A238458 Number of primes p < n with 2*P(n-p) + 1 prime, where P(.) is the partition function (A000041).

0, 0, 1, 2, 2, 3, 2, 3, 4, 2, 3, 3, 3, 5, 2, 4, 4, 5, 4, 5, 4, 4, 3, 3, 3, 4, 4, 4, 2, 4, 2, 5, 4, 4, 5, 3, 3, 6, 3, 4, 1, 3, 4, 7, 6, 4, 4, 4, 4, 4, 4, 5, 3, 5, 5, 7, 3, 3, 4, 6, 5, 8, 5, 5, 4, 4, 2, 7, 5, 7, 3, 6, 5, 7, 6, 7, 5, 5, 4, 7, 4, 5, 3, 5, 6, 8, 5, 3, 4, 6, 3, 5, 4, 5, 4, 5, 2, 6, 4, 5

Offset: 1

Author

Zhi-Wei Sun, Feb 27 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2. Also, for each n > 3 there is a prime p < n with 2*P(n-p) - 1 prime.

We have verified the conjecture for n up to 10^5.

See also A238459 for a similar conjecture involving the strict partition function.

Examples

			a(3) = 1 since 2 and 2*P(3-2) + 1 = 2*1 + 1 = 3 are both prime.
a(41) = 1 since 37 and 2*P(41-37) + 1 = 2*5 + 1 = 11 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000040, A000041, A237705, A237768, A237769, A238457, A238459.

Programs

Mathematica

p[n_,k_]:=PrimeQ[2*PartitionsP[n-Prime[k]]+1]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,100}]