Table of Contents
ISRN Discrete Mathematics
Volume 2011, Article ID 534628, 8 pages
Research Article

Some Combinatorial Interpretations and Applications of Fuss-Catalan Numbers

Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan

Received 1 August 2011; Accepted 25 August 2011

Academic Editors: L. Ji and K. Ozawa

Copyright © 2011 Chin-Hung Lin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Fuss-Catalan number is a family of generalized Catalan numbers. We begin by two definitions of Fuss-Catalan numbers and some basic properties. And we give some combinatorial interpretations different from original Catalan numbers. Finally we generalize the Jonah's theorem as its applications.

1. Introduction

Catalan numbers {𝑐𝑛}𝑛0 [1] are said to be the sequence satisfying the recursive relation𝑐𝑛+1=𝑐0𝑐𝑛+𝑐1𝑐𝑛1++𝑐𝑛𝑐0,𝑐0=1.(1.1) It is well known that the 𝑛th term of Catalan numbers is 𝑐𝑛=(1/(𝑛+1))𝑛2𝑛=(1/(2𝑛+1))𝑛2𝑛+1 and {𝑐𝑛}𝑛0={1,1,2,5,14,42,132,}. Also, one of many combinatorial interpretations of Catalan numbers is that 𝑐𝑛 is the number of shortest lattice paths from (0,0) to (𝑛,𝑛) on the 2-dimensional plane such that those paths lie beneath the line 𝑦=𝑥.

On the other hand, Fuss-Catalan numbers {𝑐𝑛(𝑠)}𝑠,𝑛0 were investigated by Fuss [2] and studied by several authors [1, 37]. Hence we have the following proposition.

Proposition 1.1. If 𝑛 and 𝑠 are nonnegative integers, the following statements are equivalent: (1)𝑐𝑛(𝑠)=1𝑛𝑠𝑛+1𝑠𝑛+1;(1.2)(2)𝑐(𝑠)𝑛+1=𝑟1+𝑟2++𝑟𝑠=𝑛𝑐𝑟(𝑠)1×𝑐𝑟(𝑠)2××𝑐𝑟(𝑠)𝑠,𝑐0(𝑠)=1;(1.3)(3)𝑐𝑛(𝑠) is the number of shortest lattice paths from (0,0) to (𝑛,(𝑠1)𝑛) on the 2-dimensional plane such that those paths lie beneath 𝑦=(𝑠1)𝑥.

It is easy to see that in the case when 𝑠=2, the sequence of Catalan numbers {𝑐𝑛} is a special case of the family of Fuss-Catalan numbers {𝑐𝑛(2)}. Although Fuss-Catalan numbers could be viewed as one kind of generalized Catalan numbers, Fuss finished this work many years before Catalan [8].

The proposition describing Fuss-Catalan numbers could be restated in the language of generating functions.

Proposition 1.2. The generating function 𝐶(𝑠)(𝑥)=𝑛0𝑐𝑛(𝑠)𝑥𝑛 satisfies the equation 𝑥(𝐴(𝑠))𝑠(𝑥)=𝐴(𝑠)(𝑥)1, where 𝐴(𝑥) is a generating function. That is, 𝑥𝐶(𝑠)𝑠(𝑥)=𝐶(𝑠)(𝑥)1.(1.4)

There are many combinatorial interpretations of Fuss-Catalan numbers, but most of them are similar to that of Catalan numbers. In order to demonstrate the importance of Fuss-Catalan number, in Section 2 we tried to find some combinatorial interpretations which is different from original Catalan numbers.

Finally, Hilton and Pedersen [9] generalized an identity called Jonah's theorem which involves Catalan numbers. So in Section 3 we restated the identity in Jonah's theorem in the form of Fuss-Catalan numbers.

2. Some Other Interpretations

It is remarkable that the interpretation in Proposition 1.1 illustrates the relation between paths in an 𝑛×𝑛 square and Catalan numbers. It is reasonable to consider whether Fuss-Catalan numbers are relevant to paths in an 𝑛×𝑛×𝑛 cube. As the cube in Figure 1, consider the shortest path in it from (0,0,0) to (𝑛,𝑛,𝑛). There are (3𝑛)!/𝑛!𝑛!𝑛! paths. But it is notable that (3𝑛)!/𝑛!𝑛!𝑛! could be also written as𝑛𝑛𝑛𝑛3𝑛2𝑛=(2𝑛+1)𝑐𝑛(3)×(𝑛+1)𝑐𝑛(2).(2.1)

Figure 1: An 𝑛×𝑛×𝑛 cube.

Maybe by giveing some constraints, shown in Figure 2, the number of paths will be precisely 𝑐𝑛(3)×𝑐𝑛(2). And here are some results, which consider a more general case on an 𝑛×𝑛×(𝑠1)𝑛 cuboid.

Figure 2: Constrained regions in Corollary 2.2 and Theorem 2.1.

Theorem 2.1. From (0,0,0) to (𝑛,𝑛,(𝑠1)𝑛) and under the following constraints: 1(𝑠1)𝑦𝑧0,𝑥s1𝑦s𝑧0,(2.2) there are 𝑐𝑛(𝑠+1)×𝑐𝑛(𝑠) shortest paths.

