Has anyone seen this combinatorial identity involving the Bernoulli and Stirling numbers?

Solution 1:

The coefficients $B_j^{(r)}$ defined by$$\sum_{j = 0}^\infty B_j^{(r)} {{x^j}\over{j!}} = \left({x\over{e^x - 1}}\right)^r$$are usually called higher order Bernoulli numbers, so your identity is a formula for $B_j^{(r)}$ for $j < r$.

Let $c(n, k) = |s(n, k)| = (-1)^{n - k}s(n, k)$. This is a fairly standard notation, used, for example, in Stanley's "Enumerative Combinatorics". Then$${{(-\log(1 - x))^k}\over{k!}} = \sum_{n = k}^\infty c(n, k) {{x^n}\over{n!}}.$$Differentiating this equation gives$${{(-\log(1 - x))^k}\over{(1 - x)k!}} = \sum_{n = k}^\infty c(n + 1, k + 1) {{x^n}\over{n!}}.\tag*{$(1)$}$$For any formal Laurent series $f = f(\alpha)$ we define the residue of $f$, denoted $\text{res}\,f$, to be the coefficient of $\alpha^{-1}$ in $f$. So the coefficient of $\alpha^k$ in $f$ is $\text{res}\,f/\alpha^{j + 1}$.

We will apply Jacobi's change of variables formula for residues, which is a form of the Lagrange inversion formula. See e.g. Gessel's survey of Lagrange inversion at https://arxiv.org/abs/1609.05988, Theorem 4.1.1.

Suppose that $f(\alpha)$ is a formal Laurent series in $\alpha$ and that $g(\alpha) = g_1 \alpha + g_2\alpha^2 + \ldots$ is a formal power series in $\alpha$ with $g_1 \neq 0$. Then Jacobi's formula says that$$\text{res}\,f(\alpha) = \text{res}\,f(g(\alpha))g'(\alpha).$$We apply Jacobi's formula with$$f(\alpha) = \left({\alpha\over{1 - e^{-\alpha}}}\right)^{n + 1} \alpha^{-j - 1}$$and$$g(\alpha) = -\log(1 - \alpha).$$Then the coefficient of $\alpha^j$ in$$\left({\alpha\over{1 - e^{-\alpha}}}\right)^{n + 1}$$is\begin{align*} \text{res}\,f(\alpha) & = \text{res}\,f(g(\alpha))g'(\alpha) \\ & = \text{res}\, {{(-\log(1 - \alpha)/\alpha)^{n + 1}}\over{(-\log(1 - \alpha))^{j + 1}(1 - \alpha)}} \\ & = \text{res}\,{{(-\log(1 - \alpha))^{n - j}}\over{\alpha^{n + 1}(1 - \alpha)}}. \end{align*}This is the coefficient of $\alpha^n$ in$${{(-\log( 1- \alpha))^{n - j}}\over{1 - \alpha}}$$which by $(1)$ is$${{(n - j)!}\over{n!}} c(n + 1, n - j + 1).$$