Question 126, Sequence $\left\{a_{n}\right\}_{n=0}^{+\infty}: a_{0}=2, a_{1}=\frac{5}{2}, a_{n+2}=a_{n+1}\left(a_{n}{ }^{2}-2\right)-\frac{5}{2}$. Try to find the general formula for $a_{n}$.
Question 126, Solution: Let $a_{n}=x^{b_{n}}+x^{-b_{n}}$, then $x=2, b_{0}=0, b_{1}=1$. And
$$
\begin{array}{l}
a_{n+2}=x^{b_{n+2}}+x^{-b_{n+2}}=\left(x^{b_{n+1}}+x^{-b_{n+1}}\right)\left(\left(x^{b_{n}}+x^{-b_{n}}\right)^{2}-2\right)-\frac{5}{2} \\
=\left(x^{b_{n+1}}+x^{-b_{n+1}}\right)\left(x^{2 b_{n}}+x^{-2 b_{n}}\right)-\frac{5}{2} \\
=x^{b_{n+1}+2 b_{n}}+x^{-\left(b_{n+1}+2 b_{n}\right)}+\left(x^{b_{n+1}-2 b_{n}}+x^{-\left(b_{n+1}-2 b_{n}\right)}\right)-\frac{5}{2}
\end{array}
$$

If we take $b_{n+2}=b_{n+1}+2 b_{n}$, then by the characteristic root method, we get $b_{n}=\frac{2^{n}-(-1)^{n}}{3}$. At this point, $b_{n+1}-2 b_{n}=$ $\frac{2^{n+1}-(-1)^{n+1}}{3}-\frac{2\left(2^{n}-(-1)^{n}\right)}{3}=(-1)^{n}$, hence $x^{b_{n+1}-2 b_{n}}+x^{-\left(b_{n+1}-2 b_{n}\right)}=x+\frac{1}{x}=\frac{5}{2}$. Substituting into (*), it holds exactly.
In summary, $a_{n}=x^{b_{n}}+x^{-b_{n}}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{2^{n}-(-1)^{n}}{3}}$.