On a Family of Generalized Pascal Triangles Defined by Exponential Riordan Arrays

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Abstract

We introduce a family of number triangles defined by exponential Riordan arrays, which generalize Pascal’s triangle. We characterize the row sums and central coefficients of these triangles, and define and study a set of generalized Catalan numbers. We establish links to the Hermite, Laguerre and Bessel polynomials, as well as links to the Narayana and Lah numbers.

1 Introduction

In [1], we studied a family of generalized Pascal triangles whose elements were defined by Riordan arrays, in the sense of [10, 13]. In this note, we use so-called “exponential Riordan arrays” to define another family of generalized Pascal triangles. These number triangles are easy to describe, and important number sequences derived from them are linked to both the Hermite and Laguerre polynomials, as well as being related to the Narayana and Lah numbers.

We begin by looking at Pascal’s triangle, the binomial transform, exponential Riordan arrays, the Narayana numbers, and briefly summarize those features of the Hermite and Laguerre polynomials that we will require. We then introduce the family of generalized Pascal triangles based on exponential Riordan arrays, and look at a simple case in depth. We finish by enunciating a set of general results concerning row sums, central coefficients and generalized Catalan numbers for these triangles.

2 Preliminaries

Pascal’s triangle, with general term C(n, k) = \binom{n}{k}, n, k ≥ 0, has fascinated mathematicians by its wealth of properties since its discovery [6]. Viewed as an infinite lower-triangular matrix, it is invertible, with an inverse whose general term is given by (−1)^{n−k}\binom{n}{k}. Invertibility follows from the fact that \binom{n}{n} = 1. It is centrally symmetric, since by definition, \binom{n}{k} = \binom{n}{n−k}. All the terms of this matrix are integers.

By a generalized Pascal triangle we shall understand a lower-triangular infinite integer matrix T = T (n, k) with T (n, 0) = T (n, n) = 1 and T (n, k) = T (n, n − k). We index all matrices in this paper beginning at the (0, 0)-th element.

We shall encounter transformations that operate on integer sequences during the course of this note. An example of such a transformation that is widely used in the study of integer sequences is the so-called Binomial transform [17], which associates to the sequence with general term a_{n} the sequence with general term b_{n} where

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}.  (1)

If we consider the sequence with general term a_{n} to be the vector a = (a_{0}, a_{1}, …) then we obtain the binomial transform of the sequence by multiplying this (infinite) vector by the lower-triangle matrix B whose (n, k)-th element is equal to \binom{n}{k}:

B = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 2 & 1 & 0 & 0 & 0 & … \\ 1 & 3 & 3 & 1 & 0 & 0 & … \\ 1 & 4 & 6 & 4 & 1 & 0 & … \\ 1 & 5 & 10 & 10 & 5 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

This transformation is invertible, with

a_{n} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}b_{k}.  (2)

We note that B corresponds to Pascal’s triangle. Its row sums are 2^{n}, while its diagonal sums are the Fibonacci numbers F (n + 1). If B^{m} denotes the m−th power of B, then the n−th term of B^{m}a where a = \{a_{n}\} is given by ∑_{k=0}^{n} m^{n−k}\binom{n}{k}a_{k}.

If A(x) is the ordinary generating function of the sequence a_{n}, then the ordinary generating function of the transformed sequence b_{n} is \frac{1}{1−x}A(\frac{x}{1−x}). Similarly, if G(x) is the exponential generating function (e.g.f.) of the sequence a_{n}, then the exponential generating function of the binomial transform of a_{n} is exp(x)G(x).

The binomial transform is an element of the exponential Riordan group, which can be defined as follows.

The exponential Riordan group [5], is a set of infinite lower-triangular integer matrices, where each matrix is defined by a pair of generating functions g(x) = 1 + g_{1}x + g_{2}x^{2} + … and f(x) = f_{1}x + f_{2}x^{2} + … where f_{1} ≠ 0. The associated matrix is the matrix whose k-th column has exponential generating function g(x)f(x)^{k}/k! (the first column being indexed by 0). The matrix corresponding to the pair f, g is denoted by (g, f) or R(g, f). The group law is then given by

(g, f) ∗ (h, l) = (g(h ∘ f), l ∘ f).

The identity for this law is I = (1, x) and the inverse of (g, f) is (g, f)^{−1} = (1/(g ∘ f̄), f̄) where f̄ is the compositional inverse of f.

If M is the matrix (g, f), and a = {a_{n}} is an integer sequence with exponential generating function A (x), then the sequence Ma has exponential generating function g(x)A(f(x)).

We note at this juncture that the exponential Riordan group, as well as the group of ‘standard’ Riordan arrays [10] can be cast in the more general context of matrices of type R^{q}(α_{n}, β_{k}; ϕ, f, ψ) as found in [7, 8, 9]. Specifically, a matrix C = (c_{nk})_{n,k=0,1,2,...} is of type R^{q}(α_{n}, β_{k}; ϕ, f, ψ) if its general term is defined by the formula

c_{nk} = \frac{β_{k}}{α_{n}} res_{x}(ϕ(x)f^{k}(x)ψ^{n}(x)x^{−n+qk−1})

where res_{x}A(x) = a_{−1} for a given formal power series A(x) = ∑_{j} a_{j}x^{j} is the formal residue of the series.

