\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.9in}

\newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}}



\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}


\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{xca}[theorem]{Exercise}
\newtheorem{examp}[theorem]{Example}
\newenvironment{example}{\begin{examp}\normalfont\quad}{\normalsize\end{examp}}




\begin{center}
\vskip 1cm{\LARGE\bf A Note on Krawtchouk Polynomials \\
\vskip .15in
and Riordan Arrays } \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\

\end{center}
\vskip .2 in

\begin{abstract} We study links between Krawtchouk polynomials and Riordan arrays of both the ordinary kind and the
exponential kind. We derive summation formulas for values of the Krawtchouk polynomials using the $A$-sequences of the
Riordan arrays.
\end{abstract}

\section{Introduction}
The Krawtchouk polynomials play an important role in various areas
of mathematics. Notable applications occur in coding theory
\cite{MS}, association schemes \cite{Cameron}, and in the theory of
group representations \cite{Vilenkin1, Vilenkin2}.

In this note, we explore links between the Krawtchouk polynomials
and Riordan arrays, of both ordinary and exponential type, and we
study integer sequences defined by evaluating the Krawtchouk
polynomials at different values of their parameters.

The link between Krawtchouk polynomials and exponential Riordan arrays is implicitly contained in
the umbral calculus approach to certain families of orthogonal polynomials. We shall look at these links
explicitly in the following.

The structure of this note is as follows. In the next section, we
shall give a brief introduction to the relevant theory of both
ordinary and exponential Riordan arrays. We then define the
Krawtchouk polynomials, using exponential Riordan arrays, and look
at some general properties of these polynomials from this
perspective. We then show that for different values of the
parameters used in the definition of the Krawtchouk polynomials,
there exist interesting families of (ordinary) Riordan arrays.

\section{Riordan arrays}

The \emph{Riordan group} \cite{Alter, SGWW, SpruSums}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$g(x)=g_0+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$. We sometimes write $f(x)=xh(x)$ where $h(0)\ne 0$. 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
$\cal{R}$$(g,f)$, and is often called the \emph{Riordan array}
defined by $g$ and $f$. When $g_0=1$, the array is called a
\emph{monic} Riordan array. The group law is given by
\begin{equation} \label{product} (g, f)*(u, v)=(g(u\circ f), v\circ
f).\end{equation} The identity for this law is $I=(1,x)$ and the
inverse of $(g, f)$ is $(g, f)^{-1}=(1/(g\circ \bar{f}), \bar{f})$
where $\bar{f}$ is the compositional inverse of $f$.

To each Riordan array as defined above is associated an integer sequence
$A=\{a_i\}$ with $a_0\ne 0$ such that every element $d_{n+1,k+1}$ of the array (not lying in column $0$ or row $0$)
can be expressed as a linear combination with coefficients in $A$ of the elements in the preceding row, starting from the
preceding column:
$$d_{n+1,k+1}=a_0d_{n,k}+a_1 d_{n,k+1}+a_2d_{n,k+2}+\cdots$$
$A=\{a_i\}$ is called the $A$-sequence of the array, and its generating function may be calculated according to
$$A(x)=[h(t)|t=xh(t)^{-1}].$$
A Riordan array of the form $(g(x),x)$, where $g(x)$ is the
generating function of the sequence $u_n$, is called the
\emph{Appell array} (or sometimes the  \emph{sequence array}) of the
sequence $u_n$. Its general term is $u_{n-k}$.
\newline\newline If $\mathbf{M}$ is the matrix $(g,f)$, and
$\mathbf{u}=(u_0,u_1,\ldots)'$ is an integer sequence with ordinary
generating function $\cal{U}$ $(x)$, then the sequence
$\mathbf{M}\mathbf{u}$ has ordinary generating function
$g(x)$$\cal{U}$$(f(x))$.

