How to solve this kind of problems?

Question

For $[z^n]\frac{z^k}{1-z^k}$ we can get $[z^n]\frac{z^k}{1-z^k}=\left( \begin{array}{ccc} n-1\\ k-1\\ \end{array} \right)$

But $[z^n](z\frac{1-z^r}{1-z})^k=?$

How to solve this kind of problems in general? What are the techniques?

Answer 1

Using the trivial splitting of the generating function: $$ g(z) = \left( z \frac{1-z^r}{1-z} \right)^k = \underbrace{\frac{z^k}{(1-z)^k} }_{g_1(z)} \cdot \underbrace{\left(1-z^r \right)^k}_{g_2(z)} $$ along with established facts $[z^n] (1-z)^{-\alpha} = \binom{\alpha+n-1}{n}$: $$ [z^n] \frac{z^k}{(1-z)^k} = [z^{n-k}] \left(1-z\right)^{-k} = \binom{k + (n-k) -1}{n-k} = \binom{n-1}{n-k} = \binom{n-1}{k-1} $$ and $$ [z^n] \left(1-z^r \right)^k = [z^n] \sum_{m=0}^k \binom{k}{m} (-1)^m z^{m r} = (-1)^{n/r}\binom{k}{n/r} [ n \bmod r = 0] $$ The series coefficients for the product of $g_1$ and $g_2$ is the Cauchy sum of series coefficients of individual generating functions: $$ [z^n] g(z) = \sum_{m=0}^n [z^{n-m}] g_1(z) \cdot [z^m] g_2(z) = \sum_{m=0}^{\left\lfloor n/r \right\rfloor} (-1)^m \binom{n-1 - m r}{k-1} \binom{k}{m} $$