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تسجيل مستخدم جديدMaximal sets of mutually orthogonal frequency squares
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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 [Electron. J. Combin. 27(3) (2020), P3.7] 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 $kle k$. 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 $0le v-uh<u/2$ for $hge1$ then we show that there exists a set of $(q^h-1)^2/(q-1)$-maxMOFS$(n;n/q)$.
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A emph{frequency square} is a matrix in which each row and column is a permutation of the same multiset of symbols. We consider only {em binary} frequency squares of order $n$ with $n/2$ zeroes and $n/2$ ones in each row and column. Two such frequenc
y squares are emph{orthogonal} if, when superimposed, each of the 4 possible ordered pairs of entries occurs equally often. In this context we say that a $k$-MOFS$(n)$ is a set of $k$ binary frequency squares of order $n$ in which each pair of squares is orthogonal. A $k$-MOFS$(n)$ must satisfy $kle(n-1)^2$, and any MOFS achieving this bound are said to be emph{complete}. For any $n$ for which there exists a Hadamard matrix of order $n$ we show that there exists at least $2^{n^2/4-O(nlog n)}$ isomorphism classes of complete MOFS$(n)$. For $2<nequiv2pmod4$ we show that there exists a $17$-MOFS$(n)$ but no complete MOFS$(n)$. A $k$-maxMOFS$(n)$ is a $k$-MOFS$(n)$ that is not contained in any $(k+1)$-MOFS$(n)$. By computer enumeration, we establish that there exists a $k$-maxMOFS$(6)$ if and only if $kin{1,17}$ or $5le kle 15$. We show that up to isomorphism there is a unique $1$-maxMOFS$(n)$ if $nequiv2pmod4$, whereas no $1$-maxMOFS$(n)$ exists for $nequiv0pmod4$. We also prove that there exists a $5$-maxMOFS$(n)$ for each order $nequiv 2pmod{4}$ where $ngeq 6$.
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