Timeline for A combinatorial triangle for the Bernoulli numbers
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 2 at 10:46 | comment | added | Peter Luschny | OEIS A363154 calls it "The Hadamard product of A173018 and A349203". | |
| Apr 24 at 17:57 | comment | added | The Amplitwist | 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" | |
| Apr 24 at 17:40 | comment | added | Andreas Holmstrom | I'm interested in this. @PeterLuschny, did you find anything? | |
| May 22, 2023 at 16:17 | comment | added | Peter Luschny | 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. | |
| May 22, 2023 at 14:15 | comment | added | Peter Taylor | "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: $$\sum_{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle {\binom {n}{m}}^{-1}=(n+1)B_{n}$$ | |
| May 22, 2023 at 10:37 | history | asked | Peter Luschny | CC BY-SA 4.0 |