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Inductive Methods and Zero-Sum Free Sequences [PDF]
AbstractA fairly long-standing conjecture is that the Davenport constant of a groupWe also prove the conjecture for groups of the form ...
Immanuel Halupczok +1 more
exaly +6 more sources
On the index of minimal zero-sum sequences over finite cyclic groups
Let \(G\) be a cyclic group of order \(n\geq 2\). A finite sequence \(S\) of not necessarily distinct elements from \(G\) with \(|S|= k\) the number of elements in \(S\) (\(k\) the length of \(S\)) will be written in the form \[ S= g_1\cdot\dots\cdot g_k= \prod^k_{i=1} g_i= \prod_{g\in G} g^{v_g(S)}, \] where \(v_g(S)\geq 0\) is called the multiplicity
Pingzhi Yuan
exaly +4 more sources
Zero-sum subsequences in bounded-sum {−r,s}-sequences [PDF]
24 pages, 1 ...
Alec Sun
exaly +4 more sources
Sequences not containing long zero-sum subsequences
For a finite abelian group \(G\), denote by \(D(G)\) (Davenport's constant) the smallest integer \(d\) such that every sequence of \(d\) elements of \(G\) contains a subsequence with sum zero. A sequence \(S\) with \(| S| \geq D(G)\) is called normal if it contains no zero-sum subsequence longer than \(| S| -D(G)+1\). The authors describe the structure
Weidong Gao 0001, Jujuan Zhuang
openaire +2 more sources
A Reciprocity on Finite Abelian Groups Involving Zero-Sum Sequences [PDF]
19 pages, accepted by SIAM Journal on Discrete ...
Hanbin Zhang
exaly +4 more sources
Extremal sequences for a weighted zero-sum constant
18 pages.
Mondal, Santanu +2 more
openaire +4 more sources
Subsequence Sums of Zero-sum-free Sequences [PDF]
Let $G$ be a finite abelian group, and let $S$ be a sequence of elements in $G$. Let $f(S)$ denote the number of elements in $G$ which can be expressed as the sum over a nonempty subsequence of $S$. In this paper, we slightly improve some results of Pixton on $f(S)$ and we show that for every zero-sum-free sequences $S$ over $G$ of length $|S|=\exp(G ...
Lev, Vsevolod F.
core +5 more sources
Zero-sum subsequences in bounded-sum {−1,1}-sequences [PDF]
The following result gives the flavor of this paper: Let $t$, $k$ and $q$ be integers such that $q\geq 0$, $0\leq t < k$ and $t \equiv k \,({\rm mod}\, 2)$, and let $s\in [0,t+1]$ be the unique integer satisfying $s \equiv q + \frac{k-t-2}{2} \,({\rm mod} \, (t+2))$. Then for any integer $n$ such that \[n \ge \max\left\{k,\frac{1}{2(t+2)}k^2 + \frac{
Yair Caro +2 more
exaly +3 more sources
Indexes of long zero-sum sequences over cyclic groups
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Xiangneng Zeng, Pingzhi Yuan
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Indexes of unsplittable minimal zero-sum sequences of length I(Cn)−1
Let G be a finite abelian group and S=g1⋯gl a minimal zero-sum sequence of elements in G. We say that S is unsplittable if there do not exist an element gi∈supp(S) and two elements x,y∈G such that x+y=gi and Sa−1xy is a minimal zero-sum sequence as well.
Pingzhi Yuan
exaly +2 more sources

