13. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=1+a_{n}+\sqrt{1+4 a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Then the general term of the sequence $a_{n}=$ $\qquad$ .
13. $1+(n-1)(n+\sqrt{5}-1)$.

Let $b_{n}=\sqrt{1+4 a_{n}}$. Then $a_{n}=\frac{b_{n}^{2}-1}{4}$.
Thus, $\frac{b_{n+1}^{2}-1}{4}=1+\frac{b_{n}^{2}-1}{4}+b_{n}$, which means $b_{n+1}^{2}=b_{n}^{2}+4 b_{n}+4=\left(b_{n}+2\right)^{2}$.

Given $b_{1}=\sqrt{1+4 a_{1}}=\sqrt{5}$ and $b_{n}>0$, we know $b_{n+1}=b_{n}+2$, i.e., $b_{n}=\sqrt{5}+2(n-1)$.
Then $a_{n}=\frac{1}{4}\left(b_{n}^{2}-1\right)$
$$
\begin{array}{l}
=\frac{1}{4}\left[5+4 \sqrt{5}(n-1)+4(n-1)^{2}-1\right] \\
=1+(n-1)(n+\sqrt{5}-1) .
\end{array}
$$