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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.

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    $\begingroup$ Does this answer your question? Solving nonlinear recursive relation $\endgroup$
    – Anne Bauval
    Commented Apr 2, 2023 at 5:02
  • $\begingroup$ 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. $\endgroup$
    – WantsToKnow
    Commented Apr 2, 2023 at 5:25
  • $\begingroup$ The answers in that duplicate (out of many) give methods (and links) to get a closed form for a general homographic sequence. $\endgroup$
    – Anne Bauval
    Commented Apr 2, 2023 at 5:29
  • $\begingroup$ 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. $\endgroup$
    – Anne Bauval
    Commented Apr 2, 2023 at 5:38
  • $\begingroup$ Thanks. I'm looking at the links and will update the question accordingly. $\endgroup$
    – WantsToKnow
    Commented Apr 2, 2023 at 5:40

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