A239214 a(n) = |{0 < k < n: p(k)*p(n)*(p(n)+1) - 1 is prime}|, where p(.) is the partition function (A000041).

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

Offset: 1

Author

Zhi-Wei Sun, Mar 12 2014

Keywords

• nonn

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.

Examples

			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.
		

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. A000040, A000041, A238457, A238509, A238516, A238393, A239207, A239209.

Programs

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,n-1}]
Table[a[n],{n,1,80}]