TITLE: Does there exist a pure recurrence formula with polynomial coefficients for Fibonacci(2^n)?
QUESTION [9 upvotes]: Let F(n) be the Fibonacci sequence as defined by F(1)=1, F(2)=1, F(n)=F(n-1)+F(n-2) for n>=3.  I'm looking for a pure recurrence formula for the function X(i)=F(2i) whose coefficients may be polynomials in i.  This is Sloane's A058635.  I also would like it to be "pure" in the sense that there is no auxiliary function involved.  Is such a formula known?
I attempted using Sister Celine's technique (as described in A=B) with the data up to 221 without success.
My motivation is that I have a fairly complicated recurrence formula for another sequence, but I am only interested in the terms whose indices are of the form 2i-3.  The existence (or non-existence) of a recursion for X(i) would be a kind of "proof of concept" as to whether or not I should explore the possibility of finding such a recursion for my sequence.

REPLY [2 votes]: Interestingly, the recursion $u_{n+1} = (u_n + 5/u_n)/2$, with $u_0=1$, gives the fractions $Lucas(2^n)/Fibonacci(2^n)$.