A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO

Capacity Calculations

Paul Barry School of Science Waterford Institute of Technology Ireland

                         pbarry@wit.ie

Aoife Hennessy Department of Computing, Mathematics and Physics Waterford Institute of Technology Ireland

                   aoife.hennessy@gmail.com

Abstract We study the Narayana triangles and related families of polynomials. We link this study to Riordan arrays and Hankel transforms arising from a special case of capacity calculation related to MIMO communication systems. A link is established between a channel capacity calculation and a series reversion.

1 Introduction

The Narayana numbers, which are closely related to the ubiquitous Catalan numbers, have an important and growing literature. Their applications are varied. In this note, we look at the mathematics of one application in the area of MIMO (multiple input, multiple output) wireless communication. For our purposes, it is useful to distinguish between three different “Narayana triangles” and their associated “Narayana polynomials”. These triangles are documented separately in Sloane’s Encyclopedia [32] along with other variants. We will find it useful in this note to use the language of Riordan arrays [30] for later sections. In the next section we provide a quick introduction to the Riordan group. We also use the notion of “Deleham array” [3], which is explained in Appendix B. Our approach to Deleham arrays is based on continued fractions [39]. We shall be interested in the Hankel transform [23, 22, 29] of a number of integer sequences that we shall encounter. We recall that if a_{n} is a given sequence, then the sequence with general term given by the determinant |a_{i+j}|_{0≤i,j≤n} is called the Hankel transform of a_{n}. We shall mention well-known orthogonal polynomials in this note. General references are [8, 18]. Links between orthogonal polynomials and Riordan arrays have been studied in [1, 2]. Techniques to calculate Hankel transforms using associated orthogonal polynomials will follow methods to be found in [10, 29].

For an integer sequence a_{n}, that is, an element of ℤ^{ℕ}, the power series f(x) = ∑_{k=0}^{∞} a_{k}x^{k} is called the ordinary generating function or g.f. of the sequence. a_{n} is thus the coefficient of x^{n} in this series. We denote this by a_{n} = [x^{n}]f(x). For instance, F_{n} = [x^{n}]\frac{x}{1−x−x^{2}} is the n-th Fibonacci number \underline{A000045}, while C_{n} = [x^{n}]\frac{1−\sqrt{1−4x}}{2x} is the n-th Catalan number \underline{A000108}. We use the notation 0^{n} = [x^{n}]1 for the sequence 1, 0, 0, 0, …, \underline{A000007}. Thus 0^{n} = [n = 0] = δ_{n,0} = \binom{0}{n}. Here, we have used the Iverson bracket notation [19], defined by [P] = 1 if the proposition P is true, and [P] = 0 if P is false.

For a power series f(x) = ∑_{n=0}^{∞} a_{n}x^{n} with f(0) = 0 we define the reversion or compositional inverse of f to be the power series f̄(x) such that f(f̄(x)) = x. We sometimes write f̄ = Revf.

2 Riordan arrays

The Riordan group [30, 34], 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 [34]. The associated matrix is the matrix whose i-th column is generated by g(x)f(x)^{i} (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. Also called the reversion of f, we will use the notation f̄ = Rev(f) as well.

A Riordan array of the form (g(x), x), where g(x) is the generating function of the sequence a_{n}, is called the sequence array of the sequence a_{n}. Its general term is a_{n−k}. Such arrays are also called Appell arrays as they form the elements of the Appell subgroup.

If M is the matrix (g, f), and a = (a_{0}, a_{1}, …)′ is an integer sequence with ordinary generating function A (x), then the sequence Ma has ordinary generating function g(x)A(f(x)). The (infinite) matrix (g, f) can thus be considered to act on the ring of integer sequences Z^{N} by multiplication, where a sequence is regarded as a (infinite) column vector. We can extend this action to the ring of power series Z[[x]] by

(g, f) : A(x) ⟶ (g, f) · A(x) = g(x)A(f(x)).

Example 1. The binomial matrix B is the element (\frac{1}{1−x}, \frac{x}{1−x}) of the Riordan group. It has general element \binom{n}{k}. More generally, B^{m} is the element (\frac{1}{1−mx}, \frac{x}{1−mx}) of the Riordan group, with general term \binom{n}{k}m^{n−k}. It is easy to show that the inverse B^{−m} of B^{m} is given by (\frac{1}{1+mx}, \frac{x}{1+mx}).

The row sums of the matrix (g, f) have generating function

(g, f) · \frac{1}{1 − x} = \frac{g(x)}{1 − f(x)}

while the diagonal sums of (g, f) have generating function g(x)/(1 − xf(x)).

Many interesting examples of Riordan arrays can be found in Neil Sloane’s On-Line Encyclopedia of Integer Sequences, [32, 33]. Sequences are frequently referred to by their OEIS number. For instance, the matrix B is \underline{A007318}.

3 The Narayana Triangles and their generating functions

In this section, we define four separate though related “Narayana triangles”, and we describe their (bi-variate) generating functions.

The number triangle N_{0} with general term

N_{0}(n, k) = \frac{1}{n + 0^{n}} \binom{n}{k}\binom{n}{k + 1}  (1)

has [40] generating function ϕ_{0}(x, y) which satisfies the equation

xyϕ_{0}^{2} + (x + xy − 1)ϕ_{0} + x = 0.

Solving for ϕ_{0}(x, y) yields

ϕ_{0}(x, y) = \frac{1 − x(1 + y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2xy} .  (2)

This triangle begins

N_{0} = ( \begin{matrix}0 & 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} ) .

