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Is there a simple expression for the generating series $$f_n(z)=\sum_{k\geq 1}\binom{kn}{n}z^k$$ for $n=1$, I know it's $\frac{z}{(1-z)^2}$, but what happens for other $n$? Thanks for any help.

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Let

$$ g_n(x) = \sum_{k \ge 0} \binom{k}{n}x^k = \frac{x^{n}}{(1 - x)^{n + 1}}. $$

Let $\zeta_n = \exp\left(2 \pi i/n \right)$. Then

$$ \frac{g_n(x) + g_n(\zeta_n x) + \cdots + g_n(\zeta_n^{n - 1} x)}{n} = \sum_{k \ge 0} \binom{kn}{n} x^{kn}. $$

So

$$ f_n(x) = \frac{g_n(x^{1/n}) + g_n(\zeta_n x^{1/n}) + \cdots + g_n(\zeta_n^{n - 1} x^{1/n})}{n}. $$

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I think the general solution will be of the form $f_n(z)=\frac{P_n(z)}{(1-z)^{n+1}}$, where $P_n(z)$ is a polynomial. This is because $\binom{nk}{k}=\frac{nk(nk-1)..(nk-n)}{n!}$, so that you need to know how to evaluate $\sum_{k=0}^\infty nk(nk-1)...(nk-n)z^k$, which can be done by computing $n$ derivatives of $\frac{1}{(1-z^{n})}$. This is not fun.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.\vphantom{\Large A}\mrm{f}_{n}\pars{z} \right\vert_{\ \verts{z}\ <\ 1} \equiv \sum_{k \geq 1}{kn \choose n}z^{k}.\quad \mbox{Note that}\quad \mrm{f}_{0}\pars{z} = {z \over 1 - z}}$.

Then, \begin{align} \left.\vphantom{\Large A}\mrm{f}_{n}\pars{z} \right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 1}} & \equiv \sum_{k \geq 1} {kn \choose n}z^{k} = \sum_{k = 0}^{\infty}{kn \choose n}z^{kn/n} = \sum_{k = n}^{\infty}{k \choose n}z^{k/n} \,{\sum_{\ell = 0}^{n - 1}r_{\ell}^{k} \over n} \end{align}

where $\ds{r_{\ell} = \exp\pars{{2\ell\pi \over n}\,\ic}.\ \ell = 0,1,2,\ldots,n - 1}$.

Then, \begin{align} \left.\vphantom{\Large A}\mrm{f}_{n}\pars{z} \right\vert_{\ n\ \in\ \mathbb{N}_{\large\ \geq\ 1}} & = {1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = n}^{\infty} {k \choose k - n}\pars{z^{1/n}r_{\ell}}^{k} = {1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = n}^{\infty} {-n - 1 \choose k - n}\pars{-1}^{k - n}\pars{z^{1/n}r_{\ell}}^{k} \\[5mm] & = {1 \over n}\sum_{\ell = 0}^{n - 1}\sum_{k = 0}^{\infty} {-n - 1 \choose k}\pars{-1}^{k}\pars{z^{1/n}r_{\ell}}^{k + n} \\[5mm] & = {1 \over n}\sum_{\ell = 0}^{n - 1}\pars{z^{1/n}r_{\ell}}^{n} \sum_{k = 0}^{\infty}{-n - 1 \choose k}\pars{-z^{1/n}r_{\ell}}^{k} = {z \over n}\sum_{\ell = 0}^{n - 1}\pars{1 - z^{1/n}r_{\ell}}^{-n - 1} \end{align}


$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{% \mrm{f}_{n}\pars{z} = \left\{\begin{array}{lcl} \ds{z \over 1 - z} & \mbox{if} & \ds{n = 0} \\[2mm] \ds{{z \over n}\sum_{\ell = 0}^{n - 1} \bracks{1 - z^{1/n}\exp\pars{-\,{2\pi\ell \over n}\,\ic}}^{-n - 1}} & \mbox{if} & \ds{n \geq 1} \end{array}\right.}} $$

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