Explicit formula for recurrence I know how to get explicit formula for simple recurrence $a_n = m_1a_{n-1} + m_2a_{n-2}\dots$ for $m_{1,2\dots}$ being constant numbers.
I'm wondering how to get explicit formula for recurrence like this $$a_n = na_{n-1} + n^2a_{n-2}$$, simply containing $n$. Can we do it using generating functions? 
I'd really appreciate some help on this.
 A: Your question is two general and broad. It is not always possible to solve for the $n$th term using conventional tools such as generating functions.
Short answer: Yes, generating functions are often one the easiest ways to find a closed formula for the $n$th term of a given sequence. Obviously, it is not the only method and it is not even guaranteed that an explicit formula for the $n$th term exists. Since your sequence is Fibonacci-like, I would first try using generating functions. Also, you need initial conditions.
Edit: The recurrence relation $a_n = n^2 a_{n - 1}$ it much easier to handle. Let us find the exponential generating function for the sequence $\{a_n\}$. To this end, let
$$E = \sum_{n \geq 0} a_n \frac{x^n}{n!}.$$
Then
\begin{align*}
E &= \sum_{n \geq 0} (n^2 a_{n - 1}) \frac{x^n}{n!}\\
&= \sum_{n \geq 0} (n a_{n - 1}) \frac{x^n}{(n - 1)!}\\
&= \sum_{n \geq 1} (n a_{n - 1}) \frac{x^n}{(n - 1)!}\\
&= x\sum_{n \geq 0} ((n + 1) a_n) \frac{x^n}{n!}\\
&= x\sum_{n \geq 0} na_n \frac{x^n}{n!} + x\sum_{n \geq 0} a_n \frac{x^n}{n!}\\
&= x^2\sum_{n \geq 0} \frac{a_n}{n!} \frac{d}{dx}x^n + xE\\
&= x^2E' + xE
\end{align*}
because $x(x^n)' = nx^n$. In other words,
$$E' = \frac{(1 - x)}{x^2}E,$$
which is a simple separable differential equation. It is routine to show that
$$E = \frac{Ce^{-1/x}}{x},$$
where $C$ is an arbitrary constant. Hence
$$E = \sum_{n \geq 0} \frac{C}{n!x^{n + 1}}.$$
Unfortunately, since this expression for $E$ is not of the form
$$E = \sum_{n \geq 0} A_n \frac{x^n}{n!},$$
we cannot equate the coefficients to deduce that $a_n = A_n$. If $E$ were of the aforesaid form, then $A_n$ would be the closed formula for $a_n$.
Also, if we consider $a_n = n a_{n - 1} + n^2 a_{n - 2}$ instead, we will run into the same problem. So, personally, I would conjecture that there is no explicit formula for $a_n$ in either case.
A: Sometimes the best way to solve a recurrence relation is to simplify it into something more tractable by multiplying by a certain function. I don't know how to solve your exact recurrence relation, but I can solve something very similar, and without using generating functions:
$$
a_n = na_n + n(n-1)a_{n-2}.
$$
To solve, divide by $n!$:
$$
\frac{a_n}{n!}= \frac{a_{n-1}}{(n-1)!}+\frac{a_{n-2}}{(n-2)!}
$$
Now let $F_n=a_n/n!$; then $F_n$ satisfies the recurrence relation
$$
F_n = F_{n-1} + F_{n-2}.
$$
Look familiar? This is the recurrence relation for the Fibonacci numbers! Assuming that $a_0=0$ and $a_1=1$, we can give an explicit solution to our original recurrence:
$$
a_n = n!\cdot F_n.
$$
(If this isn't explicit enough for you, use Binet's formula for the Fibonacci numbers.)
