Periods in missing lengths of rainbow cycles

preprint2015arXivOpen access

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 $n−cycleof-cycle of G$ is rainbow}\}$. Then S(G)\mathfrak{S}(G) is a monoid with respect to the operation n∘m=n+m−2n\circ m = n+m-2, and thus there is a least positive integer π(G)π(G), the period of S(G)\mathfrak{S}(G), such that S(G)\mathfrak{S}(G) contains the arithmetic progression {N+kπ(G)  ∣  k≥0}\{N+kπ(G)\;|\;k\ge 0\} for some sufficiently large NN. Given that n∈S(G)n\in\mathfrak{S}(G), what can be said about π(G)π(G)? Alexeev showed that π(G)=1π(G)=1 when n≥3n\ge 3 is odd, and conjectured that π(G)π(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<n∈S(G)2<n\in\mathfrak{S}(G) then π(G)π(G) is a divisor of p(n)p(n). Moreover, S(G)\mathfrak{S}(G) contains the arithmetic progression {N+kp(n)  ∣  k≥0}\{N+kp(n)\;|\;k\ge 0\} for some N=O(n2)N=O(n^2). The key observations are: If 2<n=2k∈S(G)2<n=2k\in\mathfrak{S}(G) then 3n−8∈S(G)3n-8\in\mathfrak{S}(G). If 16≠n=4k∈S(G)16\ne n=4k\in\mathfrak{S}(G) then 3n−10∈S(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 4k∈S(G)4k\in\mathfrak{S}(G), π(G)=2π(G)=2, and 4k+2∈S(H)4k+2\in\mathfrak{S}(H), π(H)=4π(H)=4.

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