Let
$f_0=x,\ f_{i+1}=f_i\pm\dfrac1{f_i}$
for $i\geq0$, i.e., $f_i=\dfrac{\sqrt{f_{i+1}^2\mp4}+f_{i+1}}2$
I have observed that
$f_1=\dfrac{x^2\pm1}x$
$f_2=\dfrac{x^4\pm3x^2+1}{x(x^2\pm1)}$
$f_3=\dfrac{x^8\pm7x^6+13x^4\pm7x^2+1}{x(x^6\pm4x^4+4x^2\pm1)}$
$f_4=\dfrac{x^{16}\pm15x^{14}+83x^{12}\pm220x^{10}+303x^8\pm220x^6+83x^4\pm15x^2+1}{x(x^{14}\pm11x^{12}+45x^{10}\pm88x^8+88x^6\pm45x^4+11x^2\pm1)}$,
and that in the positive case each denominator is the product of all previous numerators.
How can I determine the coefficients?
Edit: Note that all $\pm$ and $\mp$ here are considered to be a single instance, meaning that all $\pm$ are the opposite of all $\mp$, transcendent of the index $i$.
Sloane's A147990 appears to provide the coefficients for the numerators in the (negative) case when $\pm$ is $-$, so the absolute values should give the coefficients in the positive case. I think I might be able to establish a formula by crawling around oeis for a bit.
Edit: Oops, I didn't realize that the sequence A147985 that A147990 links to is precisely the sequence I'm looking for.
To be honest, I'm not certain how to define the coefficients based on what's given there.
And then the denominators appear to be given by A147986, which, as I said, appear to be the product of all previous numerators (in the positive case) as well.