Considering the one parameter Mittag-Leffler function, 


$$E\_{\alpha}(z)=\sum\_{k=0}^\infty\frac{z^{k}}{\Gamma(\alpha k+1)}, \Re(\alpha)>0$$


Considering then the generating function for $E\_\alpha(z^\alpha)$, we see that is


$$E\_\alpha(z^\alpha)=\frac{1}{\alpha}\sum\_{k=0}^{\alpha -1}\exp(w\_{\alpha}^kz)$$


where $w\_\alpha=\exp(2i\pi/\alpha)$.


I want to study the coefficients then of powers of this function; in other words, 


$$[E\_\alpha(z^\alpha)]^n=\sum\_{k=0}^\infty{A(n,k)\frac{z^k}{k!}}$$


I know that I can use the multinomial formula to write $[E\_\alpha(z^\alpha)]^n$ as


$$\left[\frac{1}{\alpha}\sum\_{k=0}^{\alpha -1}\exp(w\_{\alpha}^kz)\right]^n=\frac{1}{\alpha^n}\sum\_{k\_0+...+k\_{\alpha-1}=n}\binom{n}{k\_0,...,k\_{\alpha-1}}\exp((w\_{\alpha}^0k\_0+...+w\_{\alpha}^{\alpha-1}k\_{\alpha-1})x)$$
$$=\frac{1}{\alpha^n}\sum\_{k\_0+...+k\_{\alpha-1}=n}\binom{n}{k\_0,...,k\_{\alpha-1}}\sum\_{j=0}^\infty{(w\_{\alpha}^0k\_0+...+w\_{\alpha}^{\alpha-1}k\_{\alpha-1})^j}\frac{x^j}{j!}$$
$$=\sum\_{j=0}^\infty\frac{1}{\alpha^n}\sum\_{k\_0+...+k\_{\alpha-1}=n}\binom{n}{k\_0,...,k\_{\alpha-1}}{(w\_{\alpha}^0k\_0+...+w\_{\alpha}^{\alpha-1}k\_{\alpha-1})^j}\frac{x^j}{j!}$$


I could keep going, and again use the multinomial formula for $(w\_{\alpha}^0k\_0+...+w\_{\alpha}^{\alpha-1}k\_{\alpha-1})^j$ and use elementary recursive methods to get a recursive formula. But this seems like a very longwinded approach. Are there other methods to come up with a recursive formula for these coefficients or is this standard approach the preferred method? Is there a combinatoric approach that I could use?



Coefficients for Powers of the Mittag-Leffler Function
If $\alpha$ is a positive integer then one can give a combinatorial
interpretation of the coefficients of $\left(\sum\_{k\geq
0}\frac{z^k}{(\alpha k)!}\right)^n = \sum\_{k\geq 0}
B(n,k)\frac{x^k}{(\alpha k)!}$. Namely, $B(n,k)$ is the number of
multichains of sets $\emptyset=S\_0\subseteq S\_1\subseteq \cdots
\subseteq S\_n=\{1,2,\dots,\alpha k\}$, where the number of elements of
each $S\_i$ is divisible by $\alpha$.


One can also obtain a recursive formula by differentiating
$[E\_\alpha(z^\alpha)]^n$. This recurrence works for any power series $F(x)^n$ so has nothing to do with properties of $E\_\alpha(z)$ itself.


