Extracting a coefficient from a generating function Background: I am working on an exercise relating to Skolem $k-$subsets with index $k$ in Goulden and Jackson's Combinatorial Enumeration text and they broke it down to finding the coefficient of $x^n$ in the generating function
$$f(x) = \left(\sum_{i \equiv 1 \mod{p}}x^i\right)^k\sum_{r\geq 1}x^r$$
which has the closed form 
$$f(x) = \frac{x^{k+1}}{(1-x^p)^k}\frac{1}{1-x}.$$
I am a bit stuck on how someone would deduce that the answer is 
$$[x^n]f(x) = \binom{\lfloor(n+(p-1)k)\frac{1}{p}\rfloor}{k}$$
What are the intermediate steps taken in order to find this coefficient? I tried expanding the series only to find that it has the form
$$f(x) = \sum_{m\geq0}\sum_{r\geq1}\binom{k+m-1}{m}x^{mp+r+k+1}.$$
How do I proceed?
 A: Either I've lost a -1 somewhere or the answer you quote is missing one, but I think this is either the right answer or has a small mistake in it.
Note that from your first equation, $[x^n]f(x)$ equals the number of solutions $(c_1,c_2,\dots,c_{k+1})$ to the equation $$n=1+c_1p+1+c_2p+\cdots+1+c_kp+1+c_{k+1}1,$$ with nonnegative integers $c_i$. Equivalently, it’s the number of solutions $(c_1,c_2,\dots,c_{k+1})$ to
$$n-k-1=c_1p+c_2p+\cdots+c_kp+c_{k+1}1,$$ again with nonnegative integers $c_i$.
First, suppose that $n-k-1=qp+r$, $0\le r<p$, that is, $n-k-1$ has remainder $r$ when divided by $p$. The number of solutions does not depend on $r$. If there is a remainder, it can only be incorporated into the term $c'_{k+1}1$.
WLOG, then, the desired coefficient of $x^n$ is the number of solutions to
$$n'=p\lfloor (n-k-1)\frac{1}{p}\rfloor=c_1p+c_2p+\cdots+c_kp+c_{k+1}1.$$
A “stars and bars” approach can count the number of solutions. Each solution to the equation corresponds to an arrangement of $k$ “bars” within a sequence of $p\lfloor (n-k-1)\frac{1}{p}\rfloor$ “stars of value $p$.” For example, the arrangement $*||**|**$ would represent the solution $(1,0,2,2p)$, or $n'=(1+1p)+(1+0p)+(1+2p)+(2p)1$. (Note that both sides of the equation are multiples of $p$.)
Equivalently, each solution corresponds to a choice of the $k$ bar-locations in an arrangement of $p\lfloor (n-k-1)\frac{1}{p}\rfloor+k$ stars-and-bars, which is
$$p\lfloor (n-k-1)\frac{1}{p}\rfloor+k\choose k$$
For the desired result (except for a -1 that one of us missed), observe that $$p\lfloor (n-k-1)\frac{1}{p}\rfloor+k= \frac{n+(p-1)k-1}{p}=\lfloor(n+(p-1)k)\frac{1}{p}\rfloor.$$
Your question suggests you are well along in understanding these ideas, and resolving the -1 discrepancy is left as your homework.
A: Firstly there is an error in the lower limit of the sum over $r$. The sum sum should run over $r\ge 0$ since you expand the term $\frac{1}{1-x}$. Now we have:
\begin{eqnarray}
[x^n] f(x)  = \sum\limits_{m=0}^\infty \sum\limits_{r=0}^\infty \binom{k+m-1}{m} \delta_{m\cdot p+r+k+1,n} = \sum\limits_{m=0}^\infty 1_{n-k-1-m \cdot p \ge 0} \binom{k+m-1}{m} = \sum\limits_{m=0}^{\lfloor \frac{n-k-1}{p}\rfloor} \binom{k+m-1}{m} = \binom{k + \lfloor \frac{n-k-1}{p}\rfloor}{k} = \binom{\frac{k \cdot p}{p} + \lfloor \frac{n-k-1}{p}\rfloor}{k} = \binom{\lfloor \frac{n+k(p-1)-1}{p}\rfloor}{k}
\end{eqnarray}
So again, there is an issue with ``the missing minus one'' as in the answer above.
