# Coefficients for Powers of the Mittag-Leffler Function

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?

## 1 Answer

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.