# Condition for Sequence of Numbers to Have Both OGF and EGF

As an example, The Stirling Numbers of the Second Kind, denoted S(n,k), has two generating functions; an ordinary and an exponential. The ordinary is

$$\sum_{n=0}^\infty{S(n,k)x^n}=\frac{x^k}{(1-x)(1-2x)...(1-kx)}$$

and the exponential is

$$\sum_{n=0}^\infty{S(n,k)\frac{x^n}{n!}}=\frac{(e^x-1)^k}{k!}$$

Is there a way to determine whether a particular sequence of numbers has both an exponential and ordinary generating function? Not all sequences of numbers have both generating function types; Euler numbers do not appear to have both for example, and I have not found any literature stating the Bernoulli Numbers have an ordinary generating function.

• This is just a thought but it may be handy to remember the connection between egf $f_{\text{e}}$ and ogf $f_{\text{o}}$ of the same sequence $$f_{\text{o}}(x)=\int_{0}^{\infty}e^{-t}f_{\text{e}}(xt)\, \mathrm{d}t$$ perhaps this can be exploited in some way? Commented Apr 24, 2017 at 0:41
• This is interesting. It is similar to a Laplace Transform, but the argument of the exponential generating function has the change of variable argument. Is this particular transformation called anything? Commented Apr 24, 2017 at 0:58
• I'm not sure if it has a name. It exploits the gamma function property for non-negative $n$ since $$n!=\int_{0}^{\infty}e^{-t}t^n\,\mathrm{d}t$$ It's a convenient tool for expressing certain combinatorial counts, but I figured it is a nice way of connecting egfs and ogfs so it might be an indication as to whether an egf has a nice ogf and vice-versa. It was only a thought, I don't really know how to proceed from there but there may be some "bright spark" on mathSE who can help. You could probably use it to show that those two generating functions for Stirling numbers are equivalent. Commented Apr 24, 2017 at 1:06
• What do you mean "have a generating function"? Do you mean one that is convergent somewhere? Or that has a convenient formula? Every sequence has a (formal) generating function. Commented Apr 25, 2017 at 6:21
• I meant "have a convenient closed form" generating function. Yes, any sequence has a generating function, but "nice closed forms" don't always exist. Commented Apr 26, 2017 at 17:55

Just a few remarks (and interesting references).

Given a sequence $(a_n)$ we are always free to consider different types of generating functions. Amongst them are

But for a specific sequence $(a_n)$ usually not all types of generating functions will provide useful results or can be represented in a closed form.

In combinatorics we typically take ordinary generating functions to count unlabelled objects and exponential generating functions when we want to count labelled objects. See e.g. the first two chapters in Analytic Combinatorics by P. Flajolet and R. Sedgewick.

As stated in your post the ordinary generating function for Bernoulli Numbers

\begin{align*} \beta(x)=\sum_{n=0}^\infty B_n x^n=1-\frac{x}{2}+\frac{x^2}{6}-\frac{x^4}{30}+\frac{x^6}{42}-\cdots \end{align*} is rather unusual compared with the much more common exponential generating series \begin{align*} \frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_n}{n!}x^n=1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720} +\frac{x^6}{30240}-\cdots \end{align*}

In fact it is stated in Curious and Exotic Identities for Bernoulli Numbers by Don Zagier in section A.1 The other Generating Function(s) for the Bernoulli Numbers.

He states there that despite its being divergent and not being expressible as an elementary function, it fulfills the following functional equation:

The power series $\beta(x)$ is the unique solution in $\mathbb{Q}[[x]]$ of the equation \begin{align*} \frac{1}{1-x}\beta\left(\frac{1}{1-x}\right)-\beta(x)=x \end{align*}