Timeline for Is there a lop-sided permutation $\pi:\mathbb{N}\to\mathbb{N}$? [closed]
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 27, 2023 at 7:15 | vote | accept | Dominic van der Zypen | ||
| Aug 27, 2023 at 1:53 | comment | added | bof | If $\{A,B\}$ is a partition of $\mathbb N=\{1,2,3,\dots\}$ into two infinite sets with $1\in A$, then there is a permutation $\pi:\mathbb N\to\mathbb N$ such that $A=\{n:\pi(n)\gt n\}$ and $B=\{n:\pi(n)\lt n\}$. Namely, write $A=\{a_1\lt a_2\lt a_2\lt\cdots\}$ and $B=\{b_1\lt b_2\lt b_3\lt\cdots\}$ and define $\pi(a_n)=a_{n+1}$ and $\pi(b_{n+1}=b_n$ for all $n\ge1$, and $\pi(b_1)=1=a_1$. | |
| Aug 26, 2023 at 19:21 | history | closed |
YCor Denis T Andreas Blass Ville Salo Max Horn |
Not suitable for this site | |
| Aug 26, 2023 at 17:31 | history | edited | Daniel Asimov | CC BY-SA 4.0 |
excedances —> exceedances
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| S Aug 26, 2023 at 17:21 | history | suggested | CommunityBot | CC BY-SA 4.0 |
\liminf is a valid control sequence and this will have the effect of putting the subscript in the right place.
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| Aug 26, 2023 at 16:42 | review | Suggested edits | |||
| S Aug 26, 2023 at 17:21 | |||||
| Aug 26, 2023 at 15:55 | review | Close votes | |||
| Aug 26, 2023 at 19:21 | |||||
| Aug 26, 2023 at 15:36 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added 147 characters in body
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| Aug 26, 2023 at 15:33 | answer | added | Gro-Tsen | timeline score: 8 | |
| Aug 26, 2023 at 15:33 | comment | added | Dominic van der Zypen | Apologies for missing the $\neq$ sign and putting the $=$ sign instead in the original version. Of course I mean a permutation to be lop-sided if there are more excedences than... "negative excedences", or the other way round. | |
| Aug 26, 2023 at 15:31 | comment | added | Daniel Weber | Wouldn't $(1)(23)(4567)(8, 9, 10 ,11 ,12, 13, 14, 15)...$ be such a permutation? (Add 1 if it isn't one less than a power of two, add 1 and halve if it is). It has $d(pos(\pi)) = 1$ and $d(neg(\pi))=0$ | |
| Aug 26, 2023 at 15:30 | comment | added | Aleksei Kulikov | Still, slight modification of the (already deleted) comments gives us this: if on each group of 4 consecutive elements we have rotation by 1 position. Then one limit is $\frac{3}{4}$ and the other one is $\frac{1}{4}$. | |
| Aug 26, 2023 at 15:29 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added 4 characters in body
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| Aug 26, 2023 at 15:29 | comment | added | Dominic van der Zypen | Of course @gro-tsen --> will correct! | |
| S Aug 26, 2023 at 15:28 | comment | added | Dominic van der Zypen | Thanks Sam, I'll correct it in the next couple minutes. | |
| S Aug 26, 2023 at 15:28 | comment | added | Gro-Tsen | Do you perhaps mean $d\big(\text{pos}(\pi)\big) \neq d\big(\text{neg}(\pi)\big)$ is inconceivable? | |
| Aug 26, 2023 at 15:28 | comment | added | Aleksei Kulikov | I mean, both of them definitely can be zero. Is this good enough for you? | |
| Aug 26, 2023 at 15:28 | comment | added | Sam Hopkins | Terminological point: what you call $\mathrm{pos}(\pi)$ is usually called the set of excedances of the permutation $\pi$ (see also my question mathoverflow.net/questions/359684 ) | |
| Aug 26, 2023 at 15:23 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |