A Combinatorial Interpretation of the Numbers 6 (2n)! /n! (n + 2)!

Ira M. Gessel^{1} and Guoce Xin Department of Mathematics Brandeis University Waltham, MA 02454-9110 USA

                      gessel@brandeis.edu
                      guoce.xin@gmail.com

Abstract It is well known that the numbers (2m)! (2n)!/m! n! (m + n)! are integers, but in general there is no known combinatorial interpretation for them. When m = 0 these numbers are the middle binomial coefficients \binom{2n}{n}, and when m = 1 they are twice the Catalan numbers. In this paper, we give combinatorial interpretations for these numbers when m = 2 or 3.

1 Introduction

The Catalan numbers

C_{n} = \frac{1}{n + 1}\binom{2n}{n} = \frac{(2n)!}{n! (n + 1)!}

are well-known integers that arise in many combinatorial problems. Stanley [11, pp. 219–229] gives 66 combinatorial interpretations of these numbers.

In 1874 Catalan [3] observed that the numbers

\frac{(2m)! (2n)!}{m! n! (m + n)!}  (1.1)

are integers, and their number-theoretic properties were studied by several authors (see Dickson [4, pp. 265–266]). For m = 0, (1.1) is the middle binomial coefficient \binom{2n}{n}, and for m = 1 it is 2C_{n}.

1Partially supported by NSF grant DMS-0200596.

Except for m = n = 0, these integers are even, and it is convenient for our purposes to divide them by 2, so we consider the numbers

T (m, n) = \frac{1}{2} \frac{(2m)! (2n)!}{m! n! (m + n)!}.

Some properties of these numbers are given in [5], where they are called “super Catalan numbers”. An intriguing problem is to find a combinatorial interpretation to the super Catalan numbers. The following identity [5, Equation (32)], together with the symmetry property T (m, n) = T (n, m) and the initial value T (0, 0) = 1, shows that T (m, n) is a positive integer for all m and n.

∑_{n} 2^{p−2n}\binom{p}{2n}T (m, n) = T (m, m + p), p ≥ 0.  (1.2)

Formula (1.2) allows us to construct recursively a set of cardinality T (m, n) but it is difficult to give a natural description of this set. Shapiro [10] gave a combinatorial interpretation to (1.2) in the case m = 1, which is the Catalan number identity

∑_{n} 2^{p−2n}\binom{p}{2n}C_{n} = C_{p+1}.

A similar interpretation works for the case m = 0 of (1.2) (when multiplied by 2), which is the identity

∑_{n} 2^{p−2n}\binom{p}{2n}\binom{2n}{n} = \binom{2p}{p}.

Another intriguing formula for T (m, n), which does not appear in [5], is

1 + ∑_{m,n=1}^{∞} C_{m}C_{n}x^{m}y^{n} = (1 − ∑_{m,n=1}^{∞} T (m, n)x^{m}y^{n})^{−1},  (1.3)

which can easily be proved using the the generating function for 2T (m, n) given in formulas (35) and (37) of [5]. Although (1.3) suggests a combinatorial interpretation for T (m, n) based on a decomposition of pairs of objects counted by Catalan numbers, we have not found such an interpretation.

In this paper, we give a combinatorial interpretation for T (2, n) = 6 (2n)! /n! (n + 2)! for n ≥ 1 and for T (3, n) = 60 (2n)! /n! (n + 3)! for n ≥ 2. The first few values of T (m, n) for m = 2 and m = 3 are as follows:

\begin{matrix}m\n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 3 & 2 & 3 & 6 & 14 & 36 & 99 & 286 & 858 & 2652 & 8398 \\ 3 & 10 & 5 & 6 & 10 & 20 & 45 & 110 & 286 & 780 & 2210 & 6460 \end{matrix}

We show that T (2, n) counts pairs of Dyck paths of total length 2n with heights differing by at most 1. We give two proofs of this result, one combinatorial and one using generating functions. The combinatorial proof is based on the easily checked formula

T (2, n) = 4C_{n} − C_{n+1}  (1.4)

which we interpret by inclusion-exclusion.

Our interpretation for T (3, n) is more complicated, and involves triples of Dyck paths with height restrictions. Although we have the formula T (3, n) = 16C_{n} − 8C_{n+1} + C_{n+2} analogous to (1.4), we have not found a combinatorial interpretation to this formula, and our proof uses generating functions.

