Let $F\_n^{(k)}(x)= \sum\_j {\binom{n+(k-1)j}{kj} x^j}$ and $G\_n^{(k)}(x)= \sum\_j {\binom{n+j}{kj} x^j}.$


I am interested in the coefficients ${a\_{n,k,j}}$ such that
$$G\_n^{(k)}(x)=\sum\_{j\geq0 }{a\_{n,k,j}} F\_j^{(k)}((-1)^ { k}x).$$
Computations suggest the following:


Let $z\sum\_{j\geq 0}C\_{1,j}^{(k)}z^j$ be the inverse series of $\sum\_{j=1}^k (-1)^{j-1}z^j$ and let $(\sum\_{j\geq 0}C\_{1,j}^{(k)}z^j)^m=\sum\_{j\geq 0}C\_{m,j}^{(k)}z^j.$
Then $$ a\_{n,k,j}=(-1)^{k j}C\_{kj+1,n-(k-1)j}^{(k-1)}.$$
Any idea how to prove this?


**Remark:** $C\_{1,n}^{(2)}=C\_n$ are the Catalan numbers and


$( C\_{1,n}^{(3)})\_{n\geq 0}=(1,1,1,0,-4,-14,-30,-33,\dots).$ (cf. OEIS, A103779).


**Edit:** Perhaps the following observation may be useful.
A matrix inversion theorem of Gould and Hsu implies a similar result:
The coefficients $c\_{n,k,j}$ which give $$\sum\_j {c\_{n,k,j}} F\_j^{(k)}(x)=x^n$$ are $(-1)^{n-j} A\_{n-j,k,kj+1}$ where $A\_{n,k,r}=\frac{r}{kn+r}\binom{kn+r}{n}$ are Fuss-Catalan numbers.
Here we have $\sum\_{n}A\_{n,k,r}x^n=(\sum\_{n}A\_{n,k,1}x^n)^r $ and $ x\sum\_{n}A\_{n,k,1}x^{(k-1)n}$ is the inverse series of $y-y^k.$
Is there a combinatorial or other reason for the appearance of these special inverse series?



A polynomial identity related to Catalan numbers
These assertions can be proved using (formal) generating functions.
Using that for $j\geq 0, k\geq 1$
\begin{align\*} 
\sum\_{n\geq 0} {n-j+kj \choose kj} t^n &=\frac{t^j}{(1-t)^{kj+1} }\;\;\mbox{ and }\\
\sum\_{n\geq 0}{n+j \choose kj} t^n&=\frac{t^{kj-j}}{(1-t)^{kj+1}}\;\;\;,
\end{align\*}
gives that
\begin{align\*} 
\sum\_{n\geq 0} t^n F\_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt}{(1-t)^k}}\;\;\mbox{ and }\\
\sum\_{n\geq 0} t^n G\_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt^{k-1}}{(1-t)^k}}\;\;\;,
\end{align\*}


(I) consider first the (simpler) Gould-Hsu case. Here $c\_{n,k,j}= (-1)^{n-j} [t^{n-j} ]\, C\_k(t)^{kj+1}$ where
$C\_k(t)$ denotes the $k$-ary tree function, which is defined by
$$C\_k(t)=1+tC\_k(t)^k\;\;\;.$$
Thus \begin{align\*}
\sum\_{j\geq 0} c\_{n,k,j} F\_j^{(k)}(x)&=\sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{n-j}[t^{n-j}]C\_k(t)^{kj+1}\\
&=\sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{n-j}[t^{n}] t^j C\_k(t)^{kj+1}\\
&=(-1)^n [t^n] \sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{j}t^j C\_k(t)^{kj+1}\\
&=(-1)^n [t^n] \frac{C\_k(t)}{1+tC\_k(t)^k}\frac{1}{1+\frac{xtC\_k(t)^k}{(1+tC\_k(t)^k)}^k}\\
&=(-1)^n [t^n] \frac{1}{1+xt}=x^n\\
\end{align\*}


(II) now to your case above. Here $a\_{n,k,j}=(-1)^{kj}[t^{n-(k-1)j}] A\_{k-1}(t)^{kj+1}$ where $yA\_k(y)$ is the inverse of $y(z)=\sum\_{j=1}^k (-1)^{j-1} z^j$ .
A similar computation as above here gives
\begin{align\*}
\sum\_{j\geq 0} a\_{n,k,j} F\_j^{(k)}((-1)^kx)&=[t^n] \frac{A\_{k-1}(t)}{1-T(t)} \frac{1}{1- (-1)^k\frac{T(t)x}{(1-T(t))^k}}
\end{align\*}
where $T(t):=(-1)^kt^{k-1}A\_{k-1}(t)^k$. This will simplify to the generating function for the $G\_n^{(k)}$ if  

$$\frac{A\_{k-1}(t)}{1-T(t)}=\frac{1}{1-t}\;\;.$$
And this in turn follows (for $k\geq 2$) with simple steps after substituting $x=tA\_{k-1}(t)$ in the equality $$t=\frac{x+(-1)^{k-1}x^k}{1+x}$$.


(III) The Ansatz $b\_{n,k,j}=[t^n] T(t)^j Z(t)$ leads to the generating function
\begin{align\*}
\frac{Z(t)}{1-T(t)} \frac{1}{1- \frac{T(t)x}{(1-T(t))^k}}
\end{align\*}
for $R\_n(x):=\sum\_{j \geq 0} b\_{n,k,j} F\_j^{(k)}(x)$. One will expect this to be a simple function of $t$ only if $\frac{Z(t)}{1-T(t)}$ and $\frac{T(t)}{(1-T(t))^k}$ simplify
to simple functions of $t$, i.e. can be "solved" for $t$. The targetet generating functions more or less require that $Z=C\_k, T=-tC\_k^k $ in case (I), resp. that
$Z=A\_{k-1}, T=-t^{k-1}A\_{k-1}$ in case (II), this explains the appearance of these special inverse series.


