%I #10 Dec 25 2019 04:49:47
%S 1,3,7,14,22,29,41,46,62,64,67,88,167,179,207,231,239,249,263,266,286,
%T 290,309,315,322,323,326,344,350,353,354,372,392,421,444,454,458,496,
%U 505,553,560,561,571,585,613,636,647,661,669,682,745,788,790,791,815
%N Numbers m such that 180 m^2 is the average of a twin prime pair.
%C Inspired by Z. Seidov's post to the SeqFan list, cf. link. This yields A154672 as 180 a(n)^2. Indeed, if N is such that N/5 is a square, then M=5m^2 and this can't by the average of a twin prime pair unless m=6a.
%H Amiram Eldar, <a href="/A154772/b154772.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(A154672(n)/180)
%t Select[Range[10^3], And @@ PrimeQ[180#^2 + {-1, 1}] &] (* _Amiram Eldar_, Dec 25 2019 *)
%o (PARI) for(i=1,999, isprime(180*i^2+1) & isprime(180*i^2-1) & print1(i","))
%Y Cf. A014574, A037073, A154331, A154672.
%K nonn
%O 1,2
%A _M. F. Hasler_, Jan 15 2009