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Related papers: Davenport constant for semigroups II

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Let $\mathcal{S}$ be a finite commutative semigroup. The Davenport constant of $\mathcal{S}$, denoted $D(\mathcal{S})$, is defined to be the least positive integer $d$ such that every sequence $T$ of elements in $\mathcal{S}$ of length at…

Number Theory · Mathematics 2014-09-05 Haoli Wang , Lizhen Zhang , Qinghong Wang , Yongke Qu

Given a finite commutative semigroup $\mathcal{S}$ (written additively), denoted by ${\rm D}(\mathcal{S})$ the Davenport constant of $\mathcal{S}$, namely the least positive integer $\ell$ such that for any $\ell$ elements…

Combinatorics · Mathematics 2016-03-22 Guoqing Wang , Weidong Gao

Let $\mathcal{S}$ be a finite commutative semigroup written additively, and let $\exp(\mathcal{S})$ be its exponent which is defined as the least common multiple of all periods of the elements in $\mathcal{S}$. For every sequence $T$ of…

Combinatorics · Mathematics 2013-10-22 Sukumar Das Adhikari , Weidong Gao , Guoqing Wang

Let $\F_q[x]$ be the ring of polynomials over the finite field $\F_q$, and let $f$ be a polynomial of $\F_q[x]$. Let $R=\frac{\F_q[x]}{(f)}$ be a quotient ring of $\F_q[x]$ with $0\neq R\neq \F_q[x]$. Let $\mathcal{S}_R$ be the…

Combinatorics · Mathematics 2015-07-14 Lizhen Zhang , Haoli Wang , Yongke Qu

For a finite abelian group $G,$ the Davenport Constant, denoted by $D(G)$, is defined to be the least positive integer $k$ such that every sequence of length at least $k$ has a non-trivial zero-sum subsequence. A long-standing conjecture is…

Number Theory · Mathematics 2024-02-16 Anamitro Biswas , Eshita Mazumdar

For a finite group $G,$ $\mathsf{D}(G)$ is defined as the least positive integer $k$ such that for every sequence $S=g_1\bdot g_2\bdot \dotsc \bdot g_k$ of length $k$ over $G$, there exist $1 \le i_1 < i_2 <\cdots < i_m \le k $ such that…

Combinatorics · Mathematics 2025-11-25 Naveen K. Godara , Renu Joshi , Eshita Mazumdar

The Davenport constant is one measure for how "large" a finite abelian group is. In particular, the Davenport constant of an abelian group is the smallest $k$ such that any sequence of length $k$ is reducible. This definition extends…

Group Theory · Mathematics 2016-02-11 Calvin Deng

The Erd\H{o}s-Burgess constant of a semigroup $S$ is the smallest positive integer $k$ such that any sequence over $S$ of length $k$ contains a nonempty subsequence whose elements multiply to an idempotent element of $S$. In the case where…

Combinatorics · Mathematics 2018-08-21 Noah Kravitz , Ashwin Sah

Let $\mathcal{S}$ be a nonempty commutative semigroup written additively. An element $e$ of $\mathcal{S}$ is said to be idempotent if $e+e=e$. The Erd\H{o}s-Burgess constant of the semigroup $\mathcal{S}$ is defined as the smallest positive…

Combinatorics · Mathematics 2020-05-19 Guoqing Wang

Let $G$ be a finite group. By a sequence over $G$, we mean a finite unordered string of terms from $G$ with repetition allowed, and we say that it is a product-one sequence if its terms can be ordered so that their product is the identity…

Group Theory · Mathematics 2025-06-25 Jun Seok Oh

Let $G$ be a finite abelian group of exponent $n$ and let $A$ be a non-empty subset of $[1,n-1]$. The Davenport constant of $G$ with weight $A$, denoted by $D_A(G)$, is defined to be the least positive integer $\ell$ such that any sequence…

Number Theory · Mathematics 2022-02-02 Subha Sarkar

Let $G$ be a group and $A\subseteq [1,\exp(G)-1]$. We define the constant ${\sf C}_A(G),$ which is the least positive integer $\ell$ such that every sequence over $G$ of length at least $\ell$ has an $A$-weighted consecutive product-one…

Number Theory · Mathematics 2024-04-18 A. Lemos , A. O. Moura , S. Ribas , A. T. Silva

$G$ be an additive finite abelian group. The Davenport constant $\mathsf D(G)$ is the smallest integer $t$ such that every sequence (multiset) $S$ over $G$ of length $|S|\ge t$ has a non-empty zero-sum subsequence. Recently, B. Girard…

Combinatorics · Mathematics 2018-03-01 Dongchun Han

For $(G,+)$ a finite abelian group the plus-minus weighted Davenport constant, denoted $\mathsf{D}_{\pm}(G)$, is the smallest $\ell$ such that each sequence $g_1 ... g_{\ell}$ over $G$ has a weighted zero-subsum with weights +1 and -1,…

Number Theory · Mathematics 2013-08-16 Luz Elimar Marchan , Oscar Ordaz , Wolfgang Schmid

Let $G$ be a finite group written multiplicatively. By a sequence over $G$, we mean a finite sequence of terms from $G$ which is unordered, repetition of terms allowed, and we say that it is a product-one sequence if its terms can be…

Number Theory · Mathematics 2012-11-13 A. Geroldinger , D. J. Grynkiewicz

Let $R$ be a finite commutative unitary ring. An idempotent in $R$ is an element $e\in R$ with $e^2=e$. Let $\Psi$ be a subgroup of the group ${\rm Aut}(R)$ of all automorphisms of $R$. The $\Psi-$weighted Erd\H{o}s-Burgess constant ${\rm…

Combinatorics · Mathematics 2022-02-25 Guoqing Wang

Let $G$ be a finite group, written multiplicatively. The Davenport constant of $G$ is the smallest positive integer $D(G)$ such that every sequence of $G$ with $D(G)$ elements has a non-empty subsequence with product $1$. Let $D_{2n}$ be…

Number Theory · Mathematics 2017-02-01 Fabio Enrique Brochero Martínez , Sávio Ribas

Let $G = C_{n_1} \oplus \cdots \oplus C_{n_r}$ with $1 < n_1 | \cdots | n_r$ be a finite abelian group. The Davenport constant $\mathsf D(G)$ is the smallest integer $t$ such that every sequence $S$ over $G$ of length $|S|\ge t$ has a…

Combinatorics · Mathematics 2021-09-24 Chao Liu

Let $G$ be a finite group, written multiplicatively. The Davenport constant of $G$ is the smallest positive integer $d$ such that every sequence of $G$ with $d$ elements has a non-empty subsequence with product $1$. Let $C_n \simeq \mathbb…

Number Theory · Mathematics 2017-02-01 Fabio Enrique Brochero Martínez , Sávio Ribas

A generalization of the Davenport constant is investigated. For a finite abelian group $G$ and a positive integer $k$, let $D_k(G)$ denote the smallest $\ell$ such that each sequence over $G$ of length at least $\ell$ has $k$ disjoint…

Number Theory · Mathematics 2010-08-05 Michael Freeze , Wolfgang A. Schmid
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