论文标题
朝向Heim和Neuhauser在Nekrasov-Okounkov多项式上的单形成猜想
Towards Heim and Neuhauser's Unimodality Conjecture on the Nekrasov-Okounkov polynomials
论文作者
论文摘要
令$ q_n（z）$为与nekrasov -okounkov公式$$ \ sum_ \ sum_ {n \ geq 1} q_n（z）q^n：= \ prod_ {m = 1}^\ infty（1- q^m） $ q_n（z）$是单峰或更强的log-concave，用于所有$ n \ geq 1 $。通过新的递归公式，我们表明，如果$ a_ {n，k} $是$ z^k $的系数，则$ z^k $在$ q_n（z）$中，则$ a_ {n，k} $是$ k $ in $ k $ for $ k $ for $ k \ k \ ll n^{1/6}/\ log n $ and notosiss $ k gg，n $ s n $ s n $ s sq gg g g g g g g g g g g g g g。 n $。我们还提出了一个可能缩小差距的猜想。
Let $Q_n(z)$ be the polynomials associated with the Nekrasov-Okounkov formula $$\sum_{n\geq 1} Q_n(z) q^n := \prod_{m = 1}^\infty (1 - q^m)^{-z - 1}.$$ In this paper we partially answer a conjecture of Heim and Neuhauser, which asks if $Q_n(z)$ is unimodal, or stronger, log-concave for all $n \geq 1$. Through a new recursive formula, we show that if $A_{n,k}$ is the coefficient of $z^k$ in $Q_n(z)$, then $A_{n,k}$ is log-concave in $k$ for $k \ll n^{1/6}/\log n$ and monotonically decreasing for $k \gg \sqrt{n}\log n$. We also propose a conjecture that can potentially close the gap.
