对数凸性与超配分函数

摘要：

令$\overline{p}(n)$表示过配分函数。本文得到了序列$Δ^{2}\log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α}$的一个不等式，其中\begin{equation*} \log \biggl(1+\frac{3π}{4n^{5/2}}-\frac{11+5α}{n^{11/4}}\biggr) < Δ^{2} \log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α} < \log \biggl(1+\frac{3π}{4n^{5/2}}\biggr) \ \ \text{for}\ n \geq N(α), \end{equation*}是$α$的非负实数，$N(α)$是依赖于$α$的正整数，$Δ$是关于$n$的差分算子.这个不等式因此暗示了$\bigl\{\sqrt[n]{\overline{p}(n)/n}\bigr\}_{n \geq 19}$和$\bigl\{\sqrt[n]{\overline{p}(n)}\bigr\}_{n \geq 4}$的$\log$ -凸性。此外，通过证明$\underset{n \rightarrow \infty}{\lim} Δ^{2} \log \ \sqrt[n]{\overline{p}(n)/n^α} = \dfrac{3 π}{4 n^{5/2}}.$，还建立了$Δ^{2} \log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α}$的渐近增长性

英文摘要：

Let $\overline{p}(n)$ denote the overpartition function. In this paper, we obtain an inequality for the sequence $Δ^{2}\log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α}$ which states that \begin{equation*} \log \biggl(1+\frac{3π}{4n^{5/2}}-\frac{11+5α}{n^{11/4}}\biggr) < Δ^{2} \log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α} < \log \biggl(1+\frac{3π}{4n^{5/2}}\biggr) \ \ \text{for}\ n \geq N(α), \end{equation*} where $α$ is a non-negative real number, $N(α)$ is a positive integer depending on $α$ and $Δ$ is the difference operator with respect to $n$. This inequality consequently implies $\log$-convexity of $\bigl\{\sqrt[n]{\overline{p}(n)/n}\bigr\}_{n \geq 19}$ and $\bigl\{\sqrt[n]{\overline{p}(n)}\bigr\}_{n \geq 4}$. Moreover, it also establishes the asymptotic growth of $Δ^{2} \log \ \sqrt[n-1]{\overline{p}(n-1)/(n-1)^α}$ by showing $\underset{n \rightarrow \infty}{\lim} Δ^{2} \log \ \sqrt[n]{\overline{p}(n)/n^α} = \dfrac{3 π}{4 n^{5/2}}.$

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