For the exponential Riordan arrays in this note, we have α_{n} = \frac{1}{n!}, β_{k} = \frac{1}{k!}, and q = 1.

Example 1. The Binomial matrix B is the element (e^{x}, x) of the exponential Riordan group. More generally, B^{m} is the element (e^{mx}, x) of the Riordan group. It is easy to show that the inverse B^{−m} of B^{m} is given by (e^{−mx}, x).

Example 2. The exponential generating function of the row sums of the matrix (g, f) is obtained by applying (g, f) to e^{x}, the e.g.f. of the sequence 1, 1, 1, . . .. Hence the row sums of (g, f) have e.g.f. g(x)e^{f(x)}.

We shall frequently refer to sequences by their sequence number in the On-Line Encylopedia of Integer Sequences [11, 12]. For instance, Pascal’s triangle is \underline{A007318} while the Catalan numbers [18] C(n) = \binom{2n}{n}/(n + 1) are \underline{A000108}.

Example 3. An example of a well-known centrally symmetric invertible triangle is the Narayana triangle Ñ, [14, 15], defined by

Ñ(n, k) = \frac{1}{k + 1}\binom{n}{k}\binom{n + 1}{k} = \frac{1}{n + 1}\binom{n + 1}{k + 1}\binom{n + 1}{k}

for n, k ≥ 0. Other expressions for Ñ(n, k) are given by

Ñ(n, k) = \binom{n}{k}^{2} − \binom{n}{k + 1}\binom{n}{k − 1} = \binom{n + 1}{k + 1}\binom{n}{k} − \binom{n + 1}{k}\binom{n}{k + 1}.

This triangle begins

Ñ = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 1 & 6 & 6 & 1 & 0 & 0 & … \\ 1 & 10 & 20 & 10 & 1 & 0 & … \\ 1 & 15 & 50 & 50 & 15 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

Note that in the literature, it is often the triangle Ñ(n−1, k−1) = \frac{1}{n} \binom{n}{k}\binom{n}{k−1} that is referred to as the Narayana triangle. Alternatively, the triangle Ñ(n−1, k) = \frac{1}{k+1} \binom{n−1}{k}\binom{n}{k} is referred to as the Narayana triangle. We shall denote this latter triangle by N(n, k). We then have

N = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 1 & 6 & 6 & 1 & 0 & 0 & … \\ 1 & 10 & 20 & 10 & 1 & 0 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

with row sums equal to the Catalan numbers C(n).

Note that for n, k ≥ 1, N(n, k) = \frac{1}{n} \binom{n}{k}\binom{n}{k+1}. We have, for instance,

Ñ(n−1, k−1) = \frac{1}{n}\binom{n}{k}\binom{n}{k − 1} = \binom{n}{k}^{2}−\binom{n − 1}{k}\binom{n + 1}{k} = \binom{n}{k}\binom{n − 1}{k − 1}−\binom{n}{k − 1}\binom{n − 1}{k}.

The last expression represents a 2 × 2 determinant of adjacent elements in Pascal’s triangle. The Narayana triangle is \underline{A001263}.

The Hermite polynomials H_{n}(x) [19] are defined by

H_{n}(x) = (−1)^{n}e^{x^{2}} \frac{d^{n}}{dx^{n}} e^{−x^{2}}.

They obey H_{n}(−x) = (−1)^{n}H_{n}(x) and can be defined by the recurrence

H_{n+1}(x) = 2xH_{n}(x) − 2nH_{n−1}(x).  (3)

They have a generating function given by

e^{2tx−x^{2}} = ∑_{n=0}^{∞} \frac{H_{n}(t)}{n!}x^{n}.

A property that is related to the binomial transform is the following:

∑_{k=0}^{n} \binom{n}{k}H_{k}(x)(2z)^{n−k} = H_{n}(x + z).

From this, we can deduce the following proposition.

Proposition 4. For fixed x and y ≠ 0, the binomial transform of the sequence n → H_{n}(x)y^{n} is the sequence n → y^{n}H_{n}(x + \frac{1}{2y}).

Proof. Let z = \frac{1}{2y}. Then 2z = \frac{1}{y} and hence

∑_{k=0}^{n} \binom{n}{k}H_{k}(x)(y)^{k−n} = H_{n}(x + \frac{1}{2y}).

That is,

∑_{k=0}^{n} \binom{n}{k}H_{k}(x)y^{k} = y^{n}H_{n}(x + \frac{1}{2y})

as required.

The Laguerre polynomials L_{n}(x) [20] are defined by

L_{n}(x) = \frac{e^{x}}{n!} \frac{d^{n}}{dx^{n}} x^{n}e^{−x}.

They have generating function

\frac{exp(−\frac{tx}{1−x})}{1 − x} = ∑_{n=0}^{∞} \frac{L_{n}(t)}{n!}x^{n}.

