Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with permutations
Search options not deleted
user 172802
Questions related to permutations, bijections from a finite (or sometimes infinite) set to itself.
11
votes
Accepted
Min–max reversing bijections $f:\mathbb{N}\to\mathbb{N}$
$\newcommand\N{\mathbb N}$No, it is not possible to have $\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$.
Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more …
- 10.6k
42
votes
2
answers
2k
views
How decreasing can a bijection $f:\mathbb{N}\to\mathbb{N}$ be?
This is a follow-up to this question by
Dominic van der Zypen. For each bijection $f:\mathbb{N}\to\mathbb{N}$, let
$$\operatorname{rc}(f) := \liminf_{N\to\infty} \frac{\left|\left\{(m,n)\in\{1,\dots,N …
- 10.6k