THE GENERATING FUNCTION FOR THE DIFFERENCE IN EVEN AND ODD THREE-LINE LATIN RECTANGLES

Jiang Zeng

10 octobre, 1995

Résumé. On donne une fonction génératrice exponentielle pour la différence entre le nombre des rectangles latin pairs de taille 3 × n et le nombre de ceux impairs. On montre ensuite que cette formmule équivaut à une formula, qui est apparemment différente, donnée dans [4].

Abstract. We give a simple exponential generating function for the difference between the number of even reduced Latin rectangles of size 3 × n and the number of odd ones. We show that this formula is equivalent to an apparently different formula given in [4].

Let S_{n} be the set of permutations of [n] = {1, 2, …, n}. For each σ ∈ S_{n}, we denote by cyc σ (resp. fix σ) the number of cycles (resp. fixed points) of σ. Further let L_{n} be the generating polynomial for S_{n} × S_{n} defined by

L_{n}(α, β, p, q, γ) = ∑_{σ,τ∈S_{n}} α^{cyc σ}β^{cyc τ}p^{fix σ}q^{fix τ}γ^{fix στ^{−1}}.

Let (x)_{n} = x(x + 1) ⋯ (x + n − 1). Using a generalization of Cauchy’s summation theorem for Schur functions and some group representation theories Andrews et al. [1] obtained the following expression for the exponential generating function for the L_{n} :

(1)  1 + ∑_{n≥1} \frac{t^{n}}{n!}L_{n}(α, β, p, q, γ) = e^{αβt(2−p−q−γ+pqγ)}

∑_{m≥0} \frac{t^{m}}{m!} \frac{(α)_{m}(β)_{m}}{[1 + β(1 − q)t]^{m+α}[1 + α(1 − p)t]^{m+β}[1 + (1 − γ)t]^{m+αβ}} ·

In [5] the author has shown some applications of (1) and further derived a combinatorial proof of (1) by generalizing a standard method due to Gessel [3].

Recently Habsieger and Janssen [4] raised the problem to count signed Latin rectangles. Recall that a Latin rectangle of size k × n is an k × n array, such that each row is a permutation of 12 … n, and the product ^{−1} of any two distinct row permutations , is fixed-point free, i.e. is a derangement. A reduced Latin rectangle has the additional property that the first row is the identity permutation. A Latin rectangle is called row-even if the product of the signs of its row permutations is +1 and row-odd if this product is −1. Let S_{k,n} denote the difference between the number of even reduced Latin rectangles of size k × n and the number of odd ones. In [4], Habsieger and Janssen proved the following results :

(1.0)  S_{1,n} = 1,
(1.1)  S_{2,n} = (−1)^{n−1}(n − 1),
(1.2)  S_{3,n} = (−1)^{n−1} \frac{(n − 2)n!}{2}[t^{n}]e^{2(arctanh t−t)},

where [t^{n}]F (t) stands for the coefficient of t^{n} in the series expansion of F (t). Note that in [4] the series arctanh t of (1.2) was incorrectly replaced by sinh t due to an obvious error at the end of its proof [4, p. 75], where (2j + 1) was incorrectly replaced by (2j + 1)!.

Actually formulas (1.0) and (1.1) are straightforward. The purpose of this note is to show that (1.2) is a equivalent to a special case of (1).

Proposition 1 We have

(1.3)  1 + ∑_{n≥1} S_{3,n} \frac{t^{n}}{n!} = e^{2t} [\frac{(1 − t)^{2}}{1 + t} + \frac{t}{(1 + t)^{2}} ] .

Proof: Let D_{\} be the set of derangements of [n]. By definition, we have

S_{3,n} = ∑_{σ,τ,στ^{−1}∈D_{n}} ε(α)ε(β).

As ε(α)ε(β) = (−1)^{n−cyc α}(−1)^{n−cyc β} = (−1)^{cyc α}(−1)^{cyc β}, we conclude that

S_{3,n} = L_{n}(−1, −1, 0, 0, 0), (n ≥ 1).

Formula (1.3) follows then from (1).

Proposition 2 Formulas (1.2) and (1.3) are equivalent.

Proof: From (1.3) we derive

(1.4)  ∑_{n≥1}(−1)^{n−1}S_{3,n} \frac{t^{n}}{n!} = 1 + \frac{e^{−2t}}{(1 − t)^{2}} (t^{2} + t^{3} − 1).

Note that

\frac{e^{−2t}}{(1 − t)^{2}} = ∑_{n≥0} \frac{(−2)^{n}t^{n}}{n!} · ∑_{n≥0}(n + 1)t^{n}

= ∑_{n≥0} (∑_{i=0}^{n}(n − i + 1)\frac{(−2)^{i}}{i!} ) t^{n}.

