AI中文摘要
对于$0\leq k\leq n$,设$u(n,k)$是$\binom nk$的最大因子,其素因子不超过$k$。Erdős问题#684涉及特殊阈值$u(n,k)>n^2$,并询问这个小素因子部分能多早被迫变大。我们证明了每个固定幂阈值的密度一模拟。若$f_c(n)$是使得$u(n,k)>n^c$的最小$k$,则对每个固定的$c>0$,\[ f_c(n)=\left(\frac{c}{1-γ}+o(1)\right)\log n \] 对几乎所有正整数$n$成立。特别地,对于Erdős #684阈值,\[ f_2(n)=\left(\frac{2}{1-γ}+o(1)\right)\log n =(4.730544237\ldots+o(1))\log n. \] 这是一个正态阶定理,而非相应最坏情况问题的逐点解决。常数$1-γ$是算术的。Kummer定理将$\log u(n,k)$重写为进位指示器之和,完全剩余系平均给出\[ m(k)=k\sum_{p\leq k}\frac{\log p}{p-1}-\log k!=(1-γ)k+o(k). \] 该公式中的消去将典型交叉从朴素尺度$c\log n$移至$c(1-γ)^{-1}\log n$。我们在一个二进区间上对每个$k\leq A\log X$一致地证明了所需的集中性,在丢弃由整除相邻整数$n,n-1,\ldots$之一的小素数的高次幂导致的零密度例外集之后。我们还证明了对数范围中的高斯波动。若$k=k(X)\to\infty$,$k\leq A\log X$,且$n$在$[X,2X)\cap\mathbb Z$上均匀分布,则\[ \frac{\log u(n,k)-m(k)}{\sqrt{V(k)}}\Rightarrow \mathcal N(0,1), \qquad V(k)\sim (2-\log(2π))k\log k. \] 均值需要更高的素数幂,但在中心化后,它们的总和在高斯尺度上是$L^2$可忽略的;方差仅来自素数水平。
英文摘要
For $0\leq k\leq n$, let $u(n,k)$ be the largest divisor of $\binom nk$ whose prime factors are at most $k$. Erdős Problem #684 concerns the special threshold $u(n,k)>n^2$ and asks how early this small-prime part can be forced to become large. We prove the density-one analogue for every fixed power threshold. If $f_c(n)$ is the least $k$ for which $u(n,k)>n^c$, then, for each fixed $c>0$, \[ f_c(n)=\left(\frac{c}{1-γ}+o(1)\right)\log n \] for almost all positive integers $n$. In particular, \[ f_2(n)=\left(\frac{2}{1-γ}+o(1)\right)\log n =(4.730544237\ldots+o(1))\log n \] for the Erdős #684 threshold. This is a normal-order theorem, not a pointwise resolution of the corresponding worst-case problem. The constant $1-γ$ is arithmetic. Kummer's theorem rewrites $\log u(n,k)$ as a sum of carry indicators, and complete-residue averaging gives \[ m(k)=k\sum_{p\leq k}\frac{\log p}{p-1}-\log k!=(1-γ)k+o(k). \] The cancellation in this formula moves the typical crossing from the naive scale $c\log n$ to $c(1-γ)^{-1}\log n$. We prove the required concentration uniformly for every $k\leq A\log X$ on one dyadic interval, after discarding a zero-density exceptional set caused by large powers of small primes dividing one of the nearby integers $n,n-1,\ldots$. We also prove Gaussian fluctuations in the logarithmic range. If $k=k(X)\to\infty$, $k\leq A\log X$, and $n$ is uniform in $[X,2X)\cap\mathbb Z$, then \[ \frac{\log u(n,k)-m(k)}{\sqrt{V(k)}}\Rightarrow \mathcal N(0,1), \qquad V(k)\sim (2-\log(2π))k\log k. \] Higher prime powers are needed for the mean, but after centering their aggregate is $L^2$-negligible on the Gaussian scale; the variance comes only from the prime levels.