TITLE: How find $a_{n}$ if the sequence $a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: $$
a_{n}=2a_{n-1}+(2n-1)^2a_{n-2}.
$$
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 $c_n=1$, $n\in\mathbb N$.