A239209 a(n) = |{0 < k < n: k*p(n)*(p(n)-1) + 1 is prime}|, where p(.) is the partition function (A000041).

0, 1, 2, 2, 2, 1, 4, 2, 3, 3, 3, 2, 3, 4, 2, 4, 4, 4, 8, 3, 3, 4, 6, 5, 3, 5, 10, 4, 4, 7, 5, 4, 3, 8, 7, 6, 3, 4, 5, 4, 3, 7, 5, 5, 3, 4, 5, 11, 7, 10, 3, 10, 8, 12, 6, 4, 10, 4, 8, 5, 11, 7, 5, 14, 5, 7, 4, 10, 1, 10, 9, 12, 8, 5, 10, 1, 7, 7, 6, 5

Offset

1,3

Comments

Conjecture: (i) a(n) > 0 for all n > 1. If n > 1 is not equal to 25, then k*p(n)*(p(n)+1) - 1 is prime for some 0 < k < n.

(ii) Let q(.) be the strict partition function given by A000009. Then, for any integer n > 2, there is a number k among 1, ..., n with k*q(n)^2 - 1 prime. Also, we may replace k*q(n)^2 - 1 by k*q(n)^2 + 1 or k*q(n)*(q(n)+1) + 1 or k*q(n)*(q(n)+1) - 1.

We have verified that a(n) > 0 for all n = 2..10^5.

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(2) = 1 since 1*p(2)*(p(2)-1) + 1 = 1*2*1 + 1 = 3 is prime.
a(6) = 1 since 3*p(6)*(p(6)-1) + 1 = 3*11*10 + 1 = 331 is prime.
a(69) = 1 since 50*p(69)*(p(69)-1) + 1 =  50*3554345*3554344 + 1 = 631668241234001 is prime.
a(76) = 1 since 24*p(76)*(p(76)-1) + 1 = 24*9289091*9289090 + 1 = 2070892855612561 is prime.

Mathematica

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

Crossrefs

Cf. A000009, A000040, A000041, A238393, A238457, A238458, A238459, A238509, A238516, A238577, A239207, A239214.

Sequence in context: A144393 A351112 A089400 * A180824 A105777 A014572

Adjacent sequences: A239206 A239207 A239208 * A239210 A239211 A239212

Keyword

nonn

Author

Zhi-Wei Sun, Mar 12 2014

Status

approved