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arXiv:2403.19555 [math.CO]AbstractReferencesReviewsResources

On 5-cycles and strong 5-subtournaments in a tournament of odd order n

Sergey Savchenko

Published 2024-03-28Version 1

Let $T$ be a tournament of odd order $n\ge 5,$ $c_{m}(T)$ be the number of its $m$-cycles, and $s_{m}(T)$ be the number of its strongly connected $m$-subtournaments. Due to work of L.W. Beineke and F. Harary, it is well known that $s_{m}(T)\le s_{m}(RLT_{n}),$ where $RLT_{n}$ is the regular locally transitive tournament of order $n.$ For $m=3$ and $m=4,$ $c_{m}(T)$ equals $s_{m}(T),$ but it is not so for $m\ge 5.$ As J.W. Moon pointed out in his note in 1966, the problem of determining the maximum of $c_{m}(T)$ seems very difficult in general (i.e. for $m\ge 5$). In the present paper, based on the Komarov-Mackey formula for $c_{5}(T)$ obtained recently, we prove that $c_{5}(T)\le (n+1)n(n-1)(n-2)(n-3)/160$ with equality holding iff $T$ is doubly regular. A formula for $s_{5}(T)$ is also deduced. With the use of it, we show that $s_{5}(T)\le (n+1)n(n-1)(n-3)(11n-47)/1920$ with equality holding iff $T=RLT_{n}$ or $n=7$ and $T$ is regular or $n=5$ and $T$ is strong. It is also proved that for a regular tournament $T$ of (odd) order $n\ge 9,$ a lower bound $(n+1)n(n-1)(n-3)(17n-59)/3840\le s_{5}(T)$ holds with equality iff $T$ is doubly regular. These results are compared with the ones recently obtained by the author for $c_{5}(T).$

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