English

Periods in missing lengths of rainbow cycles

Combinatorics 2015-09-21 v1

Abstract

A cycle in an edge-colored graph is said to be rainbow if no two of its edges have the same color. For a complete, infinite, edge-colored graph GG, define \mathfrak{S}(G)=\{n\ge 2\;|\;\text{no ncycleof-cycle of G is rainbow}\}. Then S(G)\mathfrak{S}(G) is a monoid with respect to the operation nm=n+m2n\circ m = n+m-2, and thus there is a least positive integer π(G)\pi(G), the period of S(G)\mathfrak{S}(G), such that S(G)\mathfrak{S}(G) contains the arithmetic progression {N+kπ(G)    k0}\{N+k\pi(G)\;|\;k\ge 0\} for some sufficiently large NN. Given that nS(G)n\in\mathfrak{S}(G), what can be said about π(G)\pi(G)? Alexeev showed that π(G)=1\pi(G)=1 when n3n\ge 3 is odd, and conjectured that π(G)\pi(G) always divides 44. We prove Alexeev's conjecture: Let p(n)=1p(n)=1 when nn is odd, p(n)=2p(n)=2 when nn is divisible by four, and p(n)=4p(n)=4 otherwise. If 2<nS(G)2<n\in\mathfrak{S}(G) then π(G)\pi(G) is a divisor of p(n)p(n). Moreover, S(G)\mathfrak{S}(G) contains the arithmetic progression {N+kp(n)    k0}\{N+kp(n)\;|\;k\ge 0\} for some N=O(n2)N=O(n^2). The key observations are: If 2<n=2kS(G)2<n=2k\in\mathfrak{S}(G) then 3n8S(G)3n-8\in\mathfrak{S}(G). If 16n=4kS(G)16\ne n=4k\in\mathfrak{S}(G) then 3n10S(G)3n-10\in\mathfrak{S}(G). The main result cannot be improved since for every k>0k>0 there are GG, HH such that 4kS(G)4k\in\mathfrak{S}(G), π(G)=2\pi(G)=2, and 4k+2S(H)4k+2\in\mathfrak{S}(H), π(H)=4\pi(H)=4.

Keywords

Cite

@article{arxiv.1509.05632,
  title  = {Periods in missing lengths of rainbow cycles},
  author = {Petr Vojtěchovský},
  journal= {arXiv preprint arXiv:1509.05632},
  year   = {2015}
}