4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=0, a_{n+1}=a_{n}+4 \sqrt{a_{n}+1}+4(n \geqslant 1) \text {. }
$$

Then $a_{n}=$ $\qquad$ .
4. $4 n^{2}-4 n$.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}+4 \sqrt{a_{n}+1}+4 \\
\Rightarrow a_{n+1}+1=\left(a_{n}+1\right)+4 \sqrt{a_{n}+1}+4 \\
\quad=\left(\sqrt{a_{n}+1}+2\right)^{2} \\
\Rightarrow \sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+2 .
\end{array}
$$

Therefore, $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with the first term $\sqrt{a_{1}+1}=1$ and common difference 2.
Hence $\sqrt{a_{n}+1}=2 n-1 \Rightarrow a_{n}=4 n^{2}-4 n$.