

A187808


a(n) = {0<=k<n: 2k+3, n(nk)1, n(n+k)1 are all prime}


1



0, 1, 1, 2, 1, 3, 1, 2, 1, 3, 2, 3, 1, 4, 1, 4, 1, 3, 2, 5, 4, 4, 2, 4, 1, 4, 2, 5, 1, 4, 2, 4, 2, 6, 2, 5, 4, 4, 2, 5, 1, 7, 1, 7, 5, 3, 2, 4, 2, 4, 3, 6, 3, 6, 4, 7, 4, 8, 2, 9, 2, 8, 3, 2, 3, 7, 4, 7, 1, 7, 4, 7, 1, 7, 4, 9, 7, 8, 2, 9, 3, 6, 2, 6, 3, 7, 2, 8, 3, 7, 4, 6, 8, 9, 4, 6, 3, 9, 5, 8
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,4


COMMENTS

Conjecture: a(n)>0 for all n>1. Moreover, if n>5 is different from 9, 191, 329, 641, 711, 979, then 2k3, 2k+3, n(nk)1, n(n+k)1 are all prime for some 0<k<n.
ZhiWei Sun also made the following conjectures:
(1) For any integer n>101 there is an integer 0<k<n such that kn1 is a Sophie Germain prime.
(2) For each n=128,129,... there is an integer 0<k<n with kn1 and kn+1 both prime.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.


EXAMPLE

a(25) = 1 since 2*17+3 = 37, 25(2517)1 = 199, and 25(25+17)1 = 1049 are all prime.


MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[2k+3]==True&&PrimeQ[n(nk)1]==True&&PrimeQ[n(n+k)1]==True, 1, 0], {k, 0, n1}]
Do[Print[n, " ", a[n]], {n, 1, 100}]


CROSSREFS

Cf. A185636, A086686, A034693.
Sequence in context: A241664 A157226 A156249 * A317673 A317954 A164677
Adjacent sequences: A187805 A187806 A187807 * A187809 A187810 A187811


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 07 2013


STATUS

approved



