%I
%S 10,14,15,25,60,74,76,87,127,129,130,140,171,196,207,224,259,263,302,
%T 314,315,319,333,337,377,389,451,454,470,491,508,518,549,568,574,589,
%U 592,624,629,636,690,696,729,748,753,769,770,771,781,791,802,833,844
%N Numbers n such that 9900n^2 is the average of a twin prime pair.
%C Inspired by _Zak Seidov_'s post to the SeqFan list, cf. link: This yields A154674 as 9900 a(n)^2. Indeed, if N/11 is a square, then N=11 m^2 and this can't be the average of a twin prime pair unless m=30a (considering N+1 mod 2,3,5 and N1 mod 5).
%H Zak Seidov, <a href="http://zak08.livejournal.com/4070.html">"A154676"</a>, Jan 15 2009
%F a(n) = sqrt(A154674(n)/9900).
%t Select[Range[1000],AllTrue[9900#^2+{1,1},PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* _Harvey P. Dale_, Jan 19 2019 *)
%o (PARI) for(i=1,999, isprime(9900*i^2+1) & isprime(9900*i^21) & print1(i","))
%Y Cf. A037073, A154331, A154772A154773.
%K nonn
%O 1,1
%A _M. F. Hasler_, Jan 15 2009
