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We say that a family ${\cal S}\subseteq{\cal P}(\mathbb{N})$ is bijection-dodging if there is a bijection $\varphi:\mathbb{N}\to\mathbb{N}$ with $\varphi(T)\notin {\cal S}$ for all $T\in{\cal S}$. ...

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\...

For any set $X$, let $\newcommand{\N}{\mathbb{N}}[X]^2 = \big\{\{x,y\}:x\neq y \in X\big\}$ and set $[n]^2 = [\{0,\dotsc,n-1\}]^2$ for any positive integer $n$. For $A\subseteq [\N]^2$ we set $$\...

Let $\mathbb{N}$ denote the set of non-negative integers. We say $A\subseteq \mathbb{N}$ is a finite arithmetic progression if there are $a, n, d\in\mathbb{N}$ with $d \geq 1$ and $n \geq 2$ such that ...

Motivation. This weekend I was playing the pair-matching game Memory (also called Concentration in other parts of the world) against my youngest son, and wondered about what constitutes a "good ...