Interesting sequence with a beautiful term : $U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$ Question : 
Find the term of $U_{n}$ such that : 
$U_{0}=2$ and $U_{1}=\frac{5}{2}$ 
$U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$
My try : 
We have ; $U_{2}=\frac{5}{2}$ $U_{3}=2^{3}+2^{-3}$
After calculus ...  I think the term is : 
$U_{n}=2^{a_n}+2^{-a_n}$
Where $a_n=\frac{2^{n}-(-1)^{n}}{3}$ but I don't 
need use induction ? 
If any one have another method please tell me ? 
 A: Given any $v > 1$, construct sequence $\displaystyle\;v_n = \begin{cases} 1, & n = 0\\ v, & n = 1\\ v_{n-1}v_{n-2}^2 & n > 1\end{cases}$
Notice
$$v_{n+1} = v_n v_{n-1}^2 \iff
\frac{v_{n+1}}{v_n^2} = \frac{v_{n-1}^2}{v_n}
\implies 
\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}} =
\frac{v_{n-1}^2}{v_n} + \frac{v_n}{v_{n-1}^2}
$$
The value of expression $\displaystyle\;\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}}$ is independent of $n$ and equals to $v + \frac1v$.
Let $\displaystyle\;u_n = v_n + \frac1{v_n}$, we obtain
$$\begin{align} u_{n+1} &= v_{n+1} + \frac1{v_{n+1}}
= v_n v_{n-1}^2 + \frac{1}{v_n v_{n-1}^2}\\
&= v_n v_{n-1}^2 + \frac{1}{v_n v_{n-1}^2} +
\left[\left(\frac{v_n}{v_{n-1}^2} + \frac{v_{n-1}^2}{v_n}\right) - \left(v + \frac1v\right)\right]\\
&= \left( v_n + \frac1{v_n}\right)\left(v_{n-1}^2 + \frac{1}{v_{n-1}^2}\right)
- \left( v + \frac1v \right)\\
&= u_n \left( u_{n-1}^2 - 2 \right) - \left(v + \frac1v\right)
\end{align}
$$
Compare the recurrence relations between $U_n$ and $u_n$, we find $U_n$ is a special case of $u_n$ for $v = 2$.
To derive an explicit expression for $v_n$ and hence for $U_n$, we can use the fact
$$v_{n+1}v_n = (v_n v_{n-1})^2, \forall n \ge 1
\quad\implies\quad
v_n v_{n-1} = (v_1 v_0)^{2^{n-1}} = v^{2^{n-1}}, \forall n \ge 1$$
This leads to 
$$v_n = (v_{n-1} v_{n-2}) v_{n-2} = v^{2^{n-2}} v_{n-2}, \forall n \ge 2$$
Together with $v_0 = 1, v_1 = v$, we find 
$$\begin{align}
v_{2k} &= v^{ 2^{2k-2} + 2^{2k-4} + \cdots + 2^0}(1) = v^{\frac{2^{2k} - 1}{3}}\\
v_{2k+1} &= v^{ 2^{2k-1} + 2^{2k-3} + \cdots + 2^1}(v) = v^{\frac{2^{2k+1} + 1}{3}}
\end{align}
\quad\implies\quad v_n = v^{\frac{2^n - (-1)^n}{3}}
$$
As a special case for $v = 2$, we obtain
$$U_n = 2^{a_n} + 2^{-a_n}\quad\text{ where }\quad a_n = \frac{2^n - (-1)^n}{3}$$
A: Here's another one of my
not-quite-there
incomplete solutions.
If
$U_{n+1}
=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}
$
and
$U_n = 2^{a_n}+2^{-a_n}
$
then
$\begin{array}\\
U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}
&=(2^{a_n}+2^{-a_n})((2^{a_{n-1}}+2^{-a_{n-1}})^2-2)-\frac{5}{2}\\
&=(2^{a_n}+2^{-a_n})(2^{2a_{n-1}}+2^{-2a_{n-1}}+2-2)-\frac{5}{2}\\
&=(2^{a_n}+2^{-a_n})(2^{2a_{n-1}}+2^{-2a_{n-1}})-\frac52\\
&=(2^{a_n+2a_{n-1}}+2^{a_n-2a_{n-1}}+2^{-a_n+2a_{n-1}}+2^{-a_n-2a_{n-1}})-\frac52\\
&=(2^{a_n+2a_{n-1}}+2^{-a_n-2a_{n-1}}+2^{a_n-2a_{n-1}}+2^{-a_n+2a_{n-1}})-\frac52\\
\text{so we want}\\
2^{a_{n+1}}+2^{-a_{n+1}}
&=(2^{a_n+2a_{n-1}}+2^{-a_n-2a_{n-1}}+2^{a_n-2a_{n-1}}+2^{-a_n+2a_{n-1}})-\frac52\\
\end{array}
$
and right now
I don't see how to choose
$a_{n+1}$
as a function of
$a_n$ and $a_{n-1}$
and what the
$\frac52$
has to do with it.
Once again,
your turn.
