Is there any interesting interpretation of Taylor coefficients of $e^{-\log(1-x)}$?

Question

Consider the series expansion of $e^{-\log(1-x)} = \frac{1}{1-x}$. Using the Taylor series of $\exp(x)$ and $-\log(1-x)$, we find that

$$ 1 + \sum_{n=1}^{\infty} \Bigg( \sum_{k=1}^{\infty} \sum_{\substack{ \alpha \in \Bbb{N}_1^k \\ |\alpha| = n}} \frac{1}{k!} \frac{1}{\alpha_1 \cdots \alpha_k} \Bigg)x^n = 1 + x + x^2 + \cdots, $$

where $\alpha = (\alpha_1, \cdots, \alpha_k) \in \Bbb{N}_1^k$ is a multi-index and $|\alpha| = \alpha_1 + \cdots + \alpha_k$. Comparing both sides, we find that for any $n \geq 1$,

$$ \sum_{k=1}^{\infty} \sum_{\substack{ \alpha \in \Bbb{N}_1^k \\ |\alpha| = n}} \frac{1}{k!} \frac{1}{\alpha_1 \cdots \alpha_k} = 1. $$

Is there any known combinatorial or probabilistic interpretation on it?

We may think of this as a discrete distribution which assigns the probability $p(\alpha) = \frac{1}{k!}\frac{1}{\alpha_1 \cdots \alpha_k}$ to each $\alpha \in \Bbb{N}_1^k$ with $|\alpha| = n$, but I see no clear interpretation of this probability (if we have any).

Answer 1

There is a well-known interpretation for $e^{-\log{1-x}}$ from the theory of exponential generating functions. See Richard Stanley - Enumerative Combinatorics Vol 2, 5.1. Here it's more natural to consider the coefficient of $\frac{x^n}{n!}$ rather than $x^n$.

We have \begin{align*} \Bigg( \sum_{k=1}^{\infty} \sum_{\substack{ \alpha \in \Bbb{N}_1^k \\ |\alpha| = n}} \frac{1}{k!} \frac{1}{\alpha_1 \cdots \alpha_k} \Bigg)x^n = \Bigg( \sum_{k=1}^{\infty} \sum_{\substack{ \alpha \in \Bbb{N}_1^k \\ |\alpha| = n}} \frac{1}{k!} \binom{n}{\alpha} (\alpha_1-1)! \cdots (\alpha_k-1)! \Bigg) \frac{x^n}{n!} \end{align*}

where $\binom{n}{\alpha} = \frac{n!}{\alpha_1! \cdots \alpha_k!}$ is the multinomial coefficient. Furthermore, $(n-1)!$ is the number of ways to put an $n$-cycle on $n$ objects. Then $\binom{n}{\alpha} (\alpha_1-1)! \cdots (\alpha_k-1)!$ can be interpreted as the number of ways to break a set of $n$ objects into $k$ disjoint subsets in some order, where the $k$th subset has size $\alpha_k$, and put a cycle on each subset. This is the same as choosing a permutation with cycle type $\alpha_1, \ldots, \alpha_k$. Dividing by $k!$ means that the subsets don't come in any specific order - so we can put them in some standard order, like arranging them by their least elements. Summing over all $\alpha$ gives $$ \sum_{k=1}^{\infty} \sum_{\substack{ \alpha \in \Bbb{N}_1^k \\ |\alpha| = n}} \frac{1}{k!} \binom{n}{\alpha} (\alpha_1-1)! \cdots (\alpha_k-1)! = n!.$$

Then the summand $\frac{1}{k!} \frac{1}{\alpha_1 \cdots \alpha_k}$ can be interpreted probabilistically: it is the probability that a random permutation in $S_n$ has cycle type $\alpha_1, \ldots, \alpha_k$, so that $\alpha_1$ is the size of the cycle that has a $1$ in it, $\alpha_2$ is the size of the cycle with the next smallest element outside the first cycle, etc.