Find Nth formula of recursive formula $a_n=a_{n-1}+n(n-1)a_{n-2}$ $$a_n=a_{n-1}+n(n-1)a_{n-2}$$
$$a_0=1, a_1=-\frac{1}{2}$$
Is it possible to find explicit formula for $a_n$ just by using $a_0$ and $a_1$? 
I know how to solve this problem if $a_n=Aa_{n-1}+Ba_{n-2}$ where $A$ and $B$ is some real number(constant), but in this problem that is not a case.
 A: Let $\displaystyle\;b_n = \frac{a_n}{n!}$, the recurrence relation at hand can be transformed to
$$n ( b_n - b_{n-2} ) = b_{n-1}\quad\text{ with }\quad \begin{cases} b_0 = 1\\b_1 = -\frac12\end{cases}\tag{*1}$$
Let $f(z) = \sum_{n=0}^\infty b_n z^n$ be the OGF for $b_n$. If we multiple $(*1)$
by $z^n$ and start to sum from $n = 2$. We get
$$\begin{align}
& z\frac{d}{dz}\left[ (1-z^2) f(z) - 1 + \frac{z}{2} \right] = z(f(z)-1)\\
\iff & (1-z^2) \frac{df(z)}{dz} - (1+2z) f(z) = -\frac{3}{2}
\end{align}
$$
Solving the ODE gives us
$$f(z) = \frac{5 - 3\sin^{-1}(z)}{2(1-z)\sqrt{1-z^2}} - \frac{3}{2(1-z)}$$
Expanding $f(z)$ as a power series in $z$, we obtain following
ugly expression of $a_n$.
$$a_n = \frac{n!}{2}\left[
5\sum_{p=0}^{\lfloor \frac{n}{2}\rfloor} \frac{\binom{2p}{p}}{4^p}
- 3\sum_{s=0}^{\lfloor\frac{n-1}{2}\rfloor} \sum_{p=0}^s \frac{\binom{2p}{p}\binom{2s-2p}{s-p}}{4^s(2p+1)} - 3
\right]$$
As a double check, the first few values of $a_n$ according this formula are
$$(2a_0,2a_1,\ldots) = 2, -1, 3, -3, 33, -27, 963, -171, 53757, 41445,4879575, 9438525, \ldots$$
and they do satisfy the recurrence relation.
Update
It turns out we can simplify the mess a little bit.
The recurrence relation on $b_n$ can be rewritten as $b_n = (n+1)(b_{n+1} - b_{n-1})$. The RHS has the form of a finite difference. We can use it to get
rid of one level of summation. The end result is for all $n \ge 0$,
$$a_n = \frac{(n+1)!}{2}\left[
5\frac{\binom{2r}{r}}{4^r}
- 3\sum_{p=0}^s \frac{\binom{2p}{p}\binom{2s-2p}{s-p}}{4^s(2p+1)}
\right]
\quad\text{ with }\quad
\begin{cases}
r = \lfloor \frac{n+1}{2}\rfloor\\
s = \lfloor \frac{n}{2}\rfloor
\end{cases}
$$
A: For the second question, note that 
$$\begin{align*}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}&=\begin{bmatrix}Aa_{n-1}+Ba_{n-2}\\ a_{n-1}\end{bmatrix}=\begin{bmatrix}A &B\\ 1 & 0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}A&B\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\ a_{n-3}\end{bmatrix}\\&=\cdots=\begin{bmatrix}A&B\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_1\\a_0\end{bmatrix}\end{align*}$$
With a similar argument, we would get that 
$$\begin{align*}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}&=\begin{bmatrix}a_{n-1}+n(n-1)a_{n-2}\\ a_{n-1}\end{bmatrix}=\begin{bmatrix}1 &n(n-1)\\ 1 & 0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}\\&=\begin{bmatrix}1&n(n-1)\\1&0\end{bmatrix}\begin{bmatrix}1 &(n-1)(n-2)\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\ a_{n-3}\end{bmatrix}\\&=\cdots \\&=\left(\prod_{k=0}^{n-2} \begin{bmatrix}1&(n-k)(n-k-1)\\1&0\end{bmatrix}\right)\begin{bmatrix}a_1\\a_0\end{bmatrix}\end{align*}$$