\begin{example} The binomial matrix $\mathbf{B}$ is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ of the Riordan group. It has general
element $\binom{n}{k}$. For this matrix we have $A(x)=1+x$, which
translates the usual defining relationship for Pascal's triangle
$$\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}.$$
More generally, $\mathbf{B}^m$ is the element $(\frac{1}{1-m
x},\frac{x}{1-mx})$ of the Riordan group, with general term
$\binom{n}{k}m^{n-k}$. It is easy to show that the inverse
$\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(\frac{1}{1+mx},\frac{x}{1+mx})$.
\end{example}

\begin{example} We let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the
Catalan numbers $C_n=\frac{1}{n+1}\binom{2n}{n}$ \seqnum{A000108}.
The array $(1, xc(x))$ is the inverse of the array $(1,x(1-x))$
while the array $(1,xc(x^2))$ is the inverse of the array $(1,
\frac{x}{1+x^2})$.
\end{example}

\begin{example} The row sums of the matrix $(g, f)$ have generating function
$g(x)/(1-f(x))$ while the diagonal sums of $(g, f)$ have generating
function $g(x)/(1-xf(x))$. The row sums of the array $(1,xc(x))$, or
\seqnum{A106566}, are the Catalan numbers $C_n$ since
$\frac{1}{1-xc(x)}=c(x)$. The diagonal sums have g.f.
$\frac{1}{1-x^2 c(x)}$, \seqnum{A132364}.
\end{example}

The \emph{exponential Riordan group} \cite {PasTri,ProdMat,DeutschShap},
is a set of infinite lower-triangular integer
matrices, where each matrix is defined by a pair of generating
functions $g(x)=g_0+g_1x+g_2x^2+\ldots$ and
$f(x)=f_1x+f_2x^2+\ldots$ where $f_1\ne 0$. The associated matrix is
the matrix whose $i$-th column has exponential generating function
$g(x)f(x)^i/i!$ (the first column being indexed by 0). The matrix
corresponding to the pair $f, g$ is denoted by $[g, f]$. It is
\emph{monic} if $g_0=1$. The group law is then given by
\begin{displaymath} [g, f]*[h, l]=[g(h\circ f), l\circ
f].\end{displaymath} The identity for this law is $I=[1,x]$ and the
inverse of $[g, f]$ is $[g, f]^{-1}=[1/(g\circ \bar{f}), \bar{f}]$
where $\bar{f}$ is the compositional inverse of $f$.

If $\mathbf{M}$ is the matrix $[g,f]$, and $\mathbf{u}=\{u_n\}$ is
an integer sequence with exponential generating function
$\mathcal{U}$ $(x)$, then the sequence $\mathbf{M}\mathbf{u}$ has
exponential generating function $g(x)\mathcal{U}(f(x))$. Thus the
row sums of the array $[g,f]$ are given by $g(x)e^{f(x)}$ since the
sequence $1,1,1,\ldots$ has exponential generating function $e^x$.

As an element of the group of exponential Riordan arrays, we have
$\mathbf{B}=[e^x,x]$. By the above, the exponential generating
function of its row sums is given by $e^x e^x = e^{2x}$, as
expected.

Riordan group techniques have been used to provide alternative
proofs of many binomial identities that originally appeared in works
such as \cite{Riordan,Riordan2}. See, for instance,
\cite{SpruId,SpruSums}.

