TITLE: Show that $x_{n+2} = x_{n+1} + \frac{x_n}{2^n}$ is a bounded sequence
QUESTION [7 upvotes]: Given a sequence:
  $$
\begin{cases}
x_{n+2} = x_{n+1} + \frac{x_n}{2^n}\\
x_1 = 1\\
x_2 = 1 \\
n \in \mathbb N
\end{cases}
$$
  Show that $\{x_n\}$ is a bounded sequence.

This recurrence feels like a bad joke for precalculus level. Is it possible to show that it is bounded and find some estimations for the bounds. Clearly $x_n$ is greater than $0$. ${1\over 2^n}$ coefficient makes it hard to find closed form of the recurrence.
I've expanded a few first terms to find a pattern but it doesn't look like it exists:
$$
x_n = {1}, {1}, {3\over 2}, {7\over 4}, {31\over 16}, {131\over 64}, {1079\over 512}, {8763\over 4096}, \
{141287\over 65536}, {2269355\over 1048576} \dots
$$
Denominator is in some form of $2^{k_n}$ from which i've figured out $k_n$ is in the form:
$$
k_n = \left\lfloor {n^2\over 4}\right\rfloor
$$
So the denominator is in the form:
$$
d_n = 2^{\left\lfloor {n^2\over 4} \right\rfloor}
$$
The sequence seems to be bounded since computing a few terms shows it tends to some number:
$$
M = 2.1726687\dots
$$
I have these questions in mind regarding the given sequence:

What is the kind of the sequence to understand how to search for its kind on the internet (linear/non-linear, homogenous/non-homogenous, ...)
How do i show that it is bounded using precalculus maths only?
Is it possible to find its closed form?

Please note this problem is in precalculus even before limits are defined.

REPLY [3 votes]: Consider the auxillary sequence  $\displaystyle\;y_n = \frac{x_n}{1-2^{2-n}}$ defined for $n >2$.
It is clear all $y_n$ are positive. Notice
$$\begin{align}
y_{n+2} &= \frac{1}{1-2^{-n}}\left[(1-2^{1-n})y_{n+1} + 2^{-n}(1 - 2^{2-n}) y_n\right]\\
&\le \frac{1}{1-2^{-n}}\left[(1-2^{1-n})y_{n+1} + 2^{-n} y_n\right]\\
&\le \frac{1}{1-2^{-n}}\left[(1-2^{1-n}) + 2^{-n}\right]\max( y_n, y_{n+1} )\\
& = \max(y_n,y_{n+1})
\end{align}
$$
We find for all $n > 4$, 
$$x_n < y_n \le \max(y_{n-1},y_{n-2}) \le \max(y_{n-2},y_{n-3}) \le \cdots \le \max(y_3,y_4) 
= 3$$
Since $x_1,x_2,x_3,x_4 < 3$, we find $x_n < 3$ for all $n$.