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

Author

Zhi-Wei Sun, Mar 12 2014

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

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

Programs

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}]