A239428 Number of ordered ways to write n = k + m with 0 < k <= m such that pi(2*k) - pi(k) and pi(2*m) - pi(m) are both prime, where pi(x) denotes the number of primes not exceeding x.

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

Offset: 1

Author

Zhi-Wei Sun, Mar 20 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 9, and a(n) = 1 only for n = 8, 10, 11, 26, 33, 50, 52.

This implies that there are infinitely many positive integers n with pi(2*n) - pi(n) prime.

Recall that Bertrand's postulate proved by Chebyshev in 1850 asserts that pi(2*n) - pi(n) > 0 for all n > 0.

Examples

			a(11) = 1 since 11 = 4 + 7 with pi(2*4) - pi(4) = 4 - 2 = 2 and pi(2*7) - pi(7) = 6 - 4 = 2 both prime.
a(26) = 1 since 26 = 13 + 13 with pi(2*13) - pi(13) = 9 - 6 = 3 prime.
a(33) = 1 since 33 = 6 + 27 with pi(2*6) - pi(6) = 5 - 3 = 2 and pi(2*27) - pi(27) = 16 - 9 = 7 both prime.
a(50) = 1 since 50 = 23 + 27 with pi(2*23) - pi(23) = 14 - 9 = 5 and pi(2*27) - pi(27) = 16 - 9 = 7 both prime.
a(52) = 1 since 52 = 21 + 31 with pi(2*21) - pi(21) = 13 - 8 = 5 and pi(2*31) - pi(31) = 18 - 11 = 7 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, A108954.

Programs

Mathematica

p[n_]:=PrimeQ[PrimePi[2n]-PrimePi[n]]
a[n_]:=Sum[If[p[k]&&p[n-k],1,0],{k,1,n/2}]
Table[a[n],{n,1,80}]