login
a(n) = |{0 < k < n: p(k)*p(n)*(p(n)+1) - 1 is prime}|, where p(.) is the partition function (A000041).
4

%I #13 Mar 13 2014 01:51:30

%S 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,

%T 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,

%U 8,6,8,5,11,5,3,4,5,11,7,6,12,16,4

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

%C Conjecture: (i) a(n) > 0 for all n > 1.

%C (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.

%C (iii) For any n > 1, there is a positive integer k < n with 2*p(k)*p(n)*A000009(n)*A047967(n) + 1 prime.

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

%H Zhi-Wei Sun, <a href="/A239214/b239214.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.

%e a(2) = 1 since p(1)*p(2)*(p(2)+1) - 1 = 1*2*3 - 1 = 5 is prime.

%e a(5) = 1 since p(3)*p(5)*(p(5)+1) - 1 = 3*7*8 - 1 = 167 is prime.

%e a(21) = 1 since p(10)*p(21)*(p(21)+1) - 1 = 42*792*793 - 1 = 26378351 is prime.

%e a(45) = 1 since p(20)*p(45)*(p(45)+1) - 1 = 627*89134*89135 - 1 = 4981489349429 is prime.

%t p[n_]:=PartitionsP[n]

%t f[n_]:=p[n]*(p[n]+1)

%t a[n_]:=Sum[If[PrimeQ[p[k]*f[n]-1],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,80}]

%Y Cf. A000040, A000041, A238457, A238509, A238516, A238393, A239207, A239209.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Mar 12 2014