Timeline for Counting monomials and the Catalan numbers
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 6, 2022 at 20:09 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| Apr 26, 2022 at 20:52 | vote | accept | T. Amdeberhan | ||
| Apr 26, 2022 at 10:05 | comment | added | Martin Rubey | Oh, I feel silly now! Thank you! | |
| Apr 26, 2022 at 8:07 | comment | added | Fedor Petrov | @MartinRubey ah, I see. My theorem yields the formula $G_{\mathfrak S_{n+1}}=z_1(z_1+z_2)(z_1+z_2+z_3)\cdots (z_1+\ldots+z_n)$, is it enough for your observations? | |
| Apr 26, 2022 at 7:22 | comment | added | Martin Rubey | I think I am using the definition of @T.Amdeberhan. For example, if $n=1$ I have $G = \frac{x_1 z_1}{x_1 - x_2}$ and therefore $G_{\mathfrak S_2} = z_1$. So, I have $n+1$ variables in the $x$-alphabet and $n$ variables in the $z$-alphabet. $G_{\mathfrak S_3} = z_1^2 + z_1 z_2$, $G_{\mathfrak S_4} = z_1^3 + 2*z_1^2 z_2 + z_1 z_2^2 + z_1^2 z_3 + z_1 z_2 z_3$. In particular, the number of monomials with coefficient 1 is $1, 2, 4, \dots$. | |
| Apr 25, 2022 at 19:50 | comment | added | Fedor Petrov | hm... to be sure, do you consider all monomials, or all monomials without $z_{n+1}$, or all monomials which may occur in expanding these brackets? | |
| Apr 25, 2022 at 19:33 | comment | added | Martin Rubey | Well, also the number of monomials with coefficient 1 in my implementation is $2^n$, not Catalan. | |
| Apr 25, 2022 at 15:52 | comment | added | Fedor Petrov | (if you do not consider monomials which include $z_{n+1}$, then yes, otherwise this depends on parity of $n$, I think) | |
| Apr 25, 2022 at 15:48 | comment | added | Fedor Petrov | @MartinRubey this is quite possible, I simply did not care about the sign writing my comment | |
| Apr 25, 2022 at 14:33 | comment | added | Martin Rubey | That's strange, possibly I have a bug in my code, but my coefficients are all positive. | |
| Apr 25, 2022 at 12:30 | comment | added | Fedor Petrov | @MartinRubey the coefficient equals (upto sign) to the number of enumerations of $n+1$ vertices of the path which satisfy $n$ inequalities which in turn correspond to the $n$ edges. The inequality $\pi(x)<\pi(y)$ for an edge $xy$ corresponds to our monomial $C$ being $y$-biased in the following sense: remove the edge $xy$, let $k$ denote the number of vertices in the piece containing $y$; then the total degree of $C$ in these $k$ variables is at least $k$. So, the number of monomials with coefficient $\pm 1$ (happens when all inequalities have the same direction) is twice Catalan number. | |
| Apr 25, 2022 at 10:36 | comment | added | Martin Rubey | Does your Theorem yield a direct interpretation of the coefficients? Is it true that the coefficient equals 1 if and only if the path is a bounce path? | |
| Apr 25, 2022 at 8:04 | comment | added | Martin Rubey | This is now findstat.org/StatisticsDatabase/St001786 | |
| Apr 24, 2022 at 5:43 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Apr 23, 2022 at 16:47 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Apr 23, 2022 at 15:12 | history | answered | Fedor Petrov | CC BY-SA 4.0 |