\section{Krawtchouk polynomials}

We follow \cite{Roman} in defining the Krawtchouk polynomials. They form an important family of orthogonal polynomials \cite{Chihara, Szego, W_K}. Thus the Krawtchouk polynomials will be considered to be
the special case $\beta=-N$, $c=\frac{p}{p-1}$, $p+q=1$ of the Meixner polynomials of the first kind, which
form the Sheffer sequence for
\begin{eqnarray*} g(t)&=&\left(\frac{1-c}{1-ce^t}\right)^{\beta},\\
f(t)&=&\frac{1-e^t}{c^{-1}-e^t}.\end{eqnarray*} Essentially, this
says that the Meixner polynomials of the first kind are obtained by
operating on the vector $(1,x,x^2,x^3,\ldots)'$ by the exponential
Riordan array $[g(t),f(t)]^{-1}$, since
$$[g,f]^{-1}=\left[\frac{1}{g\circ \bar{f}}, \bar{f}\right]$$ and
$$\left[\frac{1}{g\circ \bar{f}}, \bar{f}\right]e^{xt}=\frac{1}{g\circ \bar{f}}e^{x \bar{f}(t)}$$ which is the
defining expression for the Sheffer sequence associated to $g$ and $f$.
In order to work with this expression, we calculate $[g,f]^{-1}$ as follows. Firstly,
$$\bar{f}=\log\left(\frac{t-c}{c(t-1)}\right)$$ since
\begin{eqnarray*} \frac{1-e^u}{c^{-1}-u}=x &\implies& e^u=\frac{x-c}{c(x-1)}\\
u=\log\left(\frac{x-c}{c(x-1)}\right) &\implies& \bar{f}(t)=\log\left(\frac{t-c}{c(t-1)}\right)
\end{eqnarray*}
Then we have
$$g(\bar{f}(t))=\left(\frac{1-c}{1-ce^{\bar{f}(t)}}\right)^{\beta}=\left(\frac{1-c}{1-\frac{t-c}{t-1}}\right)^{\beta}=(1-t)^{\beta}.$$
and
$$e^{x\bar{f}(t)}=e^{x\log\left(\frac{t-c}{c(t-1)}\right)}=\left(\frac{t-c}{c(t-1)}\right)^x.$$
Thus we arrive at
$$[g,f]^{-1}=\left[\frac{1}{(1-t)^{\beta}},\log\left(\frac{t-c}{c(t-1)}\right)\right]$$
and
\begin{eqnarray*}\frac{e^{x\bar{f}(t)}}{g(\bar{f}(t))}
&=&\frac{1}{(1-t)^{\beta}}\left(\frac{t-c}{c(t-1)}\right)^x\\
&=&\frac{1}{(1-t)^{\beta+x}}\left(\frac{c-t}{c}\right)^x\\
&=&(1-t)^{-\beta-x}\left(1-\frac{t}{c}\right)^x.\end{eqnarray*}
Specializing to the values $\beta=-N$ and $c=\frac{p}{p-1}=-\frac{p}{q}$, we get
$$\frac{e^{x\bar{f}(t)}}{g(\bar{f}(t))}=(1-t)^{N-x}(1+\frac{q}{p}t)^x.$$
Extracting the coefficient of $t^k$ in this expression, we obtain
\begin{eqnarray*} [t^k]\frac{e^{x\bar{f}(t)}}{g(\bar{f}(t))}&=&[t^k]\sum_{i=0}\binom{N-x}{i}(-1)^it^i\sum_{j=0}\binom{x}{j}\left(\frac{q}{p}\right)^jt^j\\
&=&\sum_{j=0}^k\binom{N-x}{k-j}\binom{x}{j}(-1)^{k-j}q^jp^{-j}.\end{eqnarray*}
Scaling by $p^k$, we thus obtain
$$p^k[x^k]\frac{e^{x\bar{f}(t)}}{g(\bar{f}(t))}=\sum_{j=0}^k\binom{N-x}{k-j}\binom{x}{j}(-1)^{k-j}q^jp^{k-j}.$$
We use the notation
$$\kappa^{(p)}_n(x,N)=\sum_{j=0}^n\binom{N-x}{n-j}\binom{x}{j}(-1)^{n-j}q^jp^{n-j}$$
for the Krawtchouk polynomial with parameters $N$ and $p$.
This can be expressed in hypergeometric form as
$$\kappa^{(p)}_n(x,N)=(-1)^n\binom{N}{n}p^n {}_2F_1(-n,-x;-N;1/p).$$
The form of $[g, f]^{-1}$ allows us to make some interesting deductions. For instance, if we write
$$[g(t), f(t)]^{-1}=\left[\frac{1}{(1-t)^{\beta}}, \log\left(\frac{1-\frac{t}{c}}{1-t}\right)\right]$$ then
setting $\beta=-N$ and $c=\frac{p}{p-1}$, we get
$$ [g(t),f(t)]^{-1}=\left[\frac{1}{(1-t)^{-N}}, \log\left(\frac{1-\frac{p-1}{p}t}{1-t}\right)\right].$$
Now we let $t=ps$, giving
\begin{eqnarray*}[g(t), f(t)]^{-1}&=&\mathrm{Diag}(1/p^n)*\left[(1-ps)^N, \log\left(\frac{1-(p-1)s}{1-ps}\right)\right]\\
&=&\mathrm{Diag}(1/p^n)*[(1-ps)^N,s]*\left[1,\frac{s}{1-(p-1)s}\right]*\left[1, \log\left(\frac{1}{1-s}\right)\right]\\
&=&\mathrm{Diag}(1/p^n)*\mathbf{P}[p]^{-N}*\mathbf{Lah}[p-1]* \mathbf{s}.\end{eqnarray*}
where we have used the notation of \cite{Lah} and where for instance $\mathbf{s}=\left[1, \log\left(\frac{1}{1-s}\right)\right]$ is the
Stirling array of the first kind.


