A Congruence Connecting Latin Rectangles and Partial Orthomorphisms

Abstract

A partial orthomorphism of ${\mathbb{Z}_{n}}$ is an injective map ${\sigma : S \rightarrow \mathbb{Z}_{n}}$ such that ${S \subseteq \mathbb{Z}_{n}}$ and σ(i)–i ≢ σ(j)− j (mod n) for distinct ${i, j \in S}$. We say σ has deficit d if ${|S| = n - d}$. Let ω(n, d) be the number of partial orthomorphisms of ${\mathbb{Z}_{n}}$ of deficit d. Let χ(n, d) be the number of partial orthomorphisms σ of ${\mathbb{Z}_n}$ of deficit d such that σ(i) ∉ {0, i} for all ${i \in S}$. Then ω(n, d) = χ(n, d)n ²/d ² when ${1\,\leqslant\,d < n}$ . Let R _{k, n} be the number of reduced k × n Latin rectangles. We show that

$$R_{k, n} \equiv \chi (p, n - p)\frac{(n - p)!(n - p - 1)!^{2}}{(n - k)!}R_{k-p,\,n-p}\,\,\,\,(\rm {mod}\,p)$$

when p is a prime and ${n\,\geqslant\,k\,\geqslant\,p + 1}$ . In particular, this enables us to calculate some previously unknown congruences for R _{n, n} . We also develop techniques for computing ω(n, d) exactly. We show that for each a there exists μ _a such that, on each congruence class modulo μ _a , ω(n, n-a) is determined by a polynomial of degree 2a in n. We give these polynomials for ${1\,\leqslant\,a\,\leqslant 6}$ , and find an asymptotic formula for ω(n, n-a) as n → ∞, for arbitrary fixed a.