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arXiv math/0503095math.COmath.NT
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关于$p$ -零和自由序列的结构及其对Erdos-Ginzburg-Ziv定理的一个变体的应用

On the structure of $p$-zero-sum free sequences and its application to a variant of Erdos--Ginzburg--Ziv theorem

W D Gao, A Panigrahi, R Thangadurai

摘要:

设$p$为任意奇素数。令$k$为任何正整数,使得$2\leq k\leq [\frac{p+1}3]+1$。令$S = (a_1,a_2,...,a_{2p-k})$是${\Bbb Z}_p$中的任意序列,使得在$\zp$中不存在$S$的长度为$p$且其和为零的子序列。然后我们证明我们可以安排序列$S$如下:$ S = (\underbrace{a, a, ..., a}_{u {\rm times}}, \underbrace{b, b, >..., b}_{v {\rm times}}, a_1', a_2', >..., a_{2p-k-u-v}') $其中$u\geq v$,$u+v\geq 2p-2k+2$和$a-b$生成$\zp$。这将\cite{gao10}中的结果扩展到满足$(p+1)/4+3\leq k\leq (p+1)/3+1$的所有素数$p$和$k$。此外,我们证明,如果$g$表示不同的剩余类模$p$的数量出现在序列$S$在$\zp$的长度$2p-k$ $(2\leq k\leq [(p+1)/4]+1)$,和$g\geq 2\sqrt{2}\sqrt{k-2}$,那么存在一个子序列的$S$的长度$p$的总和为零的$\zp$。

英文摘要:

Let $p$ be any odd prime number. Let $k$ be any positive integer such that $2\leq k\leq [\frac{p+1}3]+1$. Let $S = (a_1,a_2,...,a_{2p-k})$ be any sequence in ${\Bbb Z}_p$ such that there is no subsequence of length $p$ of $S$ whose sum is zero in $\zp$. Then we prove that we can arrange the sequence $S$ as follows: $ S = (\underbrace{a, a, ..., a}_{u {\rm times}}, \underbrace{b, b, >..., b}_{v {\rm times}}, a_1', a_2', >..., a_{2p-k-u-v}') $ where $u\geq v$, $u+v\geq 2p-2k+2$ and $a-b$ generates $\zp$. This extends a result in \cite{gao10} to all primes $p$ and $k$ satisfying $(p+1)/4+3\leq k\leq (p+1)/3+1$. Also, we prove that if $g$ denotes the number of distinct residue classes modulo $p$ appearing in the sequence $S$ in $\zp$ of length $2p-k$ $(2\leq k\leq [(p+1)/4]+1)$, and $g\geq 2\sqrt{2}\sqrt{k-2}$, then there exists a subsequence of $S$ of length $p$ whose sum is zero in $\zp$.

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