The triangle N_{1} with general term

N_{1}(n, k) = 0^{n+k} + \frac{1}{n + 0^{n}} \binom{n}{k}\binom{n}{k + 1} = \frac{1}{k + 1}\binom{n − 1}{k}\binom{n}{k}  (3)

which begins

N_{1} = ( \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} ) ,

clearly has generating function

ϕ_{1}(x, y) = 1 + ϕ_{0}(x, y) = \frac{1 − x(1 − y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2xy} .  (4)

The triangle N_{2}, which is the reversal of N_{1}, has general term

N_{2}(n, k) = [k ≤ n]N_{1}(n, n − k) = 0^{n+k} + \frac{1}{n + 0^{nk}} \binom{n}{k}\binom{n}{k − 1}  (5)
= \frac{1}{n − k + 1}\binom{n − 1}{n − k}\binom{n}{k},  (6)

and begins

N_{2} = ( \begin{matrix}1 & 0 & 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 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This triangle has generating function

ϕ_{2}(x, y) = 1 + yϕ_{0}(x, y) = \frac{1 + x(1 − y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2x} .  (7)

Finally the “Pascal-like” variant N_{3} with general term

N_{3}(n, k) = N_{0}(n + 1, k) = \frac{1}{n + 1}\binom{n + 1}{k}\binom{n + 1}{k + 1}  (8)

which begins

N_{3} = ( \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} ) ,

has generating function

ϕ_{3}(x, y) = \frac{ϕ_{0}(x, y)}{x} = \frac{1 − x(1 + y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2x^{2}y} .  (9)
T riangle A-number  Generating function
N_{1}  \underline{A131198} ϕ_{1}(x, y) = ^{\frac{1−x(1−y)−\sqrt{1−2x(1+y)+x^{2}(1−y)^{2}}}{2xy}}
N_{2}  \underline{A090181} ϕ_{2}(x, y) = ^{\frac{1+x(1−y)−\sqrt{1−2x(1+y)+x^{2}(1−y)^{2}}}{2x}}
N_{3}  \underline{A001263} ϕ_{3}(x, y) = ^{\frac{1−x(1+y)−\sqrt{1−2x(1+y)+x^{2}(1−y)^{2}}}{2x^{2}y}}

4 Narayana Triangles and series reversion

Using the generating functions obtained in the last section, we can relate the Narayana triangles to the process of reverting sequences. Proposition 2. We have

ϕ_{1}(x, y) = \frac{1}{x}Rev_{x} \frac{x(1 − xy)}{1 − x(y − 1)}.  (10)

Proof. We calculate the reversion of the expression

\frac{x(1 − xy)}{1 − x(y − 1)},

considered as a function in x, with parameter y. This amounts to solving the equation

\frac{u(1 − uy)}{1 − u(y − 1)} = x

for the unknown u. We obtain

u = xϕ_{1}(x, y).

Thus we have

ϕ_{1}(x, y) = \frac{1}{x}Rev_{x} \frac{x(1 − xy)}{1 − x(y − 1)},

as required.

In like manner, we have

Proposition 3.

ϕ_{2}(x, y) = \frac{1}{x}Rev_{x} \frac{x(1 − x)}{1 − x(1 − y)}.  (11)

Proof. We calculate the x-reversion of the expression \frac{x(1−x)}{1−x(1−y)}. Thus we wish to solve for u, where

\frac{u(1 − u)}{1 − u(1 − y)} = x.

We obtain

u = xϕ_{2}(x, y).

Thus

ϕ_{2}(x, y) = \frac{1}{x}Rev_{x} \frac{x(1 − x)}{1 − x(1 − y)}.

In similar fashion, we can establish that

Proposition 4.

ϕ_{3}(x, y) = \frac{1}{x}Rev_{x} \frac{x}{1 + (1 + y)x + yx^{2}} .  (12)

We note that the Narayana triangles are not Riordan arrays.

5 The Narayana Triangles and continued fractions

In this section, we develop continued fraction versions for each of the generating functions ϕ_{1}, ϕ_{2}, ϕ_{3}. In this case of ϕ_{3}, we give two distinct (but equivalent) continued fraction expressions.

Proposition 5. [4, Section 3.5] We have the following continued fraction expansion of ϕ_{1}(x, y):

ϕ_{1}(x, y) = \frac{1}{1 − \frac{x}{1 − \frac{xy}{1 − \frac{x}{1 − \frac{xy}{1 − ⋯}}}}}.  (13)

Proof. It is easy to see that ϕ_{1}(x, y) obeys the equation [4]

xyϕ_{1}^{2} − (xy − x + 1)ϕ_{1} + 1 = 0.  (14)

Thus

ϕ_{1}(1 − x − xyϕ_{1}) = 1 − xyϕ_{1}

and hence

ϕ_{1} = \frac{1 − xyϕ_{1}}{1 − xyϕ_{1} − x}
= \frac{1}{1 − \frac{x}{1−xyϕ_{1}}} .

We thus obtain the result that ϕ_{1}(x, y) can be expressed as the continued fraction

ϕ_{1}(x, y) = \frac{1}{1 − \frac{x}{1 − \frac{xy}{1 − \frac{x}{1 − \frac{xy}{1 − ⋯}}}}}.

Corollary 6. We have

N_{1} = [1, 0, 1, 0, 1, 0, 1, …] Δ [0, 1, 0, 1, 0, 1, …].

Proposition 7. We have the following continued fraction expansion of ϕ_{2}(x, y):

ϕ_{2}(x, y) = \frac{1}{1 − \frac{xy}{1 − \frac{x}{1 − \frac{xy}{1 − \frac{x}{1 − ⋯}}}}}.  (15)

Proof. It is easy to establish that

xϕ_{2}^{2} − (1 + x − xy)ϕ_{2} + 1 = 0  (16)

from which we deduce

ϕ_{2}(1 − xϕ_{2} − xy) = 1 − xϕ_{2}

and hence

ϕ_{2} = \frac{1 − xϕ_{2}}{1 − xϕ_{2} − xy}
= \frac{1}{1 − \frac{xy}{1−xϕ_{2}}} .

The result follows from this.

