(13) Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=1+a_{n}+\sqrt{1+4 a_{n}}(n \in$ $\mathbf{N}^{*}$ ), then the general term of the sequence $a_{n}=$ $\qquad$ .
Let $b_{n}=\sqrt{1+4 a_{n}}$, then $a_{n}=\frac{b_{n}^{2}-1}{4}$, hence we have
$$
\frac{b_{n+1}^{2}-1}{4}=1+\frac{b_{n}^{2}-1}{4}+b_{n},
$$

which simplifies to
$$
b_{n+1}^{2}=b_{n}^{2}+4 b_{n}+4=\left(b_{n}+2\right)^{2} .
$$

Given $b_{1}=\sqrt{1+4 a_{1}}=\sqrt{5}$ and $b_{n}>0$, we know $b_{n+1}=b_{n}+2$, which means $b_{n}=\sqrt{5}+2(n-1)$. Therefore, we get
$$
\begin{aligned}
a_{n} & =\frac{1}{4}\left(b_{n}^{2}-1\right) \\
& =\frac{1}{4}\left[5+4 \sqrt{5}(n-1)+4(n-1)^{2}-1\right] \\
& =1+\sqrt{5}(n-1)+(n-1)^{2} \\
& =1+(n-1)(n+\sqrt{5}-1) .
\end{aligned}
$$