Maximal sets of mutually orthogonal frequency squares

Abstract

A frequency square is a square matrix in which each row and column is a permutation of the same multiset of symbols. A frequency square is of type $(n;\lambda )$ if it contains $n/\lambda$ symbols, each of which occurs $\lambda$ times per row and $\lambda$ times per column. In the case when $\lambda =n/2$ we refer to the frequency square as binary. A set of k-MOFS$(n;\lambda )$ is a set of k frequency squares of type $(n;\lambda )$ such that when any two of the frequency squares are superimposed, each possible ordered pair occurs equally often. A set of k-maxMOFS$(n;\lambda )$ is a set of k-MOFS$(n;\lambda )$ that is not contained in any set of $(k+1)$-MOFS$(n;\lambda )$. For even n, let $\mu (n)$ be the smallest k such that there exists a set of k-maxMOFS(n; n/2). It was shown in Britz et al. (Electron. J. Combin. 27(3):#P3.7, 26 pp, 2020) that $\mu (n)=1$ if n/2 is odd and $\mu (n)>1$ if n/2 is even. Extending this result, we show that if n/2 is even, then $\mu (n)>2$. Also, we show that whenever n is divisible by a particular function of k, there does not exist a set of $k'$-maxMOFS(n; n/2) for any $k'\leqslant k$. In particular, this means that $\limsup \mu (n)$ is unbounded. Nevertheless we can construct infinite families of maximal binary MOFS of fixed cardinality. More generally, let $q=p^u$ be a prime power and let $p^v$ be the highest power of p that divides n. If $0\leqslant v-uh<u/2$ for $h\geqslant 1$ then we show that there exists a set of $(q^h-1)^2/(q-1)$-maxMOFS(n; n/q).