A187808 a(n) = |{0<=k

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

Author

Zhi-Wei Sun, Jan 07 2013

Keywords

• nonn

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

Zhi-Wei Sun also made the following conjectures:

(1) For any integer n>101 there is an integer 0

(2) For each n=128,129,... there is an integer 0

Examples

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

Crossrefs

Cf. A185636, A086686, A034693.

Programs

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}]

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