Timeline for Coincidences between average Catalan tableaux
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 27, 2020 at 12:26 | history | edited | Igor Pak | CC BY-SA 4.0 |
sorting probability of Catalan posets paper is done
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| May 19, 2019 at 6:44 | comment | added | Christian Stump | @IgorPak: I believe the above formula for $p_{21}-p_{12}$ is correct, well, after dividing by $C_n$. Indeed, $C_n \cdot p_{12} = 2C_{n} + C_{n-1}$ and $C_n \cdot p_{21} = C_{n+1}$. This also immediately give $p_{21}-p_{12} \rightarrow 4 - 2 - 1/4 = 7/4$. | |
| May 18, 2019 at 21:32 | history | edited | Igor Pak | CC BY-SA 4.0 |
minor edit
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| May 18, 2019 at 21:25 | history | edited | Igor Pak | CC BY-SA 4.0 |
update
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| May 18, 2019 at 21:18 | history | edited | Igor Pak | CC BY-SA 4.0 |
update
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| May 18, 2019 at 20:09 | comment | added | Igor Pak | @ChristianStump: A side note - it follows from Richard's formulas below that $p_{21}-p_{12}\to 7/4$ as you originally observed. This doesn't answer my question, but a good thing to know. | |
| May 15, 2019 at 23:15 | answer | added | Richard Stanley | timeline score: 7 | |
| May 15, 2019 at 17:09 | comment | added | Christian Stump | Sorry, I was reading big O all the time... Please ignore my last two comments, neither of the two pairs are witnesses for small $o(1)$. | |
| May 15, 2019 at 16:39 | comment | added | Sam Hopkins | @ChristianStump: $a_{12}$ will be greater than $a_{21}$ (by only 1) about $1/4$ of the time, because there are $C_{n-1}$ tableaux where the first column is $1,2$. But $a_{21}$ will be greater than $a_{12}$ by at least 1, and in fact sometimes much more, about $3/4$ of the time. So $p_{21}-p_{12}$ is gonna be at least $1/2$, not $o(1)$. | |
| May 15, 2019 at 16:36 | comment | added | Christian Stump | If I am not mistaken, $p_{21}-p_{12} = C_{n+1} - 2C_{n} - C_{n-1}$. Since $C_{n+1} / C_n \rightarrow 4$, $(p_{12},p_{21})$ is a witness. Or am I too rusty computing limits? | |
| May 15, 2019 at 16:20 | comment | added | Sam Hopkins | @ChristianStump: I seriously doubt $(p_{12},p_{21})$ could work. | |
| May 15, 2019 at 16:08 | comment | added | Christian Stump | Observation 2: $|p_{14} - p_{22}| \leq 1$ and $|p_{12}-p_{21}| \leq 2$ for all $4 \leq n \leq 13$, so either of these pairs might be a candidate for a witness for an affirmative answer to your question. | |
| May 15, 2019 at 16:07 | comment | added | Christian Stump | Observation 1: for $n \leq 12$: the minimum is obtained at a unique pair $(i,j) \neq (k,l)$ and its 180°-rotation counterpart. | |
| May 15, 2019 at 15:48 | comment | added | Sam Hopkins | Another trivial observation that might be useful is that $p_{i,j} \geq p_{i-1,j} + 1$ and $p_{i,j} \geq p_{i,j-1} + 1$. | |
| May 15, 2019 at 11:35 | history | asked | Igor Pak | CC BY-SA 4.0 |