They are governed by the following recurrence relationship:

(n + 1)L_{n+1}(t) = (2n + 1 − t)L_{n}(t) − nL_{n−1}(t)  (4)

3 Introducing the family of centrally symmetric invertible triangles

We recall that the Binomial matrix B, or Pascal’s triangle, is the element (e^{x}, x) of the Riordan group. For a given integer r, we shall denote by B_{r} the element (e^{x}, x(1 + rx)) of the Riordan group. We note that B = B_{0}. We can characterize the general element of B_{r} as follows.

Proposition 5. The general term B_{r}(n, k) of the matrix B_{r} is given by

B_{r}(n, k) = \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!}.

Proof. We have

B_{r}(n, k) = \frac{n!}{k!}[x^{n}](e^{x}(x(1 + rx)^{k})
= \frac{n!}{k!}[x^{n}] ∑_{i=0}^{∞} \frac{x^{i}}{i!} x^{k} ∑_{j=0}^{k} \binom{k}{j}r^{j}x^{j}

= \frac{n!}{k!}[x^{n−k}] ∑_{i=0}^{∞} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{i!} x^{i+j}

= \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!}.

From the above expression we can easily establish that B_{r}(n, k) = B_{r}(n, n − k) and B_{r}(n, 0) = B_{r}(n, n) = 1.

An alternative derivation of these results can be obtained be observing that the matrix B_{r} may be defined as the array R^{1}(\frac{1}{n!}, \frac{1}{k!}; e^{x}, (1 + rx), 1). Then we have

B_{r}(n, k) = \frac{1/k!}{1/n!}res_{x}(e^{x}(1 + rx)^{k}x^{−n+k−1})

= \frac{n!}{k!}res_{x}(∑_{i}^{∞} \frac{x^{i}}{i!} ∑_{j}^{k} \binom{k}{j}r^{j}x^{j}x^{−n+k−1})

= \frac{n!}{k!}res_{x}(∑_{i}^{∞} ∑_{j}^{k} \binom{k}{j}\frac{r^{j}}{i!} x^{i+j−n+k−1})

= \frac{n!}{k!} ∑_{j}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!}.

Thus B_{r} is a centrally symmetric lower-triangular matrix with B_{r}(n, 0) = B_{r}(n, n) = 1. In this sense B_{r} can be regarded as a generalized Pascal matrix. Note that by the last property, this matrix is invertible.

Proposition 6. The inverse of B_{r} is the element (e^{−u}, u) of the Riordan group, where

u = \frac{\sqrt{1 + 4rx} − 1}{2r}.

Proof. Let (g^{∗}, f̄) be the inverse of (e^{x}, x(1 + rx)). Then

(g^{∗}, f̄)(e^{x}, x(1 + rx)) = (1, x) ⇒ f̄(1 + rf̄) = x.

Solving for f̄ we get

f̄ = \frac{\sqrt{1 + 4rx} − 1}{2r}.

But g^{∗} = \frac{1}{g∘f̄} = e^{−f̄}.

This result allows us to easily characterize the row sums of the inverse B_{r}^{−1}. Corollary 7. The row sums of the inverse triangle B_{r}^{−1} are given by 0^{n} = 1, 0, 0, 0, . . .. Proof. We have B_{r}^{−1} = (e^{−u}, u) as above. Hence the e.g.f. of the row sums of B_{r}^{−1} is e^{−u}e^{u} = 1. The result follows from this. Example 8. B_{1} = (e^{x}, x(1 + x)) is given by

B_{1} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 4 & 1 & 0 & 0 & 0 & … \\ 1 & 9 & 9 & 1 & 0 & 0 & … \\ 1 & 16 & 42 & 16 & 1 & 0 & … \\ 1 & 25 & 130 & 130 & 25 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

The row sums of B_{1} are

1, 2, 6, 20, 76, 312, 1384, 6512, 32400, …

or \underline{A000898}.

From the above, the terms of this sequence are given by

s_{1}(n) = ∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{1}{(n − k − j)!}.

with e.g.f. g(x)e^{f(x)} = e^{x}e^{x(1+x)} = e^{2x+x^{2}}. What is less evident is that

s_{1}(n) = H_{n}(−i)i^{n}

where i = \sqrt{−1}. This follows since

e^{2x+x^{2}} = e^{2(−i)(ix)−(ix)^{2}}
= ∑_{n=0} \frac{H_{n}(−i)}{n!}(ix)^{n}
= ∑_{n=0} \frac{H_{n}(−i)i^{n}}{n!} x^{n}

and hence e^{2x+x^{2}} is the e.g.f. of H_{n}(−i)i^{n}. We therefore obtain the identity

H_{n}(−i)i^{n} = ∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{1}{(n − k − j)!}.

We can characterize the row sums of B_{1} in terms of the diagonal sums of another related special matrix. For this, we recall [16] that

Bessel(n, k) = \frac{(n + k)!}{2^{k}(n − k)!k!} = \binom{n + k}{2}(2k − 1)!!

defines the triangle \underline{A001498} of coefficients of Bessel polynomials that begins

Bessel = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 3 & 0 & 0 & 0 & … \\ 1 & 6 & 15 & 15 & 0 & 0 & … \\ 1 & 10 & 45 & 105 & 105 & 0 & … \\ 1 & 15 & 105 & 420 & 945 & 945 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

We then have

Proposition 9. The row sums of the matrix B_{1} are equal to the diagonal sums of the matrix with general term Bessel(n, k)2^{n}. That is

H_{n}(−i)i^{n} = ∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{1}{(n − k − j)!} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k)2^{n−k}.

