General formula for a(n+1)=a(n)^2+2*a(n)2 and a(1)=k+1 is a(n)=Floor[((k + Sqrt[k^2 + 4])/2)^(2^((n+1)  1))
From Peter Bala, Nov 12 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.
Let x > 2 and let alpha := {x + sqrt(x^2  4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(2^(n1)) + (1/alpha)^(2^(n1))  1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n)  2 with the initial condition a(1) = x  1.
A second recurrence is a(n) = (a(1) + 2)*{product {k = 1..n1} a(k)}  2.
The following algebraic identity is valid for x > 2:
(x + 1)/sqrt(x^2  4) = (1 + 1/(x  1))*(y + 1)/sqrt(y^2  4), where y  1 = (x  1)^2 + 2*(x  1)  2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2  4) = product {n = 1..inf} (1 + 1/a(n)).
A second expansion is product {n = 1..inf} (1 + 2/(a(n) + 1) = sqrt((x + 2)/(x  2)). For an alternative approach to these identities see the Bala link.
(End)
