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Motivation: We informally call an infinite lower triangular matrix $\operatorname{T}(n, k)$ of integers a combinatorial triangle of a sequence of integers or rational numbers if it can be obtained from it by some simple operation (e.g., by summation or alternating summation of the rows), in the case of a rational sequence with an additional normalization.

For example, the triangle of the Stirling set numbers is a combinatorial triangle for the Bell numbers, Euler's triangle (Eulerian numbers) for the factorial numbers, or Pascal's triangle for the powers of 2. Of course, there can be a multitude of triangles that give a combinatorial interpretation of a sequence in this way. Are there such triangles for the Bernoulli numbers?

Construction: The first-order Eulerian numbers are defined as

\begin{equation} \left\langle n\atop k \right\rangle = (k+1) \left\langle n-1\atop k \right\rangle + (n-k) \left\langle n-1\atop k-1 \right\rangle, \end{equation}

with boundary conditions $\left\langle 0\atop 0 \right\rangle=1$, $\left\langle n\atop k \right\rangle =0$ for $k<0$ or $k > n$.

Further we set, for integer $0 \le k \le n$:

\begin{equation} \left[ n \atop k \right] = \frac{\operatorname{lcm}_{j=0}^n \binom{n}{j}}{\binom{n}{k}} \end{equation}

\begin{equation} \operatorname{T}_{n, k} = \left[ n \atop k \right] \left\langle n \atop k \right\rangle \end{equation}

The first few values of the triangle $\operatorname{T}(n,k)$ are:

\begin{equation} \begin{matrix} 1 \\ 1 \quad 0 \\ 2 \quad 1 \quad 0 \\ 3 \quad 4 \quad 1 \quad 0 \\ 12 \quad 33 \quad 22 \quad 3 \quad 0 \\ 10 \quad 52 \quad 66 \quad 26 \quad 2 \quad 0 \\ \ldots \end{matrix} \end{equation}

The Bernoulli numbers ($ \operatorname{B}_1 = 1/2 $) can be obtained for $n \ge 0$ with: \begin{equation} \operatorname{B}_n = \frac{\sum_{k=0}^n (-1)^k \operatorname{T}_{n,k}} { \operatorname{lcm}_{k=0}^n (k + 1)}. \end{equation} \begin{equation} \operatorname{B}_n = \frac11, \frac12, \frac16, 0, -\frac{2}{60}, 0, \frac{10}{420}, 0, -\frac{84}{2520}, 0, \frac{2100}{27720}, 0, -\frac{91212}{360360}, \ldots \end{equation}

Which combinatorial objects does $\operatorname{T}(n,k)$ count?

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  • $\begingroup$ "Are there such triangles for the Bernoulli numbers?" appears not to be the main question, but a quick glance at Wikipedia turns up $$B_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {k!}{k+1}} \left\{ {n+1\atop k+1} \right\}$$ in terms of weighted Stirling numbers of the second kind. An identity given in A002944 shows that your expression is equivalent to another one from Wikipedia: $$\sum_{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle {\binom {n}{m}}^{-1}=(n+1)B_{n}$$ $\endgroup$ Commented May 22, 2023 at 14:15
  • $\begingroup$ Thank you Peter, however I'm not interested in this identity per se (I described it in a blog post more than ten years ago as one of the motivations for choosing B(1) = 1/2) or in its proof, which Ira Gessel gave here on MO. My concern is exactly what the request in the last line says. $\endgroup$ Commented May 22, 2023 at 16:17
  • $\begingroup$ I'm interested in this. @PeterLuschny, did you find anything? $\endgroup$ Commented Apr 24, 2024 at 17:40
  • $\begingroup$ Reposting a link mentioned in a previous comment so that it appears in the "Linked" questions list: Ira Gessel's answer to "Eulerian number identity" $\endgroup$ Commented Apr 24, 2024 at 17:57
  • $\begingroup$ OEIS A363154 calls it "The Hadamard product of A173018 and A349203". $\endgroup$ Commented May 2, 2024 at 10:46

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