Proof. We shall prove this in two steps. First, we shall show that

∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k)2^{n−k} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \frac{(2k)!}{k!}\binom{n}{2k}2^{n−2k} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋}(2k − 1)!!\binom{n}{2k}2^{n−k}.

We shall then show that this is equal to H_{n}(−i)i^{n}. Now

∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k)2^{n−k} = ∑_{k=0}^{n} Bessel(n − \frac{k}{2}, \frac{k}{2})2^{n− \frac{k}{2}} (1 + (−1)^{k})/2

= ∑_{k=0}^{n} \frac{(n − \frac{k}{2} + \frac{k}{2})!2^{n− \frac{k}{2}}}{2^{\frac{k}{2}} (n − \frac{k}{2} − \frac{k}{2})!(\frac{k}{2})!}(1 + (−1)^{k})/2

= ∑_{k=0}^{n} \frac{n!}{(n − k)!} \frac{2^{n−k}}{(\frac{k}{2})!} (1 + (−1)^{k})/2

= ∑_{k=0}^{n} \frac{k!}{(\frac{k}{2})!}\binom{n}{k}2^{n−k}(1 + (−1)^{k})/2

= ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \frac{(2k)!}{k!}\binom{n}{2k}2^{n−2k}

= ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \frac{(2k)!}{2^{k}k!}\binom{n}{2k}2^{n−k}

= ∑_{k=0}^{⌊ \frac{n}{2} ⌋}(2k − 1)!!\binom{n}{2k}2^{n−k}.

establishes the first part of the proof. The second part of the proof is a consequence of the following more general result, when we set a = 2 and b = 1.

Proposition 10. The sequence with e.g.f. e^{ax+bx^{2}} has general term u_{n} given by

u_{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}a^{n−2k}b^{k} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}C(k)(k + 1)!a^{n−2k}b^{k}.

Proof. We have

n![x^{n}]e^{ax+bx^{2}} = n![x^{n}]e^{ax}e^{bx^{2}}
= n![x^{n}] ∑_{i=0}^{∞} \frac{a^{i}x^{i}}{i!} ∑_{k=0}^{∞} \frac{b^{k}x^{2k}}{k!}
= n![x^{n}] ∑_{i=0}^{∞} ∑_{k=0}^{∞} \frac{a^{i}b^{k}}{i!k!} x^{i+2k}
= n! ∑_{k=0}^{∞} \frac{a^{n−2k}b^{k}}{(n − 2k)!k!}
= ∑_{k=0}^{∞} \frac{n!}{(n − 2k)!(2k)!} \frac{(2k)!}{k!}a^{n−2k}b^{k}
= ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}a^{n−2k}b^{k}.

Corollary 11.

H_{n}(−\frac{a}{2\sqrt{b}}i)(\sqrt{b}i)^{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}a^{n−2k}b^{k}.

Corollary 12. Let u_{n} be the sequence with e.g.f. e^{ax+bx^{2}}. Then u_{n} satisfies the recurrence

u_{n} = au_{n−1} + 2(n − 1)bu_{n−2}

with u_{0} = 1, u_{1} = a. Proof. Equation 3 implies that

H_{n}(x) = 2xH_{n−1}(x) − 2(n − 1)H_{n−2}(x).

Thus

H_{n}(−\frac{a}{2\sqrt{b}}i) = −2\frac{a}{2\sqrt{b}}iH_{n−1}(−\frac{a}{2\sqrt{b}}i) − 2(n − 1)H_{n−2}(−\frac{a}{2\sqrt{b}}i).

Now multiply both sides by (\sqrt{b}i)^{n} to obtain

u_{n} = au_{n−1} + 2(n − 1)bu_{n−2}.

Since

u_{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}a^{n−2k}b^{k}

we obtain the initial values u_{0} = 1, u_{1} = a.

Corollary 13. The binomial transform of ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k} \frac{(2k)!}{k!}a^{n−2k}b^{k} is given by

∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}(a + 1)^{n−2k}b^{k}.

Proof. The e.g.f. of the binomial transform of the sequence with e.g.f. e^{ax+cx^{2}} is e^{x}e^{ax+bx^{2}} = e^{(a+1)x+bx^{2}}.

Equivalently, the binomial transform of ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}C(k)(k + 1)!a^{n−2k}b^{k} is given by

∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}C(k)(k + 1)!(a + 1)^{n−2k}b^{k}.

We note that in [1], it was shown that the binomial transform of ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}C(k)a^{n−2k}b^{k} is given by

∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}C(k)(a + 1)^{n−2k}b^{k}.

Corollary 14. The row sums of B_{1} satisfy the recurrence equation

u_{n} = 2u_{n−1} + 2(n − 1)u_{n−2}

with u_{0} = 1, u_{1} = 2.

