Timeline for Recurrence relation for number of reduced words of longest element in $S_n$
Current License: CC BY-SA 4.0
15 events
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| Apr 30, 2020 at 18:34 | history | edited | Matt Samuel | CC BY-SA 4.0 |
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| Apr 30, 2020 at 14:48 | comment | added | Bipolar Minds | @SamHopkins thanks for the reference, this exactly the direction I'm aiming for | |
| Apr 30, 2020 at 14:44 | comment | added | Matt Samuel | @BipolarMinds Not sure if it will be useful, but I combinatorially described a recurrence for $\frac{\binom{n+1}2!}{|R(w_0(n))|}$. | |
| Apr 30, 2020 at 14:42 | history | edited | Matt Samuel | CC BY-SA 4.0 |
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| Apr 30, 2020 at 14:39 | comment | added | Sam Hopkins | @BipolarMinds: it sounds like you might be interested in higher Bruhat orders in the sense of Manin--Schechtman/Ziegler (doi.org/10.1016/0040-9383(93)90019-R). In that paper Ziegler says that, beyond the case covered by Stanley's work, little is known in terms of enumeration. But of course that paper is 30 years old so perhaps more is known now. | |
| Apr 30, 2020 at 14:26 | history | edited | Matt Samuel | CC BY-SA 4.0 |
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| Apr 30, 2020 at 14:22 | comment | added | Bipolar Minds | Yes, I actually did the same thing at first :) | |
| Apr 30, 2020 at 14:21 | comment | added | Matt Samuel | @BipolarMinds That's fair, but are you sure it's not always a natural number? | |
| Apr 30, 2020 at 14:16 | comment | added | Bipolar Minds | Well yes, fair enough :) My problem with this is that the quotient is not always a natural number, so it doesn't count anything. The recurrence relation should come from a bijection on sets, as for example $S_n= S_{n-1}\times [n]$ gives $n! =n(n-1)!$. In some sense, it should be a higher analogue of this identity, but I don't expect it to be easy. Sorry for my imprecise formulation of the question. | |
| Apr 30, 2020 at 13:04 | history | edited | Matt Samuel | CC BY-SA 4.0 |
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| Apr 27, 2020 at 18:36 | comment | added | Bipolar Minds | I found it in a paper of Richard Stanley | |
| Apr 27, 2020 at 18:29 | comment | added | Matt Samuel | @BipolarMinds The formula is well known, for example it's in Combinatorics of Coxeter groups by Bjorner and Brenti, though I'm not sure of the original source. | |
| Apr 27, 2020 at 18:26 | comment | added | Bipolar Minds | Thx! I actually forgot to write that I mean recurrence relation in $n$ but now that there is a closed formula, all the better | |
| Apr 27, 2020 at 18:23 | vote | accept | Bipolar Minds | ||
| Apr 30, 2020 at 10:20 | |||||
| Apr 27, 2020 at 18:18 | history | answered | Matt Samuel | CC BY-SA 4.0 |