赖纳和斯坦顿1998年单模猜想的推广

IF 0.4 Q4 MATHEMATICS, APPLIED Journal of Combinatorics Pub Date : 2017-11-27 DOI:10.4310/joc.2020.v11.n1.a5
R. Stanley, Fabrizio Zanello
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引用次数: 3

摘要

赖纳和斯坦顿的一个有趣的、仍然广泛开放的猜想预言$q$ -二项式系数的某些“奇怪的”对称差总是非负的和单峰的。我们将他们的猜想扩展到一个更广泛的,也许更自然的框架,通过推测,对于每个$k\ge 5$,多项式$$f(k,m,b)(q)=\binom{m}{k}_q-q^{\frac{k(m-b)}{2}+b-2k+2}\cdot\binom{b}{k-2}_q$$是非负的,并且对于所有$m\gg_k 0$和$b\le \frac{km-4k+4}{k-2}$都是单峰的,例如$kb\equiv km$ (mod 2),除了$b=\frac{km-4k+2}{k-2}$是整数时的唯一例外。使用KOH定理,我们组合地展示了$k=5$这种情况。事实上,对于$k\le 5$,我们完全刻画了$f(k,m,b)$的非负性和单模性。(这也为Reiner-Stanton的猜想提供了一个孤立的反例,当$k=3$。)进一步,我们证明,对于每个$k$和$m$,它足以证明我们对$b$的最大$2k-6$值的猜想。
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A generalization of a 1998 unimodality conjecture of Reiner and Stanton
An interesting, and still wide open, conjecture of Reiner and Stanton predicts that certain "strange" symmetric differences of $q$-binomial coefficients are always nonnegative and unimodal. We extend their conjecture to a broader, and perhaps more natural, framework, by conjecturing that, for each $k\ge 5$, the polynomials $$f(k,m,b)(q)=\binom{m}{k}_q-q^{\frac{k(m-b)}{2}+b-2k+2}\cdot\binom{b}{k-2}_q$$ are nonnegative and unimodal for all $m\gg_k 0$ and $b\le \frac{km-4k+4}{k-2}$ such that $kb\equiv km$ (mod 2), with the only exception of $b=\frac{km-4k+2}{k-2}$ when this is an integer. Using the KOH theorem, we combinatorially show the case $k=5$. In fact, we completely characterize the nonnegativity and unimodality of $f(k,m,b)$ for $k\le 5$. (This also provides an isolated counterexample to Reiner-Stanton's conjecture when $k=3$.) Further, we prove that, for each $k$ and $m$, it suffices to show our conjecture for the largest $2k-6$ values of $b$.
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来源期刊
Journal of Combinatorics
Journal of Combinatorics MATHEMATICS, APPLIED-
自引率
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发文量
21
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