$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$? 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 ? 
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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 ? 
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Update
See also
About $a_n = \frac{a_{n-1}(a_{n-1} + C)}{a_{n-2}} , t(12,13) = \frac{3}{2}$

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.
