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A239209 a(n) = |{0 < k < n: k*p(n)*(p(n)-1) + 1 is prime}|, where p(.) is the partition function (A000041). 4

%I #15 Mar 13 2014 11:08:36

%S 0,1,2,2,2,1,4,2,3,3,3,2,3,4,2,4,4,4,8,3,3,4,6,5,3,5,10,4,4,7,5,4,3,8,

%T 7,6,3,4,5,4,3,7,5,5,3,4,5,11,7,10,3,10,8,12,6,4,10,4,8,5,11,7,5,14,5,

%U 7,4,10,1,10,9,12,8,5,10,1,7,7,6,5

%N a(n) = |{0 < k < n: 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. If n > 1 is not equal to 25, then k*p(n)*(p(n)+1) - 1 is prime for some 0 < k < n.

%C (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.

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

%H Zhi-Wei Sun, <a href="/A239209/b239209.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 1*p(2)*(p(2)-1) + 1 = 1*2*1 + 1 = 3 is prime.

%e a(6) = 1 since 3*p(6)*(p(6)-1) + 1 = 3*11*10 + 1 = 331 is prime.

%e a(69) = 1 since 50*p(69)*(p(69)-1) + 1 = 50*3554345*3554344 + 1 = 631668241234001 is prime.

%e a(76) = 1 since 24*p(76)*(p(76)-1) + 1 = 24*9289091*9289090 + 1 = 2070892855612561 is prime.

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

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

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

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

%Y Cf. A000009, A000040, A000041, A238393, A238457, A238458, A238459, A238509, A238516, A238577, A239207, A239214.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Mar 12 2014

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Last modified March 29 10:59 EDT 2024. Contains 371277 sequences. (Running on oeis4.)