当前位置: X-MOL 学术J. Algebraic Comb. › 论文详情
Our official English website, www.x-mol.net, welcomes your feedback! (Note: you will need to create a separate account there.)
On permutations of $$\{1,\ldots ,n\}$$ { 1 , … , n } and related topics
Journal of Algebraic Combinatorics ( IF 0.8 ) Pub Date : 2021-03-25 , DOI: 10.1007/s10801-021-01028-8
Zhi-Wei Sun

In this paper, we study combinatorial aspects of permutations of \(\{1,\ldots ,n\}\) and related topics. In particular, we prove that there is a unique permutation \(\pi \) of \(\{1,\ldots ,n\}\) such that all the numbers \(k+\pi (k)\) (\(k=1,\ldots ,n\)) are powers of two. We also show that \(n\mid {\mathrm{per}}[i^{j-1}]_{1\leqslant i,j\leqslant n}\) for any integer \(n>2\). We conjecture that if a group G contains no element of order among \(2,\ldots ,n+1\) then any \(A\subseteq G\) with \(|A|=n\) can be written as \(\{a_1,\ldots ,a_n\}\) with \(a_1,a_2^2,\ldots ,a_n^n\) pairwise distinct. This conjecture is confirmed when G is a torsion-free abelian group. We also prove that for any finite subset A of a torsion-free abelian group G with \(|A|=n>3\), there is a numbering \(a_1,\ldots ,a_n\) of all the elements of A such that all the n sums

$$\begin{aligned}&a_1+a_2+a_3,\quad a_2+a_3+a_4,\ldots , a_{n-2}+a_{n-1}+a_n,\\&\quad a_{n-1}+a_n+a_1,\quad a_n+a_1+a_2 \end{aligned}$$

are pairwise distinct, and conjecture that this remains valid if G is cyclic.



中文翻译:

关于$$ \ {1,\ ldots,n \} $$ {1,…,n}的排列和相关主题

在本文中,我们研究\(\ {1,\ ldots,n \} \)和相关主题的组合的组合方面。特别是,我们证明存在\(\ {1,\ ldots,n \} \)的唯一置换\(\ pi \),使得所有数字\(k + \ pi(k)\)\( k = 1,\ ldots,n \))是2的幂。我们还显示\(n \ mid {\ mathrm {per}} [i ^ {j-1}] _ {1 \ leqslant i,j \ leqslant n} \)对于任何整数\(n> 2 \)。我们推测,如果组G\(2,\ ldots,n + 1 \)中不包含顺序元素,则任何具有\(| A | = n \)的\(A \ subseteq G \)都可以写为\ (\ {a_1,\ ldots,a_n \} \)\(a_1,a_2 ^ 2,\ ldots,a_n ^ n \)成对区分。当G是无扭转的阿贝尔群时,这一猜想得到了证实。我们还证明,对于任何有限子集自由扭转的阿贝尔群的ģ\(| A | = N> 3 \),有一个编号\(A_1,\ ldots,A_N \)所有的元件的这样所有的n个

$$ \ begin {aligned}&a_1 + a_2 + a_3,\ quad a_2 + a_3 + a_4,\ ldots,a_ {n-2} + a_ {n-1} + a_n,\\&\ quad a_ {n-1 } + a_n + a_1,\ quad a_n + a_1 + a_2 \ end {aligned} $$

是成对的,并且猜想如果G是循环的,那么这仍然有效。

更新日期:2021-03-26
down
wechat
bug