We can use Proposition 4 to study the inverse binomial transform of s_{1}(n). By that proposition, the inverse binomial transform of H_{n}(−i)i^{n} is given by i^{n}H_{n}(−i + \frac{1}{2i}) = H_{n}(−\frac{i}{2})i^{n}. This is the sequence

1, 1, 3, 7, 25, 81, 331, 1303, 5937, …

with e.g.f. e^{x+x^{2}}. This is \underline{A047974} which satisfies the recurrence a_{n} = a_{n−1} +2(n−1)a_{n−2}. It is in fact equal to ∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k)2^{k}. The second inverse binomial transform of s_{1}(n) is the sequence

1, 0, 2, 0, 12, 0, 120, 0, 1680, 0, 30240, …

with e.g.f. e^{x^{2}}. This is an “aerated” version of the quadruple factorial numbers C(n)(n+1)! = \frac{(2n)!}{n!}, or \underline{A001813}.

We now look at the central coefficients B_{1}(2n, n) of B_{1}. We have

B_{1}(2n, n) = \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j)!}

= C(n)(n + 1)! ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j)!}

= C(n)(n + 1) ∑_{j=0}^{n} \binom{n}{j}^{2}j!
= C(n)(n + 1)!L_{n}(−1).

Hence

\frac{B_{1}(2n, n)}{C(n)(n + 1)!} = L_{n}(−1).

We note that this is the rational sequence 1, 2, \frac{7}{2}, \frac{17}{3}, \frac{209}{24}, . . .. Two other ratios are of interest.

1. \frac{B_{1}(2n,n)}{C(2n,n)} = n!L_{n}(−1) is \underline{A002720}. It has e.g.f. \frac{1}{1−x} exp(\frac{x}{1−x}). It is equal to the number of partial permutations of an n-set, as well as the number of matchings in the bipartite graph K(n, n). Using Equation (4) we can show that these numbers obey the following recurrence:

u_{n} = 2nu_{n−1} − (n − 1)^{2}u_{n−2}

with u_{0} = 1, u_{1} = 2.

2. \frac{B_{1}(2n,n)}{C(n)} = (n + 1)!L_{n}(−1) is \underline{A052852}(n + 1). It has e.g.f. given by

\frac{d}{dx} \frac{x}{1 − x} exp(\frac{x}{1 − x}) = \frac{1}{(1 − x)^{3}} exp(\frac{x}{1 − x}).

Again using Equation (4) we can show that these numbers obey the following recurrence:

v_{n} = 2(n + 1)v_{n−1} − (n^{2} − 1)v_{n−2}

with v_{0} = 1, v_{1} = 4.

This sequence counts the number of (121, 212)-avoiding n-ary words of length n. Specifically,

\frac{B_{1}(2n, n)}{C(n)} = f_{121,212}(n + 1, n + 1)

where

f_{121,212}(n, k) = ∑_{j=0}^{k} \binom{k}{j}\binom{n − 1}{j − 1}j!

is defined in [4].

From this last point, we find the following expression

B_{1}(2n, n) = C(n) ∑_{j=0}^{n+1} \binom{n + 1}{j}\binom{n}{j − 1}j!  (5)

Based on the fact that C(n) = \binom{2n}{n} − \binom{2n}{n−1} we define

C_{1}(n) = B_{1}(2n, n) − B_{1}(2n, n − 1) = B_{1}(2n, n) − B_{1}(2n, n + 1)

to be the generalized Catalan numbers associated with the triangle B_{1}. We calculate B_{1}(2n, n− 1) as follows:

B_{1}(2n, n − 1) = \frac{(2n)!}{(n − 1)!} ∑_{j=0}^{n−1} \binom{n − 1}{j}\frac{1}{(n − j + 1)!}

= \frac{(2n)!}{(n − 1)!} ∑_{j=0}^{n−1} \binom{n}{j}\frac{n − j}{n} \frac{1}{(n − j + 1)!}

= \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j)!} \frac{n − j}{n − j + 1}.

Hence

B_{1}(2n, n) − B_{1}(2n, n − 1) = \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j)!}(1 − \frac{n − j}{n − j + 1})

= \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j)!} \frac{1}{n − j + 1}

= \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j + 1)!}.

Starting from the above, we can find many expressions for C_{1}(n). For example,

C_{1}(n) = \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{1}{(n − j + 1)!}

= C(n) ∑_{j=0}^{n} \binom{n}{j}\frac{(n + 1)!}{(n + 1 − j)!}

= C(n) ∑_{j=0}^{n} \binom{n}{j}\binom{n + 1}{j}j!

= C(n) ∑_{j=0}^{n} \binom{n}{j}^{2} \frac{n + 1}{n − j + 1}j!

= C(n) ∑_{j=0}^{n} \binom{n}{j}\binom{n + 1}{j + 1}\frac{(j + 1)!}{n − j + 1}.

where we have used the fact that \frac{(2n)!}{n!} = C(n)(n + 1)!. This is the sequence \underline{A001813} of quadruple factorial numbers with e.g.f. \frac{1}{\sqrt{1−4x}}.

