Timeline for Number of collinear ways to fill a grid
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Sep 20 at 13:12 | answer | added | Martin.s | timeline score: 0 | |
| Apr 7, 2021 at 19:10 | comment | added | darij grinberg | And indeed, a solution has appeared in Crux Mathematicorum 47/3 (pp. 141--142). From a quick look, the tactic seems to be "generalize and induct on the new parameters". And sure, this is not the original source for the problem, but it is a new appearance with a potentially new solution. | |
| Feb 10, 2021 at 19:55 | comment | added | Sebastien Palcoux | @darijgrinberg: what you mentioned (here) is in 2020. It also appeared in Miklos Schweitzer Competition (here) in 2019. But as far as I know, it appeared first as this MathOverflow post (and was solved) in 2018. | |
| Feb 10, 2021 at 2:17 | comment | added | darij grinberg | FYI: This problem has appeared as problem OC500 in the Olympiad Corner of Crux Mathematicorum ( cms.math.ca/publications/crux ) recently. I would expect a solution to appear around issue 47/3. | |
| Feb 5, 2021 at 19:36 | answer | added | atenao | timeline score: 3 | |
| Nov 16, 2019 at 11:47 | answer | added | Fedor Petrov | timeline score: 5 | |
| Sep 28, 2018 at 15:09 | vote | accept | Sebastien Palcoux | ||
| Jul 15, 2018 at 19:11 | answer | added | Yibo Gao | timeline score: 12 | |
| May 23, 2018 at 2:09 | answer | added | Sam Hopkins | timeline score: 3 | |
| Apr 20, 2018 at 2:03 | answer | added | Richard Stanley | timeline score: 13 | |
| Apr 12, 2018 at 2:38 | answer | added | j.c. | timeline score: 2 | |
| Apr 9, 2018 at 22:02 | comment | added | Sebastien Palcoux | About the Generalized Hockey Stick Identities: pdfs.semanticscholar.org/5a96/… | |
| Apr 9, 2018 at 21:24 | comment | added | user44191 | @SebastienPalcoux Also, I just recognized the sum as a generalized hockeystick. $\sum_{\ell=1}^m \frac{m!}{(m-\ell)!} \ell (2m-\ell-1)! = m! \sum_{\ell=1}^m \frac{(2m-\ell-1)!}{(m-\ell)!} \ell = m! (m - 1)! \sum_{\ell = 1}^m \binom{l}{1} \binom{2m - \ell - 1}{m - 1}$ $= m! (m - 1)! \binom{2m}{m + 1} = m! \frac{(2m)!}{(m + 1)!} = \frac{(2m)!}{m + 1}$ | |
| Apr 9, 2018 at 21:11 | comment | added | Yuzhou Gu | I wrote some little code and verified the formula for $nm \le 1000$. | |
| Apr 9, 2018 at 20:25 | comment | added | user44191 | It may be useful to study $\tilde{g}(m, n, i) = \frac{g(m, n, i)}{m! n!}$, which represents the number of collinear orderings that reach each row and column in order. This satisfies the simpler recursion relation $\tilde{g}(m, n, i+1) = (mn - i) \tilde{g}(m, n, i) + n \tilde{g}(m-1, n, i) + m \tilde{g}(m, n-1, i)$, and also keeps the numbers smaller. The conjecture is then that $\tilde{g}(m, n, mn) = \frac{(mn)!}{(m + n - 1)!}$. | |
| Apr 9, 2018 at 19:29 | comment | added | j.c. | I've just posted a question inspired by this one here mathoverflow.net/questions/297411/… | |
| Apr 9, 2018 at 16:47 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
About the first case in the proof: clarification to avoid confusion.
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| Apr 9, 2018 at 16:10 | comment | added | Sebastien Palcoux | @HughThomas: yes, with the use of WolframAlpha. A direct proof would be better, and useful for the general case. | |
| Apr 9, 2018 at 15:57 | comment | added | Hugh Thomas | Let me note that the case $m=2$ is solved in the linked question. | |
| Apr 9, 2018 at 15:37 | comment | added | Per Alexandersson | Side note: The formula is a bit similar to en.wikipedia.org/wiki/Catalan_number#Generalizations the super-Catalan numbers (or one of the super versions). | |
| Apr 9, 2018 at 14:18 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |