English

An inequality for Kruskal-Macaulay functions

Combinatorics 2009-04-27 v2

Abstract

Given integers k1k\geq1 and n0n\geq0, there is a unique way of writing nn as n=(nkk)+(nk1k1)+...+(n11)n=\binom{n_{k}}{k}+\binom{n_{k-1}}{k-1}+...+\binom{n_{1}}{1} so that 0n1<...<nk1<nk0\leq n_{1}<...<n_{k-1}<n_{k}. Using this representation, the \emph{Kruskal-Macaulay function of}nn is defined as \partial^{k}(n) =\binom{n_{k}-1}{k-1}+\binom{n_{k-1}-1}{k-2}+...+\binom{n_{1}-1}% {0}. We show that if a0a\geq0 and a<k+1(n)a<\partial^{k+1}(n) , then k(a)+k+1(na)k+1(n).\partial^{k}(a) +\partial^{k+1}(n-a) \geq \partial^{k+1}(n) . As a corollary, we obtain a short proof of Macaulay's Theorem. Other previously known results are obtained as direct consequences.

Keywords

Cite

@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}
}

Comments

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

R2 v1 2026-06-21T11:22:30.424Z