

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.


3



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
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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

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei 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[nk], 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

ZhiWei Sun, Mar 20 2014


STATUS

approved



