15. (1993 China National Training Team Test Question) Let $a_{1}=a_{2}=\frac{1}{3}$, and for $n=3,4,5, \cdots$, $a_{n}=$ $\frac{\left(1-2 a_{n-2}\right) a_{n-1}^{2}}{2 a_{n-1}^{2}-4 a_{n-2} a_{n-1}^{2}+a_{n-2}}$, find the general term formula of the sequence $\left\{a_{n}\right\}$.
15. Slightly explained: By mathematical induction, it is easy to prove $00\right)$, then $b_{n}=\frac{\left(b_{n-1}^{2}+2\right)}{b_{n-2}}$. From this, we can get $b_{n} b_{n-2}=b_{n-1}^{2}+2 \cdots$ (1)
In (1), replace $n$ with $n-1$ to get $b_{n-1} b_{n-3}=b_{n-2}^{2}+2 \cdots$ (2)
From (1) and (2), we have $b_{n} b_{n-2}-b_{n-1}^{2}=b_{n-1} b_{n-3}-b_{n-2}^{2}=2$, which means $b_{n-2}\left(b_{n}+b_{n-2}\right)=b_{n-1}\left(b_{n-3}+b_{n-1}\right)$.
Therefore, $\frac{b_{n}+b_{n-2}}{b_{n-1}}=\frac{b_{n-1}+b_{n-3}}{b_{n-2}}=\cdots=\frac{b_{3}+b_{1}}{b_{2}}$.
And $b_{1}=\sqrt{\frac{1}{a_{1}}-2}=1, b_{2}=\sqrt{\frac{1}{a_{2}}-2}=1, b_{3}=\frac{b_{2}^{2}+2}{b_{1}}=3$.
So $\frac{b_{n}+b_{n-2}}{b_{n-1}}=\frac{3+1}{1}=4$, which means $b_{n}=4 b_{n-1}-b_{n-2} \cdots$ (3)
It is easy to know that the characteristic equation of (4) is $r^{2}=4 r-1$, solving which gives $r_{1}=2+\sqrt{3}, r_{2}=2-\sqrt{3}$ as the characteristic roots.
Therefore, we set $b_{n}=c_{1}(2+\sqrt{3})^{n}+c_{2}(2-\sqrt{3})^{n}$.
We define $b_{0}$ to satisfy $b_{2}=4 b_{1}-b_{0}$, i.e., $b_{0}=4 b_{1}-4 b_{2}=3$, then
$\left\{\begin{array}{l}c_{1}+c_{2}=3 \\ c_{1}(2+\sqrt{3})+c_{2}(2-\sqrt{3})=1\end{array}\right.$, solving which gives $c_{1}=\frac{3}{2}-\frac{5}{6} \sqrt{3}, c_{2}=\frac{3}{2}+\frac{5}{6} \sqrt{3}$.
Therefore, $\dot{b}_{n}=\left(\frac{3}{2}-\frac{5}{6} \sqrt{3}\right)(2+\sqrt{3})^{n}+\left(\frac{3}{2}+\frac{5}{6} \sqrt{3}\right)(2-\sqrt{3})^{n}$.
Thus, we have $b_{n}=\left(b_{n}^{2}+2\right)^{-1}$
$=\left\{\left[\left(\frac{3}{2}-\frac{5}{6} \sqrt{3}\right)(2+\sqrt{3})^{n}+\left(\frac{3}{2}+\frac{5}{6} \sqrt{3}\right)(2-\sqrt{3})^{n}\right]^{2}+2\right\}^{-1}\left(n \in \mathbf{N}^{*}\right)$.