Generating function for sequence $a_i={n+im-1 \choose im}$ It is well known that $\sum_{i=0}^\infty {n+i-1 \choose i}x^i=\frac{1}{(1-x)^n},$ i.e. $\frac{1}{(1-x)^n}$ is  generating function for sequence $a_i={n+i-1 \choose i.}$
But I want to find generating function for sequence $a_i={n+im-1 \choose im.}$
Using The On-Line Encyclopedia of Integer Sequences, I understood that generating function for sequence $a_i={n+2i-1 \choose 2i.}$ is $\frac{1+(n-1)x}{(1-x)^2}.$ But I cann't recieve generating function if $m>2.$  I found two formulas   for multiset formula. But they didn't help me.
Thanks a lot in advance for any help!
 A: We can simplify our problem:
$$ \binom{n+im-1}{im} = (-1)^{im}\binom{-n}{im} $$
We are looking for
$$ A_p(x) = \sum\limits_{i=0}^\infty \binom{-n}{im+p}x^i $$
Where $p\in \{ 0, 1, \dots, m-1 \} $ (particularly, $p=0$).
We also have:
$$ S(x) = \sum\limits_{i=0}^\infty \binom{-n}{i} x^i = \frac{1}{(1+x)^n} $$
Let $\{ z, z^2, \dots, z^m \}$ be different complex solutions to the equation $z^m=1$. 
$$ S(z^l \cdot x ) = \sum\limits_{i=0}^\infty \binom{-n}{i} (z^l\cdot x)^i = A_0(x^m) +z^l \cdot xA_1(x^m) + \dots + z^{l(m-1)} \cdot x^{m-1}A_{m-1}(x^m) $$
We created a linear system of equations, where the variables are functions $A_p(x)$. Now we will have, because $z+z^2+\dots+z^m=0$:
$$ A_0(x^m) = \frac{1}{m} \sum\limits_{j=1}^m S(z^j \cdot x) = \frac{1}{m} (\frac{1}{(1+zx)^n}+\frac{1}{(1+z^2x)^n}+\dots+\frac{1}{(1+z^nx)^n}) $$
When $m=2$
$$ A_0(x^2) = \frac{1}{2}(\frac{1}{(1+x)^n}+\frac{1}{(1-x)^n}) $$
Derivation for any $p$ is possible, but needs more work with solving the linear equation.
A: The $m$-th roots of unity $\omega_j=\exp\left(\frac{2\pi ij}{m}\right), 0\leq j<m$ have the nice property to filter elements. For $m,N>0$ we have
\begin{align*}
\frac{1}{m}\sum_{j=0}^{m-1}\exp\left(\frac{2\pi ij N}{m}\right)=
\begin{cases}
1&\qquad m\mid N\\
0& \qquad otherwise
\end{cases}
\end{align*}
If  $A(x)=\sum_{k=0}^\infty a_kx^k$  then
\begin{align*}
A(x)+A(\omega_1x)+\cdots+A(\omega_{m-1}x)=\sum_{k=0}^\infty  a_{mk}x^{mk}\tag{1}
\end{align*}

We set
  \begin{align*}
A_1(x)=\sum_{k=0}^\infty \binom{n+k-1}{k}x^k=\frac{1}{(1-x)^n}
\end{align*}
According to  (1)   we   obtain
  \begin{align*}
\color{blue}{A_m(x)}&\color{blue}{=\sum_{k=0}^\infty\binom{n+mk-1}{mk}x^k}\\
&=A_1\left(x^{1/m}\right)+A_1\left(\omega_1 x^{1/m}\right)+\cdots+A_{m-1}\left(\omega_{m-1}  x^{1/m}\right)\\
&=\sum_{j=0}^{m-1}A_1\left(\omega_j  x^{1/m}\right)\\
&\,\,\color{blue}{=\sum_{j=0}^{m-1}\frac{1}{\left(1-\omega_j  x^{1/m}\right)^n}}
\end{align*}

Some special  cases:
Case  $m=2$:
We have $\{\omega_0,\omega_1\}=\{1,e^{\pi i}\}=\{1,-1\}$
\begin{align*}
A_2(x)&=\sum_{k=0}^\infty \binom{n+2k-1}{2k}x^k\\
&=\frac{1}{(1-\sqrt{x})^n}+\frac{1}{(1+\sqrt{x})^n}
\end{align*}
Case  $m=3$:
We have $\{\omega_0,\omega_1,\omega_2\}=\{1,e^{\frac{2\pi   i}{3}},e^{\frac{4\pi  i}{3}}\}=\{1,\frac{1}{2}(-1+\sqrt{3}),\frac{1}{2}\left(-1-\sqrt{3}\right)\}$
\begin{align*}
A_3(x)&=\sum_{k=0}^\infty \binom{n+3k-1}{3k}x^k\\
&=\frac{1}{(1-\sqrt[3]{x})^n}+\frac{1}{(1+\frac{1}{2}(1-\sqrt{3})\sqrt[3]{x})^n}+\frac{1}{(1+\frac{1}{2}(1+\sqrt{3})\sqrt[3]{x})^n}
\end{align*}
Case  $m=4$:
We have $\{\omega_0,\omega_1,\omega_2,\omega_3\}=\{1,e^{\frac{\pi   i}{2}},e^{\pi   i},e^{\frac{3\pi i}{2}}\}=\{1,i,-1,-i\}$
\begin{align*}
A_4(x)&=\sum_{k=0}^\infty \binom{n+4k-1}{4k}x^k\\
&=\frac{1}{(1-\sqrt[4]{x})^n}+\frac{1}{(1-i\sqrt[4]{x})^n}+\frac{1}{(1+\sqrt[4]{x})^n}+\frac{1}{(1+i\sqrt[4]{x})^n}
\end{align*}
