帕斯卡三角形行中最小公倍数的极值问题
Extremal Least Common Multiples in Rows of Pascal's Triangle
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中文总结 AI 辅助
研究帕斯卡三角形行中选取n个项的最小公倍数的极值，给出渐近公式与结构描述，并验证了n=15和n=41处的有限结构转变。
中文摘要 AI 辅助
对于 $r \geq 0$，令 $\mathcal P_r=\{\binom{r}{0},\binom{r}{1},\ldots,\binom{r}{\lfloor r/2\rfloor}\}$ 为帕斯卡三角形第 $r$ 行中不同项的集合。我们研究从某一行中选取 $n$ 个项的最小可能的最小公倍数，其中行本身也是自由的：\\[ a(n)=\min_{\substack{r\geq 0,\\ S\subseteq \mathcal P_r\\\\ |S|=n}}\operatorname{lcm}(S). \\] 我们首先将固定行问题精确地转化为一个加权素数次幂排斥问题。这一结构描述解释了为什么最优支撑集可能出现空洞，并为有限情形提供了一种精确的验证方法。我们的主要渐近结果是 \\[ \log a(n)=2n+O\\!\left(n\exp\\!\left(-c\frac{(\log n)^{3/5}}{(\log\log n)^{1/5}}\right)\right) \\] 对于某个绝对常数 $c>0$，因此 $a(n)^{1/n}\to e^2$。一个双带细化进一步表明，每个最优行满足 \\[ r_n=2n+O\\!\left(n\exp\\!\left(-c\frac{(\log n)^{3/5}}{(\log\log n)^{1/5}}\right)\right)，\\] 且最优支撑集的最小前缀缺陷为 $o(n)$。最后，两个独立的精确实现验证了最初的有限结构转变：$n=15$ 是第一个非前缀最优情形，而 $n=41$ 是第一个最小前缀缺陷大于一的情形。
英文摘要
For $r \ge 0$, let $\mathcal P_r=\{\binom{r}{0},\binom{r}{1},\ldots,\binom{r}{\lfloor r/2\rfloor}\}$ be the set of distinct entries in row $r$ of Pascal's triangle. We study the least possible least common multiple of $n$ entries chosen from one row, with the row itself also free: $a(n)=\min_{r\ge0,\ S\subseteq\mathcal P_r,\ |S|=n}\operatorname{lcm}(S)$. Whereas the least common multiple of an entire row is given by a classical identity of Farhi, allowing both the row and the selected coefficients to vary creates a different optimization problem. We first recast the fixed-row problem exactly as a weighted prime-power exclusion problem, which explains how omitting a few coefficients can remove expensive prime-power contributions. Our main asymptotic result is $\log a(n)=2n+O\!\left(n\exp\!\left(-c(\log n)^{3/5}(\log\log n)^{-1/5}\right)\right)$ for some absolute $c>0$, so $a(n)^{1/n}\to e^2$. A two-band refinement further shows that every optimal row satisfies $r_n=2n+O\!\left(n\exp\!\left(-c(\log n)^{3/5}(\log\log n)^{-1/5}\right)\right)$, and that the minimum prefix defect of an optimal support is $o(n)$.
