10. The sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}$ satisfies $a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}}\left(\left[a_{n}\right],\left\{a_{n}\right\}\right.$ represent the integer part and fractional part of $a_{n}$, respectively), then $a_{2016}=$ $\qquad$ .
$$
a_{1}=1+\frac{1}{\sqrt{3}-1}=2+\frac{\sqrt{3}-1}{2}, a_{2}=2+\frac{2}{\sqrt{3}-1}=3+\sqrt{3} \text {. }
$$

Conjecture: $a_{2 k-1}=3 k-1+\frac{\sqrt{3}-1}{2}, a_{2 k}=3 k+\sqrt{3}$. Assuming the conclusion holds for $n=k$, then for $n=k+1$,
$$
a_{2 k+1}=3 k+1+\frac{1}{\sqrt{3}-1}=3 k+2+\frac{\sqrt{3}-1}{2}, a_{2 k+2}=3 k+2+\frac{2}{\sqrt{3}-1}=3 k+3+\sqrt{3} \text {, }
$$

which means the conclusion also holds for $n=k+1$, thus the above conjecture is correct. Therefore, $a_{2016}=3024+\sqrt{3}$.