How to find the $n$- th term of the generating function $\frac{1}{(1-x-x^2)}^k$?

Question

I know how to find the $n$ -th term for $k = 1$, of $\frac{1}{\left(1-x^2-x\right)^k}$ but I want to know if there's a general formula for $k > 1$.

Answer 1

First, notice that the auxiliar function $R(s)=\dfrac{1}{1-s}$ satisfies the curious formula: $$ \dfrac{1}{(1-s)^k}=(-1)^k\dfrac{R^{(k)}(s)}{k!} $$

Thus, one can simply find the $nth$ term of the EGF progression $$ \dfrac{1}{(1-s)^k}=\sum_{n=0}^\infty a_{n,k} s^n $$ by simply compute the $kth$ derivative of the geometric series expansion $$ \dfrac{1}{1-s}=\sum_{n=0}^\infty s^n.$$

That is, $$ \dfrac{1}{(1-s)^k}=(-1)^k\dfrac{R^{(k)}(s)}{k!}=\sum_{n=k}^\infty (-1)^k\left(\begin{array}{c}n \\ k\end{array}\right)s^{n-k}=\sum_{n=0}^\infty (-1)^k\left(\begin{array}{c}n+k \\ k\end{array}\right)s^{n}. $$

Thus, $$a_{n,k}=(-1)^k\left(\begin{array}{c}n+k \\ k\end{array}\right).$$

In order to conclude our computation, one needs to perform the substitution $s=x^2+x$ on the above series expansion.

A straightorward application of the binomial identity to $(x^2+x)^n$ will allows you to determine a close formula for the coefficients of the aforementioned EGF.