3.2.15 * Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=a_{2}=1$, and $a_{n+2}=\frac{1}{a_{n+1}}+a_{n}, n=1,2, \cdots$, find $a_{2004}$.
From the given, we have $a_{n+2} a_{n+1} - a_{n+1} a_{n} = 1$, so $\left\{a_{n+1} a_{n}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. Therefore, $a_{n+1} a_{n} = n, n=1$, $2, \cdots$, thus $a_{n+2} - \frac{n+1}{a_{n+1}} = \frac{n+1}{\frac{n}{a_{n}}} = \frac{n+1}{n} a_{n}, n=1,2, \cdots$, so
$$
\begin{aligned}
a_{2004} & = \frac{2003}{2002} a_{2002} = \frac{2003}{2002} \cdot \frac{2001}{2000} a_{2000} \\
& = \cdots = \frac{2003}{2002} \cdot \frac{2001}{2000} \cdot \cdots \cdot \frac{3}{2} a_{2} = \frac{3 \cdot 5 \cdot \cdots \cdot 2003}{2 \cdot 4 \cdot \cdots \cdot 2002}\left(= \frac{2003!!}{2002!!}\right) .
\end{aligned}
$$