中文

Kruskal-Macaulay函数的一个不等式

组合数学 2009-04-27 v2

摘要

给定整数k1k\geq1n0n\geq0,存在唯一的方式将nn表示为n=(nkk)+(nk1k1)+...+(n11)n=\binom{n_{k}}{k}+\binom{n_{k-1}}{k-1}+...+\binom{n_{1}}{1},使得0n1<...<nk1<nk0\leq n_{1}<...<n_{k-1}<n_{k}。利用此表示,nn的\emph{Kruskal-Macaulay函数}定义为\partial^{k}(n) =\binom{n_{k}-1}{k-1}+\binom{n_{k-1}-1}{k-2}+...+\binom{n_{1}-1}% {0}. 我们证明,若a0a\geq0a<k+1(n)a<\partial^{k+1}(n) ,则k(a)+k+1(na)k+1(n).\partial^{k}(a) +\partial^{k+1}(n-a) \geq \partial^{k+1}(n) . 作为推论,我们得到了Macaulay定理的一个简短证明。其他已知结果也作为直接推论得到。

关键词

引用

@article{arxiv.0809.3549,
  title  = {An inequality for Kruskal-Macaulay functions},
  author = {Bernardo M. Ábrego and Silvia Fernández-Merchant and Bernardo Llano},
  journal= {arXiv preprint arXiv:0809.3549},
  year   = {2009}
}

备注

February 9th, 2009 version. The introduction was improved. Theorem 1 now establishes equality for some $n$. Corollary 2 (Bj\"{o}rner and Vre\'{c}ica Theorem) was added. Acknowledgements were added