What is this mystery function that wolfram alpha says my exponential generating function is equal too?

Question

The $E_n(x^n)$ is the mystery function

$$\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}=E_n(x^n)$$

Here are the first 3 values of the function

Answer 1

This is the Mittag-Leffer generalization of the exponential function.

Answer 2

For $$E_n=\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}$$ the first terms are simple $$\left( \begin{array}{cc} n & E_n \\ 1 & e^x \\ 2 & \cosh (x) \\ 3 & \frac{2}{3} e^{-x/2} \cos \left(\frac{\sqrt{3} x}{2}\right)+\frac{e^x}{3} \\ 4 & \frac{\cos (x)}{2}+\frac{\cosh (x)}{2} \end{array} \right)$$ For $n>4$, they are hypergeometric functions $$E_5=\, _0F_4\left(;\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{x^5}{5^5}\right)$$ $$E_6=\, _0F_5\left(;\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{x^6}{6^6 }\right)$$ $$E_7=\, _0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}; \frac{x^7}{7^7}\right)$$ and the pattern is quite clear.