计算有缺陷的停车功能
Counting Defective Parking Functions
摘要：
假设$n$驾驶员各自在具有$m$空间的线性停车场中选择优选的停车空间。每个司机去选择的空间和公园那里，如果它是免费的，否则采取第一个可用的空间与较大的数字（如果有的话）。如果所有驾驶员都成功停车，则选择的序列称为停车功能。一般来说，如果$k$驾驶员未能停车，我们有一个\emph{defective parking function}的\emph{defect} $k$。令$\cp(n,m,k)$为此类函数的数量。本文建立了数$\cp(n,m,k)$的一个递推关系，并将其表示为一个三元母函数的方程。我们使用核方法求解这个方程，并显式地提取系数：结果表明，累积总数是阿贝尔二项式恒等式中的部分和。最后，我们计算了$\cp(n,m,k)$的渐近性。特别地，对于$m=n$的情况，如果随机地独立地做出选择，则缺陷的极限分布（未停车的驾驶员的数量）（由$n$的平方根缩放）是瑞利分布。另一方面，在$m=ω(n)$的情况下，所有空间被占据的概率渐近趋于1。
英文摘要：
Suppose that $n$ drivers each choose a preferred parking space in a linear car park with $m$ spaces. Each driver goes to the chosen space and parks there if it is free, and otherwise takes the first available space with larger number (if any). If all drivers park successfully, the sequence of choices is called a parking function. In general, if $k$ drivers fail to park, we have a \emph{defective parking function} of \emph{defect} $k$. Let $\cp(n,m,k)$ be the number of such functions. In this paper, we establish a recurrence relation for the numbers $\cp(n,m,k)$, and express this as an equation for a three-variable generating function. We solve this equation using the kernel method, and extract the coefficients explicitly: it turns out that the cumulative totals are partial sums in Abel's binomial identity. Finally, we compute the asymptotics of $\cp(n,m,k)$. In particular, for the case $m=n$, if choices are made independently at random, the limiting distribution of the defect (the number of drivers who fail to park), scaled by the square root of $n$, is the Rayleigh distribution. On the other hand, in case $m=ω(n)$, the probability that all spaces are occupied tends asymptotically to one.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
