If your initial conditions are $a_0$ and $a_1$, $p(x)=c(x)=(a_1-a_0)x+a_0$. Look at it this way. You have $$a_n=a_{n-1}+a_{n-2}\tag{1}$$ for $n\ge 2$. Assume that $a_n=0$ for $n<0$. Then $(1)$ is valid for all integers $n$ except possibly $0$ and $1$. To make it valid for all integers, add a couple of terms using the Iverson bracket to get $$a_n=a_{n-1}+a_{n-2}+(a_1-a_0)[n=1]+a_0[n=0]\;.\tag{2}$$
Note that while $a_0$ is straightforward, you have to be careful for $n>0$, since the earlier initial values are automatically built into the basic recurrence.
Now multiply $(2)$ through by $x^n$ and sum:
$$\begin{align*}
\sum_na_nx^n&=\sum_na_{n-1}x^n+\sum_na_{n-2}x^n+(a_1-a_0)\sum_n[n=1]x^n+a_0\sum_n[n=0]x^n\\
&=x\sum_na_nx^n+x^2\sum_na_nx^n+(a_1-a_0)x+a_0\;.
\end{align*}$$
Thus, if your generating function is $A(x)=\displaystyle\sum_na_nx^n$, you have $$A(x)=xA(x)+x^2A(x)+(a_1-a_0)x+a_0\;,$$ and hence $$A(x)=\frac{(a_1-a_0)x+a_0}{1-x-x^2}\;.$$
This obviously generalizes to higher-order recurrences and other starting points for the initial values. For example, a third-order recurrence with initial values $a_0,a_1,a_2$ would have
$$\begin{align*}
p(x)=c(x)&=a_0+(a_1-a_0)x+\left(a_2-\big(a_0+(a_1-a_0)\big)\right)x^2\\
&=a_0+(a_1-a_0)x+(a_2-a_1)x^2\;.
\end{align*}$$
In general with initial values $a_0,\dots,a_m$ you’ll get Iverson terms
$$a_0[n=0]+(a_1-a_0)[n=1]+(a_2-a_1)[n=2]+\cdots+(a_m-a_{m-1})[n=m]$$
in the recurrence and
$$c(x)=p(x)=a_0+(a_1-a_0)x+\cdots+(a_m-a_{m-1})x^m\;.$$
Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Then
\begin{eqnarray}
A(x)&=&1+\sum_{n=1}^\infty a_{n}x^n\\
&=&1+\sum_{n=1}^\infty (3a_{n-1}+\frac{1}{2}n(n-1))x^n\\
&=&1+3xA(x)+\frac{1}{2}\sum_{n=1}^\infty n(n-1)x^n.
\end{eqnarray}
Note $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ for $|x|<1$. Differentiating this twice, you can give
$$ \sum_{n=2}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}. $$
Thus
$$ A(x)=1+3xA(x)+\frac{x^2}{(1-x)^3} $$
from which you can get $A(x)$.
Best Answer
If $$f(x) = \frac{P(x)}{Q(x)}$$ where $P,Q$ are polynomials, then we have that
$$f(x)Q(x) = P(x)$$
Now use Leibniz's formula for differentiating a product
$$\sum_{r=0}^{n} {n \choose r} f^{r}(x) Q^{n-r}(x) = P^{n}(x)$$
which you can evaluate at $0$ for $n=1,2,\dots$ and you can obtain $(n+1)^{th}$ derivative of $f$ at $0$ from the first $n$ derivatives using the above formula.
The coefficient you need would be $\frac{f^{n}(0)}{n!}$.
This should be easily doable by a computer. No integration etc needed.