<s>[INST] Solve recurring sequence using a generating function I have the sequence $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known terms, and I want to find a non-recursive equation for $a_n$ using a generating function.
What I have done:
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
  = a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
  -3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
  +x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}\\
& = 2+2x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)\\
& = \frac{2+2x+4x^2}{1-3x+3x^2-x^3}\\
& = \frac{4x^2+2x+2}{(1-x)^3}
\end{align}
$$
As shown above, I have reached a solution for $A(x)$, but I'm not sure how to use it to find a solution for $a_n$. Any tips pointing me in the right direction would be greatly appreciated.
 [/INST] Using Martin's hint, recall that $$ \frac{d^2}{dx^2} \left( \frac{1}{1-x} \right ) = \frac{2}{(1-x)^3}. $$
From a table (erm, Wikipedia), we see that the generating function for the sequence $a_n = \binom{n+2}{2}$ is the function $g(x) = \frac{1}{(1-x)^3}$. Doing some rearrangements we see that the intended solution is just $\frac{1}{4}\binom{n+2}{2}.$
</s>