A Common Generating Function for Catalan Numbers and Other Integer Sequences

G. E. Cossali Università di Bergamo 24044 Dalmine Italy

                       cossali@unibg.it

Abstract Catalan numbers and other integer sequences (such as the triangular numbers) are shown to be particular cases of the same sequence array g(n, m) = \frac{(2n+m)!}{m!n!(n+1)!} . Some features of the sequence array are pointed out and a unique generating function is proposed.

1 Introduction

Catalan numbers can be found in many different combinatorial problems, as shown by Stanley [1], and exhaustive information about this sequence can be found in [2]. In this note I show that the Catalan numbers (A000108) and other known sequences (triangular numbers A000217, A034827, A001700, A002457, A002802, A002803, A007004, A024489) can be derived by the same generating function and are related to the same polynomial set.

2 The polynomials j_{m}(y)

Consider the following recurrence relation defining the polynomials j_{m}(y):

\begin{matrix}j_{0}(y) = 1; \\ j_{m+1} (y) = yj_{m} (y) + ∑_{s=0}^{m} j_{s} (y) j_{m−s} (y) . \end{matrix}  (1)

It may immediately be noticed that for y = 0 this formula coincides with the recursive definition of the Catalan numbers:

C_{s+1} = ∑_{m=0}^{s} C_{m}C_{s−m}  (2)

where

C_{n} = \frac{2n!}{n!(n + 1)!} = \frac{1}{n + 1}\binom{2n}{n}.  (3)

This means that C_{n} is the zero-order coefficient of the nth-order polynomial j_{n} (y), i.e.,

j_{m}(y) = ∑_{q=0}^{m} e(m, q) y^{q}  (4)

and e(m, 0) = C_{m}. The first few values of e(m, q) are shown in Table 1.

m ⧹ q 0 1 2 3 4 5 6 7
0  1
1  1  1
2  2  3  1
3  5 10  6  1
4  14 35 30 10  1
5  42 126 140 70 15 1
6  132 462 630 420 140 21 1
7  429 1716 2772 2310 1050 252 28 1

Table 1: Some of the coefficients e(m, q)

Equation (4) can be introduced into (1) to obtain

∑_{q=0}^{m+1} e(m + 1, q) y^{q} = ∑_{q=0}^{m} e(m, q) y^{q+1} + ∑_{s=0}^{m} ∑_{p=0}^{m−s} e(m − s, p) ∑_{r=0}^{s} e(s, r) y^{p+r}

and transformed as follows: ∑_{q=0}^{m+1} e(m + 1, q) y^{q} = ∑_{q=1}^{m+1} e(m, q − 1) y^{q} + ∑_{s=0}^{m} ∑_{p=0}^{m−s} e(m − s, p) ∑_{r=0}^{s} e(s, r) y^{p+r} =

= ∑_{q=1}^{m+1} e(m, q − 1) y^{q} + ∑_{p=0}^{m} ∑_{r=0}^{m−p} y^{p+r} [∑_{s=r}^{m−p} e(m − s, p) e(s, r)] = = ∑_{q=1}^{m+1} e(m, q − 1) y^{q} + ∑_{p=0}^{m} ∑_{q=p}^{m} y^{q} [∑_{s=q−p}^{m−p} e(m − s, p) e(s, q − p)] = = ∑_{q=1}^{m+1} e(m, q − 1) y^{q} + ∑_{q=0}^{m} ∑_{p=0}^{q} [∑_{s=q−p}^{m−p} e(m − s, p) e(s, q − p)] y^{q} = = ∑_{q=1}^{m+1} e(m, q − 1) y^{q} + ∑_{q=0}^{m} ∑_{p=0}^{q} [∑_{l=q}^{m} e(m − l + p, p) e(l − p, q − p)] y^{q}.

