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).

甲频率正方形是正方形矩阵，其中每行和列是码元的相同的多集的排列。如果频率平方包含\（n / \ lambda \）符号，则频率平方的类型为 \（（n; \ lambda）\），每个符号每行出现\（\ lambda \）次和\（\ lambda \）次每列。在\（\ lambda = n / 2 \）的情况下，我们将频率平方称为二进制。一组k -MOFS \（（n; \ lambda）\）是一组类型为\（（n; \ lambda）\）的k个频率平方这样，当任何两个频率平方重叠时，每个可能的有序对就会频繁出现。一组k -maxMOFS \（（n; \ lambda）\）是一组不包含在\（（k + 1）\）中的k -MOFS \（（n; \ lambda ）\） -MOFS \（（（n; \ lambda）\）。对于偶数n，令\（\ mu（n）\）是最小的k，从而存在一组k -maxMOFS（n ;  n / 2）。在Britz等人的文章中对此进行了说明。（Electron。J.Combin.27（3）：＃P3.7，26 pp，2020 ）如果n / 2为奇数且\（\ mu（n）= 1 \）\（\ mu（n）> 1 \）如果n / 2是偶数。扩展该结果，我们表明如果n / 2是偶数，则\（\ mu（n）> 2 \）。而且，我们表明，只要n被k的特定函数整除，对于任何\（k'\ leqslant k \）都不存在\（k'\）- maxMOFS（n ;  n / 2）的集合。特别是，这意味着\（\ limsup \ mu（n）\）是无界的。但是，我们可以构造固定基数的最大二进制MOFS的无限族。更一般而言，令\（q = p ^ u \）为素数幂，令\（p ^ v \）为p的最高幂除以n。如果\（0 \ leqslant v-uh <u / 2 \）为\（h \ geqslant 1 \），那么我们表明存在一组\（（q ^ h-1）^ 2 /（q-1） \）- maxMOFS（n ;  n / q）。