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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

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

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  • 2
    $\begingroup$ There's some superficial similarity to the sequences (Somos, Gale-Robinson, etc.) which (in part) motivated the initial study of cluster algebras. $\endgroup$ – Sam Hopkins Dec 17 '19 at 17:33
  • $\begingroup$ You switch at some point from $t$ to $f$; are they the same? Also, what does "when $A B a_n$ is an integer sequence" mean? $\endgroup$ – LSpice Dec 17 '19 at 18:11
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    $\begingroup$ Finally, why not ask your mentor how to prove the asserted equality? $\endgroup$ – LSpice Dec 17 '19 at 18:11
  • $\begingroup$ It means that when you multiply $A$ and $B$ and $a_n$ that for every n its value is an integer. And oh yes sorry t = f. $\endgroup$ – mick Dec 17 '19 at 19:09
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Sometimes it's helpful to look at problems of this sort in a slightly more general geometric framework, especially if you want to work over other fields. You can model your problem as a map of $\mathbb A^2$ using rational functions, but I find it better to look at a model in $\mathbb P^2$ and use homomgeneous coordinates. So consider the function $$ F(x,y,z) = [xy,y(y+Cz),xz],\quad F:\mathbb P^2\to\mathbb P^2. $$ Start with the point $[A,B,1]$, and write the $n$th iterate of $F$ as $$ F^{\circ n}([A,B,1]) = [U_n,V_n,W_n]. $$ Then, if I haven't made an algebra error, your sequence is $$ U_n/W_n = a_{n+1}\quad\text{and}\quad V_n/W_n = a_{n+2}. $$ The map $F$ is a rational map of degree 2 on $\mathbb P^2$, with three indeterminacy points, $[0,0,1]$, $[1,0,0]$ and $[0,-C,1]$. None of this answers your question, of course, but it may point you in the direction of the general theory of rational maps (of degree $2$) on $\mathbb P^2$, for which there is a voluminous literature.

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