Condition for Sequence of Numbers to Have Both OGF and EGF

130 Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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*}