论文标题
关于具有不同部分总和的循环基团的序列
On Sequences in Cyclic Groups with Distinct Partial Sums
论文作者
论文摘要
如果有一个元素的$（x_1，\ ldots，x_k）的订购$（y_0，y_0，y_1，y_1，\ ldots，y_k）$，则是{\ em序列可}的子集，是{\ em序列}，如果有$（x_1，\ ldots，x_k）$ \ leq k $是不同的，除了我们可能具有$ y_k = y_0 = 0 $的例外。我们演示了$ \ mathbb {z} _n \ setMinus \ {0 \} $的$ k $ of $ k $的子集的测序性，当时$ n = mt $在许多情况下，包括$ m $是prime或所有主要因素时，包括$ n的所有素数大于$ k！ /2 $ for $ k \ leq 11 $和$ t \ leq 5 $以及$ k = 12 $和$ t \ t \ leq 4 $。我们获得类似但部分的结果，售价为$ 13 \ leq k \ leq 15 $。这代表了有关阿贝尔群体子集的测序性的文献中的各种问题和猜想的进展，我们将其结合并总结为猜想，即如果阿贝尔基团的子集不包含0，则可以进行测序。
A subset of an abelian group is {\em sequenceable} if there is an ordering $(x_1, \ldots, x_k)$ of its elements such that the partial sums $(y_0, y_1, \ldots, y_k)$, given by $y_0 = 0$ and $y_i = \sum_{j=1}^i x_i$ for $1 \leq i \leq k$, are distinct, with the possible exception that we may have $y_k = y_0 = 0$. We demonstrate the sequenceability of subsets of size $k$ of $\mathbb{Z}_n \setminus \{ 0 \}$ when $n = mt$ in many cases, including when $m$ is either prime or has all prime factors larger than $k! /2$ for $k \leq 11$ and $t \leq 5$ and for $k=12$ and $t \leq 4$. We obtain similar, but partial, results for $13 \leq k \leq 15$. This represents progress on a variety of questions and conjectures in the literature concerning the sequenceability of subsets of abelian groups, which we combine and summarize into the conjecture that if a subset of an abelian group does not contain 0 then it is sequenceable.