Interpretations of the number T (2, n) in terms of trees, related to each other, but not, apparently, to our interpretation, have been found by Schaeffer [9], and by Pippenger and Schleich [7, pp. 34]. A combinatorial interpretation of (1.2) for m = 2, using Pippenger and Schleich’s interpretation of T (2, n), has been given by Callan [2].

2 The main theorem

All paths in this paper have steps (1, 1) and (1, −1), which we call up steps and down steps. A step from a point u to a point v is denoted by u → v. The level of a point in a path is defined to be its y-coordinate. A Dyck path of semilength n (or of length 2n) is a path that starts at (0, 0), ends at (2n, 0), and never goes below level 0. It is well-known that the number of Dyck paths of semilength n equals the Catalan number C_{n}. The height of a path P, denoted by h(P), is the highest level it reaches.

Every nonempty Dyck path R can be factored uniquely as UP DQ, where U is an up step, D is a down step, and P and Q are Dyck paths. Thus the map R ↦ (P, Q) is a bijection from nonempty Dyck paths to pairs of Dyck paths. Let B_{n} be the set of pairs of Dyck paths (P, Q) of total semilength n. This bijection gives |B_{n}| = C_{n+1}, so by (1.4), we have T (2, n) = 4C_{n} − |B_{n}|.

Our interpretation for T (2, n) is a consequence of the following Lemma 2.1. We give two proofs of this lemma, one combinatorial and the other algebraic. The algebraic proof will be given in the next section.

Lemma 2.1. For n ≥ 1, C_{n} equals the number of pairs of Dyck paths (P, Q) of total semilength n, with P nonempty and h(P) ≤ h(Q) + 1.

Proof. Let D_{n} be the set of Dyck paths of semilength n, and let E_{n} be the set of pairs of Dyck paths (P, Q) of total semilength n, with P nonempty and h(P) ≤ h(Q) + 1. We first establish a bijection from E_{n} to D_{n}.

For a given pair (P, Q) in E_{n}, since P is nonempty, the last step of P must be a down step, say, u → v. By replacing u → v in P with an up step u → v′, we get a path F_{1}. Now raising Q by two levels, we get a path F_{2}. Thus F := F_{1}F_{2} is a path that ends at level 2 and never goes below level 0. The point v′ belongs to both F_{1} and F_{2}, but we treat it as a point only in F_{2}, even if F_{2} is the empty path. The condition that h(P) ≤ h(Q) + 1 yields h(F_{1}) < h(F_{2}), which implies that the highest point of F must belong to F_{2}. See Figure 1 below.

Now let y be the leftmost highest point of F (which is in F_{2}), and let x → y be the step in F leading to y. Then x → y is an up step. By replacing x → y with a down step x → y′, and lowering the part of F_{2} after y by two levels, we get a Dyck path D ∈ D_{n}. See Figure 2 below.

P u v  Q  F_{1} u Fv′  F_{2}

Figure 1: First step of the bijection

u Fv′ F_{1}  x  y F_{2}  u D v′  y′ x

Figure 2: Second step of the bijection

With the following two key observations, it is easy to see that the above procedure gives a bijection from E_{n} to D_{n}. First, x in the final Dyck path D is the rightmost highest point. Second, u in the intermediate path F is the rightmost point of level 1 in both F and F_{1}.

Theorem 2.2. For n ≥ 1, the number T (2, n) counts pairs of Dyck paths (P, Q) of total semilength n with |h(P) − h(Q)| ≤ 1.

Proof. Let F be the set of pairs of Dyck paths (P, Q) with h(P) ≤ h(Q) + 1, and let G be the set of pairs of Dyck paths (P, Q) with h(Q) ≤ h(P) + 1. By symmetry, we see that |F| = |G|. Now we claim that the cardinality of F is 2C_{n}. This claim follows from Lemma 2.1 and the fact that if P is the empty path, then h(P) ≤ h(Q) + 1 for every Q ∈ D_{n}.

Clearly we have that F ∪ G = B_{n}, and that F ∩ G is the set of pairs of Dyck paths (P, Q), with |h(P) − h(Q)| ≤ 1. The theorem then follows from the following computation:

|F ∩ G| = |F| + |G| − |F ∪ G| = 4C_{n} − |B_{n}| = 4C_{n} − C_{n+1}.