The following set of equations can then be obtained for any natural number m:

(1) for q = 0:

e(m + 1, 0) = ∑_{s=0}^{m} e(m − s, 0) e(s, 0),

whose solution is

e(m, 0) = C_{m} = \frac{1}{m + 1}\binom{2m}{m}.  (5)

(2) for 0 < q ≤ m:

e(m + 1, q) = e(m, q − 1) + ∑_{p=0}^{q} ∑_{l=q}^{m} e(m − l + p, p) e(l − p, q − p).  (6)

(3) for q = m + 1:

e(m + 1, m + 1) = e(m, m) = ⋯ = 1.  (7)

It is useful to introduce the modified matrix g(n, k) defined as follows:

\begin{matrix}g(n, k) = e(n + k, k) \\ e(n, k) = g(n − k, k) \end{matrix}  (8)

Table 2 reports the first few values:

\begin{matrix}m ⧹ q & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 \\ 2 & 2 & 10 & 30 & 70 & 140 & 252 & 420 & 660 \\ 3 & 5 & 35 & 140 & 420 & 1050 & 2310 & 4620 & 8580 \\ 4 & 14 & 126 & 630 & 2310 & 6930 & 18018 & 42042 & 90090 \\ 5 & 42 & 462 & 2772 & 12012 & 42042 & 126126 & 336336 & 816816 \\ 6 & 132 & 1716 & 12012 & 60060 & 240240 & 816816 & 2450448 & 6651216 \\ 7 & 429 & 6435 & 51480 & 291720 & 1312740 & 4988412 & 16628040 & 49884120 \end{matrix}

Table 2: Some values of the coefficients g(m, q)

Equations (5), (6), (7) then become:

(1) for q = 0:

g(n, 0) = C_{n}.  (9)

(2) for 0 < q ≤ n + 1:

g(n + 1, q) = g(n + 1, q − 1) + ∑_{p=0}^{q} ∑_{l=0}^{n} g(n − l, p) q(l, q − p),  (10)

where m − q was replaced by n = m − q. Moreover, from (7) we get g(0, n) = 1. The solution of (10) can be written as follows:

g(n, q) = \frac{(2n + q)!}{q!n!(n + 1)!} = C_{n}\binom{2n + q}{q}.  (11)

In fact, (9) is satisfied, and substituting (11) into (10), we get

C_{n+1}\binom{2(n + 1) + q}{q} = C_{n+1}\binom{2(n + 1) + q − 1}{q − 1} +  (12)

+ ∑_{l=0}^{n} C_{n−l}C_{l} [∑_{p=0}^{q} \binom{2(n − l) + p}{p}\binom{2l + q − p}{q − p}]

It is possible to show that (see appendix)

∑_{p=0}^{q} \binom{2(n − l) + p}{p}\binom{2l + q − p}{q − p} = \binom{2n + q + 1}{q},

and

\begin{matrix}\binom{2(n + 1) + q}{q} − \binom{2(n + 1) + q − 1}{q − 1} = \binom{2(n + 1) + q − 1}{q − 1} [\frac{2(n + 1) + q}{q} − 1] \\ = \{\frac{[2(n + 1) + q − 1]!}{[q − 1]! [2(n + 1)]!}\} [\frac{2(n + 1)}{q}] = \{\frac{(2n + q + 1)!}{q! (2n + 1)!}\} = \binom{2n + q + 1}{q}. \end{matrix}

By using the recursive definition (2) of Catalan numbers, (12) becomes an indentity.

3 Some features of the array g(n, q)

The sequence

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

can be seen as a generalization of the Catalan sequence, as it reduces to the Catalan sequence for q = 0. There are also some other interesting features. In Table 3 I report the known names of the integer sequences, referenced in The On-line Encyclopedia of Integer Sequences [2], that can be extracted from the matrix g(n, q).

A000108 A001700 A002457 A002802 A002803 none
m ⧹ q  0  1  2  3  4  5
-  0  1  1  1  1  1  1
A000217  1  1  3  6  10  15  21
A034827  2  2  10  30  70  140  252
none  3  5  35  140  420  1050 2310
none  4  14  126  630  2310  6930 18018
-  5  42  462  2772  12012  42042 126126
-  6  132  1716  12012  60060 240240 816816

Table 3: g(m, q) numbers and names of known sequences

The first five columns correspond to the known sequences: A000108 (Catalan), A001700, A002457, A002802, A002803. The first two rows correspond to the sequences A000217 (g(1, q), triangular numbers) and A034827. For the other rows and columns no reference was found by the author. Also the sequence on the main diagonal g(k, k) (1,3,30,420,6930,126126,…) is known as A007004 and the sequence on the diagonal g(k, k + 1) (1,6,70,1050,18018, …) is known as A024489.

