OFFSET
1,1
COMMENTS
Inspired by Zak Seidov's post to the SeqFan list, cf. link: This yields A154675 as 468 a(n)^2. Indeed, if N/13 is a square, then N=13 k^2 and this can't be the average of a twin prime pair unless k=6m.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
Zak Seidov, "A154676", Jan 15 2009
FORMULA
a(n) = sqrt(A154675(n)/468).
MATHEMATICA
okQ[n_]:=Module[{av=468n^2}, PrimeQ[av-1]&&PrimeQ[av+1]]; Select[Range[1000], okQ] (* Harvey P. Dale, Jan 21 2011 *)
PROG
(PARI) for(i=1, 999, isprime(468*i^2+1) & isprime(468*i^2-1) & print1(i", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jan 15 2009
STATUS
approved