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%I #15 Dec 25 2019 04:49:59
%S 2,4,5,42,46,49,59,82,84,100,119,128,137,182,185,187,192,233,264,301,
%T 303,340,376,390,395,422,438,446,471,472,494,518,527,570,598,609,611,
%U 633,667,688,714,716,726,728,733,744,831,837,865,875,896,926,940,948
%N Numbers k such that 13*(6*k)^2 is the average of a twin prime pair.
%C 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.
%H Amiram Eldar, <a href="/A154775/b154775.txt">Table of n, a(n) for n = 1..10000</a>
%H Zak Seidov, <a href="http://zak08.livejournal.com/4070.html">"A154676"</a>, Jan 15 2009
%F a(n) = sqrt(A154675(n)/468).
%t okQ[n_]:=Module[{av=468n^2},PrimeQ[av-1]&&PrimeQ[av+1]]; Select[Range[1000],okQ] (* _Harvey P. Dale_, Jan 21 2011 *)
%o (PARI) for(i=1,999, isprime(468*i^2+1) & isprime(468*i^2-1) & print1(i","))
%Y Cf. A014574, A037073, A154331, A154772, A154773, A154774, A154675.
%K nonn
%O 1,1
%A _M. F. Hasler_, Jan 15 2009
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