2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1$, and $a_{2 n}=a_{n}, a_{2 n+1}=a_{n}+1\left(n \in \mathbf{Z}_{+}\right)$.
Then $a_{2013}=$ . $\qquad$
2.9.

From the problem, we know
$$
\begin{array}{l}
a_{2013}=a_{1006}+1=a_{503}+1=a_{251}+2 \\
=a_{125}+3=a_{62}+4=a_{31}+4=a_{15}+5 \\
=a_{7}+6=a_{3}+7=a_{1}+8=9 .
\end{array}
$$