4 Generating function

Consider the algebraic equation in J:

−xJ^{2} + (1 − yx)J − 1 = 0,  (13)

and its solutions

J(x, y) = \frac{(1 − yx) ± \sqrt{(1 − yx)^{2} − 4x}}{2x}.  (14)

Let now suppose that J(x, y) admits a Taylor expansion in x (which excludes the solution J(x, y) = ^{\frac{(1−yx)+\sqrt{(1−yx)^{2}−4x}}{2x}} unlimited in x = 0)

J(x, y) = ∑_{m=0}^{∞} j_{m} (y) x^{m}.  (15)

Substituting (15) into equation (13) we get 0 = −x ∑_{m=0}^{∞} j_{m} (y) x^{m} ∑_{m=0}^{∞} j_{n} (y) x^{n} + ∑_{m=0}^{∞} j_{m} (y) x^{m} − yx ∑_{m=0}^{∞} j_{m} (y) x^{m} − 1 = = −x ∑_{s=0}^{∞} j_{s} (y) ∑_{m=s}^{∞} j_{s−m} (y) x^{s} + ∑_{m=0}^{∞} j_{m} (y) x^{m} − y ∑_{m=0}^{∞} j_{m} (y) x^{m+1} − 1 = = − ∑_{s=0}^{∞} ∑_{m=0}^{s} j_{s} (y) j_{s−m} (y) x^{s+1} + ∑_{s=0}^{∞} j_{s} (y) x^{s} − y ∑_{s=0}^{∞} j_{s} (y) x^{s+1} − 1 =

= j_{0}(y) − 1 + ∑_{s=0}^{∞} [− ∑_{m=0}^{s} j_{s} (y) j_{s−m} (y) + j_{s+1} (y) − yj_{s−1} (y)] x^{s+1}.

Then

\begin{matrix}j_{0}(y) = 1 \\ j_{s+1} (y) = y j_{s} (y) + ∑_{m=0}^{s} j_{s} (y) j_{s−m} (y) , \end{matrix}  (16)

which is the recursive definition given by (1). This means that

j_{m} (y) = lim_{x→0} \frac{1}{m!} \frac{d^{m}J(x, y)}{dx^{m}},

and for y = 0 the function J(x, 0) is the generating function of the Catalan sequence

\begin{matrix}j_{m}(0) = C_{m} \\ J(x, 0) = \frac{1−\sqrt{1−4x}}{2x} = C_{a}(x). \end{matrix}

Now, using Equations (4) and (15), we get

\begin{matrix}J(x, y) = ∑_{m=0}^{∞} ∑_{q=0}^{m} e(m, q) y^{q}x^{m} = ∑_{q=0}^{∞} ∑_{m=q}^{∞} e(m, q) y^{q}x^{m} = \\ = ∑_{q=0}^{∞} y^{q} ∑_{m=0}^{∞} e(m + q, q) x^{m+q} = ∑_{q=0}^{∞} (yx)^{q} ∑_{m=0}^{∞} g(m, q) x^{m}. \end{matrix}  (17)

Then, the function

L(x, z) = \frac{(1 − z) − \sqrt{(1 − z)^{2} − 4x}}{2x} = J(x, z/x)

can be expanded to get (see equation (17))

L(x, z) = ∑_{q=0}^{∞} ∑_{m=0}^{∞} g(m, q) x^{m}z^{q},  (18)

and this can be seen to be the generating function of g(m, q):

g(m, q) = lim_{x,z→0} \frac{1}{m!q!} \frac{∂^{m+q}L(x, z)}{∂x^{m}∂z^{q}}.

