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Statement

Section 4 (pp. 234--236) is a sketch: the paper says it will "merely sketch the results" (p. 234). Notation as in Theorem 1: $N$ counts the ways to add a $(k+1)$-st row to a given $k$-row Latin rectangle on $1,\ldots,n$, $f(n,k)$ counts $k$-row Latin rectangles, $(n)_t$ is the falling factorial $n(n-1)\cdots(n-t+1)$, and ${}_kC_j=\binom kj$.

• (p. 234) A more careful argument shows that the error term in (7) is of the order of $k^2n^{-1}$; keeping $B(r,1)$ as well as $B(r,0)$ reduces it to the order of $k^4n^{-2}$, and continuing gives successive terms of an asymptotic series, whose existence the paper credits as a conjecture to Jacob.
• (p. 235, (20)) Correct up to $n^{-2}$,

$$\frac{Ne^k}{n!}=1-\frac{n\binom k2}{(n)_2}+\frac{2n\binom k3}{(n)_3} +\frac{\binom n2\binom k2^2}{(n)_4}+\cdots =1-\frac{\binom k2}{n}+\frac{\binom k2(k+4)(3k-7)}{12n^2}+\cdots.$$

The text above (20) gives $F(2,3)=3n\binom k3$, while the middle term of (20) is printed with $2n\binom k3$. The two agree once the term $-F(3,3)/(n)_3$ of (19) is counted, which (19) leaves in its dots and the paper does not mention: $F(3,3)=n\binom k3$ (three equal entries in distinct columns, with all three of their pairs). The closed form on the second line agrees with $2n\binom k3$. The coefficient of $n^{-3}$ already involves the number $X$ of pairs of integers occurring together in two different columns, which depends on the rectangle; the paper bounds $X\le n\binom k2(k-1)$.

• (p. 235, (21)) Multiplying (20) from $1$ to $k-1$,

$$f(n,k)\,(n!)^{-k}\exp\binom k2 =1-\frac{\binom k3}{n}+\frac{\binom k3(k^3-3k^2+8k-30)}{12n^2}+\cdots.$$

For $k=3$ the right side is $1-1/n-1/(2n^2)+\cdots$, which the paper compares in a table with Kerawala's exact values for $n=5,10,15,20,25$.

The paper adds (p. 236) that the form of (21) "strongly suggests that at about $k = n^{1/3}$ the expression ceases to be valid", which it cannot prove; the introduction (p. 230) likewise calls $(\log n)^{3/2}$ an apparent "natural boundary" of the method and says the authors believe the actual break occurs at $k=n^{1/3}$.

Proof pointer

Pp. 234--235, a sketch only. Run the two sieves of Theorem 1 without truncation, drop the remainder $\theta$ of the exponential series, which gives $Ne^k/n!=1-F(1,2)/(n)_2+F(2,3)/(n)_3+F(2,4)/(n)_4-\cdots$ (19), and count $F(1,2)=n\binom k2$, $F(2,3)=3n\binom k3$ and $F(2,4)$ up to $X$ by hand. No error bounds are proved for (20) or (21).

Read depth

Claims checked: the expansions (19)--(21) and the remarks of pp. 234--236 were read against the page images of the print. The closed form of (20) was checked against its first line, and the $n^{-1}$ and $n^{-2}$ coefficients of (21) against the product of (20) over $1,\ldots,k-1$, by exact arithmetic for $k\le11$; the paper itself proves no error bound.

Dependencies

• Theorem 1 (p. 232), whose sieve this extends.

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: sketched further terms, to order $n^{-2}$, of the formula of Theorem 2, with the authors' unproved expectation that the formula $f(n,k)\sim(n!)^ke^{-\binom k2}$ stops holding near $k=n^{1/3}$; it proves nothing beyond Theorem 2's range.