6. If the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=2$, and
$$
\begin{aligned}
a_{n+2}= & 2(2 n+3)^{2} a_{n+1}- \\
& 4(n+1)^{2}(2 n+1)(2 n+3) a_{n} \\
& (n \geqslant 0),
\end{aligned}
$$

then the general term formula of the sequence $\left\{a_{n}\right\}$ is $\qquad$ .
$$
\begin{array}{l}
\text { 6. } a_{n}=\left\{\begin{array}{ll}
0, & n=0 ; \\
(2 n)!\sum_{i=1}^{n} \frac{1}{i}, & n \geqslant 1 .
\end{array}\right. \\
\text { Let } v_{n}=\frac{a_{n}}{(2 n)!} .
\end{array}
$$

Then $v_{0}=0, v_{1}=1$, and
$$
\begin{array}{l}
(n+2) v_{n+2}=(2 n+3) v_{n+1}-(n+1) v_{n} \\
\Rightarrow(n+2)\left(v_{n+2}-v_{n+1}\right) \\
\quad=(n+1)\left(v_{n+1}-v_{n}\right)=\cdots=v_{1}-v_{0}=1 \\
\Rightarrow v_{n}=\sum_{i=1}^{n} \frac{1}{i} \Rightarrow a_{n}=(2 n)!\sum_{i=1}^{n} \frac{1}{i} .
\end{array}
$$