5 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 $a_{2004}$. (Wu Weizhao)
From the given condition, we have
$$
a_{n+2} a_{n+1}-a_{n+1} a_{n}=1,
$$

Therefore, $\left\{a_{n+1} a_{n}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1, thus
$$
a_{n+1} a_{n}=n, n=1,2, \cdots,
$$

Hence, $\quad a_{n+2}=\frac{n+1}{a_{n+1}}=\frac{n+1}{\frac{n}{a_{n}}}=\frac{n+1}{n} a_{n}, n=1,2, \cdots$,
So $a_{2004}=\frac{2003}{2002} a_{2002}=\frac{2003}{2002} \cdot \frac{2001}{2000} a_{2000}$
$$
\begin{array}{l}
=\cdots=\frac{2003}{2002} \cdot \frac{2001}{2000} \cdot \cdots \cdot \frac{3}{2} a_{2} \\
=\frac{3 \cdot 5 \cdot \cdots \cdot 2003}{2 \cdot 4 \cdot \cdots \cdot 2002}\left(=\frac{2003!!}{2002!!}\right)
\end{array}
$$