Corollary 8. We have

N_{2} = [0, 1, 0, 1, 0, 1, …] Δ [1, 0, 1, 0, 1, 0, 1, …].

In order to find an expression for ϕ_{3}, we first note that

ϕ_{3} = \frac{ϕ_{1} − 1}{x} ⇒ ϕ_{1} = 1 + xϕ_{3}.

Substituting into Eq. (14) and simplifying, we find that

ϕ_{3}(1 − xy − x^{2}yϕ_{3}) = 1 + xϕ_{3}  (17)

and hence

ϕ_{3} = \frac{1 + xϕ_{3}}{1 − xy(1 + xϕ_{3})}
= \frac{1}{−xy + \frac{1}{1+xϕ_{3}}}
= \frac{1}{−xy + \frac{1}{ϕ_{1}}} .

But

\frac{1}{ϕ_{1}} = 1 − \frac{x}{1 − \frac{xy}{1 − \frac{x}{1 − ⋯}}}.

Hence we obtain: Proposition 9. We have the following continued fraction expansion of ϕ_{3}(x, y):

ϕ_{3}(x, y) = \frac{1}{1 − xy − \frac{x}{1 − \frac{xy}{1 − \frac{x}{1 − ⋯}}}}.  (18)

Corollary 10. We have

N_{3} = [0, 1, 0, 1, 0, 1, …] Δ^{(1)} [1, 0, 1, 0, 1, 0, 1, …].

We end this section by expressing the g.f. of N_{3} in another way. Proposition 11. The generating function of N_{3} has the following continued fraction expression

ϕ_{3}(x, y) = \frac{1}{1 − x − xy − \frac{x^{2}y}{1 − x − xy − \frac{x^{2}y}{1 − x − xy − \frac{x^{2}y}{1 − ⋯}}}}.

Proof. This can be seen by solving the equation

u = \frac{1}{1 − x − xy − x^{2}yu}

and comparing the solution u(x, y) with ϕ_{3}(x, y).

6 Narayana polynomials and moment sequences

To each of the above triangles, there is a family of “Narayana” polynomials [35, 37], where the triangles take on the role of coefficient arrays. Thus we get the polynomials

N_{1,n}(y) = ∑_{k=0}^{n} N_{1}(n, k)y^{k}

N_{2,n}(y) = ∑_{k=0}^{n} N_{2}(n, k)y^{k}

N_{3,n}(y) = ∑_{k=0}^{n} N_{3}(n, k)y^{k}.

Note that since N_{2} is the reversal of N_{1}, we have

N_{2,n}(y) = ∑_{k=0}^{n} N_{2}(n, k)y^{k} = ∑_{k=0}^{n} N_{1}(n, k)y^{n−k}.

Example 12. The first terms of the sequence (N_{1,n}(y))_{n≥0} are:

1, 1, 1 + y, 1 + 3y + y^{2}, 1 + 6y + 6y^{2} + y^{3}, 1 + 10y + 20y^{2} + 10y^{3} + y^{4}, …

Using the results of section 3, we see that

N_{1,n}(y) = [x^{n+1}]Rev_{x} \frac{x(1 − xy)}{1 − (y − 1)x}
N_{2,n}(y) = [x^{n+1}]Rev_{x} \frac{x(1 − x)}{1 − (1 − y)x}
N_{3,n}(y) = [x^{n+1}]Rev_{x} \frac{x}{1 + (1 + y)x + yx^{2}} .

Values of these polynomials are often of significant combinatorial interest. Sample values for these polynomials are tabulated below.

y N_{1,0}(y), N_{1,1}(y), N_{1,2}(y), … A-number  Name
1  1, 1, 2, 5, 14, 42, …  \underline{A000108}  Catalan numbers
2  1, 1, 3, 11, 45, 197 …  \underline{A001003} little Schröder numbers
3  1, 1, 4, 19, 100, 562, …  \underline{A007564}
4 1, 1, 5, 29, 185, 1257, …  \underline{A059231}

y N_{2,0}(y), N_{2,1}(y), N_{2,2}(y), … A-number  Name
1  1, 1, 2, 5, 14, 42, …  \underline{A000108}  Catalan numbers
2  1, 2, 6, 22, 90, 394, …  \underline{A006318} Large Schröder numbers
3 1, 3, 12, 57, 300, 1686, …  \underline{A047891}
4 1, 4, 20, 116, 740, 5028, … \underline{A082298}
y N_{3,0}(y), N_{3,1}(y), N_{3,2}(y), … A-number  Name
1 1, 2, 5, 14, 42, 132, … \underline{A000108(n+1)} shifted Catalan numbers
2 1, 3, 11, 45, 197, 903, … \underline{A001003(n+1)} shifted little Schröder numbers
3 1, 4, 19, 100, 562, 3304, … \underline{A007564(n+1)}
4 1, 5, 29, 185, 1257, 8925, … \underline{A059231(n+1)}

We can derive a moment representation for these polynomials using the generating functions above and the Stieltjes transform. We obtain the following : Proposition 13. The families of polynomials (N_{1,n}(y))_{n≥0}, (N_{2,n}(y))_{n≥0}, (N_{3,n}(y))_{n≥0}, are each a family of moments corresponding to an associated family of orthogonal functions. Proof. Using the established generating functions ϕ_{1}(x, y), ϕ_{2}(x, y) and ϕ_{3}(x, y), and the Stieltjes-Perron transform (see Appendix C), we can establish the following moment representations, for the densities shown.

