

A238597


Number of primes p < 2*n with 2*pi(p) + 1 and p*(2n1)  2 both prime, where pi(.) is given by A000720.


2



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

1,3


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 for no n > 195.
(ii) For any integer n > 1, there is a prime p < 2*n with 2*pi(p) + 1 (or 2*pi(p)  1) and 2*n + p both prime.
Part (i) of this conjecture is an extension of the conjecture in A238580.


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(9) = 1 since 5, 2*pi(5) + 1 = 2*3 + 1 = 7 and 5*(2*91)  2 = 5*17  2 = 83 are all prime.
a(28) = 1 since 3, 2*pi(3) + 1 = 2*2 + 1 = 5 and 3*(2*281)  2 = 3*55  2 = 163 are all prime.
a(195) = 1 since 71, 2*pi(71) + 1 = 2*20 + 1 = 41 and 71*(2*1951)  2 = 27617 are all prime.


MATHEMATICA

p[n_, k_]:=p[n, k]=PrimeQ[k]&&PrimeQ[2*PrimePi[k]+1]&&PrimeQ[k*(2n1)2]
a[n_]:=a[n]=Sum[If[p[n, k], 1, 0], {k, 1, 2n1}]
Table[a[n], {n, 1, 80}]


CROSSREFS

Cf. A000040, A000720, A237284, A238580.
Sequence in context: A116560 A103784 A153916 * A045870 A229037 A036863
Adjacent sequences: A238594 A238595 A238596 * A238598 A238599 A238600


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 01 2014


STATUS

approved



