Timeline for Show that this ratio of factorials is always an integer
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2014 at 9:36 | answer | added | user45639 | timeline score: 2 | |
| Oct 16, 2013 at 3:40 | comment | added | Thomas | What is the generating function of the sequence? | |
| Jun 4, 2013 at 4:49 | history | edited | GH from MO |
edited tags
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| May 13, 2013 at 15:02 | comment | added | JSE | Now I'm curious; if {L_1, ... L_r} and {M_1, .. M_s} are linear forms in k variables, and the sum of the L_i is the same as the sum of the M_j, what further conditions guarantee that prod L_i ! / prod M_j ! is always an integer? | |
| May 12, 2013 at 15:00 | answer | added | Ira Gessel | timeline score: 16 | |
| May 11, 2013 at 17:48 | history | edited | Ryan Reich | CC BY-SA 3.0 |
improve title; added 2 characters in body
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| May 11, 2013 at 16:52 | comment | added | cardinal | This is just an observation: We can write the quantity as ${2m \choose m}{2n \choose n}/{m+n \choose m}$. The numerator is the number of random walk paths of length $2(m+n)$ such that the random walk is zero at time $2m$ and $2(m+n)$. The denominator hints at quotienting out the locations of pairs of steps instead of considering the first $2m$ and the subsequent $2n$. But, I haven't made that work out and suspect it won't. | |
| May 11, 2013 at 16:39 | comment | added | user9072 | @Roland Bacher: "Thus the question for a nice counting argument seems to be a valuable one." Perhaps. However this is not the question that was asked here. (I did not vote.) | |
| May 11, 2013 at 16:17 | history | reopened |
Douglas Zare user6976 Joseph O'Rourke Roland Bacher Seva |
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| May 11, 2013 at 16:14 | comment | added | Roland Bacher | I do not understand the closing of this question. A combinatorial interpretation of these integers seems to be lacking and this is surely a (nice, in my opinion) research question. Thus the question for a nice counting argument seems to be a valuable one. | |
| May 11, 2013 at 15:41 | history | closed |
Qiaochu Yuan Steven Landsburg Gerry Myerson Fernando Muro Andreas Blass |
off topic | |
| May 11, 2013 at 14:54 | comment | added | Douglas Zare | Also oeis.org/A068555. | |
| May 11, 2013 at 14:45 | answer | added | Douglas Zare | timeline score: 20 | |
| May 11, 2013 at 14:36 | comment | added | Douglas Zare | See also: mathoverflow.net/questions/26336/… | |
| May 11, 2013 at 14:07 | answer | added | Barry Cipra | timeline score: 14 | |
| May 11, 2013 at 8:51 | answer | added | user33772 | timeline score: 46 | |
| May 11, 2013 at 8:18 | vote | accept | karan | ||
| May 11, 2013 at 7:37 | answer | added | Yuichiro Fujiwara | timeline score: 45 | |
| May 11, 2013 at 7:21 | comment | added | Tony Huynh | @Qiaochu and Karan. Sorry, I think I spoke too soon. I was thinking of the case m=n in which the embedding was clear to me. I see that it's not clear what to do if m≠n. | |
| May 11, 2013 at 7:03 | comment | added | karan | @Tony : Can you add some detail on how the latter is a subgroup of the former ? | |
| May 11, 2013 at 7:01 | comment | added | Qiaochu Yuan | @Tony: is it? That isn't clear to me. What embedding do you have in mind? | |
| May 11, 2013 at 6:58 | comment | added | Tony Huynh | I am not sure this is appropriate for MO. That being said, the numerator is the size of the group $S_{2m} \times S_{2n}$ while the denominator is the size of the group $S_{m} \times S_{n} \times S_{m+n}$. The latter is easily seen to be a subgroup of the former. | |
| May 11, 2013 at 6:37 | history | edited | karan | CC BY-SA 3.0 |
added 3 characters in body
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| May 11, 2013 at 6:32 | history | asked | karan | CC BY-SA 3.0 |