N_{1,n}(y) = \frac{y − 1}{y}0^{n} + \frac{1}{2π} ∫_{y−2\sqrt{y}+1}^{y+2\sqrt{y}+1} x^{n} \frac{\sqrt{−x^{2} + 2x(1 + y) − (1 − y)^{2}}}{2y} dx,

N_{2,n}(y) = \frac{1}{2π} ∫_{y−2\sqrt{y}+1}^{y+2\sqrt{y}+1} x^{n} \frac{\sqrt{−x^{2} + 2x(1 + y) − (1 − y)^{2}}}{x} dx,

N_{3,n}(y) = \frac{1}{2π} ∫_{y−2\sqrt{y}+1}^{y+2\sqrt{y}+1} x^{n} \frac{\sqrt{−x^{2} + 2x(1 + y) − (1 − y)^{2}}}{y} dx.

The associated orthogonal polynomials are determined by the densities shown.

Using the theory developed in [1, 2], we can exhibit these families of polynomials as the first columns of three related Riordan arrays. More precisely, we have Proposition 14. The elements of the three families of polynomials (N_{1,n}(y))_{n≥0}, (N_{2,n}(y))_{n≥0}, (N_{3,n}(y))_{n≥0} are given by the first column of the inverse Riordan arrays given by (\frac{1}{1+x}, \frac{x}{(1+x)(1+yx)}),

(\frac{1}{1+yx}, \frac{x}{(1+x)(1+yx)}), and (\frac{1}{(1+x)(1+yx)}, \frac{x}{(1+x)(1+yx)}), respectively. These Riordan arrays are the coefficient arrays of the corresponding families of orthogonal polynomials. Thus

N_{1,n}(y) is given by the first column of (\frac{1}{1 + x}, \frac{x}{(1 + x)(1 + yx)})^{−1} ,
N_{2,n}(y) is given by the first column of (\frac{1}{1 + yx}, \frac{x}{(1 + x)(1 + yx)})^{−1}
N_{3,n}(y) is given by the first column of (\frac{1}{(1 + x)(1 + yx)}, \frac{x}{(1 + x)(1 + yx)})^{−1} .

Proof. We look at the case of N_{1,n}, as the other cases are proved in similar manner. Thus we let

(\frac{1}{1 + x}, \frac{x}{(1 + x)(1 + yx)}) = (g, f).

We wish then to show that

ϕ_{1}(x, y) = \frac{1}{g(Rev_{x}f(x, y))}.

For f(x, y) = \frac{x}{(1+x)(1+yx)}, we find that

Rev_{x}f(x, y) = \frac{1 − x(1 + y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2xy} .

Then since g(x) = \frac{1}{1+x}, we find that

\frac{1}{g(Rev_{x}f(x, y))} = 1+Rev_{x}f(x, y) = 1+\frac{1 − x(1 + y) − \sqrt{1 − 2x(1 + y) + x^{2}(1 − y)^{2}}}{2xy} = ϕ_{1}(x, y)

as required. Now

(\frac{1}{1 + x}, \frac{x}{(1 + x)(1 + yx)}) = (\frac{1 + yx}{(1 + x)(1 + yx)}, \frac{x}{(1 + x)(1 + yx)})
= (\frac{1 + yx}{1 + (1 + y)x + yx^{2}} , \frac{x}{1 + (1 + y)x + yx^{2}} ) ,

and hence [2] (\frac{1}{1+x}, \frac{x}{(1+x)(1+yx)}) is the coefficient array of a family of orthogonal polynomials.

7 An investigation inspired by a MIMO application of the Narayana numbers

The role of the Catalan numbers and more recently the Narayana polynomials in the elucidation of the behaviour of certain families of random matrices, along with applications to areas such as MIMO wireless communication, is an active field of research. See for instance [15, 16, 20, 25, 26, 31, 38]. Other areas where Narayana polynomials and their generalizations find applications include that of associahedra [6, 17, 28] and secondary RNA structures [14]. The investigations in this section and those that follow are inspired by MIMO (multiple input, multiple output) data-communication applications in [25] and [38]. The reader is referred to Appendix A for the link with MIMO capacity calculations. We let

G_{β}(z) = −\frac{1}{2} + \frac{β − 1}{2z} + \sqrt{\frac{(1 − β)^{2}}{4z^{2}} + \frac{1}{4} − \frac{1 + β}{2z}} .

In terms of wireless transmission,

β = \frac{T}{R}

where we have T transmit antennas and R receive antennas (see Appendix A). In this section β can be treated as a parameter. Then the function

g_{β}(x) = −\frac{1}{x}G_{β}(\frac{1}{x})

satisfies

g_{β}(x) = \frac{1 + (1 − β)x − \sqrt{1 − 2x(1 + β) + (1 − β)^{2}x^{2}}}{2x}

and generates the sequence

1, β, β(β + 1), β(β^{2} + 3β + 1), β(β^{3} + 6β^{2} + 6β + 1), …

In other words, g_{β}(x) is the generating function of the sequence

a_{n}^{(β)} = ∑_{k=0}^{n} N_{2}(n, k)β^{k} = N_{2,n}(β).

Thus

g_{β}(x) = ϕ_{2}(x, β).

We have the following moment representation:

a_{n}^{(β)} = \frac{1}{2π} ∫_{1+β−2\sqrt{β}}^{1+β+2\sqrt{β}} x^{n} \frac{\sqrt{−x^{2} + 2x(1 + β) − (1 − β)^{2}}}{x} dx

= \frac{1}{2π} ∫_{(1−\sqrt{β})^{2}}^{(1+\sqrt{β})^{2}} x^{n} \frac{\sqrt{((1 − \sqrt{β})^{2} − x)(x − (1 + \sqrt{β})^{2})}}{x}dx

= \frac{\sqrt{β}}{π} ∫_{(1−\sqrt{β})^{2}}^{(1+\sqrt{β})^{2}} x^{n} \frac{\sqrt{1 − (\frac{1+β−x}{2\sqrt{β}})^{2}}}{x} dx

= \frac{\sqrt{β}}{π} ∫_{(1−\sqrt{β})^{2}}^{(1+\sqrt{β})^{2}} x^{n} \frac{w_{U} (\frac{1+β−x}{2\sqrt{β}})}{x}dx

where w_{U}(x) = \sqrt{1 − x^{2}} is the weight function for the Chebyshev polynomials of the second kind. This is an example of the well-known Marčenko-Pastur [24] distribution.

