 is prime.
a(45) = 1 since p(20)*p(45)*(p(45)+1)  1 = 627*89134*89135  1 = 4981489349429 is prime.


MATHEMATICA

p[n_]:=PartitionsP[n]
f[n_]:=p[n]*(p[n]+1)
a[n_]:=Sum[If[PrimeQ[p[k]*f[n]1], 1, 0], {k, 1, n1}]
Table[a[n], {n, 1, 80}]


CROSSREFS

Cf. A000040, A000041, A238457, A238509, A238516, A238393, A239207, A239209.
Sequence in context: A230294 A104483 A080717 * A200181 A121062 A045831
Adjacent sequences: A239211 A239212 A239213 * A239215 A239216 A239217


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 12 2014


STATUS

approved