Recognizing that the terms after C(n) represent convolutions, we can also write

C_{1}(n) = C(n) ∑_{j=0}^{n} \binom{n}{j}\frac{(n + 1)!}{(j + 1)!}

= C(n) ∑_{j=0}^{n} \binom{n}{j}\binom{n + 1}{j + 1}(n − j)!

= C(n) ∑_{j=0}^{n} \binom{n}{j}^{2} \frac{n + 1}{j + 1}(n − j)!

We note that the first expression immediately above links C_{1}(n) to the Lah numbers \underline{A008297}. The ratio \frac{C_{1}(n)}{C(n)}, or ∑_{j=0}^{n} \binom{n}{j} \frac{(n+1)!}{(k+1)!}, is the sequence

1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, …

or \underline{A000262}(n + 1). This is related to the number of partitions of [n] = \{1, 2, 3, …, n\} into any number of lists, where a list means an ordered subset. It also has applications in quantum physics [2]. The sequence has e.g.f.

\frac{d}{dx}e^{\frac{x}{1−x}} = \frac{e^{\frac{x}{1−x}}}{(1 − x)^{2}} .

We can in fact describe this ratio in terms of the Narayana numbers Ñ(n, k) as follows:

\frac{C_{1}(n)}{C(n)} = ∑_{j=0}^{n} \binom{n}{j}\binom{n + 1}{j + 1}(n − j)!

= ∑_{j=0}^{n} \frac{n − j + 1}{n + 1}\binom{n + 1}{j}\binom{n + 1}{j + 1}(n − j)!

= ∑_{j=0}^{n} \frac{1}{n + 1}\binom{n + 1}{j}\binom{n + 1}{j + 1}(n − j + 1)!

= ∑_{j=0}^{n} Ñ(n, j)(n − j + 1)!

= ∑_{j=0}^{n} Ñ(n, n − j)(j + 1)!

= ∑_{j=0}^{n} Ñ(n, j)(j + 1)!

Hence we have

\frac{C_{1}(n)}{C(n)} = ∑_{j=0}^{n} Ñ(n, j)(n − j + 1)! = ∑_{j=0}^{n} Ñ(n, j)(j + 1)! = ∑_{j=0}^{n} \binom{n}{j}\frac{(n + 1)!}{(j + 1)!}

4 The General Case

We shall now look at the row sums, central coefficients and generalized Catalan numbers associated with the general matrix B_{r}. In what follows, proofs follow the methods developed in the last section. Proposition 15. The row sums s_{r}(n) of B_{r} are given by H_{n}(−\frac{i}{\sqrt{r}})(\sqrt{r}i)^{n}. Proof. The row sums of B_{r} are given by the sequence

∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!}

with e.g.f. g(x)e^{f(x)} = e^{x}e^{x(1+rx)} = e^{2x+rx^{2}}. Now

e^{2x+rx^{2}} = e^{2( \frac{−i}{\sqrt{r}})(i\sqrt{r}x)−(i\sqrt{r}x)^{2}}
= ∑_{n=0}^{∞} \frac{H_{n}(−\frac{i}{\sqrt{r}})}{n!}(i\sqrt{r}x)^{n}
= ∑_{n=0}^{∞} \frac{H_{n}(−\frac{i}{\sqrt{r}})(i\sqrt{r})^{n}}{n!} x^{n}

Corollary 16. We have the identity

∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!} = H_{n}(−\frac{i}{\sqrt{r}})(\sqrt{r}i)^{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}2^{n−2k}r^{k}.

As before, we can rewrite this using the fact that \frac{(2k)!}{k!} = C(k)(k + 1)! = 2^{k}(2k − 1)!!. We note that the second inverse binomial transform of s_{r}(n) has e.g.f. e^{rx^{2}}.

Proposition 17. The row sums of B_{r} are equal to the diagonal sums of the matrix with general term Bessel(n, k)2^{n}r^{k}. That is,

∑_{k=0}^{n} \frac{n!}{k!} ∑_{j=0}^{k} \binom{k}{j}\frac{r^{j}}{(n − k − j)!} = H_{n}(−\frac{i}{\sqrt{r}})(\sqrt{r}i)^{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k)2^{n−k}r^{k}.

Proposition 18. The row sums of B_{r} obey the recurrence

u_{n} = 2u_{n−1} + 2r(n − 1)u_{n−2}

with u_{0} = 1, u_{1} = 2.

We now turn our attention to the central coefficients of B_{r}.

Proposition 19. B_{r}(2n, n) = C(n)(n + 1)! ∑_{j=0}^{n} \binom{k}{j}^{2}j!r^{j}

Proof. The proof is the same as the calculation for B_{1}(2n, n) in Example 8, with the extra factor of r^{j} to be taken into account.

Corollary 20.

\frac{B_{r}(2n, n)}{C(n)(n + 1)!} = r^{n}L_{n}(−\frac{1}{r}) = ∑_{j=0}^{n} \binom{n}{j}\frac{r^{j}}{(n − j)!}

for r ≠ 0.

We note that the above expressions are not integers in general.

