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Statement

Setting (p. 113). f(n,k)f(n,k) is the number of nn by kk Latin rectangles in the symbols 1,2,…,n1,2,\ldots,n, that is, of kk rows of length nn with no symbol repeated in a row or in a column; the rows, columns and symbols are labeled.

Theorem 2 (p. 118, quoted). "For k<n1/3−δk<n^{1/3-\delta}, δ\delta being positive constant, we have the asymptotic relation

f(n,k)∼(n!)kexp⁡(−k(k−1)/2)f(n,k)\sim(n!)^k\exp(-k(k-1)/2)

for the number f(n,k)f(n,k) of nn by kk Latin rectangles."

The relation is as n→∞n\to\infty, uniformly in the kk allowed: the proof gives (1−cn−2ε)k<exp⁡(k(k−1)/2) (n!)−kf(n,k)<(1+cn−2ε)k(1-cn^{-2\varepsilon})^k<\exp(k(k-1)/2)\,(n!)^{-k}f(n,k)<(1+cn^{-2\varepsilon})^k with ε=1/6+δ\varepsilon=1/6+\delta for all sufficiently large nn, and both bounds tend to 11.

Remark (p. 119). The paper says the generalization made for Theorem 1 is immediate here: δ\delta may be a positive function of nn with n−δ→0n^{-\delta}\to0 as n→∞n\to\infty. The introduction (p. 113) states the result in this form, for δ\delta a positive constant "or more generally, may be a positive-valued function of nn tending to zero such that" n−δ→0n^{-\delta}\to0.

Erdős and Kaplansky had proved the same relation for k<(log⁡n)3/2−εk<(\log n)^{3/2-\varepsilon} and conjectured it for kk up to nearly n1/3n^{1/3} (p. 113); Theorem 2 confirms that conjecture. The paper adds (p. 119) that the range k<n1/3−δk<n^{1/3-\delta} is the limit of its method, as seen from (27), and that it seems likely to be the natural boundary of the problem, as Erdős and Kaplansky observed.

Proof pointer

Pp. 118--119. Apply Theorem 1 with ε=1/6+δ\varepsilon=1/6+\delta to the extensions of nn by ii Latin rectangles for i=0,1,…,k−1i=0,1,\ldots,k-1, where i<n1/3−δ=n1/2−εi<n^{1/3-\delta}=n^{1/2-\varepsilon}: every nn by ii Latin rectangle has between n! e−i(1−cn−2ε)n!\,e^{-i}(1-cn^{-2\varepsilon}) and n! e−i(1+cn−2ε)n!\,e^{-i}(1+cn^{-2\varepsilon}) extensions, so 1−cn−2ε<ei(n!)−1f(n,i+1)/f(n,i)<1+cn−2ε1-cn^{-2\varepsilon}<e^i(n!)^{-1}f(n,i+1)/f(n,i)<1+cn^{-2\varepsilon}, with f(n,0)=1f(n,0)=1. Multiplying these kk inequalities gives the bounds above, and klog⁡(1+cn−2ε)<cn−2(1/6+δ)n1/3−δ=cn−3δk\log(1+cn^{-2\varepsilon})<cn^{-2(1/6+\delta)}n^{1/3-\delta}=cn^{-3\delta} (27) shows they tend to 11.

Read depth

Claims checked: the setting, Theorem 2, the remark after it and the introduction's statement were read clause by clause on the page images of the print, and the proof on pp. 118--119 was followed. Nothing here is independently reviewed.

Dependencies

  • Theorem 1 (p. 118), the per-row extension estimate.

Source. K. Yamamoto, On the asymptotic number of Latin rectangles, Jpn. J. Math. 21 (1951), 113--119, doi:10.4099/jjm1924.21.0_113; the edition read is named on the source card.

Bears on

  • Problem 725: Theorem 2 gives the asymptotic number f(n,k)∼(n!)kexp⁡(−k(k−1)/2)f(n,k)\sim(n!)^k\exp(-k(k-1)/2) of nn by kk Latin rectangles, the problem's k×nk\times n Latin rectangles, for k<n1/3−δk<n^{1/3-\delta}, and by the remark for δ\delta a positive function of nn with n−δ→0n^{-\delta}\to0. It says nothing about larger kk, which the problem also asks about.