The matrix $\mathbf{P}[p]^{-N}*\mathbf{Lah}[p-1]*
\mathbf{s}=\left[(1-pt)^N,\log\left(\frac{1-(p-1)t}{1-pt}\right)\right]$
is of course a monic exponential Riordan array. If its general term
is $T(n,k)$, then that of the corresponding array $[g,f]^{-1}$ is
given by $T(n,k)/p^n$.

The above matrix factorization indicates that the Krawtchouk polynomials can be expressed as combinations of the
Stirling polynomials of the first kind $1, x, x(x + 1), x(x^2 + 3x + 2), x(x^3 + 6x^2 + 11x + 6),\ldots$.
\begin{example} Taking $N=-1$ and $p=2$ we exhibit an interesting property of the matrix $\left[(1-pt)^N,\log\left(\frac{1-(p-1)t}{1-pt}\right)\right]$, which
in this case is the matrix $\left[\frac{1}{1-2t},
\log\left(\frac{1-t}{1-2t}\right)\right]$. An easy calculation shows
that
$$\left[\frac{1}{1-2t}, \log\left(\frac{1-t}{1-2t}\right)\right]^{-1}=\left[\frac{1}{2e^t-1}, \frac{e^t-1}{2e^t-1}\right].$$
We recall that the Binomial matrix with general term $\binom{n}{k}$
is the Riordan array $[e^t,t]$. Now
$$\left[\frac{1}{1-2t}, \log\left(\frac{1-t}{1-2t}\right)\right][e^t,t]\left[\frac{1}{2e^t-1}, \frac{e^t-1}{2e^t-1}\right]=\left[\frac{1-t}{1-2t},t\right].$$
Hence the matrices $[e^t,t]$ and $\left[\frac{1-t}{1-2t},t\right]$
are similar, with $\left[\frac{1}{1-2t},
\log\left(\frac{1-t}{1-2t}\right)\right]$ serving as matrix of
change of basis for the similarity.
\end{example}
\section{Krawtchouk polynomials and Riordan arrays}
In this section, we shall use the following notation, where we
define a variant on the polynomial family $\kappa^{(p)}_n(x,N)$.
Thus we let
$$K(n,k,x,q)=\sum_{j=0}^k (-1)^j
\binom{x}{j}\binom{n-x}{k-j}(q-1)^{k-j}.$$ We then have
$$K(n,k,x,q)=[t^k](1-t)^x (1+(q-1)t)^{n-x},$$
which implies that
$$K(N,k,N-x,q)=[t^k](1-t)^{N-x}(1+(q-1)t)^x.$$
Letting $P=1/q$ and thus $(1-P)/P=q-1$ we obtain
$$K(N,k,N-x,q)=\frac{1}{q^n}\kappa_n^{(P)}(x,N).$$
We shall see in the sequel that by varying the parameters $n,k,x$
and $q$, we can obtain families of (ordinary) Riordan arrays defined
by the corresponding Krawtchouk expressions.
\begin{example} We first look at the term $K(k,n-k,r,q)$. We have
\begin{eqnarray*} K(k,n-k,r,q)&=&\sum_{j=0}^{n-k} (-1)^j\binom{r}{j}\binom{k-r}{n-k-j}(q-1)^{n-k-j}
\\&=&\sum_{j=0}^{r} (-1)^j\binom{r}{j}\binom{k-r}{n-k-j}(q-1)^{n-k-j}.\end{eqnarray*}
But this last term is the general term of the Riordan array
\begin{equation}\label{R1} \left(\left(\frac{1-x}{1+(q-1)x}\right)^r,x(1+(q-1)x)\right). \end{equation}
The term $(-1)^{n-k}K(k,n-k,r,q)$ then represents the general term of the inverse of
this Riordan array, which is given by
$$\left(\left(\frac{1+x}{1-(q-1)x}\right)^r,x(1-(q-1)x)\right).$$
The $A$-sequence of the array (\ref{R1}) is given by
$$A(x)=\frac{1+\sqrt{1+4(q-1)x}}{2}.$$ Thus
$$a_0=1, \qquad a_n=(-1)^{n-1}(q-1)^n C_{n-1}.$$
With these values, we therefore have
