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 $c_n$.


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

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  • $\begingroup$ 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

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