A239330 Number of odd primes p <= n with pi(n*(p+1)/2) - pi(n*(p-1)/2) prime, where pi(x) denotes the number of primes not exceeding x.

0, 0, 0, 1, 1, 2, 2, 3, 3, 2, 4, 2, 5, 3, 3, 4, 1, 3, 4, 4, 5, 4, 4, 4, 4, 3, 3, 5, 5, 5, 3, 6, 8, 5, 5, 3, 5, 6, 4, 4, 7, 6, 4, 4, 3, 5, 3, 4, 3, 5, 4, 4, 3, 3, 4, 2, 4, 2, 4, 4, 3, 4, 9, 3, 7, 4, 6, 4, 5, 5, 7, 4, 9, 9, 7, 7, 11, 7, 8, 8

Offset: 1

Author

Zhi-Wei Sun, Mar 16 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 3, and a(n) = 1 only for n = 4, 5, 17.

We have verified this for n up to 3*10^5.

Examples

			a(4) = 1 since 3 and pi(4*(3+1)/2) - pi(4*(3-1)/2) = pi(8) - pi(4) = 4 - 2 = 2 are both prime.
a(5) = 1 since 5 and pi(5*(5+1)/2) - pi(5*(5-1)/2) = pi(15) - pi(10) = 6 - 4 = 2 are both prime.
a(17) = 1 since 11 and pi(17*(11+1)/2) - pi(17*(11-1)/2) = pi(102) - pi(85) = 26 - 23 = 3 are both 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, A000720, A237578, A238278, A239328.

Programs

Mathematica

p[n_,k_]:=PrimeQ[PrimePi[n*(Prime[k]+1)/2]-PrimePi[n*(Prime[k]-1)/2]]
a[n_]:=Sum[If[p[n,k],1,0],{k,2,PrimePi[n]}]
Table[a[n],{n,1,80}]