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

1,12

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.

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.

Example

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.

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}]

Crossrefs

Cf. A000040, A000720, A108954.

Sequence in context: A106696 A131839 A143299 * A257497 A266225 A248596

Adjacent sequences: A239425 A239426 A239427 * A239429 A239430 A239431

Keyword

nonn

Author

Zhi-Wei Sun, Mar 20 2014

Status

approved