Let $A,B,C > 0$. Put $a_1 = A$ and $a_2 = B$ and, for integer $n > 2$, $$a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}}$$ and $$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

Notice the limit seems to depend on 3 variables, but actually only depends on $A/C$, $B/C$. You can check that if you multiply $A$, $B$, $C$ by $D>0$ you get the same limit value, hence the above is true.

Let’s rewrite it as a function. $$T = t(A/C,B/C) = t(B/C,A/C) $$ So this function is symmetric. ( may I say commutative?? And scale-invariant ??)

So for simplicity we can reduce to taking $C = 1$.

So from now on we take $C= 1$ and we prefer $B \ge A$.

So we investigate

Let $ B \ge A > 0$. Put $a_1 = A$ and $a_2 = B$, and, for integer $n > 2$,

$$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$ and $$T = t(A, B) = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

We notice

$$t(A,B) = t(a_2,a_3) = t(a_3,a_4) = t(a_4,a_5)= \dotsb $$

And when $B \gg A > 1 $ then $t(A,B)$ is close to $B/A$.

Some identities are known and some are proven.

\begin{gather*} t(1,1) = t(1,2) = t(2,6) = 2 + \sqrt 3 \\ t(4,4) = 2 \\ t(3,6) = 1 + \phi \\ \lim_{h \to \infty} t(h, vh) = v. \end{gather*}

In particular when $A B a_n$ is an integer sequence , it often satisfies a Fibonacci type recursion or almost satisfies it. And that helps to find and prove the value of $t(A,B)$.

But there are many open questions.

Such as extending to complex numbers. Or investigating if the function is analytic. How the function looks on the complex plane.

Or just simply getting closed form solutions for a given $A$, $B$.

That is a pretty long intro so I will ask the main question.

My mentor tommy1729 said

$$ t(12,13) = \frac{3}{2}. $$

How to prove this?

See also

https://math.stackexchange.com/questions/3396793/a-n-fraca-n-1a-n-1-1a-n-2-and-t-3-73205080

P.S. A similar identity. Let $a_1 = a_2 = 1$ and, for integer $ n > 2$, $$a_n = \frac{(a_{n-1} + \frac{1}{6})^2}{a_{n-2}}$$ and $$T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$ Then $$T = t^2(12,13) = \frac{9}{4}.$$

——-

Edit

Extended conjecture.

Let $n>0$ be an integer and let $F(n)$ be the n’th triangular number. So - for clarity $F(1) = 1,F(2)=3,F(3)= 6 , ... $

Then we have

$$ t( 4 F(n), 4 F(n) ) = \frac{n+1}{n} $$