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A187808 a(n) = |{0<=k<n: 2k+3, n(n-k)-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 2k-3, 2k+3, n(n-k)-1, n(n+k)-1 are all prime for some 0<k<n.

Zhi-Wei Sun also made the following conjectures:

(1) For any integer n>101 there is an integer 0<k<n such that kn-1 is a Sophie Germain prime.

(2) For each n=128,129,... there is an integer 0<k<n with kn-1 and kn+1 both prime.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

EXAMPLE

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

MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[2k+3]==True&&PrimeQ[n(n-k)-1]==True&&PrimeQ[n(n+k)-1]==True, 1, 0], {k, 0, n-1}]

Do[Print[n, " ", a[n]], {n, 1, 100}]

CROSSREFS

Cf. A185636, A086686, A034693.

Sequence in context: A241664 A157226 A156249 * A164677 A001511 A265331

Adjacent sequences:  A187805 A187806 A187807 * A187809 A187810 A187811

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Jan 07 2013

STATUS

approved

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Last modified February 24 17:22 EST 2018. Contains 299624 sequences. (Running on oeis4.)