8 Riordan arrays, orthogonal polynomials and N_{2}

We now note that xg_{β}(x) is in fact the series reversion of the function

\frac{x(1 − x)}{1 + (β − 1)x}.

This simple form leads us to investigate the nature of the coefficient array of the orthogonal polynomials P_{n}^{(β)}(x) associated to the weight function

w(x) = \frac{1}{2π} \frac{\sqrt{−x^{2} + 2x(1 + β) − (1 − β)^{2}}}{x} = \frac{1}{2π} \frac{\sqrt{4β − (x − 1 − β)^{2}}}{x} dx

for which the elements

a_{n}^{(β)} = ∑_{k=0}^{n} N_{2}(n, k)β^{k}

are the moments. Put otherwise, these are the family of orthogonal polynomials associated to the Narayana polynomials N_{2,n}. These polynomials can be expressed in terms of the Hankel determinants associated to the sequence a_{n}^{(β)}. We find that the coefficient array of the polynomials P_{n}^{(β)}(x) is given by the Riordan array

(\frac{1}{1 + βx}, \frac{x}{1 + (1 + β)x + βx^{2}} )

whose inverse is given by

L = (g_{β}(x), \frac{g_{β}(x) − 1}{β}) = (ϕ_{2}(x, β), \frac{ϕ_{2}(x, β) − 1}{β}) .

The Jacobi-Stieltjes array [11, 12, 27] for L is found [1, 2] to be

( \begin{matrix}β & 1 & 0 & 0 & 0 & 0 & … \\ β & β + 1 & 1 & 0 & 0 & 0 & … \\ 0 & β & β + 1 & 1 & 0 & 0 & … \\ 0 & 0 & β & β + 1 & 1 & 0 & … \\ 0 & 0 & 0 & β & β + 1 & 1 & … \\ 0 & 0 & 0 & 0 & β & β + 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

indicating that the Hankel transform of the sequence a_{n}^{(β)} is β^{\binom{n+1}{2}}, and that

g_{β}(x) = \frac{1}{1 − βx − \frac{βx^{2}}{1 − (β + 1)x − \frac{βx^{2}}{1 − (β + 1)x − \frac{βx^{2}}{1 − ⋯}}}}.

We note that the coefficient array L^{−1} can be factorized as follows:

L^{−1} = (\frac{1}{1 + βx}, \frac{x}{1 + (1 + β)x + βx^{2}} ) = (1, \frac{x}{1 + x}) · (\frac{1 − x}{1 + (β − 1)x}, \frac{x(1 − x)}{1 + (β − 1)x}) . (19)

Hence

L = (\frac{1 − x}{1 + (β − 1)x}, \frac{x(1 − x)}{1 + (β − 1)x})^{−1} · (1, \frac{x}{1 + x})^{−1}
= (g_{β}(x), xg_{β}(x)) · (1, \frac{x}{1 − x}) .

The general term of the matrix

(\frac{1 − x}{1 + (β − 1)x}, \frac{x(1 − x)}{1 + (β − 1)x})^{−1} = (g_{β}(x), xg_{β}(x))

is given by

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

For instance, when β = 1, which is the case of the matrix (1 − x, x(1 − x))^{−1}, we get the expression

\frac{k + 1}{n + 1}\binom{2n − k}{n − k}

for the general term. Now the general term of the matrix (1, \frac{1}{1−x}) is given by

\binom{n − 1}{k − 1} + 0^{n}(−1)^{k}.

Hence the general term of L is given by

∑_{j=0}^{n} ∑_{i=0}^{n−j} \frac{j + 1}{i + j + 1}\binom{n}{i + j}\binom{n + i}{i}(β − 1)^{n−j−i}(\binom{j − 1}{k − 1} + 0^{j}(−1)^{k}).  (20)

It is interesting to note that

L^{−1} = (\frac{1 + x}{(1 + x)(1 + βx)}, \frac{x}{(1 + x)(1 + βx)}) .  (21)

We can use the factorization in Eq. (19) to express the orthogonal polynomials P_{n}^{(β)}(x) in terms of the Chebyshev polynomials of the second kind U_{n}(x). Thus we recognize [2] that the Riordan array

(\frac{1}{1 + (1 + β)x + βx^{2}} , \frac{x}{1 + (1 + β)x + βx^{2}} )

is the coefficient array of the modified Chebyshev polynomials of the second kind β^{\frac{n}{2}} U_{n}(\frac{x−(β+1)}{2\sqrt{β}}). Hence by the factorization in Eq. (19) we obtain

P_{n}^{(β)}(x) = β^{\frac{n}{2}} U_{n} (\frac{x − (β + 1)}{2\sqrt{β}}) + β^{\frac{n−1}{2}} U_{n−1} (\frac{x − (β + 1)}{2\sqrt{β}}) .  (22)

We state this as a proposition.

Proposition 15. The family {P_{n}(x)} of orthogonal polynomials for which the Narayana polynomials a_{n}^{(β)} = N_{2,n}(x) are moments is given by

P_{n}^{(β)}(x) = β^{\frac{n}{2}} U_{n} (\frac{x − (β + 1)}{2\sqrt{β}}) + β^{\frac{n−1}{2}} U_{n−1} (\frac{x − (β + 1)}{2\sqrt{β}}) .

9 On the Hankel transform of the row sums of L

We recall that

L = (g_{β}(x), \frac{g_{β}(x) − 1}{β})

is the matrix whose first column is given by terms of the Narayana polynomial sequence N_{2,n}(β). We now wish to calculate the Hankel transform of the row sums s_{n}^{β} of the matrix L. The generating function of these row sums is given by

g_{s}(x) = \frac{(β + 1)\sqrt{1 − 2x(β + 1) + (1 − β)^{2}x^{2}} + (β − 1)(x(β + 1) + 1)}{2(1 − 2x(1 + β))}.

