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a(0) = 1; for n>1, a(n) is the denominator of b(n) = Product_{i=1..n} (prime(i)^2 + 1)/(prime(i)^2 - 1).
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%I #16 Apr 22 2017 23:03:29

%S 1,3,12,144,3456,41472,3483648,501645312,18059231232,4767637045248,

%T 400481511800832,38446225132879872,26297217990889832448,

%U 4417932622469491851264,4082169743161810470567936,4506715396450638759507001344,486725262816668986026756145152

%N a(0) = 1; for n>1, a(n) is the denominator 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="/A285603/b285603.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) = 3.

%e b(2) = 5/3 * (3^2 + 1)/(3^2 - 1) = 25/12, so a(2) = 12.

%e b(3) = 25/12 * (5^2 + 1)/(5^2 - 1) = 325/144, so a(3) = 144.

%e b(4) = 325/144 * (7^2 + 1)/(7^2 - 1) = 8125/3456, so a(4) = 3456.

%t a[n_]:= If[n==0, 1, Denominator[Product[(Prime[k]^2 + 1)/(Prime[k]^2 - 1), {k, n}]]]; Table[a[n], {n, 0, 16}] (* _Indranil Ghosh_, Apr 22 2017 *)

%o (PARI) a(n) = if (n==0, 1, denominator(prod(k=1, n, (prime(k)^2 + 1)/(prime(k)^2 - 1)))); \\ _Michel Marcus_, Apr 22 2017

%Y Cf. A000040, A285602.

%K nonn,frac

%O 0,2

%A _Seiichi Manyama_, Apr 22 2017