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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 12 2014
STATUS
approved