2. A sequence of numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is constructed according to the following rule: $a_{1}=7, a_{k}=a_{k-1}^{2}$'s digit sum $+1(k=2,3, \cdots)$. For example, $a_{2}=14, a_{3}=17$, and so on.
Then $a_{2013}=$ $\qquad$ .
2.8.

From the problem, we know
$$
\begin{array}{l}
a_{1}=7, a_{2}=14, a_{3}=17, a_{4}=0, \\
a_{5}=5, a_{6}=8, a_{7}=11, a_{8}=5 .
\end{array}
$$

Thus, $a_{8}=a_{5}$, meaning from $a_{5}$ onwards, the sequence repeats with a period of 3.
Therefore, $a_{2013}=a_{6+3 \times 669}=8$.