Proof. Let 𝑃 be a shortest path constrained by the conditions in Theorem 2.1. First consider the projection of 𝑃 on the 𝑦𝑧-plane. The projection could be thought as a shortest path in an 𝑛×(𝑠1)𝑛 right triangle in Proposition 1.1. So there are 𝑐𝑛(𝑠) ways to decide a path on this triangle. Fix one path and use this path to cut the cuboid with the positive direction of the 𝑥-axis. The graph is like a ladder in the right side of Figure 3 and could be put on a plane, and then it becomes an 𝑛×𝑠𝑛 right triangle. So in this situation, there are 𝑐𝑛(𝑠+1) ways to be chosen. Finally since we may choose the paths in the 𝑛×(𝑠1)𝑛 triangle and that in the 𝑛×𝑠𝑛 triangle independently, there are 𝑐𝑛(𝑠+1)×𝑐𝑛(𝑠) paths totally.

Figure 3: Auxiliary graphs for Theorem 2.1.

Corollary 2.2. From (0,0,0) to (𝑛,𝑛,𝑛) and under the following constraints: 1𝑦𝑧0,𝑥21𝑦2𝑧0,(2.3) there are 𝑐𝑛(3)×𝑐𝑛(2) shortest paths.

Proof. This is a special case of Theorem 2.1.

If now we loosen the condition “the base of the cuboid must be square”, we can get some more general result. And the proof of this theorem is similar to that of Theorem 2.1.

Theorem 2.3. Let 𝑚, 𝑛, 𝑠, 𝑡 be positive integers and 𝑠𝑛=(𝑡1)𝑚. From (0,0,0) to (𝑚,𝑛,(𝑠1)𝑛) and under the following constraints: 𝑚(𝑠1)𝑦𝑧0,𝑥𝑚𝑠𝑛𝑦𝑠𝑛𝑧0,(2.4) there are 𝑐𝑚(𝑡)×𝑐𝑛(𝑠) shortest paths.

3. Generalized Jonah's Theorem

Jonah's theorem [9] is the identity𝑚=𝑛+1𝑖0𝑐𝑖𝑛2𝑖𝑚𝑖,𝑛0,𝑚0,(3.1) where 0 is the set of nonnegative integers and 𝑐𝑖 is the 𝑖th term of Catalan numbers. Hilton and pedersen [9] proved the new identity𝑚=𝑛+1𝑖0𝑐𝑖𝑛2𝑖𝑚𝑖,𝑛,𝑚0,(3.2) where is the set of real numbers. The theorem is proven by lattice paths. Here we try to generalize the identity (3.2) as follows showing the connection with Fuss-Catalan numbers 𝑐𝑛(𝑠):𝑚=𝑛+1𝑖0𝑐𝑖(𝑠)𝑛𝑠𝑖𝑚𝑖,𝑛,𝑚0.(3.3)

The following lemmas will be needed to prove the identity (3.3).

Lemma 3.1. For any generating function 𝑓(𝑥) with 𝑓(0)=0, the equation 𝑓(𝑥)𝐵𝑠(𝑥)=𝐵(𝑥)1(3.4) has at most one solution of generating function (abbreviated SGF). That is, if 𝑓(𝑥) is a generating function with 𝑓(0)=0, there is at most one generating function 𝑔(𝑥) satisfying 𝑓(𝑥)𝑔𝑠(𝑥)=𝑔(𝑥)1.(3.5)

Proof. The cases 𝑠=0 and 𝑠=1 is easy since (3.4) could be solved immediately. So we assume that 𝑠2. If 𝑔(𝑥) is one SGF of (3.4), we have the identity 𝑓𝐵𝑠𝐵+1=(𝐵𝑔)𝑓𝐵𝑠1+𝑓𝑔𝐵𝑠2++𝑓𝑔𝑠2𝐵+𝑓𝑔𝑠1,1(3.6) where 𝑓, 𝐵, and 𝑔 are the abbreviations of 𝑓(𝑥), 𝐵(𝑥), and 𝑔(𝑥). Since the ring of formal power series is an integral domain, if 𝐵(𝑥) has any SGF other than 𝑔(𝑥) then the second term in the right hand side must be identically zero. However, if 𝐵(𝑥) is a generating function, we have 𝐵(0)<+ and so 𝑓(0)𝐵𝑖(0)=𝑓(0)𝑔𝑠1(0)=0 for all proper integer 𝑖. Since the second term on the right equals −1 but not 0 as 𝑥=0, it could not be identically zero. That is, 𝑔(𝑥) is the only SGF.

