2. The sequence $a_{0}, a_{1}, \cdots$ satisfies:
$$
a_{0}=\sqrt{5}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}},
$$

where, $[x]$ denotes the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $a_{2019}=$ $\qquad$
$2.8076+\sqrt{5}$.
By calculating the first few terms, we find
$$
a_{n}=4 n+\sqrt{5}(n \in \mathbf{N}) \text {. }
$$

We will prove this by mathematical induction.
Check $a_{0}=\sqrt{5}, a_{1}=4+\sqrt{5}$.
Assume that when $n=k$, $a_{k}=4 k+\sqrt{5}$.
Then when $n=k+1$,
$$
\begin{array}{l}
a_{k+1}=\left[a_{k}\right]+\frac{1}{\left\{a_{k}\right\}}=4 k+2+\frac{1}{\sqrt{5}-2} \\
=4(k+1)+\sqrt{5} .
\end{array}
$$

Therefore, $a_{2019}=8076+\sqrt{5}$.