3 An algebraic proof and further results

In this section we give an algebraic proof of Lemma 2.1. Although not as simple or elegant as the proof given in section 2, this proof generalizes to a larger class of paths with bounded height, while the combinatorial proof of Lemma 2.1 does not seem to generalize easily.

Let c(x) be the generating function for the Catalan numbers, so that

c(x) = ∑_{n=0}^{∞} \frac{1}{n + 1}\binom{2n}{n}x^{n} = \frac{1 − \sqrt{1 − 4x}}{2x}.

Then c(x) satisfies the functional equation c(x) = 1 + xc(x)^{2}. Let C = xc(x)^{2} = c(x) − 1 and let G_{k} be the generating function for Dyck paths of height at most k. Although G_{k} is a rational function, and explicit formulas for it as a quotient of polynomials are well-known, a formula for G_{k} in terms of C will be of more use to us. An equivalent formula can be found in [1, Equation (16)].

Lemma 3.1. For k ≥ −1,

G_{k} = (1 + C)\frac{1 − C^{k+1}}{1 − C^{k+2}} .  (3.1)

Proof. Let P be a path of height at most k ≥ 1. If P is nonempty then P can be factored as UP_{1}DP_{2}, where U is an up step, P_{1} is a Dyck path of height at most k − 1 (shifted up one unit), D is a down step, and P_{2} is a Dyck path of height at most k. Thus G_{k} = 1+xG_{k−1}G_{k}, so G_{k} = 1/(1 − xG_{k−1}). Equation (3.1) clearly holds for k = −1 and k = 0. Now suppose that for some k ≥ 1,

G_{k−1} = (1 + C)\frac{1 − C^{k}}{1 − C^{k+1}} .

Then the recurrence, together with the formula x = C/(1 + C)^{2}, gives

G_{k} = [1 − x(1 + C)\frac{1 − C^{k}}{1 − C^{k+1}} ]^{−1} = [1 − \frac{C}{1 + C} \frac{1 − C^{k}}{1 − C^{k+1}} ]^{−1}

= [\frac{1 − C^{k+2}}{(1 + C)(1 − C^{k+1})}]^{−1} = (1 + C)\frac{1 − C^{k+1}}{1 − C^{k+2}} .

We can prove Lemma 2.1 by showing that ∑_{n=0}^{∞} G_{n+1}(G_{n} − G_{n−1}) = 1 + 2C; this is equivalent to the statement that the number of pairs (P, Q) of Dyck paths of semilength m > 0 with h(P) ≤ h(Q) + 1 is 2C_{m}.

Theorem 3.2.

∑_{n=0}^{∞}(G_{n} − G_{n−1})G_{n+1} = 1 + 2C.

Proof. Let

Ψ_{k} = ∑_{n=k}^{∞} \frac{C^{n}}{1 − C^{n}} .

Thus if j < k then

Ψ_{j} = ∑_{n=j}^{k−1} \frac{C^{n}}{1 − C^{n}} + Ψ_{k}.

We have

G_{n}G_{n+1} = (1 + C)^{2} \frac{1 − C^{n+1}}{1 − C^{n+3}}

and

G_{n−1}G_{n+1} = (1 + C)^{2} \frac{(1 − C^{n})(1 − C^{n+2})}{(1 − C^{n+1})(1 − C^{n+3})}.

Let

S_{1} = ∑_{n=0}^{∞} (\frac{1 − C^{n+1}}{1 − C^{n+3}} − 1)

and

S_{2} = ∑_{n=0}^{∞} (\frac{(1 − C^{n})(1 − C^{n+2})}{(1 − C^{n+1})(1 − C^{n+3})} − 1) .

Then ∑_{n=0}^{∞}(G_{n} − G_{n−1})G_{n+1} = (1 + C)^{2}(S_{1} − S_{2}). We have

\frac{1 − C^{n+1}}{1 − C^{n+3}} − 1 = −\frac{(1 − C^{2})C^{n+1}}{1 − C^{n+3}} ,

so S_{1} = −(1 − C^{2})C^{−2}Ψ_{3}, and

\frac{(1 − C^{n})(1 − C^{n+2})}{(1 − C^{n+1})(1 − C^{n+3})} − 1 = −\frac{(1 − C)C^{n}}{(1 + C)(1 − C^{n+1})} − \frac{(1 − C^{3})C^{n+1}}{(1 + C)(1 − C^{n+3})},

so

S_{2} = −\frac{1 − C}{1 + C} C^{−1}Ψ_{1} − \frac{1 − C^{3}}{1 + C} C^{−2}Ψ_{3}.

