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

TL;DR

This paper investigates whether the integers can be partitioned into 4-element subsets with identical gap sequences, providing new evidence that such partitions are possible when the gap sequence length is three.

Contribution

It proves that for any two positive integers, sufficiently large integers can be partitioned into 4-sets with a specified gap sequence of length three.

Findings

Partition exists for large enough integers with given gap sequences

Extends known results from gap sequences of length two to length three

Provides constructive proof for the existence of such partitions

Abstract

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 $\{x_1, \ldots, x_n\}$ of integers where $x_1<\cdots<x_n$, let the {\it gap sequence} of this set be the nondecreasing sequence $d_1, \ldots, d_{n-1}$ where $\{d_1, \ldots, d_{n-1}\}$ equals $\{x_{i+1}-x_i:i\in\{1,\ldots, n-1\}\}$ as a multiset. 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 $r\geq r_0$, the set of integers can be partitioned into $4$-sets with gap sequence $p, q$, $r$.