It is interesting to observe that

L(x, z) = ∑_{q=0}^{∞} ∑_{m=0}^{∞} g(m, q) x^{m}z^{q} = ∑_{m=0}^{∞} C_{m} ∑_{q=0}^{∞} \binom{2m + q}{q}z^{q}x^{m} =
= ∑_{m=0}^{∞} C_{m}T_{m}(z)x^{m}

with

T_{m}(z) = ∑_{q=0}^{∞} \binom{2m + q}{q}z^{q}.

It can be proven that

T_{m}(z) = \frac{1}{(1 − z)^{2m+1}}

as

\frac{1}{q!} \frac{d^{q}T_{m}(z)}{dz^{q}} = \frac{(2m + 1) ⋯ (2m + q)}{q! (1 − z)^{2m+1+q}}

and

lim_{z→0} \frac{1}{q!} \frac{d^{q}T_{m}(z)}{dz^{q}} = \frac{(2m + q)!}{q! 2m!} = \binom{2m + q}{q}.

The generating function L(x, z) can then be written also in the form

L(x, z) = \frac{1}{(1 − z)} ∑_{n=0}^{∞} C_{n} [\frac{x}{(1 − z)^{2}} ]^{n} = \frac{1}{(1 − z)}C_{a} [\frac{x}{(1 − z)^{2}} ]

that better shows the strong link existing between the sequence array g(n, q) and the Catalan numbers.

5 Appendix

The following binomial identity:

∑_{p=0}^{q} \binom{m + p}{m}\binom{n + q − p}{n} = \binom{m + n + q + 1}{q}  (19)

holds for any non-negative integers m, n, q. Proof. We define

M(m, n, q) = ∑_{p=0}^{q} \binom{m + p}{m}\binom{n + q − p}{n}.

Then the proposition (19) is equivalent to

M(m, n, q) = \binom{m + n + q + 1}{q}

The proof is based on the use of the binomial identity

∑_{k=r}^{n} \binom{k}{r} = \binom{n + 1}{r + 1}  (20)

that can also be written as

∑_{k=0}^{m} \binom{r + k}{r} = \binom{m + r + 1}{r + 1}.

The identity (19) holds for n = 0 and any m, q. In fact,

M(m, 0, q) = ∑_{p=0}^{q} \binom{m + p}{m}\binom{q − p}{0} = ∑_{p=0}^{q} \binom{m + p}{m} = \binom{m + q + 1}{m + 1} = \binom{m + q + 1}{q},

where the binomial identity (20) was used.

For any n ≠ 0, using (20) again, and with q, m natural numbers,

M(m, n, q) = ∑_{p=0}^{q} \binom{m + p}{m}\binom{n + q − p}{n} = ∑_{p=0}^{q} \binom{m + p}{m}\binom{(n − 1) + q − p + 1}{(n − 1) + 1} =

= ∑_{p=0}^{q} \binom{m + p}{m} ∑_{k=0}^{q−p} \binom{n − 1 + k}{n − 1} = ∑_{k=0}^{q} ∑_{p=0}^{q−k} \binom{m + p}{m}\binom{n − 1 + k}{n − 1} =

= ∑_{k=0}^{q} \binom{q − k + m + 1}{m + 1}\binom{n − 1 + k}{n − 1} = M(m + 1, n − 1, q).

By repeatedly applying the rule M(m, n, q) = M(m + 1, n − 1, q), it is easy to obtain

M(m, n, q) = M(m + n, 0, q) = \binom{m + n + q + 1}{q}.

References

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

[2] N. J. A. Sloane. On-Line Encyclopedia of Integer Sequences. published electronically at

   http://www.research.att.com/∼njas/sequences.

2000 Mathematics Subject Classification: Primary 11B83; Secondary 05A15, 11Y55, 11B65. Keywords: Generating function, Catalan numbers, binomial identity, polynomials

(Concerned with sequences \underline{A000108} \underline{A000217} \underline{A034827} \underline{A001700} \underline{A002802} \underline{A002803} \underline{A007004} \underline{A024489} \underline{A002457}.)

Received October 6, 2002; revised version received April 17, 2003. Published in Journal of Integer Sequences May 2, 2003.

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