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Statement

Setting (p. 230, Section 2). An nn by kk Latin rectangle has kk rows, each an arrangement of 1,…,n1,\ldots,n, with distinct integers in each column (the printed definition says nn rows and kk columns, but puts 1,…,n1,\ldots,n in each row and adds rows one at a time; see Theorem 1). The paper writes kC2{}_kC_2 for (k2)\binom k2.

Theorem 2 (p. 234). Let f(n,k)f(n,k) be the number of nn by kk Latin rectangles and suppose k<(log⁡n)3/2−ϵk<(\log n)^{3/2-\epsilon}. Then

f(n,k) (n!)−kexp⁡(k2)→1as n→∞.(18)f(n,k)\,(n!)^{-k}\exp\binom k2\to1 \quad\text{as } n\to\infty. \qquad (18)

So f(n,k)∼(n!)ke−(k2)f(n,k)\sim(n!)^k e^{-\binom k2} in that range, with kk fixed or growing with nn; ϵ\epsilon is a fixed positive number, as in Theorem 1. The introduction (p. 230) calls this formula an easy heuristic conjecture, previously proved for k=3k=3, and says the paper proves it for kk fixed and for k<(log⁡n)3/2−ϵk<(\log n)^{3/2-\epsilon}.

Proof pointer

P. 234. By Theorem 1, applied to each ii-row rectangle, f(n,i+1)f(n,i+1) lies between f(n,i) n! e−i(1±n−c)f(n,i)\,n!\,e^{-i}(1\pm n^{-c}). Multiplying from i=1i=1 to k−1k-1 puts f(n,k)f(n,k) between (n!)kexp⁡(−(k2))(1±n−c)k(n!)^k\exp(-\binom k2)(1\pm n^{-c})^k, and both (1+n−c)k(1+n^{-c})^k and (1−n−c)k(1-n^{-c})^k tend to 11 because kk is at most a power of log⁡n\log n.

Read depth

Claims checked: Theorem 2 and its proof were read clause by clause on the page images of the print. Nothing here is independently reviewed.

Dependencies

Source. P. Erdős and I. Kaplansky, The asymptotic number of Latin rectangles, Amer. J. Math. 68 (1946), no. 2, 230--236, doi:10.2307/2371834; the edition read is named on the source card.

Bears on

  • Problem 725: the problem asks for an asymptotic formula for the number of k×nk\times n Latin rectangles without restricting kk; Theorem 2 gives f(n,k)∼(n!)ke−(k2)f(n,k)\sim(n!)^k e^{-\binom k2} for k<(log⁡n)3/2−ϵk<(\log n)^{3/2-\epsilon} and says nothing about larger kk. The problem's claim page for this paper records that partial answer.