$$K(k+1,n-k,r,q)=K(k,n-k,r,q)+a_1K(k+1,n-k-1,r,q)+a_2K(k+2,n-k-2,r,q)+\ldots$$
\end{example}
\begin{example} We next look at the family defined by $(-1)^k
K(n,k,k,q)$. We have
\begin{eqnarray*} (-1)^kK(n,k,k,q)&=&(-1)^k\sum_{j=0}^k (-1)^j
\binom{k}{j}\binom{n-k}{k-j}(q-1)^{k-j}\\
&=&\sum_{j=0}^k\binom{k}{j}\binom{n-k}{k-j}(1-q)^{k-j}\\
&=&\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{j}(1-q)^j.\end{eqnarray*}
Using the results of \cite{Tri}, we see that these represent the
family of Riordan arrays
$$\left(\frac{1}{1-x},\frac{x(1-qx)}{1-x}\right).$$
The $A$-sequence for this array is given by
$$A(x)=\frac{1+x+\sqrt{1+2x(1-2q)+x^2}}{2}.$$
For example, the matrix with general term $T(n,k)=(-1)^k
K(n,k,k,-3)$ is the Riordan array
$\left(\frac{1}{1-x},\frac{x(1+3x)}{1-x}\right)$, \seqnum{A081578}
or
$$\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 &
1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 5 & 1 & 0 & 0 & 0 & \ldots \\ 1 &
9 & 9 & 1 & 0 & 0 & \ldots \\ 1 &13 & 33 & 13 & 1 & 0 & \ldots
\\1 & 17 & 72 & 73 & 7 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)$$
The $A$-sequence for this array has g.f. $\frac{1+x+\sqrt{1+14x+x^2}}{2}$ which expands to
$$1, 4, -12, 84, -732, 7140, -74604,\ldots$$
Thus
\begin{eqnarray*}(-1)^{k+1} K(n+1,k+1,k+1,-3)&=&(-1)^k K(n,k,k,-3)\\
& &+4(-1)^{k+1}K(n,k+1,k+1,-3)\\& &-12(-1)^{k+2}K(n,k+2,k+2,-3)+\ldots\end{eqnarray*}
The matrix with general term $(-1)^kK(n,k,k,2)$ is the Riordan array
$\left(\frac{1}{1-x},\frac{x(1-2x)}{1-x}\right)$ or
$$\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 &
1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 0 & 1 & 0 & 0 & 0 & \ldots \\ 1 &
-1 & -1 & 1 & 0 & 0 & \ldots \\ 1 &-2 & -2 & -2 & 1 & 0 & \ldots
\\1 & -3 & -2 & -2 & -3 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)$$
The rows of this matrix \seqnum{A098593} are the anti-diagonals (and a signed version of the diagonals) of the so-called
Krawtchouk matrices \cite{Feinsilver1, Feinsilver2} which are defined as the family of $(N+1) \times (N+1)$ matrices
with general term
$$K_{ij}^{(N)}=\sum_{k} (-1)^k\binom{j}{k}\binom{N-j}{i-k}.$$
The matrix with general term $T(n,k)=(-1)^k K(n,k,k,-1)$
is the well-known Delannoy number triangle
$\left(\frac{1}{1-x},\frac{x(1+x)}{1-x}\right)$ \seqnum{A008288} given by
$$\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 &
1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 1 & 0 & 0 & 0 & \ldots \\ 1 &
5 & 5 & 1 & 0 & 0 & \ldots \\ 1 &7 & 13 & 7 & 1 & 0 & \ldots
\\1 & 9 & 25 & 25 & 9 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)$$ Thus in particular $(-1)^n
K(2n,n,n,-1)$ is the general term of the sequence of Delannoy
numbers $1,3,13,63,\ldots$ \seqnum{A001850}. We have
\begin{proposition}
The array with general term $T(n,k)=[k\le n](-1)^k K(n,k,k,q)$ is
the Riordan array $\left(\frac{1}{1-x},\frac{x(1-qx)}{1-x}\right)$.
\end{proposition}

