<s>[INST] Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3 
Thanks in advance!
 [/INST] Let $2x_{n+2} = 3 \, x_{n+1} + 8 \, x_{n} + 3 \, x_{n-1}$, with $x_{0} = -1,
x_{1} = 3$ and $x_{2} = 3$ for which letting $x_{n} = p^{n}$ the equation 
$$2 \, p^{3} - 3 \, p^{2} - 8 \, p - 3 = 0$$ 
is obtained. This equation can be factored to $(p-3)(p+1)(2p+1) = 0$ and leads to the roots $p \in \{ 3, -1, -1/2 \}$. From this the general form of $x_{n}$ is
$$x_{n} = a_{0} \, 3^{n} + a_{1} \, (-1)^{n} + a_{2} \, \left( - \frac{1}{2} \right)^{n}.$$ 
Applying the initial conditions yields:
\begin{align}
-1 &= a_{0} + a_{1} + a_{2} \\
3 &= 3 \, a_{0} - a_{1} - \frac{a_{2}}{2} \\
3 &= 9 \, a_{0} + a_{1} + \frac{a_{2}}{4}
\end{align}
which leads to $a_{0} = \frac{1}{2}$, $a_{1} = - \frac{3}{2}$, $a_{2} = 0$.
The resulting sequence is generated by
\begin{align}
x_{n} = \frac{3}{2} \, \left( 3^{n-1} + (-1)^{n-1} \right)  
\end{align} 
</s>