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arxiv: 0809.3549 · v2 · submitted 2008-09-21 · 🧮 math.CO

An inequality for Kruskal-Macaulay functions

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keywords binompartialgeq0kruskal-macaulayconsequencescorollarydefineddirect
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Given integers $k\geq1$ and $n\geq0$, there is a unique way of writing $n$ as $n=\binom{n_{k}}{k}+\binom{n_{k-1}}{k-1}+...+\binom{n_{1}}{1}$ so that $0\leq n_{1}<...<n_{k-1}<n_{k}$. Using this representation, the \emph{Kruskal-Macaulay function of}$n$ 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 $a\geq0$ and $a<\partial^{k+1}(n) $, then $\partial^{k}(a) +\partial^{k+1}(n-a) \geq \partial^{k+1}(n) .$ As a corollary, we obtain a short proof of Macaulay's Theorem. Other previously known results are obtained as direct consequences.

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