科研速览 · Science Skim继续刷下去 · Keep skimming →
◇ arXiv2026-09-19· math.GM

A Combinatorial Problem in Cinema Seating

Madjid Mirzavaziri, Daniel Yaqubi

原始摘要(英文原文)· Original abstract
We address a problem concerning cinema audiences: ``A cinema has $n$ seats numbered from $1$ to $n$, and there are $n$ people with tickets numbered from $1$ to $n$. People enter the cinema in order. If someone has the ticket number $i$, they can choose seats whose numbers are multiples of $i$. They should exit the cinema if the permitted seats are occupied by previous audience members. In how many ways can they be seated under these conditions?" We give an algorithm to create the list of situations that meet these conditions. We also focus on finding the number of situations in two special cases: when exactly one seat is unoccupied whose total number is denoted by $ω(n)$, and when all audiences $1, \ldots, n-1$ are seated, whose total number is denoted by $ψ(n)$. Giving the recursive formula $ψ(n)=1+\sum_{d|n, d\neq n}ψ(d)$ with the initial value $ψ(1)=1$, we provide an explicit formula for $ψ(p^αq^β)$, where $p$ and $q$ are distinct prime numbers. Furthermore, we show that $ω(n)=-n+\sum_{i=1}^nψ(i)$.
读原文 · Read the paper ↗

AI 追问PRO

登录后使用 AI 追问

讨论区

登录后参与讨论

相关论文 · Related

A Combinatorial Problem in Cinema Seating — 科研速览 Science Skim