Substituting this into (1.4) and equating the coefficients of t^{n} yield

(1.5)  (−1)^{n−1} \frac{S_{3,n}}{n!} = ∑_{i=0}^{n−3}(n − 4 − i)\frac{(−2)^{i}}{i!} + \frac{(−2)^{n−1}}{n(n − 3)!}.

On the other hand, since

arctanh x = ∑_{n≥0} \frac{x^{2n+1}}{2n + 1} = \frac{1}{2} ln (\frac{1 + x}{1 − x})

therefore

[x^{n}]e^{2 arctanh x−2x} = [x^{n}]\frac{1 + x}{1 − x}e^{−2x} = 2 ∑_{i=0}^{n−1} \frac{(−2)^{i}}{i!} + \frac{(−2)^{i}}{n!} ,

and it follows from (1.2) that

(1.6)  (−1)^{n−1} \frac{S_{3,n}}{n!} = (n − 2) ∑_{i=0}^{n−1} \frac{(−2)^{i}}{i!} + (−1)^{n}(n − 2)\frac{2^{n−1}}{n!} ·

It is easy to check that the right hand-sides of formulas (1.5) and (1.6) are equal. Indeed, if we subtract the right-hand side of (1.6) from that of (1.5), we obtain

∑_{i=0}^{n−3}(2 + i)\frac{(−2)^{i}}{i!} + \frac{(−2)^{n−2}}{(n − 3)!} = 2 ∑_{i=0}^{n−3} \frac{(−2)^{i}}{i!} − 2 ∑_{i=0}^{n−3} \frac{(−2)^{i−1}}{(i − 1)!} + \frac{(−2)^{n−2}}{(n − 3)!}
= 2\frac{(−2)^{n−3}}{(n − 3)!} + \frac{(−2)^{n−2}}{(n − 3)!}
= 0,

which was to be proved.

Finally, by applying Darboux’s method [Co, p. 121] we get from (1.4) a good asymptotic estimate for (−1)^{n−1}S_{3,n} :

(−1)^{n−1}S_{3,n} ∼ \sqrt{2π} n^{n+3/2}e^{−n−2} (n ⟶ +∞).

1 Résumé substantiel en français

Un rectangle latin est pair selon les lignes si le produit des signes des permutation de lignes vaut +1 et impair selon les lignes si ce produit vaut −1. Notons S_{k,n} la différence entre le nombre de rectangle réduits pairs de taille k ×n et le nombre de ceux impairs. Dans [4], Habsieger et Janssen ont établi des formules explicites pour S_{k,n} lorsque k = 1, 2, 3 :

(1.0)  S_{1,n} = 1,
(1.1)  S_{2,n} = (−1)^{n−1}(n − 1),
(1.2)  S_{3,n} = (−1)^{n−1} \frac{(n − 2)n!}{2}[t^{n}]e^{2(arctanh t−t)},

Le but de cette note est de signaler une formule simple pour la fonction génératrice exponentielle de la suite {S_{3,n}}_{n≥1} :

(1.3)  1 + ∑_{n≥1} S_{3,n} \frac{t^{n}}{n!} = e^{2t} [\frac{(1 − t)^{2}}{1 + t} + \frac{t}{(1 + t)^{2}} ] .

Cette formule est en fait un cas particulier d’une formule plus générale due à Gessel [3] ou de celle de Andrews et al. [1]. On montre ensuite que les deux formules (1.2) et (1.3) sont equivalentes.

References

[1] G. Andrews, I. P. Goulden and D. M. Jackson: Generalization of Cauchy’s summation theorem for Schur functions, Trans. Amer. Math. Soc., 310(1988), 805–820.

[2] L. Comtet Analyse combinatoire, Vol. 2, Press Univ. de France, Paris, 1970.

[3] I. M. Gessel: Counting three-line Latin rectangles, Combinatoire Énumérative (Montréal, 1985), Lecture Notes in Math., vol. 1234, Springer-Verlag, Nex-York, 106–111.

[4] L. Habsieger and J. Janssen The difference in even and odd latin rectangles for small cases, Ann. Sci. Math. Québec, 19(1995), 69-77.

[5] J. Zeng: Counting a pair of permutations and the linearization coefficients for Jacobi polynomials, Actes de l’Atelier de Combinatoire Franco-Québécois (Bordeaux, 6-7 mai 1991), Publications du LACIM, vol. 10, Université du Québec à Montréal, Montréal, 1992, 243-257.

Jiang ZENG UFR de Mathématiques et Informatique Université Louis Pasteur 7, rue René Descartes 67084 Strasbourg Cedex, France e-mail : jzeng@math.u-strasbg.fr