We infer from this (using Stieltjes-Perron) that the row sum elements s_{n}^{(β)} are the moments for the following weight function :

w_{s}(x) = \frac{1}{2π} \frac{\sqrt{−x^{2} + 2x(1 + β) − (β − 1)^{2}}(β + 1)}{x(2(1 + β) − x)}

with support on the interval [1 + β − 2\sqrt{β}, 1 + β + 2\sqrt{β}]. From this we can determine that the Hankel transform of s_{n}^{(β)} is given by

(β + 1)^{n}β^{\binom{n}{2}}.

In fact, if we let

H_{s} = L_{s}D_{s}L_{s}^{t}

be the LDU decomposition [27] of the Hankel matrix associated with s_{n}^{(β)}, then we have

L_{s} = (g_{s}(x), \frac{1 − (1 + β)x − \sqrt{1 − 2x(1 + β) + (1 − β)^{2}x^{2}}}{2βx} ) .

Equivalently,

L_{s}^{−1} = (\frac{1 − x^{2}}{1 + (1 + β)x + βx^{2}} , \frac{x}{1 + (1 + β)x + βx^{2}} )

is the coefficient array of the orthogonal polynomials associated to the sequence s_{n}^{(β)}. This is so since the Stieltjes-Jacobi matrix associated to s_{n}^{(β)} takes the form

( \begin{matrix}β + 1 & 1 & 0 & 0 & 0 & 0 & … \\ β + 1 & β + 1 & 1 & 0 & 0 & 0 & … \\ 0 & β & β + 1 & 1 & 0 & 0 & … \\ 0 & 0 & β & β + 1 & 1 & 0 & … \\ 0 & 0 & 0 & β & β + 1 & 1 & … \\ 0 & 0 & 0 & 0 & β & β + 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

10 The Hankel transform of the row sums of (g_{β}(x), xg_{β}(x))

We have seen that

L = (g_{β}(x), xg_{β}(x)) · (1, \frac{x}{1 − x}) .

In this section, we look at the Hankel transform of the row sums of the factor (g_{β}(x), xg_{β}(x)) of L. The row sums in question have generating function

\frac{g_{β}(x)}{1 − xg_{β}(x)} = \frac{1 − (1 + β)x − \sqrt{1 − 2(1 + β)x + (1 − β)^{2}x^{2}}}{2βx^{2}} .

From this we can infer that the row sums of (g_{β}(x), xg_{β}(x)) are the moments for the weight function

w(x) = \frac{1}{2π} \frac{\sqrt{−x^{2} + 2(1 + β)x − (1 − β)^{2}}}{β} .

This then allows us to prove that the Hankel transform sought is β^{\binom{n+1}{2}}.

11 Formulas for the row sums of (g_{β}(x), xg_{β}(x))

We can characterize the row sums of (g_{β}(x), xg_{β}(x)) by observing that

\frac{xg_{β}(x)}{1 − xg_{β}(x)} = \frac{1 − (1 + β)x − \sqrt{1 − 2(1 + β)x + (1 − β)^{2}x^{2}}}{2βx}

is the series reversion of the function

\frac{x}{1 + (1 + β)x + βx^{2}} .

Hence the row sums are given by the (n + 1)-st term of the series reversion of \frac{x}{1+(1+β)x+βx^{2}} . Thus the row sums of (g_{β}(x), xg_{β}(x)) are precisely

N_{3,n}(β) = ∑_{k=0}^{n} N_{3}(n, k)β^{k}.

This may also be expressed by

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

or by

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

We have seen that the general term of the matrix L is given by Eq. (20) and hence the general term s_{n}^{(β)} of the row sums of L is given by

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

12 Narayana polynomials and hypergeometric functions

We recall that the hypergeometric function _{2}F_{1}(α, β; γ; x) is defined by

2F_{1}(α, β; γ; x) = ∑_{k=0}^{∞} \frac{(α)_{k}(β)_{k}}{(γ)_{k}} \frac{x^{k}}{k!} ,

where

(α)_{k} = ∏_{j=0}^{k−1}(α + j).

For n ∈ ℤ, we have (n)_{k} = (−1)^{k}k!\binom{−n}{k}. Thus we have

2F_{1}(−n, −n − 1; 2; x) = ∑_{k=0}^{∞} \frac{(−n)_{k}(−n − 1)_{k}}{(2)_{k}} \frac{x^{k}}{k!}
= ∑_{k=0}^{∞} \frac{(−1)^{k}k!\binom{n}{k}(−1)^{k}k!\binom{n+1}{k}}{(n + 1)!} \frac{x^{k}}{k!}
= ∑_{k=0}^{n} \frac{1}{k + 1}\binom{n}{k}\binom{n + 1}{k}x^{k}
= ∑_{k=0}^{n} N_{3}(n, k)x^{k}
= N_{3,n}(x).

In two recent articles [5, 21], a link between N_{3,n}(x) and the Jacobi polynomials has been established. This is that

N_{3,n}(x) = \frac{1}{n + 1}(1 − x)^{n}P_{n}^{(1,1)} (\frac{1 + x}{1 − x}) .  (23)

Now

P_{n}^{α,β}(x) = \binom{n + α}{n}_{2}F_{1}(−n, n + α + β + 1; α + 1; \frac{1}{2}(1 − x)),

and so

N_{3,n}(x) = (1 − x)^{n} _{2}F_{1} (−n, n + 3; 2; \frac{x}{x − 1}) .  (24)

We can modify Eqn. (23) to obtain the following expression for N_{1,n}(x) :

N_{1,n}(x) = \frac{1 − x · 0^{n}}{n + 0^{n}} (1 − x)^{n−1}P_{n+0^{n}−1}^{(1,1)} (\frac{1 + x}{1 − x}) .

