$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$?

Question

Consider the following sequence : Let $a_1 = a_2 = 1.$ For integer $ n > 2 : $

$$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$

$$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

$$T = ??$$

What is the value of $T$ ?

Is there a closed form or integral for $T$?

I get

$$ T = 3.73205080..$$

The convergeance is fast.

Does anyone recognize this ?

——

Edit

So apparantly $ T = 2 + \sqrt 3 $

Let us generalize.

Take $a_1= 1, a_2 > a_1$

And now the whole sequence depends on $y = a_2$.

We thus define

$$ T(y) = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$

We know $T(1) = T(2) = 2 + \sqrt 3 $.

$$T(3) = 4.4415184401122.. $$

Apparantly $T(3) = \frac{7 + 2 \sqrt 10}{3} $ as found ( no proof ) by lhf.

How about a closed form for $T(y)$ ?

Can all of these rational recursions be transformed into a linear recursion ?

——— Update

See also

About $a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}} , t(12,13) = \frac{3}{2}$

Answer 1

Empirical answer to the last question:

$T(y)$ seems to be $\dfrac{A(y)+\sqrt {B(y+2)}}{y}$, where $A(y)$ is OEIS/A000124 and $B(y)$ is OEIS/A327319.

Tested for $y=1,2,\dots,6$.

Answer 2

EDIT: By already knowing the solution to your recurrence a priori, I am able to solve your problem. However, a more interesting question might be why your recurrence is equivalent to $$a_n=5a_{n-1}-5a_{n-2}+a_{n-3}$$ and how to see this a priori.

Your recurrence is solved by the sequence (as can be checked by direct calculations) $$a_{n+1} = \frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^n\cdot(3+\sqrt 3)}{12}.$$

So \begin{split}\frac{a_{n+1}}{a_n}&=\frac{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^n + (2-\sqrt 3)^{n}\cdot(3+\sqrt 3)}{6+(3-\sqrt 3)\cdot(2+\sqrt 3)^{n-1} + (2-\sqrt 3)^{n-1}\cdot(3+\sqrt 3)} \\ &= (2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+3-\sqrt 3+3+\sqrt 3}{\frac{6}{(2+\sqrt3)^{n-1}}+3-\sqrt 3+3+\sqrt 3}\\ &=(2+\sqrt 3)\cdot\frac{\frac{6}{(2+\sqrt3)^{n}}+6}{\frac{6}{(2+\sqrt3)^{n-1}}+6} \\&\xrightarrow{n\to\infty}(2+\sqrt 3). \end{split}

I got the closed form for $a_{n+1}$ from A101879 of the OEIS.