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Periods in missing lengths of rainbow cycles

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 $G$, define $\mathfrak{S}(G)=\{n\ge 2\;|\;\text{no $n$-cycle of $G$ is rainbow}\}$. Then $\mathfrak{S}(G)$ is a monoid with respect to the operation $n\circ m = n+m-2$, and thus there is a least positive integer $π(G)$, the period of $\mathfrak{S}(G)$, such that $\mathfrak{S}(G)$ contains the arithmetic progression $\{N+kπ(G)\;|\;k\ge 0\}$ for some sufficiently large $N$. Given that $n\in\mathfrak{S}(G)$, what can be said about $π(G)$? Alexeev showed that $π(G)=1$ when $n\ge 3$ is odd, and conjectured that $π(G)$ always divides $4$. We prove Alexeev's conjecture: Let $p(n)=1$ when $n$ is odd, $p(n)=2$ when $n$ is divisible by four, and $p(n)=4$ otherwise. If $2<n\in\mathfrak{S}(G)$ then $π(G)$ is a divisor of $p(n)$. Moreover, $\mathfrak{S}(G)$ contains the arithmetic progression $\{N+kp(n)\;|\;k\ge 0\}$ for some $N=O(n^2)$. The key observations are: If $2<n=2k\in\mathfrak{S}(G)$ then $3n-8\in\mathfrak{S}(G)$. If $16\ne n=4k\in\mathfrak{S}(G)$ then $3n-10\in\mathfrak{S}(G)$. The main result cannot be improved since for every $k>0$ there are $G$, $H$ such that $4k\in\mathfrak{S}(G)$, $π(G)=2$, and $4k+2\in\mathfrak{S}(H)$, $π(H)=4$.

preprint2015arXivOpen access

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