How to define a recursive non-constant geometric sequence non-recursively (if that even makes sense) The title, basically.
I have a sequence that I've defined as $a_n=a_{n-1}(4n^2-8n+3)$, where $a_0=1$ and $a_1=-1$. I want to find a way to define it non-recursively, yet my mind has failed me.
I've tried the standard method for geometric sequences, giving $\frac{(4n^2-8n+3)^n}{(4n^2-8n+3)}$, but that leads to problems with $n^n$, and it did not generate the same sequence as the recursive one.
Out of desperation I even tried a weird leap of logic: I thought since $a_n$ is $a_{n-1}(4n^2-8n+3)$, I could maybe try replacing the $a_{n-1}$ with $a_{n-1}(4n^2-8n+3)$, just with $n-1$ instead of $n$. That lead to a needlessly complicated polynomial which also failed to reproduce the sequence.
I've been scouring the internet for any resources that could help me solve this one single type of problem, then I came across this site and figured posting a request to help won't be that bad. Thank you for any answer you can give me!
 A: Note
$$4n^2 - 8n + 3 = (2n - 3)(2n - 1) \tag{1}\label{eq1A}$$
as well as that $2(n + 1) - 3 = 2n - 1$. This means as we start from $a_1$ and consecutively calculate up to $a_n$, each factor of $2n - 1$ for $a_n$ will be repeated as $2(n + 1) - 3$ for $a_{n+1}$ and, thus, are squared in the final result. Thus, the only factors of the RHS of \eqref{eq1A} that are not squared would be the initial first factor, i.e., $2(2) - 3 = 1$ (but it can be considered to be squared since $1^2 = 1$) and last second factor, i.e., $2n - 1$. Thus, for $n \ge 1$, this results in
$$\begin{equation}\begin{aligned}
& a_1 = -1 \\
& a_2 = a_1(2(2) - 3)(2(2) - 1) = -(2(2) - 1) \\
& a_3 = a_2(2(3) - 3)(2(3) - 1) = -(2(2) - 1)^2(2(3) - 1) \\
& a_4 = a_3(2(4) - 3)(2(4) - 1) = -(2(2) - 1)^2(2(3) - 1)^2(2(4) - 1) \\
& \qquad \qquad \qquad \qquad \qquad \vdots \\
& a_n = -\left(\prod_{i=1}^{n-1}(2i - 1)^2\right)(2n - 1)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

As Robert Israel's comment states, this can also be written in terms of factorials, with it resulting in
$$\begin{equation}\begin{aligned}
a_n & = -\left(\prod_{i=1}^{n-1}(2i - 1)\right)\left(\prod_{i=1}^{n}(2i - 1)\right) \\
& = -\left(\frac{\prod_{i=1}^{2n-2}i}{\prod_{i=1}^{n-1}2i}\right)\left(\frac{\prod_{i=1}^{2n}i}{\prod_{i=1}^{n}2i}\right) \\
& = -\left(\frac{1}{2^{2n-1}}\right)\left(\frac{\prod_{i=1}^{2n-2}i}{\prod_{i=1}^{n-1}i}\right)\left(\frac{\prod_{i=1}^{2n}i}{\prod_{i=1}^{n}i}\right) \\
& = -\frac{(2n-2)!(2n)!}{2^{2n-1}(n-1)!(n!)}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Also, as Claude Leibovici's comment indicates, this can be made shorter using the Gamma function, where
$$(n-1)! = \Gamma(n) \tag{4}\label{eq4A}$$
and the Legendre duplication formula of
$$\Gamma(z)\Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z}\sqrt{\pi}\;\Gamma(2z) \; \; \to \; \; \frac{\Gamma(2z)}{\Gamma(z)} = \frac{2^{2z - 1}\Gamma\left(z + \frac{1}{2}\right)}{\sqrt{\pi}} \tag{5}\label{eq5A}$$
Using $z = n - 1$ and $z = n$ in \eqref{eq5A}, plus using \eqref{eq4A}, gives
$$\frac{(2n-3)!}{(n-2)!} = \frac{2^{2n-3}\Gamma\left(n - \frac{1}{2}\right)}{\sqrt{\pi}}, \; \; \frac{(2n-1)!}{(n-1)!} = \frac{2^{2n-1}\Gamma\left(n + \frac{1}{2}\right)}{\sqrt{\pi}} \tag{6}\label{eq6A}$$
Using these values in \eqref{eq3A} then gives
$$\begin{equation}\begin{aligned}
a_n & = -\frac{2^{2n-3}(2n-2)(2n)\Gamma\left(n - \frac{1}{2}\right)\Gamma\left(n + \frac{1}{2}\right)}{(n-1)(\sqrt{\pi})(n)(\sqrt{\pi})} \\
& = -\frac{2^{2n-1}\Gamma\left(n - \frac{1}{2}\right)\Gamma\left(n + \frac{1}{2}\right)}{\pi}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