Lemma 3.2. Let 𝑝(𝑥)=1+𝑥,𝑞(𝑥)=𝐶(𝑠)(𝑥(1+𝑥)𝑠)=𝑖0𝑐𝑖(𝑠)𝑥𝑖(1+𝑥)𝑖𝑠,(3.7) where 𝐶(𝑠) is the generating function of 𝑐𝑛(𝑠) where 𝑠 is fixed. Then for 𝑓(𝑥)=𝑥(1+𝑥)𝑠, both 𝑝(𝑥) and 𝑞(𝑥) are SGFs of (3.4). Hence 𝑝(𝑥)=𝑞(𝑥). That is, 1+𝑥=𝑖0𝑐𝑖(𝑠)𝑥𝑖(1+𝑥)𝑖𝑠.(3.8)

Proof. Naturally 𝑝(𝑥) is a generating function; 𝑞(𝑥) is also a generating function since it is the linear combinartion of power of the generating function 𝑓(𝑥).
Observe that
(1)𝑥(1+𝑥)𝑠𝑝𝑠(𝑥)=𝑥(1+𝑥)𝑠(1+𝑥)𝑠=𝑥=𝑝(𝑥)1; (2)𝑥(1+𝑥)𝑠𝑞𝑠(𝑥)=𝑥(1+𝑥)𝑠(𝐶(𝑠)(𝑥(1+𝑥)𝑠))𝑠=𝐶(𝑠)(𝑥(1+𝑥)𝑠)1=𝑞(𝑥)1, since 𝑥𝐶(𝑠)(𝑥)𝑠=𝐶(𝑠)(𝑥)1(3.9) by Proposition 1.2. So both 𝑝(𝑥) and 𝑞(𝑥) are SGFs of (3.4). Finally by Lemma 3.1, there is at most one SGF. Hence 𝑝(𝑥)=𝑞(𝑥).

Theorem 3.3. For any real number 𝑛 and integer 𝑚, the following identity holds: 𝑚=𝑛+1𝑖0𝑐𝑖(𝑠)𝑛𝑠𝑖𝑚𝑖,𝑛,𝑚0.(3.10)

Proof. By multiplying both sides of (3.8) by (1+𝑥)𝑛, we have (1+𝑥)𝑛+1=𝑐0(𝑠)(1+𝑥)𝑛+𝑐1(𝑠)𝑥(1+𝑥)𝑛𝑠++𝑐𝑚(𝑠)𝑥𝑚(1+𝑥)𝑛𝑚𝑠+.(3.11) Then we get (3.10) by comparing the coefficients.

The following are the special cases of Theorem 3.3:

(i) 𝑠=0Pascal's theorem𝑚=𝑛𝑚+𝑛𝑛+1𝑚1,(3.12)

(ii) 𝑠=1Chu Shih-Chieh's theorem𝑚=𝑛𝑚+0𝑛+1𝑛1𝑚1++𝑛𝑚.(3.13)

(iii) 𝑠=2Jonah's theorem.

On the other hand, even when 𝑚𝑛+1 the identity holds.

Example 3.4. Recall that (𝑎𝑎)=1 and (𝑎𝑏)=0 when 𝑏>𝑎.
(i) When 𝑠=3, 𝑛=4, 𝑚=5, 551==𝑐0(3)45+𝑐1(3)14+𝑐2(3)32+𝑐3(3)25+𝑐4(3)18+𝑐5(3)011=1×0+1×0+3×(4)+12×15+55×(8)+273×1=1.(3.14)
(ii) When 𝑠=3, 𝑛=3, 𝑚=5, 450==𝑐0(3)35+𝑐1(3)04+𝑐2(3)33+𝑐3(3)26+𝑐4(3)19+𝑐5(3)012=1×0+1×0+3×(10)+12×21+55×(9)+273×1=0.(3.15)

Note 1. Theorem 3.3 can also be proved by lattice paths when 𝑛 is nonnegative integer and 𝑛𝑚+1(𝑠1)𝑚 (see Figure 4).
Consider the number of shortest path from (0,0) to (𝑚,𝑛𝑚+1), which is 𝑚𝑛+1. On the other hand, consider the auxiliary line 𝐿𝑦=(𝑠1)𝑥. Then every path must pass through 𝐿 in order to reach the ending point (𝑚,𝑛𝑚+1). So we can classify all the paths by the points they pass 𝐿 for the “first time”. Then there are 𝑐𝑖(𝑠)𝑛𝑠𝑖𝑚𝑖 paths passing by point (𝑖,(𝑠1)𝑖), because before (𝑖,(𝑠1)𝑖) the path lies beneath 𝐿, and thus there are 𝑐𝑖(𝑠) ways; after (𝑖,(𝑠1)𝑖) the path must go upward to (𝑖,(𝑠1)𝑖+1) and then finally reach (𝑚,𝑛𝑚+1) without any constraints, and thus there are 𝑛𝑠𝑖𝑚𝑖 ways. So the total number of paths will be the summation of that of each point.

Figure 4: The lattice for the proof of Theorem 3.3.


This paper was written in the Summer Research Program of Taiwan Academia Sinica in 2009. The author would like to thank Taiwan Academia Sinica for providing the financial support. He would like to thank Professor Peter Shiue very much for several useful ideas, suggestions, and comments to improve this paper. He would also like to thank Professor Sen-Peng Eu for encouragements and giving advices.


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