Question: Here is the question:

Define a sequence by $x\_0=5$ and
$$
x\_{n+1}=\frac{x\_{n}^{2}+5 x\_{n}+4}{x\_{n}+6}.
$$
Let $m$ be the least positive integer such that 
$$
x\_{m} \leq 4+\frac{1}{2^{20}},
$$
In which of the following intervals does $m$ lie?
$$
[9,26],[27,80],[81,242],[243,728],[729,+\infty].
$$

I noticed that these intervals involve powers of $3$ so I am thinking of using some approximations to $x\_n$. This sequence seems to have no general formula, which makes me struggle to find a formula that could approximate $m$.

Is this the correct way or are there better solutions to this problem?

Here is the plot of the sequence.
[![enter image description here](https://i.stack.imgur.com/cicl0.png)](https://i.stack.imgur.com/cicl0.png)

Answer: 
Write $x\_n:=4+y\_n$ $(n\geq0)$ with $y\_0:=1$. Then you obtain the recursion
$$y\_{n+1}={9y\_n+y\_n^2\over 10+y\_n}=y\_n\left({9\over10}+{y\_n\over100+10 y\_n}\right)\qquad(n\geq0)\ .\tag{1}$$
This shows that the $y\_n$ are approximately multiplied with ${9\over10}$ at each step. Solving $$\left({9\over10}\right)^m=2^{-20}$$
for $m$ gives $m\approx131.576$, so that we conjecture the true $m\_\*$ to lie in the interval $[81,242]$. 

For a *proof* we write $(1)$ in the form
$$y\_{n+1}=\lambda\_n y\_n\quad(n\geq0),\qquad\lambda\_n:={9+y\_n\over 10+y\_n}\ .$$ 
As $0<y\_n\leq1$ for all $n\geq0$ we have 
$${9\over10}\leq\lambda\_n\leq {10\over11}\qquad(n\geq0)\ .$$
This implies $$\left({9\over10}\right)^n\leq y\_n\leq\left({10\over11}\right)^n\qquad(n\geq0)\ .$$
Solving $\left({9\over10}\right)^m=2^{-20}$ and $\left({10\over11}\right)^m=2^{-20}$ then gives you a lower and an upper estimate for the true $m\_\*$, namely $132\leq m\_\*\leq146$.
Seems like I've got an approximation.

Let $y\_n=x\_n-4$, then 
$$
y\_{n}=y\_n\left(1-\frac{1}{y\_{n-1}+10}\right)<\frac{9}{10}y\_{n-1}<\cdots<\left(\frac{9}{10}\right)^ny\_0=\left(\frac{9}{10}\right)^n.
$$
Let $\left(\frac{9}{10}\right)^n\leqslant \frac{1}{2^{20}}$, we get 
$$
n\leqslant \log\_{9/10} \frac{1}{2^{20}}\approx 131.5762695792117.
$$
Seems like I should choose the interval $[81,242]$...? I still don't understand why the choices are divided by powers of 3.