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Henry Robert Thackeray

Publications and source records attributed to Henry Robert Thackeray.

5 recordsLinked to original sources

Power residue symbols and the exponential local-global principle

The exponential local-global principle, or Skolem conjecture, says: Suppose that \(b\) is a positive integer, and that the sequence \((u_{n})_{n = -\infty}^{\infty}\) is such that every term is in \(\mathbb{Z}[1/b]\), the linear recurrence \(u_{n + d} = a_{1}u_{n + d - 1} + \cdots + a_{d}u_{n}\) holds for all integers \(n\), and every root of \(x^{d} - a_{1}x^{d - 1} - a_{2}x^{d - 2} - \cdots - a_{d}\) is nonzero and simple; then there is no zero term \(u_{n}\) if and only if, for some integer \(m\) that is larger than \(1\) and relatively prime to \(b\), every term \(u_{n}\) is not in \(m\mathbb{Z}[1/b]\). Particular cases of the conjecture are known, but the general conjecture is open. This paper proves some apparently new quadratic and degenerate cubic cases of the exponential local-global principle via power residue symbols. This work was presented at the Stellenbosch Number Theory Conference 2025 in January 2025 at Stellenbosch University; much of the work was also presented at the 67th Annual Congress of the South African Mathematical Society in December 2024 at the University of Pretoria.

math.NT

Each friend of 10 has at least 10 nonidentical prime factors

For each positive integer n, if the sum of the factors of n is divided by n, then the result is called the abundancy index of n. If the abundancy index of some positive integer m equals the abundancy index of n but m is not equal to n, then m and n are called friends. A positive integer with no friends is called solitary. The smallest positive integer that is not known to have a friend and is not known to be solitary is 10. It is not known if the number 6 has odd friends, that is, if odd perfect numbers exist. In a 2007 article, Nielsen proved that the number of nonidentical prime factors in any odd perfect number is at least 9. A 2015 article by Nielsen, which was more complicated and used a computer program that took months to complete, increased the lower bound from 9 to 10. This work applies methods from Nielsen's 2007 article to show that each friend of 10 has at least 10 nonidentical prime factors. This is a formal write-up of results presented at the Southern Africa Mathematical Sciences Association Conference 2023 at the University of Pretoria.

math.NT

Solution to a BCC 2022 problem

For positive integers $n$ and $k$ such that $k$ is at most $n$, we find an explicit one-to-one correspondence between the following two sets: the set of words consisting of $k$ $R$s, $k$ $U$s, and $n - k$ $D$s, where the first letter of the word is not $D$; and the set of subgraphs $H$ of a cycle of length $2n$ (where that cycle has differently labelled vertices) such that $H$ has $n$ edges and $k$ connected components. This solves a problem of Thomas Selig from the 29th British Combinatorial Conference held at Lancaster University in July 2022.

math.CO

The cap set problem: 41-cap 5-flats

An s-cap n-flat, or an n-dimensional cap of size s, is a pair (S,F) where F is an n-dimensional affine space over Z/3Z and the size-s subset S of F contains no triple of collinear points. The cap set problem in dimension n asks for the largest s for which an s-cap n-flat exists. This series of articles investigates the cap set problem in dimensions up to and including 7. This is the second paper in the series. By applying and adapting methods from the first paper in the series, we systematically classify all 5-dimensional caps of size at least 41.

math.CO

The cap set problem: Up to dimension 7

An s-cap n-flat is given by a set of s points, no three of which are on a common line, in an n-dimensional affine space over the field of three elements. The cap set problem in dimension n is: what is the maximum s such that there is an s-cap n-flat? The first two papers in this series of articles considered the cap set problem in dimensions up to and including 5. In this paper, which is the third in the series, we consider dimensions 6 and 7: we prove that every 110-cap 6-flat is a 112-cap 6-flat minus two cap points, and that there are no 289-cap 7-flats.

math.CO