On tiling the integers with 4-sets of the same gap sequence

Partitioning a set into similar, if not, identical, parts is a fundamental research topic in combinatorics. The question of partitioning the integers in various ways has been considered throughout history. Given a set $\{x_1,\dots ,x_n\}$ of integers where $x_1<?<x_n$, let the gap sequence of this set be the unordered multiset $\{d_1,\dots ,d_{n?1}\}=\{x_{i+1}?x_i:i\in \{1,\dots ,n?1\}\}$. This paper addresses the following question, which was explicitly asked by Nakamigawa: can the set of integers be partitioned into sets with the same gap sequence? The question is known to be true for any set where the gap sequence has length at most two. This paper provides evidence that the question is true when the gap sequence has length three. Namely, we prove that given positive integers $p$ and $q$, there is a positive integer $r_0$ such that for all $r\ge r_0$, the set of integers can be partitioned into 4-sets with gap sequence $p,q$, $r$.     

Maximum average degree and relaxed coloring

We say a graph is $(d,d,\dots ,d,0,\dots ,0)$-colorable with $a$ of $d$’s and $b$ of $0$’s if $V\left(G\right)$ may be partitioned into $b$ independent sets $O_1,O_2,\dots ,O_b$ and $a$ sets $D_1,D_2,\dots ,D_a$ whose induced graphs have maximum degree at most $d$. The maximum average degree, $mad\left(G\right)$, of a graph $G$ is the maximum average degree over all subgraphs of $G$. In this note, for nonnegative integers $a,b$, we show that if $mad\left(G\right)<\frac{4}{3}a+b$, then $G$ is $(1_1,1_2,\dots ,1_a,0_1,\dots ,0_b)$-colorable.