%I #18 Apr 22 2017 23:03:22
%S 1,5,25,325,8125,99125,8425625,1221715625,44226105625,11719917990625,
%T 986817094810625,94931804520782125,65028286096735755625,
%U 10937757721470954096125,10117425892360632538915625,11179755611058498955501765625
%N a(0) = 1; for n>1, a(n) is the numerator of b(n) = Product_{i=1..n} (prime(i)^2 + 1)/(prime(i)^2 - 1).
%C The limit of b(n) is 5/2.
%D Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005.
%H Seiichi Manyama, <a href="/A285602/b285602.txt">Table of n, a(n) for n = 0..284</a>
%e b(1) = (2^2 + 1)/(2^2 - 1) = 5/3, so a(1) = 5.
%e b(2) = 5/3 * (3^2 + 1)/(3^2 - 1) = 25/12, so a(2) = 25.
%e b(3) = 25/12 * (5^2 + 1)/(5^2 - 1) = 325/144, so a(3) = 325.
%e b(4) = 325/144 * (7^2 + 1)/(7^2 - 1) = 8125/3456, so a(4) = 8125.
%t a[n_]:= If[n==0, 1, Numerator[Product[(Prime[k]^2 + 1)/(Prime[k]^2 - 1), {k, n}]]]; Table[a[n], {n, 0, 15}] (* _Indranil Ghosh_, Apr 22 2017 *)
%o (PARI) a(n) = if (n==0, 1, numerator(prod(k=1, n, (prime(k)^2 + 1)/(prime(k)^2 - 1)))); \\ _Michel Marcus_, Apr 22 2017
%Y Cf. A000040, A285603.
%K nonn,frac
%O 0,2
%A _Seiichi Manyama_, Apr 22 2017