Find the value of $\sum_{d=0}^\infty\binom{d+m-1}{m-1}\frac{x^d}{d!}$.

Question

It is known that the generating function of the number of multisets is given by $$ \sum_{d=0}^\infty\binom{d+m-1}{m-1}{x^d}=\frac{1}{(1-x)^m}. $$ Then, does a similar series $$ \sum_{d=0}^\infty\binom{d+m-1}{m-1}\frac{x^d}{d!} $$ to the generating function have an explicit formula ?

Answer 1

The series is a confluent hypergeometric function, or a generalised hypergeometric function $_1F_1$: $$\sum_{d = 0}^\infty {d+m-1 \choose m-1} \frac{x^d}{d!} = \sum_{d=0}^\infty \frac{(m)_d}{(1)_d} \frac{x^d}{d!} = {}_1F_1(m;1;x) = M(m,1,x),$$ where $(a)_n := a(a+1)\dots (a+n-1)$ denotes the rising factorial.

These functions are also related to the Laguerre polynomials $L_n$: $${}_1F_1(m;1;x) = e^x L_{m-1}(-x).$$