A239207 a(n) = |{0 < k <= n: k*p(n)*q(n)*r(n) - 1 is prime}|, where p(.), q(.) and r(.) are given by A000041, A000009 and A047967 respectively.

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

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)*q(n)*r(n) + 1 is prime for some k = 1, ..., n.

(ii) For any integer n > 1, there is a number k among 1, ..., n with k*p(n)*q(n) - 1 (or k*p(n)*q(n) + 1) prime.

(iii) For each n > 1, there is a positive integer k < n with k*p(n) + 1 (or k*q(n) + 1) prime. If n > 1, then k*p(n) - 1 is prime for some k = 1, ..., n. If n > 2, then k*q(n) - 1 is prime for some 0 < k < n.

We have verified that a(n) > 0 for all n = 2, ..., 83000.

See also A239209 and A239214 for related conjectures.

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 2*p(2)*q(2)*r(2) - 1 = 2*2*1*1 - 1 = 3 is prime.
a(23) = 1 since 12*p(23)*q(23)*r(23) - 1 = 12*1255*104*1151 - 1 = 1802742239 is prime.

Mathematica

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

Crossrefs

Cf. A000009, A000040, A000041, A047967, A238393, A238577, A239209, A239214.

Sequence in context: A159636 A023647 A332413 * A329157 A082978 A005536

Adjacent sequences: A239204 A239205 A239206 * A239208 A239209 A239210

Keyword

nonn

Author

Zhi-Wei Sun, Mar 12 2014

Status

approved