Five, let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a_{2}=1$, and
$$
a_{n+2}=\frac{1}{a_{n+1}}+a_{n}, n=1,2, \cdots \text {. }
$$

Find $a_{2004}$.
(Supplied by Wu Weizhao)
$$
\begin{array}{l}
\text { From the given, } a_{n+2} a_{n+1}-a_{n+1} a_{n}=1, \text { so, } \left\{a_{n+1} a_{n}\right\} \text { is an arithmetic sequence with the first term 1 and common difference 1. } \\
\text { Therefore, } a_{n+1} a_{n}=n, n=1,2, \cdots . \\
\text { Thus, } a_{n+2}=\frac{n+1}{a_{n+1}}=\frac{n+1}{\frac{n}{u_{n}}}=\frac{n+1}{n} \cdot a_{n}, n=1,2, \cdots . \\
\text { Hence, } a_{2004}=\frac{2003}{2002} \cdot a_{2002}=\frac{2003}{2002} \cdot \frac{2001}{2000} \cdot a_{2000} \\
\quad=\cdots=\frac{2003}{2000} \cdot \frac{2001}{2000} \cdots \cdot \frac{3}{2} \cdot a_{2} \\
\quad=\frac{3 \cdot 5 \cdots \cdots 2003}{2 \cdot 4 \cdots \cdots 2002}\left(=\frac{2003!!}{2002!!}\right) .
\end{array}
$$