A classic proof of a recurrence for a very classical sequence

Dominique Foata^{1} and Doron Zeilberger^{2}

To Marco Schützenberger, in memoriam.

Richard Stanley [St96] has recently narrated the fascinating story of how the classical Schröder [Sch1870] numbers s(n) are even more classical than has been believed before. They (at least s(10) = 103049) have been known to Hipparchus (190-127 B.C.). Stanley recalled the three-term linear recurrence ([Co64]; [Co74], p. 57)

(1) 3(2n − 1)s(n) = (n + 1)s(n + 1) + (n − 2)s(n − 1) (n ≥ 2), s(1) = s(2) = 1,

and stated that “no direct combinatorial proof of this formula seems to be known.” The purpose of this note is to fill this gap.

The present proof reflects the ideas of our great master, Marcel-Paul Schützenberger (1920-1996), who taught us that every algebraic relation is to be given a combinatorial counterpart and vice versa. The methodology has been vigorously and successfully pursued by the École bordelaise (e.g., [Cor75], [Vi85].)

The recurrence (1) is tantalizingly similar to the linear recurrence

(2)  2(2n − 1)c(n) = (n + 1)c(n + 1) (n ≥ 1),  c(1) = 1,

that is obviously satisfied by the Catalan numbers c(n) = (1/n)\binom{2n−2}{n−1}. Our proof is inspired by Rémy’s elegant combinatorial proof [Re85] of (2) shown to us by Viennot [Vi82].

Recall ([Co74], pp. 56-57) that a Schröder tree T is either the tree consisting of its root alone T = r, or an ordered tuple [r; T_{1}, …, T_{l}], where l ≥ 2 and T_{1}, …, T_{l} are smaller Schröder trees. The first symbol r is called the root of T and the roots of T_{1}, …, T_{l} are called the sons of T. A sonless node is called a leaf. The number of Schröder trees with n leaves is denoted by s(n).

The first values of the s(n)’s appear in Table 1.

n  1 2 3 4 5 6  7  8  9  10
s(n)  1 1 3 11 45 197 903 4279 20793 103049

Table 1

1 Département de mathématique, Université Louis Pasteur, 7, rue René-Descartes, F-67084 Strasbourg, France. (foata@math.u-strasbg.fr)

2 Department of mathematics, Temple University, Philadelphia, Pennsylvania 19122.

(zeilberg@math.temple.edu)

Our proof will be based on another combinatorial model for the Schröder numbers, the well-weighted binary plane trees. Recall that a plane tree is said to be binary, if each node has either no sons or exactly two sons. As is well-known (see, e.g., [Co74], pp. 52-53), the number of such trees with exactly n leaves is the Catalan number c(n). A node is said to be interior, if it is not a leaf. The binary trees we will be using are weighted. This means that each interior node is given a weight equal to 1 or equal to 2. A node is said to be well-weighted, if, whenever it has weight 2, its right son is not a leaf.

A binary plane tree is well-weighted if it is weighted and if all its interior nodes are well-weighted. In short, we will speak of a well-weighted tree. As shown in Table 2, there are c(4) = 5 binary plane trees with four leaves, and s(4) = 11 well-weighted trees with four leaves. The nodes that can get either weight 1 or weight 2 are indicated by the symbol “◯.”

[figure: 1; 1; ◯; 1; 1; 1; 1; 1; ◯]

[figure: 1; ◯; 1; ◯; ◯; 1]

Table 2 A well-weighted tree can be also defined recursively as follows. It

is either a single unweighted node (serving both as root and leaf), or a triple [r; T_{1}, T_{2}], where r is either 1 or 2, and where T_{1} and T_{2} are smaller well-weighted trees, with the provision that if T_{2} is a mere leaf, then r must be 1.

Define a mapping from the set of Schröder trees to the set of wellweighted trees as follows: if T = r, then Φ(T) := r; if the root of T has exactly two sons, i.e., T = [r; T_{1}, T_{2}], then Φ(T) := [1; Φ(T_{1}), Φ(T_{2})]; if the root of T has more than two sons, i.e., T = [r; T_{1}, …, T_{l}] with l > 2, then Φ(T) := [2; Φ(T_{1}), Φ([r; T_{2}, …, T_{l}]) ]. It is clear that Φ is a bijection that preserves the number of leaves.

