Why to ignore the factorials as part of the coefficients in the exponential generating function?

Question

I just perceived that, for the ordinary generating function $e^x$, it's coefficients are:

$$\left(1,1,\frac{1}{2!},\frac{1}{3!},\frac{1}{4!},\dots \right)$$

But for the exponential generating function $e^x$, it's coefficients are:

$$(1,1,1,1,1,1,\dots)$$

That is, the factorials are ignored for the EGF's. Why is that? I don't know if that is done by definition or if it's useful for something or if it's defined in a convenient way to achieve something else.

Answer 1

Here are the definitions for the two types of generating function, according to Wikipedia:

$${\rm G}(a_n;x)=\sum_{n=0}^\infty a_nx^n\\ {\rm EG}(a_n;x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}\\$$

As can be seen, the exponential generating function has terms of the form $a_n\frac{x^n}{n!}$, where $a_n$ is the coefficient. $n!$ is already there, so $(a_n)=(1,1,1,\dots)$ for $e^x$.

Answer 2

There are different types of generating functions, the ordinary gf and the exponential gf being the most famous representatives (but there are also others, e.g. Dirichlet generating function).

Different types of generating functions are useful for different problems, and while ordinary gfs are mostly used to count unlabeled structures, exponential gfs are used for labeled ones.

Why is that so? Well, that would certainly be too long for this answer and I recommend reading the standard literature, e.g. Flajolet's and Sedgewick's Analytic Combinatorics, but note that the generating function $$ A(z) = \sum_{n \ge 0} n! z^n $$ has a radius of convergence $0$. So it cannot be analyzed analytically (which is one of the aims of generatingfunctionology (which, by the way, is also a name of a good book)).
But by dividing every coefficient by $n!$, we get the nice function $B(z)= \frac{1}{1-z}$.