Timeline for An identity for the ratio of two partial Bell polynomials
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2023 at 6:34 | answer | added | qifeng618 | timeline score: 0 | |
| Oct 20, 2023 at 2:58 | history | edited | qifeng618 | CC BY-SA 4.0 |
added 386 characters in body
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| Oct 19, 2023 at 23:33 | comment | added | qifeng618 | @PeterTaylor I tested numerically for many $(\ell,m)$, but I neglected those such that the identity (o-o) is not valid. Your findings are true, I made stupid mistakes in the so-called guess. I will modify this question very soon. Thank you very much. | |
| Oct 19, 2023 at 16:17 | answer | added | Peter Taylor | timeline score: 1 | |
| Oct 19, 2023 at 15:40 | comment | added | Abdelmalek Abdesselam | ...is the article by Krattenthaler sciencedirect.com/science/article/pii/S0097316596900588 | |
| Oct 19, 2023 at 15:39 | comment | added | Abdelmalek Abdesselam | @qifeng618: sorry I was just floating ideas and recommending interpreting the $(2n-1)!!$ as counting perfect matchings of $2n$ elements en.wikipedia.org/wiki/Perfect_matching (this requires introducing, e.g., $\ell$ extra fictitious elements) and interpreting the Bell polynomials as counting set partitions en.wikipedia.org/wiki/Bell_polynomials Since you have squares of $(2n-1)!!$ that means you have not just one but two perfect matchings, say with red and blue edges. A reference which can help for understanding this... | |
| Oct 19, 2023 at 15:17 | comment | added | Peter Taylor | How far have you tested it numerically? My Sage code may be buggy, but it only seems to hold for $\ell \in \{1, 2\}$, $m > 0$. | |
| Oct 19, 2023 at 2:10 | comment | added | qifeng618 | @AbdelmalekAbdesselam Are you combinatorially interpreting partial Bell polynomials $B_{m,k}$? My problem is to prove the identity (o-o)! Anyway, thank you very much! But I really need a complete proof of the identity (o-o) in details. | |
| Oct 18, 2023 at 21:30 | comment | added | qifeng618 | @AbdelmalekAbdesselam I am not a combinatorist, so I cannot understand what you sketched. | |
| Oct 18, 2023 at 21:17 | comment | added | Abdelmalek Abdesselam | Sounds like you are counting partial matchings. Writing the sums in terms of combinatorial objects instead of sums with factorials is probably a better way to go. Try colored perfect matchings with $\ell$ colors on a set of of $\ell+2m+2\ell$ elements with $\ell$ distinguished elements forced to end up with different colors inherited from edges. | |
| Oct 18, 2023 at 20:39 | comment | added | qifeng618 | @MichaelHardy Yes, you are right. \begin{equation*} B_{m,k}(x_1,x_2,\dotsc,x_{m-k+1})=\sum_{\substack{1\le j\le m-k+1\\ \ell_j\in\{0\}\cup\mathbb{N}\\ \sum_{j=1}^{m-k+1}j\ell_j=m\\ \sum_{j=1}^{m-k+1}\ell_j=k}}\frac{m!}{\prod_{j=1}^{m-k+1}\ell_j!} \prod_{j=1}^{m-k+1}\biggl(\frac{x_j}{j!}\biggr)^{\ell_j}. \end{equation*} | |
| Oct 18, 2023 at 18:21 | comment | added | Michael Hardy | To be clear about which polynomials you're talking about: would $$B_{6,2}(x_1,x_2,x_3,x_4,x_5) = 6x_5x_1 + 15x_4x_2 + 10x_3^2$$ be an example of what you have in mind? $\qquad$ | |
| Oct 18, 2023 at 16:53 | history | asked | qifeng618 | CC BY-SA 4.0 |