Being binary, every well-weighted tree with n leaves, possesses exactly (n−1) interior nodes and hence (2n−1) nodes altogether. A well-weighted tree with n leaves is said to be pointed, if exactly one of its (2n − 1) nodes is pointed; it is leaf-pointed if exactly one of its n leaves is pointed; it is interior-pointed if exactly one of its (n − 1) interior nodes is pointed. Let P T (n), (resp. LT (n), resp. IT (n)) be the set of pointed (resp. leaf-pointed, resp. interior-pointed) well-weighted trees with n leaves.

As (2n − 1)s(n) = #P T (n), (n + 1)s(n + 1) = #LT (n + 1) and (n − 2)s(n − 1) = #IT (n − 1), formula (1) will be proved combinatorially (or bijectively, as some people say to-day), if we can construct a bijection σ of {1, 2, 3} × P T (n) onto the disjoint union LT (n + 1) ∪ IT (n − 1).

The construction of such a bijection will consist of adding a new leaf to each pointed well-weighted tree of P T (n), in three different ways denoted by L_{1}, L_{2}, R_{1}. We shall get all of LT (n+1) plus a set B(n+1) of leaf-pointed weighted trees, but not well-weighted, which is in one-to-one correspondence with IT (n − 1).

To construct the bijection σ we proceed as follows. Start with t in P T (n) and let s be the subtree of t whose root is the pointed node of t. For j = 1, 2 define σ′(L_{j}, t) to be the leaf-pointed weighted tree with (n + 1) leaves obtained from t, by performing the following replacement:

• s  ↦  [figure: •; s; j]

Notice that the new interior node receives weight j and the point gets moved from the root of s to the new leaf.

In the same manner σ′(R_{1}, t) is defined by performing the replacement:

• s  ↦  [figure: s; •; 1]

Clearly, σ′ is an injection of {L_{1}, L_{2}, R_{1}} × P T (n) into the set of leaf-pointed weighted trees with (n + 1) leaves. Furthermore, σ′(L_{1}, t) and σ′(R_{1}, t) are always well-weighted and σ′(L_{2}, t) is well-weighted when the subtree s is not a leaf. Define σ = σ′ in all those cases.

When s is a leaf, the subtree of t whose root is the father of the pointed node is one of the following forms

[figure: t′; •; 1; s; case (a); •; t′; s; 1; case (b); •; t′′; s; 2; case (c)]

where the subtree t′ may be any tree, while the subtree t′′ must not be a leaf. When applying σ′ to (L_{2}, t), we will get

[figure: t′; 2; s; 1; •; case (a); t′; 2; 1; •; s; case (b); t′′; 2; 2; •; s; case (c)]

Alas, none of those trees is well-weighted (since s is a leaf). To remedy cases (a) and (b) replace them by:

[figure: t′; 1; 2; •; s; case (a); t′; 1; 2; •; s; case (b)]

and define σ(L_{2}, t) to be the tree thereby obtained. It now belongs to LT (n + 1). It is straightforward to verify that σ is a bijection of the set of all pairs in {L_{1}, L_{2}, R_{1}} × P T (n) that do not belong to case (c) onto LT (n + 1).

Finally, to obtain σ(L_{2}, t) in case (c), replace the portion depicted above that belongs to σ′(L_{2}, t) by just the subtree t′′ (losing two leaves) and make the root of t′′ be the pointed node of the new tree. We get an interior-pointed well-weighted plane tree with (n−1) leaves. Furthermore, the restriction of σ to case (c) is a bijection onto IT (n − 1).

References

[Co64] Louis Comtet. Calcul pratique des coefficients de Taylor d’une fonction algébrique, Enseignement Math., vol. 10, 1964, p. 267–270.

[Co74] Louis Comtet. Advanced Combinatorics. Dordrecht-Holland/Boston, 1974.

[Cor75] Robert Cori. Un code pour les graphes planaires et ses applications, Astérisque, vol. 27, 1975.

[Re85] Jean-Luc Rémy. Un procédé itératif de dénombrement d’arbres binaires et son application à leur génération aléatoire, RAIRO Inform. Théor., vol. 19, 1985, p. 179–195.

[Sch1870] Ernst Schröder. Vier combinatorische Probleme, Z. für Math. Physik, vol. 15,

1870, p. 361-370.

[St96] Richard P. Stanley. Hipparchus, Plutarch, Schröder and Hough, to appear in Amer. Math. Monthly, 1996, also available from:

http://www-math.mit.edu/∼rstan/papers.html.

[Vi82] Gérard-Xavier Viennot. Oral communication, Oberwolfach, 1982.

[Vi85] Gérard-Xavier Viennot. Problèmes combinatoires posés par la physique statistique, Séminaire Bourbaki, 36ième année, 1983/1984, no. 626, Astérisque, vol. 225-226, 1985.