For instance, \frac{B_{2}(2n,n)}{C(2n,n)} = n!2^{n}L_{n}(−\frac{1}{2}) is \underline{A025167}, and \frac{B_{3}(2n,n)}{C(2n,n)} = n!3^{n}L_{n}(−\frac{1}{3}) is \underline{A102757}. In general, we have

Proposition 21. \frac{B_{r}(2n,n)}{C(2n,n)} = n!r^{n}L_{n}(−1/r) has e.g.f. \frac{1}{1−rx} exp(\frac{x}{1−rx}), and satisfies the recurrence relation

u_{n} = ((2n − 1)r + 1)u_{n−1} − r^{2}(n − 1)^{2}u_{n−2}

with u_{0} = 1, u_{1} = r + 1.

Proof. We have

n![x^{n}]\frac{e^{\frac{x}{1−rx}}}{(1 − rx)} = n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!} \frac{x^{i}}{(1 − rx)^{i}} (1 − rx)^{−1}

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!}x^{i}(1 − rx)^{−i}(1 − rx)^{−1}

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!}x^{i}(1 − rx)^{−(i+1)}

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!}x^{i} ∑_{j=0} \binom{−(i + 1)}{j}(−1)^{j}r^{j}x^{j}

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!} ∑_{j=0} \binom{i + j}{j}r^{j}x^{i+j}

= n! ∑_{j=0}^{n} \binom{n}{j}\frac{r^{j}}{(n − j)!}.

To prove the second assertion, we use Equation (4) with t = −\frac{1}{r}. Multiplying by n!r^{n+1}, we obtain

(n + 1)!r^{n+1}L_{n+1}(−\frac{1}{r}) = (2n + 1 + \frac{1}{r})r^{n+1}n!L_{n}(−\frac{1}{r}) − r^{n+1}n^{2}(n − 1)!L_{n−1}(−\frac{1}{r}).

Simplifying, and letting n → n − 1, gives the result.

Corollary 22. \frac{B_{r}(2n,n)}{C(n)} = (n + 1)!r^{n}L_{n}(−1/r) has e.g.f. \frac{d}{dx} \frac{x}{1−rx} exp(\frac{x}{1−rx}), and satisfies the recurrence

w_{n} = ((2n − 1) + r)\frac{n + 1}{n}w_{n−1} − r^{2}(n^{2} − 1)w_{n−2}

for n > 1, with w_{0} = 1 and w_{1} = 2r + 2.

We can generalize Equation (5) to get

B_{r}(2n, n) = C(n) ∑_{j=0}^{n+1} \binom{n + 1}{j}\binom{n}{j − 1}j!r^{j−1}.

We define the generalized Catalan numbers associated with the triangles B_{r} to be the numbers

C_{r}(n) = B_{r}(2n, n) − B_{r}(2n, n − 1).

Using the methods of Example 8, we have Proposition 23. We have the following equivalent expressions for C_{r}(n):

C_{r}(n) = \frac{(2n)!}{n!} ∑_{j=0}^{n} \binom{n}{j}\frac{r^{j}}{(n − j + 1)!}
= C(n) ∑_{j=0}^{n} \binom{n}{j}\frac{(n + 1)!}{(j + 1)!}r^{n−j}
= C(n) ∑_{j=0}^{n} \binom{n}{j}\binom{n + 1}{j + 1}(n − j)!r^{n−j}
= C(n) ∑_{j=0}^{n} \frac{n + 1}{j + 1}\binom{n}{j}^{2}(n − j)!r^{n−j}
= C(n) ∑_{j=0}^{n} Ñ(n, j)(j + 1)!r^{j}.

For instance, C_{2}(n)/C(n) is \underline{A025168}. Proposition 24. \frac{C_{r}(n)}{C(n)} has e.g.f.

\frac{d}{dx}e^{\frac{x}{1−rx}} = \frac{e^{\frac{x}{1−rx}}}{(1 − rx)^{2}} .

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

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!}x^{i}(1 − rx)^{−i}(1 − rx)^{−2}

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

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!}x^{i} ∑_{j=0}^{∞} \binom{−(i + 2)}{j}(−1)^{j}r^{j}x^{j}

= n![x^{n}] ∑_{i=0}^{∞} \frac{1}{i!} ∑_{j=0}^{∞} \binom{i + j + 1}{j}r^{j}x^{i+j}

= n! ∑_{j=0}^{∞} \binom{n + 1}{j}\frac{r^{j}}{(n − j)!}

= (n + 1)! ∑_{j=0}^{n} \binom{n}{j}\frac{r^{j}}{(n − j + 1)!}.