Therefore

S_{1} − S_{2} = \frac{1 − C}{1 + C} C^{−1}Ψ_{1} + (\frac{1 − C^{3}}{1 + C} − (1 − C^{2})) C^{−2}Ψ_{3}

= \frac{1 − C}{1 + C} C^{−1}(Ψ_{1} − Ψ_{3})

= \frac{1 − C}{1 + C} C^{−1} (\frac{C}{1 − C} + \frac{C^{2}}{1 − C^{2}} ) = \frac{1 + 2C}{(1 + C)^{2}} .

Thus (1 + C)^{2}(S_{1} − S_{2}) = 1 + 2C.

By similar reasoning, we could prove Theorem 2.2 directly: The generating function for pairs of paths with heights differing by at most 1 is

∑_{n=0}^{∞}(G_{n} − G_{n−1})(G_{n+1} − G_{n−2}),

where we take G_{−1} = G_{−2} = 0, and a calculation like that in the proof of Theorem 3.2 shows that this is equal to

1 + 2C − C^{2} = 4c(x) − c(x)^{2} − 2 = 4c(x) − \frac{c(x) − 1}{x} − 2

= 1 + ∑_{n=1}^{∞}(4C_{n} − C_{n+1})x^{n} = 1 + ∑_{n=1}^{∞} T (2, n)x^{n}.

Although the fact that the series in Theorem 3.2 telescopes may seem surprising, we shall see in Theorem 3.4 that it is a special case of a very general result on sums of generating functions for Dyck paths with restricted heights.

In the following lemma, the fact that C = c(x) − 1 is not used, and in fact C may be completely arbitrary, as long as the series involved converge.

Lemma 3.3. Let R(z, C) be a rational function of z and C of the form

\frac{zN(z, C)}{∏_{i=1}^{m}(1 − zC^{a_{i}})},

where N(z, C) is a polynomial in z of degree less than m with coefficients that are rational functions of C, and the a_{i} are distinct positive integers. Let L = − lim_{z→∞} R(z, C). Then

∑_{n=0}^{∞} R(C^{n}, C) = Q(C) + LΨ_{1},

where Q(C) is a rational function of C, and Ψ_{1} = ∑_{n=1}^{∞} C^{n}/(1 − C^{n}).

Proof. First we show that the lemma holds for R(z, C) = z/(1 − zC^{a}). In this case, L = − lim_{z→∞} R(z, C) = C^{−a} and

∑_{n=0}^{∞} R(C^{n}, C) = ∑_{n=0}^{∞} \frac{C^{n}}{1 − C^{n+a}} = C^{−a} ∑_{n=a}^{∞} \frac{C^{n}}{1 − C^{n}} = − ∑_{n=0}^{a−1} \frac{C^{n−a}}{1 − C^{n}} + C^{−a}Ψ_{1}.

Now we consider the general case. Since R(z, C)/z is a proper rational function of z, it has a partial fraction expansion

\frac{1}{z}R(z, C) = ∑_{i=1}^{m} \frac{U_{i}(C)}{1 − zC^{a_{i}}}

for some rational functions U_{i}(C), so

R(z, C) = ∑_{i=1}^{m} U_{i}(C)\frac{z}{1 − zC^{a_{i}}} .

The general theorem then follows by applying the special case to each summand.

Theorem 3.4. Let i_{1}, i_{2}, …, i_{m} be distinct integers and let j_{1}, j_{2}, …, j_{m} be distinct integers. Then

∑_{n=0}^{∞} (G_{n+i_{1}}G_{n+i_{2}} ⋯ G_{n+i_{m}} − G_{n+j_{1}}G_{n+j_{2}} ⋯ G_{n+j_{m}})  (3.2)

is a rational function of C.

Proof. By (3.1),

\frac{G_{n+i_{1}}G_{n+i_{2}} ⋯ G_{n+i_{m}}}{(1 + C)^{m}} − 1 = R(C^{n}, C),

where

R(z, C) = \frac{1 − zC^{i_{1}+1}}{1 − zC^{i_{1}+2}} ⋯ \frac{1 − zC^{i_{m}+1}}{1 − zC^{i_{m}+2}} − 1.

