%I #10 Nov 04 2013 21:52:29
%S 6,6726,13765255184676885126
%N Numerators in a product expansion for sqrt(2).
%C a(3) has 96 digits and a(4) has 479 digits.
%C Iterating the algebraic identity sqrt(1 + 4/x) = (1 + 2*(x + 2)/(x^2 + 3*x + 1)) * sqrt(1 + 4/(x*(x^2 + 5*x + 5)^2)) produces a rapidly converging product expansion sqrt(1 + 4/x) = product {n = 0..inf} (1 + 2*a(n)/b(n)), where a(n) and b(n) are integer sequences when x is a positive integer.
%C The present case is when x = 4. The denominators b(n) are in A219015. See also A219010 (x = 1) and A219012 (x = 2).
%F Let tau = 3 + 2*sqrt(2). Then a(n) = tau^(5^n) + 1/tau^(5^n).
%F Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 6.
%e The first two terms of the product give 18 correct decimal places of sqrt(2): (1 + 2*6/29)*(1 + 2*6726/45232349) = 1.41421 35623 73095 048(5...).
%Y Cf. A219010, A219012, A219015.
%K nonn,easy,bref
%O 0,1
%A _Peter Bala_, Nov 09 2012