1
$\begingroup$

Does there exist a sequence $c_n$ such that $$S(n, k) = \frac{c_n}{c_k c_{n - k}}$$ for $0 \leq k \leq n$, where $S(n, k)$ are the Stirling numbers of the second kind?

I ask because I'm trying to create a type of generating function that conveniently expresses the Stirling transform $$a_n \mapsto \sum_k S(n, k) a_k.$$ If there were such a $c_n$, then the generating functions $$\sum_{k \geq 0} \frac{a_k}{c_k} x^k$$ would satisfy the useful product rule

$$\left( \sum_{k \geq 0} \frac{a_k}{c_k} x^k \right) \left( \sum_{k \geq 0} \frac{b_k}{c_k} x^k \right) = \sum_{n \geq 0} \frac{x^n}{c_n} \sum_k S(n, k) a_k b_{n - k}.$$

I know that the Stirling transform itself can be expressed using exponential generating functions, but I don't immediately see how that gives us the above product rule or tells us how to find $c_n$.

$\endgroup$
2
$\begingroup$

That would imply that $$S(n,n-k)=\frac{c_n}{c_{n-k}c_k}=S(n,k).$$ That's just not true.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah, such a simple observation! Thanks for pointing this out. This dashes my hopes for a "Stirling generating function," at least a straightforward one. $\endgroup$ – Robert D-B Jun 8 '19 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.