Straight-forward calculation also establishes that

N_{2}(n, k) = [x^{n−k}]_{2}F_{1}(k + 1, k; 2; x).

Note also that the triangle with general term

T (n, k) = [x^{n−k}]_{2}F_{1}(k + 1, k; 1; x)

is the triangle (see \underline{A103371}) with general term

T (n, k) = 0^{n+k} + \binom{n − 1}{k − 1}\binom{n}{k},

which begins

( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & … \\ 0 & 1 & 0 & 0 & 0 & 0 & … \\ 0 & 2 & 1 & 0 & 0 & 0 & … \\ 0 & 3 & 6 & 1 & 0 & 0 & … \\ 0 & 4 & 18 & 12 & 1 & 0 & … \\ 0 & 5 & 40 & 60 & 20 & 1 & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

This matrix and N_{2} are related as follows:

N_{2}(n, k) = \frac{T (n, k)}{(n − k + 1)} = \frac{0^{n+k} + \binom{n−1}{k−1}\binom{n}{k}}{n − k + 1}.

For the matrix N_{1}, we have the following:

N_{1,n}(x) = _{2}F_{1}(−n, −n + 1, 2, x).

Since N_{2} is the reversal of N_{1}, we obtain

N_{2,n}(x) = x^{n} _{2}F_{1}(−n, −n + 1, 2, 1/x).

We can also note the following. We have seen (section 5) that N_{3} has generating function

\frac{1}{1 − x − xy − \frac{x^{2}y}{1 − x − xy − \frac{x^{2}y}{1 − x − xy − \frac{x^{2}y}{1 − ⋯}}}}.

N_{3} is thus seen [3] to be the binomial transform of the array with generating function

\frac{1}{1 − xy − \frac{x^{2}y}{1 − xy − \frac{x^{2}y}{1 − xy − \frac{x^{2}y}{1 − ⋯}}}}.

This array begins

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

This is \underline{A107131}, which is the coefficient array of the polynomials given by

x^{n} _{2}F_{1} (\frac{1}{2} − \frac{n}{2}, −\frac{n}{2}; 2; \frac{4}{x}) .

It also counts the number of Motzkin paths of length n with k steps U = (1, 1) or H = (1, 0). The general term of this array is

[k ≤ n]\binom{n}{2n − 2k}C_{n−k}.

Thus

N_{3}(n, k) = ∑_{j=0}^{n} \binom{n}{j}\binom{j}{2(j − k)}C_{j−k}.  (25)

Since the row sums of N_{3} yield the shifted Catalan numbers, we arrive at the identity

C_{n+1} = ∑_{k=0}^{n} ∑_{j=0}^{n} \binom{n}{j}\binom{j}{2(j − k)}C_{j−k}.  (26)

13 Appendix A - calculation of MIMO capacity

We follow [20] to derive an expression for MIMO capacity in a special case. This is a form of transmission technology which increases the transmission channel capacity by taking advantage of the multipath nature of transmission when many antennas transmit to many receivers. Thus we assume that we have R receive antennas and T transmit antennas, modeled by

r = Hs + n

where r is the receive signal vector, s is the source signal vector, n is an additive white Gaussian noise (AWGN) vector, which is a realization of a complex normal distribution N(0, σ^{2}I_{R}), and the channel is represented by the complex matrix H ∈ ℂ^{R×T}. We have the eigenvalue decomposition

H^{H}H = \frac{1}{T} QΛQ^{H}.

We assume T < R. Then the capacity of the uncorrelated MIMO channels is given by [25]

C_{MIMO} = \frac{1}{R} log_{2} det(I_{T} + H^{H}(σ^{2}I_{R})^{−1}H)
= \frac{1}{R} log_{2} det(I_{T} + \frac{1}{σ^{2}} H^{H}H)
= \frac{1}{R} log_{2} det(I_{T} + \frac{1}{σ^{2}} \frac{1}{T} QΛQ^{H})
= \frac{1}{R} log_{2} det(I_{T} + \frac{1}{σ^{2}T} Λ)
= \frac{T}{R} \frac{1}{T} ∑_{i=1}^{T} log_{2}(1 + \frac{1}{σ^{2}T} λ_{i})

= \frac{β}{ln 2} \frac{1}{T} ∑_{i=1}^{T} ln(1 + \frac{1}{σ^{2}T} λ_{i})

where we have set

β = \frac{T}{R}.

Now

ln(1 + x) = ln(1 + x_{0}) + ∑_{k=1}^{N}(−1)^{k−1} \frac{(x − x_{0})^{k}}{k(1 + x_{0})^{k}} ,  |x − x_{0}| < 1

= ln(1 + x_{0}) + ∑_{k=1}^{N} \frac{(−1)^{k−1}}{k(1 + x_{0})^{k}} ∑_{j=0}^{k} \binom{k}{j}x^{j}(−1)^{k−j}x_{0}^{k−j}

= ln(1 + x_{0}) + ∑_{k=1}^{n} ∑_{j=0}^{k} \binom{k}{j}(−1)^{j−1} \frac{x_{0}^{k−j}}{k(1 + x_{0})^{k}} x^{j}

= ∑_{k=0}^{N} p_{k}x^{k},

where it is appropriate to take x_{0} = \frac{1}{σ^{2}} . We thus obtain

C_{MIMO} = \frac{β}{ln 2} \frac{1}{T} ∑_{i=1}^{T} ∑_{k=0}^{N} p_{k} (\frac{λ_{i}}{σ^{2}T} )^{k}

= \frac{β}{ln 2} ∑_{k=0}^{N} \frac{p_{k}}{(σ^{2}T)^{k}} (\frac{1}{T} ∑_{i=1}^{T} λ_{i}^{k})

= \frac{β}{ln 2} ∑_{k=0}^{N} \frac{p_{k}}{(σ^{2}T)^{k}} m_{k}

= \frac{β}{ln 2} ∑_{k=0}^{N} \frac{p_{k}}{(σ^{2}T)^{k}} ∑_{j=0}^{k} N_{2}(k, j)β^{j}.