5 The case r = \frac{1}{2}

The assumption so far has been that r is an integer. In this section, we indicate that r = \frac{1}{2} also produces a generalized Pascal triangle. We have B_{\frac{1}{2}} = (e^{x}, x(1 + x/2)). This begins

B_{\frac{1}{2}} = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 1 & 1 & 0 & 0 & 0 & 0 & … \\ 1 & 3 & 1 & 0 & 0 & 0 & … \\ 1 & 6 & 6 & 1 & 0 & 0 & … \\ 1 & 10 & 21 & 10 & 1 & 0 & … \\ 1 & 15 & 55 & 55 & 15 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

This is triangle \underline{A100862}. Quoting from \underline{A100862}, B_{\frac{1}{2}} (n, k) “is the number of k-matchings of the corona K′(n) of the complete graph K(n) and the complete graph K(1); in other words, K′(n) is the graph constructed from K(n) by adding for each vertex v a new vertex v′ and the edge vv′”. The row sums of this triangle, \underline{A005425}, are given by

1, 2, 5, 14, 43, 142, 499, 1850, 7193, …

These have e.g.f. e^{2x+x^{2}/2} and general term

H_{n}(−\sqrt{2}i)(i/\sqrt{2})^{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} \binom{n}{2k}\frac{(2k)!}{k!}2^{n−3k}.

They obey the recurrence

u_{n} = 2u_{n−1} + (n − 1)u_{n−2}

with u_{0} = 1, u_{1} = 2.

[3] provides an example of their use in quantum physics. Using Proposition 4 or otherwise, we see that the inverse binomial transform of this sequence, with e.g.f. e^{x+x^{2}/2}, is given by

H_{n}(−\sqrt{2}i + \frac{i}{\sqrt{2}})(i/\sqrt{2})^{n} = H_{n}(−\frac{i}{\sqrt{2}})(i/\sqrt{2})^{n}.

This is the sequence

1, 1, 2, 4, 10, 26, 76, 232, 765, …

or \underline{A000085}. It has many combinatorial interpretations, including for instance the number of matchings in the complete graph K(n). These numbers are the diagonal sums of the Bessel triangle Bessel:

H_{n}(−\frac{i}{\sqrt{2}})(i/\sqrt{2})^{n} = ∑_{k=0}^{⌊ \frac{n}{2} ⌋} Bessel(n − k, k).

The row sums of B_{\frac{1}{2}} are the second binomial transform of the sequence

1, 0, 1, 0, 3, 0, 15, 0, 105, 0, …

with e.g.f. e^{x^{2}/2}. This is an “aerated” version of the double factorial numbers (2n − 1)!!, or \underline{A001147}. These count the number of perfect matchings in the complete graph K(2n). The row sums count the number of 12 − 3 and 214 − 3-avoiding permutations, as well as the number of matchings of the corona K′(n) of the complete graph K(n) and the complete graph K(1).

This example prompts us to define a new family B̃_{r} where B̃_{r} is the element (e^{x}, x(1+ \frac{rx}{2})) of the exponential Riordan group. Then we have B̃_{0} = B, B̃_{1} = B_{\frac{1}{2}} , B̃_{2} = B_{1} etc. We can then show that B̃_{r} is the product of the binomial matrix B and the matrix with general term Bessel(k, n − k)r^{n−k}. We have

B̃_{r}(n, k) = \frac{n!}{k!} ∑_{j=0}^{k} \frac{1}{2^{j}} \binom{n}{k}\frac{r^{j}}{(n − k − j)!} = ∑_{j=0}^{n} \binom{n}{j}\frac{j!r^{j−k}}{(2k − j)!2^{j−k}(j − k)!}.

Thus

B̃_{r}(n, k) = \binom{n}{k} ∑_{j=0}^{n} \binom{n − k}{n − j}\frac{k!}{(2k − j)!} \frac{r^{j−k}}{2^{j−k}}

and in particular

B̃_{r}(2n, n) = \binom{2n}{n} ∑_{j=0}^{2n} \binom{n}{j − n}\frac{n!}{(2n − j)!} \frac{r^{j−n}}{2^{j−n}} .

Finally,

B̃_{r}(n, k) = B̃_{r}(n − 1, k − 1) + B̃_{r}(n − 1, k) + r(n − 1)B̃_{r}(n − 2, k − 1).

6 Conclusion

The foregoing has shown that the triangles B_{r}, and more generally B̃_{r}, defined in terms of exponential Riordan arrays, are worthy of further study. Many of the sequences linked to them have significant combinatorial interpretations. B_{\frac{1}{2}} as documented in \underline{A100862} by Deutsch has a clear combinatorial meaning. This leaves us with the challenge of finding combinatorial interpretations for the general arrays B̃_{r}, r ∈ Z.

7 Acknowledgments

The author wishes to thank an anonymous referee for their careful reading of the first version of this paper, and for their pertinent comments and suggestions.

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2000 Mathematics Subject Classification: Primary 11B83; Secondary 05A19, 33C45, 11B37, 11B65. Keywords: Pascal’s triangle, Narayana numbers, Catalan numbers, Lah numbers, Hermite polynomials, Laguerre polynomials, Bessel polynomials.

(Concerned with sequences \underline{A000085}, \underline{A000108}, \underline{A000262}, \underline{A000898}, \underline{A001147}, \underline{A001263}, \underline{A001498}, \underline{A001813}, \underline{A002720}, \underline{A005425}, \underline{A007318}, \underline{A008297}, \underline{A025167}, \underline{A025168}, \underline{A047974}, \underline{A052852}, \underline{A100862}, and \underline{A102757}.)

Received January 16 2006; revised version received March 27 2007. Published in Journal of Integer Sequences, March 28 2007.

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