5. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{0}=\sqrt{6}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}} \text {, }
$$

where, $[a]$ denotes the greatest integer not exceeding the real number $a$, and $\{a\}=a-[a]$. Then $a_{2020}=$ $\qquad$ .
5. $6060+\sqrt{6}$.

Notice,
$$
\begin{array}{l}
a_{0}=\sqrt{6}=2+(\sqrt{6}-2), \\
a_{1}=2+\frac{1}{\sqrt{6}-2}=4+\frac{\sqrt{6}-2}{2}, \\
a_{2}=4+\frac{2}{\sqrt{6}-2}=8+(\sqrt{6}-2), \\
a_{3}=8+\frac{1}{\sqrt{6}-2}=10+\frac{\sqrt{6}-2}{2}, \\
a_{4}=10+\frac{2}{\sqrt{6}-2}=14+(\sqrt{6}-2) .
\end{array}
$$

By mathematical induction, it is easy to get
$$
\begin{array}{l}
a_{2 k}=6 k+\sqrt{6}, \\
a_{2 k+1}=6 k+4+\frac{\sqrt{6}-2}{2}(k \in \mathbf{N}) .
\end{array}
$$

Therefore, $a_{2020}=6060+\sqrt{6}$.