On the number of orthogonal latin squares

Haim Hanani
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引用次数: 23

Abstract

Let N(n) be the maximal number of mutually orthogonal Latin squares of order n and let nr be the smallest integer such that N(n)≥r for every n>nr. It is known that N(n)→∞ as n→∞ and that n2=6. A proof is given for n3≤51, n5≤62 and n29≤34, 115, 553.

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关于正交拉丁方的个数
设N(N)为N阶相互正交的拉丁平方的最大个数,设nr为对每个N >nr使N(N)≥r的最小整数。已知N(N)→∞为N→∞,且n2=6。给出了n3≤51,n5≤62,n29≤34,115,553的证明。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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