How find $a_{n}$ if the sequence $a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$ Question:

let sequence $\{a_{n}\}$,such $a_{-1}=1,a_{0}=1$
and
$$a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$$

Find the $a_{n}$
I  find
$$a_{0}=1,a_{1}=3,a_{2}=15,a_{3}=105$$
and I found
$$a_{0}=1$$
$$a_{1}=1\cdot 3$$
$$a_{2}=1\cdot 3\cdot 5$$
$$a_{3}=1\cdot 3\cdot 5\cdot 7$$
so I guess
$$a_{n}=1\cdot 3\cdot 5\cdots (2n+1)$$
and It is easy use Mathematical induction
to prove it.
Now My question: can you someone have other methods?
why I want to see other methods,because I think this problem reslut is interesting.and this sequence form is seem not ugly,so I think this problem have without mathematical indution.
 A: Taking $$ g_n = \frac{a_{n+1}}{a_n} $$
This gives us $$ g_{n-1}(g_n-2)=(2n+1)^2$$ 
This gives, $g_n = 2n+3$
Now , $$\frac{a_{n+1}}{a_{-1}} = g_{-1}g_1g_2\cdots g_n $$
$$ a_n = g_{n-1} \cdot g_{n-2} \cdots \cdot g_{-1} $$
$$ a_n = (2n+1)!! \Box $$
A: Starting with
$$
a_n=2a_{n-1}+(2n-1)^2a_{n-2}\tag{1}
$$
Let $a_n=2^nb_n$, then we have
$$
\begin{align}
b_n&=b_{n-1}+(n-\tfrac12)^2b_{n-2}\tag{2a}\\
b_{n-1}&=b_{n-2}+(n-\tfrac32)^2b_{n-3}\tag{2b}\\
b_n-b_{n-1}&=(n-\tfrac12)^2b_{n-2}\tag{2c}\\
(n-\tfrac12)b_{n-1}&=(n-\tfrac12)b_{n-2}+(n-\tfrac12)(n-\tfrac32)^2b_{n-3}\tag{2d}\\
b_n-(n+\tfrac12)b_{n-1}&=(n-\tfrac12)(n-\tfrac32)b_{n-2}-(n-\tfrac12)(n-\tfrac32)^2b_{n-3}\tag{2e}\\
&=(n-\tfrac12)(n-\tfrac32)\left[b_{n-2}-(n-\tfrac32)b_{n-3}\right]\tag{2f}
\end{align}
$$
Explanation:
$\mathrm{(2a)}$: $a_n=2^nb_n$ applied to $(1)$
$\mathrm{(2b)}$: substitute $n\mapsto n-1$ in $\mathrm{(2a)}$
$\mathrm{(2c)}$: subtract $b_{n-1}$ from $\mathrm{(2b)}$
$\mathrm{(2d)}$: multiply $\mathrm{(2b)}$ by $(n-\frac12)$
$\mathrm{(2e)}$: subtract $\mathrm{(2d)}$ from $\mathrm{(2c)}$
$\mathrm{(2f)}$: factor the right hand side of $\mathrm{(2e)}$
Noting that $b_0-\frac12b_{-1}=1-\frac12\cdot2=0$ and $b_1-\frac32b_0=\frac32-\frac32\cdot1=0$, equation $\mathrm{(2f)}$ ensures that $b_n=(n+\frac12)b_{n-1}$ for $n\ge0$, which is equivalent to
$$
a_n=(2n+1)a_{n-1}\tag{3}
$$
Since $a_{-1}=1$, $(3)$ implies
$$
a_n=(2n+1)!!\tag{4}
$$
A: From your results, it is clear that $a_n=(2n+1)!!$ (which is sequence A001147 at OEIS) and which verifies the recurrence relation. 
Another way to write it is $$a_n= \frac{2^{n+1} \left(n+\frac{1}{2}\right)!}{\sqrt{\pi }}$$ 
For the time being, I did not find any generating function except 
$$\sum_{i=0}^\infty \frac{x^n}{a_n}={\sqrt{\frac{\pi }{2x}} e^{x/2}
   \text{erf}\left({\sqrt{\frac{x}{2}}}\right)}$$
In OEIS, they give $$a_n=\int_0^\infty \frac{e^{-x/2} x^{n}}{\sqrt{2 \pi x}}dx$$
A: $$
a_{n}=2a_{n-1}+(2n-1)^2a_{n-2}.
$$
Let $b_n=a_n/a_{n-1}$. Then $\{b_n\}$ satisfies
$$
b_{n}=2+(2n-1)^2/b_{n-1}, \quad b_1=3.
$$
Let $c_n=b_n/(2n+1)$, then $\{c_n\}$ satisfies
$$
(2n+1)c_{n}=2+\frac{2n-1}{c_{n-1}},\quad c_1=1,
$$
which is satisfied only by the sequence $c_n=1$, $n\in\mathbb N$.
