Analytic solution to recursive functions

Let's say we have a function $$F( n \in \mathbb{Z}^{+} )$$ where

$$F(1) = \mathrm{const.}$$ $$F( n \geq 2 ) = C\frac{ F(n-1) + a }{ F(n-1) + b }$$

Where C, a and b are constants. Is there a clear-cut process to go about finding a generic solution to F(n)?

I realize that being a constant F(1) can be rewritten as $$F(1) = C\frac{ A + a }{ A + b }$$ but it doesn't help much.

Edit: please note that the function F appears in both the numerator and the denominator.

I have tried rewriting F as $$F(n) = C F_1(n) F_2(n)$$ $$F_1(n) = F(n-1) + a$$ $$F_2(n) = \frac{1}{ F(n-1) + b }$$

But again it doesn't help, because upon expansion of those functions F(n) appears again.

• Does this answer your question? Solving nonlinear recursive relation Commented Apr 2, 2023 at 5:02
• The example only has the function in the denominator, so it can be easily simplified, whereas in this instance the function F appears in both the numerator and the denominator. Commented Apr 2, 2023 at 5:25
• The answers in that duplicate (out of many) give methods (and links) to get a closed form for a general homographic sequence. Commented Apr 2, 2023 at 5:29
• People are not supposed to read the comments to understand what you mean by "The example" in your edit. Moreover, my previous comment answers that objection. Commented Apr 2, 2023 at 5:38
• Thanks. I'm looking at the links and will update the question accordingly. Commented Apr 2, 2023 at 5:40