Then by Lemma 3.3,

∑_{n=0}^{∞} (\frac{G_{n+i_{1}}G_{n+i_{2}} ⋯ G_{n+i_{m}}}{(1 + C)^{m}} − 1)

is a rational function of C. Similarly,

∑_{n=0}^{∞} (\frac{G_{n+j_{1}}G_{n+j_{2}} ⋯ G_{n+j_{m}}}{(1 + C)^{m}} − 1)

is a rational function of C, and the result follows easily.

4 A combinatorial interpretation for T (3, n)

It is natural to ask whether the super Catalan numbes T (m, n) for m > 2 have combinatorial interpretations similar to that of Theorem 2.2. Using the partial fraction procedure described in the proof of Lemma 3.3, it is straightforward (with the help of a computer algebra system) to evaluate as rational functions of C the sums that count k-tuples of paths with height restrictions when Theorem 3.4 applies. By rationalizing the denominator, we can express any rational function of C in the form

A(x) + B(x)\sqrt{1 − 4x},  (4.1)

where A(x) and B(x) are rational functions. Such a formula can be related to the super Catalan numbers with the help of the following formula.

Lemma 4.1.

(1 − 4x)^{m−1/2} = ∑_{k=0}^{m−1}(−4)^{k}\binom{m − 1/2}{k}x^{k} + 2(−1)^{m} ∑_{n=0}^{∞} T (m, n)x^{m+n}.  (4.2)

Proof. It is easily verified that T (m, n) = \frac{1}{2}(−1)^{n}4^{m+n}\binom{m−1/2}{m+n}. Thus

(1 − 4x)^{m−1/2} = ∑_{k=0}^{m−1}(−4)^{k}\binom{m − 1/2}{k}x^{k} + ∑_{k=m}^{∞}(−4)^{k}\binom{m − 1/2}{k}x^{k}

= ∑_{k=0}^{m−1}(−4)^{k}\binom{m − 1/2}{k}x^{k} + ∑_{n=0}^{∞}(−4)^{m+n}\binom{m − 1/2}{m + n}x^{m+n}

= ∑_{k=0}^{m−1}(−4)^{k}\binom{m − 1/2}{k}x^{k} + 2(−1)^{m} ∑_{n=0}^{∞} T (m, n)x^{m+n}.

Thus if in (4.1) we expand the numerator of B_{1}(x) in powers of 1 − 4x, we will get an expression involving rational functions and generating functions for super Catalan numbers.

For example, the sum

∑_{n=0}^{∞}(G_{n} − G_{n−1})(G_{n+2} − G_{n+1})(G_{n+4} − G_{n+3})  (4.3)

counts triples of paths whose heights are n, n + 2, and n + 4 for some n. It can easily be expressed in terms of sums to which Theorem 3.4 applies, and we find that (4.3) is equal to

\frac{C^{6}(1 + C^{2})(1 + 2C^{2} + C^{4} − C^{5} + C^{6} − 2C^{7} + C^{8})}{(1 + C^{2})^{2}(1 + C + C^{2})(1 + C + C^{2} + C^{3} + C^{4})^{2}}

= \frac{(1 − 3x)(1 − 13x + 63x^{2} − 140x^{3} + 142x^{4} − 56x^{5} + 6x^{6})}{2(1 − x)^{2}(1 − 2x)^{2}(1 − 3x + x^{2})^{2}}

− \frac{(1 − 4x)^{5/2}}{(1 − x)(1 − 2x)(1 − 3x + x^{2})}. (4.4)

The rational functions that appear in (4.4) can be simplified by partial fraction expansion, and we can write down an explicit formula involving T (3, n) for the coefficients of (4.4). What we obtain is far from a combinatorial interpretation of T (3, n), but the computation suggests that perhaps some modification of this set of paths might lead to the desired interpretation. We note also that sums with m paths instead of three empirically give similar expressions with (1 − 4x)^{m−1/2} instead of (1 − 4x)^{5/2}.

Our strategy for finding a combinatorial interpretation for T (3, n) is to consider more general paths that give us counting formulas that generalize (3.1), in the hope that this additional generality may lead us to a combinatorial interpretation for T (3, n). In this we are partially successful: we do find in Theorem 4.3 a set of triples of paths, not too different from the set counted by (4.3) whose generating function is ∑_{n=0}^{∞} T (3, n+1)x^{n} plus a rational function, and moreover we can interpret the coefficients of the rational function in terms of paths. Although the set of paths is not very natural, the result does suggest that there is some hope for this approach to give a nice combinatorial interpretation of T (3, n) and perhaps even for T (m, n).