Here, we have replaced the expression m_{k} = \frac{1}{T} ∑_{i=1}^{T} λ_{i}^{k} by the k-th moment of the limiting eigenvalue distribution function, which following [31] can to be shown to have Stieltjes transform

G_{β}(z) = −\frac{1}{2} + \frac{β − 1}{2z} + \sqrt{\frac{(1 − β)^{2}}{4z^{2}} + \frac{1}{4} − \frac{1 + β}{2z}} .

Thus by our preceding results (see Sections 3 and 6) we arrive at the new expression

C_{MIMO} = \frac{β}{ln 2} ∑_{k=0}^{N} \frac{p_{k}}{(σ^{2}T)^{k}} [x^{k+1}]Rev_{x} [\frac{x(1 − x)}{1 − (1 − β)x}] .  (27)

14 Appendix B - the Deleham construction

For the purposes of this note, we define the Deleham construction [3] as follows. Given two sequences r_{n} and s_{n},we use the notation

r Δ s = [r_{0}, r_{1}, r_{2}, …] Δ [s_{0}, s_{1}, s_{2}, …]

to denote the number triangle whose bi-variate generating function is given by

\frac{1}{1 − \frac{(r_{0}x + s_{0}xy)}{1 − \frac{(r_{1}x + s_{1}xy)}{1 − \frac{(r_{2}x + s_{2}xy)}{1 − ⋯}}}}.

We furthermore define

r Δ^{(1)} s = [r_{0}, r_{1}, r_{2}, …] Δ^{(1)} [s_{0}, s_{1}, s_{2}, …]

to denote the number triangle whose bi-variate generating function is given by

\frac{1}{1 − (r_{0}x + s_{0}xy) − \frac{(r_{1}x + s_{1}xy)}{1 − \frac{(r_{2}x + s_{2}xy)}{1 − ⋯}}}.

See \underline{A084938} for the original definition.

15 Appendix C - The Stieltjes transform of a measure

The Stieltjes transform of a measure μ on ℝ is a function G_{μ} defined on ℂ \ ℝ by

G_{μ}(z) = ∫_{ℝ} \frac{1}{z − t}μ(t).

If f is a bounded continuous function on ℝ, we have

∫_{ℝ} f(x)μ(x) = − lim_{y→0^{+}} ∫_{ℝ} f(x)ℑG_{μ}(x + iy)dx.

If μ has compact support, then G_{μ} is holomorphic at infinity and for large z,

G_{μ}(z) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} ,

where a_{n} = ∫_{ℝ} t^{n}μ(t) are the moments of the measure. If μ(t) = dψ(t) = ψ′(t)dt then (Stieltjes-Perron)

ψ(t) − ψ(t_{0}) = −\frac{1}{π} lim_{y→0^{+}} ∫_{t_{0}}^{t} ℑG_{μ}(x + iy)dx.

If now g(x) is the generating function of a sequence a_{n}, with g(x) = ∑_{n=0}^{∞} a_{n}x^{n}, then we can define

G(z) = \frac{1}{z}g (\frac{1}{z}) = ∑_{n=0}^{∞} \frac{a_{n}}{z^{n+1}} .

By this means, under the right circumstances we can retrieve the density function for the measure that defines the elements a_{n} as moments.

Example 16. We let g(z) = \frac{1−\sqrt{1−4z}}{2z} be the g.f. of the Catalan numbers. Then

G(z) = \frac{1}{z}g (\frac{1}{z}) = \frac{1}{2} (1 − \sqrt{\frac{x − 4}{x}}) .

Then

ℑG_{μ}(x + iy) = −\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} ,

and so we obtain

ψ′(x) = −\frac{1}{π} lim_{y→0^{+}} \{−\frac{\sqrt{2}\sqrt{\sqrt{x^{2} + y^{2}}\sqrt{x^{2} − 8x + y^{2} + 16} − x^{2} + 4x − y^{2}}}{4\sqrt{x^{2} + y^{2}}} \}

= \frac{1}{2π} \frac{\sqrt{x(4 − x)}}{x}.

16 Acknowledgements

The second-named author was supported by a Council of Directors Strand I grant during this work.

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[37] R. A. Sulanke, Generalizing Narayana and Schröder numbers to higher dimensions, Electron. J. Combin. 11 (2004), #R54.

[38] A. Tulino and S. Verdú, Random Matrix Theory and Wireless Communications, Now Publishers Inc., Hanover, MA, 2004.

[39] H. S. Wall, Analytic Theory of Continued Fractions, AMS Chelsea Publishing, 2000.

[40] D. Zeilberger, Six études in generating functions, Internat. J. Comput. Math., 29 (1989), 201–215.

2000 Mathematics Subject Classification: Primary 11B83; Secondary 05A10, 05A19, 94A05, 94A11. Keywords: Integer sequence, Narayana triangle, Narayana polynomial, Riordan array, Hankel transform, orthogonal polynomials, multiple-input multiple-output (MIMO) systems, Marčenko-Pastur.

(Concerned with sequences \underline{A000007}, \underline{A000045}, \underline{A000108}, \underline{A001003}, \underline{A001263}, \underline{A006318}, \underline{A007318}, \underline{A007564}, \underline{A047891}, \underline{A059231}, \underline{A082298}, \underline{A084938}, \underline{A090181}, \underline{A103371}, \underline{A107131}, and \underline{A131198}.)

Received April 1 2009; revised version received June 30 2010; September 15 2010; March 10 2011. Published in Journal of Integer Sequences, March 26 2011.

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