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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
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.
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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 07 2013
STATUS
approved