For our interpretation of T (3, n), we need to consider paths that end at levels greater than 0. Let us define a ballot path to be a path that starts at level 0 and never goes below level 0.

In the previous section all our paths had an even number of steps, so it was natural to assign a path with n steps the weight x^{n/2}. We shall continue to weight paths in this way, even though some of our paths now have odd lengths.

Let G_{k}^{(j)} be the generating function for ballot paths of height at most k that end at level j.

Lemma 4.2. For 0 ≤ j ≤ k + 1 we have

G_{k}^{(j)} = C^{j/2}(1 + C)\frac{1 − C^{k−j+1}}{1 − C^{k+2}}  (4.5)

Proof. The case j = 0 is Lemma 3.1. Now let W be a ballot path counted by G_{k}^{(j)}, where j > 0, so that W is of height at most k and W ends at level j. Then W can be factored uniquely as W_{1}UW_{2}, where W_{1} is a path of height at most k that ends at level 0 and W_{2} is a path from level 1 to level j that never goes above level k nor below level 1. Using \sqrt{x} = \sqrt{C/(1 + C)^{2}} = \sqrt{C}/(1 + C), we obtain

G_{k}^{(j)} = G_{k}\sqrt{x}G_{k−1}^{(j−1)} = (1 + C)\frac{1 − C^{k+1}}{1 − C^{k+2}} · \frac{\sqrt{C}}{1 + C} · G_{k−1}^{(j−1)} = \sqrt{C} \frac{1 − C^{k+1}}{1 − C^{k+2}} G_{k−1}^{(j−1)},

and (4.5) follows by induction on j.

We note an alternative formula that avoids half-integer powers of C,

G_{k}^{(j)} = x^{j/2}(1 + C)^{j+1} \frac{1 − C^{k−j+1}}{1 − C^{k+2}} ,

which follow easily from (4.5) and the formula \sqrt{x} = \sqrt{C}/(1 + C).

There is a similar formula for the generating function G_{k}^{(i,j)} for paths of height at most k that start at level i, end at level j, and never go below level 0: for 0 ≤ i ≤ j ≤ k + 1 we have

G_{k}^{(i,j)} = C^{(j−i)/2}(1 + C)\frac{(1 − C^{i+1})(1 − C^{k−j+1})}{(1 − C)(1 − C^{k+2})},  (4.6)

with G_{k}^{(i,j)} = G_{k}^{(j,i)} for i > j. Although we will not use (4.6) in this paper, it may be helpful in further applications of this method. We have not found (4.5) or (4.6) in the literature, although they may be derived from the known rational generating function for G_{k}^{(i,j)} described below.

We note two variants of (4.6), also valid for 0 ≤ i ≤ j ≤ k + 1:

G_{k}^{(i,j)} = x^{(j−i)/2}(1 + C)^{j−i+1} \frac{(1 − C^{i+1})(1 − C^{k−j+1})}{(1 − C)(1 − C^{k+2})},

= x^{−1/2}C^{(j−i+1)/2} \frac{(1 − C^{i+1})(1 − C^{k−j+1})}{(1 − C)(1 − C^{k+2})}.

It is well known that G_{k}^{(i,j)} is x^{(j−i)/2} times a rational function of x, and it is useful to have an explicit formula for it as a quotient of polynomials. (See Sato and Cong [8] and Krattenthaler [6].) Let us define polynomials p_{n} = p_{n}(x) by

p_{n}(x) = ∑_{0≤k≤n/2}(−1)^{k}\binom{n − k}{k}x^{k}.

The first few values are

p_{0} = 1
p_{1} = 1
p_{2} = 1 − x
p_{3} = 1 − 2x
p_{4} = 1 − 3x + x^{2}
p_{5} = 1 − 4x + 3x^{2}
p_{6} = 1 − 5x + 6x^{2} − x^{3}

These polynomials can be expressed in terms of the Chebyshev polynomials of the second kind U_{n}(x) by

p_{n}(x) = x^{n/2}U_{n} (\frac{1}{2\sqrt{x}}) .

It is not difficult to show that

p_{n} = \frac{1 − C^{n+1}}{(1 − C)(1 + C)^{n}} ,

and thus we obtain

G_{k}^{(i,j)} = x^{(j−i)/2} \frac{p_{i}p_{k−j}}{p_{k+1}} ,

for 0 ≤ i ≤ j ≤ k, and in particular, G_{k}^{(j)} = x^{j/2}p_{k−j}/p_{k+1} and G_{k} = p_{k}/p_{k+1}.

We can now describe our combinatorial interpretation of T (3, n): T (3, n) counts triples of ballot paths whose heights are k, k − 2, and k − 4 for some k, ending at levels 4, 3, and 2, together with some additional paths of height at most 5. (Note that if a path of height k − 4 ends at level 2, then k must be at least 6.) More precisely, let H_{k}^{(j)} be the generating function for ballot paths of height k that end at level j, so that H_{k}^{(j)} = G_{k}^{(j)} − G_{k−1}^{(j)}. Then we have:

Theorem 4.3.

1 + ∑_{n=0}^{∞} T (3, n + 1)x^{n} = \sqrt{x} ∑_{k=6}^{∞} H_{k}^{(4)}H_{k−2}^{(3)}H_{k−4}^{(2)} + 2G_{1} + 2G_{2} + G_{3} + G_{5}.  (4.7)

Proof. With the help of Lemma 4.1 we find that

−\frac{(1 − 4x)^{5/2}}{2x^{4}} − \frac{10}{x} + \frac{15}{x^{2}} − \frac{5}{x^{3}} + \frac{1}{2x^{4}} = ∑_{n=0}^{∞} T (3, n + 1)x^{n}.  (4.8)

Using the method described in Lemma 3.3 and Theorem 3.4, we find, with the help of Maple, that the sum

\sqrt{x} ∑_{k=6}^{∞} H_{k}^{(4)}H_{k−2}^{(3)}H_{k−4}^{(2)}

is equal to

−\frac{(1 − 4x)^{5/2}}{2x^{4}} − \frac{10}{x} + \frac{15}{x^{2}} − \frac{5}{x^{3}} + \frac{1}{2x^{4}} +1− \frac{2}{1 − x} −2\frac{1 − x}{1 − 2x} − \frac{1 − 2x}{1 − 3x + x^{2}} − \frac{1 − 4x + 3x^{2}}{1 − 5x + 6x^{2} − x^{3}} .

Then (4.7) follows from (4.8), the formula G_{k} = p_{k}/p_{k+1}, and the formulas for p_{k}, k = 1, …, 6.

References

[1] N. G. de Bruijn, D. E. Knuth, and S. O. Rice, “The average height of planted plane trees”, in Graph Theory and Computing, R. C. Read, ed., Academic Press, New York-London, 1972, pp. 15–22.

[2] D. Callan, A combinatorial interpretation for a super-Catalan recurrence, J. Integer Seq. 8 (2005), no. 1, Article 05.1.8, 7 pp. (electronic).

[3] E. Catalan, Question 1135, Nouvelles Annales de Mathématiques (2) 13 (1874), 207.

[4] L. E. Dickson, History of the Theory of Numbers, New York: Chelsea. Vol. 1, 1966. Originally published in 1919 by the Carnegie Institute of Washington.

[5] I. Gessel, Super ballot numbers, J. Symb. Comput. 14 (1992), 179–194.

[6] C. Krattenthaler, Permutations with restricted patterns and Dyck paths, Adv. Appl. Math. 27 (2001), 510–530.

[7] N. Pippenger and K. Schleich, Topological characteristics of random surfaces generated

   by cubic interactions, arXiv:physics.gr-qc/0306049.

[8] M. Sato and T. T. Cong, The number of minimal lattice paths restricted by two parallel lines, Disc. Math. 43 (1983), 249–261.

[9] G. Schaeffer, A combinatorial interpretation of super-Catalan numbers of order two,

   http://www.loria.fr/~schaeffe/Pub/Conjugacy/superCat.ps , (2001).

[10] L. W. Shapiro, A short proof of an identity of Touchard’s concerning Catalan numbers, J. Combin. Theory Ser. A 20 (1976), 375–376.

[11] R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 1999.

2000 Mathematics Subject Classification: Primary 05A10; Secondary 05A15. Keywords: Dyck paths, super Catalan numbers

(Concerned with sequences \underline{A007054} and \underline{A007272}.)

Received March 5 2005; revised version received March 16 2005. Published in Journal of Integer Sequences, April 27 2005.

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