\end{example}

\begin{example}
We now turn our attention to the expression
$(-1)^{n-k}K(n-k,n-k,n,q)$. We have
\begin{eqnarray*}
(-1)^{n-k}K(n-k,n-k,n,q)&=&(-1)^{n-k}\sum_{j=0}^{n-k}(-1)^j\binom{n}{j}\binom{n-k-n}{n-k-j}(q-1)^{n-k-j}\\
&=&(-1)^{n-k}\sum_{j=0}^{n-k}(-1)^j\binom{n}{j}\binom{-k}{n-k-j}(q-1)^{n-k-j}\\
&=&\sum_{j=0}^{n-k}(-1)^{n-k-j}\binom{n}{j}\binom{n-j-1}{n-k-j}(q-1)^{n-k-j}\\
&=&\sum_{j=0}^{n-k}\binom{n-j-1}{n-k-j}q^{n-k-j}\\
&=&[x^n]\frac{1}{1-x}\left(\frac{x}{1-qx}\right)^k.\end{eqnarray*}
Thus the matrix with the general term
$T(n,k;q)=(-1)^{n-k}K(n-k,n-k,n,q)$ is the Riordan array
$\left(\frac{1}{1-x},\frac{x}{1-qx}\right)$. Taking the $q$-th
inverse binomial transform of this array, we obtain
$$\left(\frac{1}{1+qx},\frac{x}{1+qx}\right)*\left(\frac{1}{1-x},\frac{x}{1-qx}\right)=\left(\frac{1}{1+(q-1)x},x\right).$$
Reversing this equality gives us
$$\left(\frac{1}{1-x},\frac{x}{1-qx}\right)=\left(\frac{1}{1-qx},\frac{x}{1-qx}\right)*\left(\frac{1}{1+(q-1)x},x\right).$$
Thus
$$(-1)^{n-k}K(n-k,n-k,n,q)=\sum_{j=k}^n
\binom{n}{j}q^{n-j}(1-q)^{j-k}.$$ The row sums of the Riordan array
$\left(\frac{1}{1-x},\frac{x}{1-qx}\right)$ have generating function
$$\frac{\frac{1}{1-x}}{1-\frac{x}{1-qx}}=\frac{1-qx}{(1-x)(1-(q+1)x)}.$$
This is thus the generating function of the sum
$$\sum_{k=0}^n (-1)^{n-k}K(n-k,n-k,n,q)=\sum_{k=0}^n \sum_{j=k}^n
\binom{n}{j}q^{n-j}(1-q)^{j-k}=\frac{(1+q)^n-(1-q)}{q}.$$ We remark
that $(-1)^k K(k,k,n,q)$ is a triangle given by the reverse of the
Riordan array $\left(\frac{1}{1-x},\frac{x}{1-qx}\right)$, and will
thus have the same row sums and central coefficients.

The $A$-sequence of this array is simply $1+qx$, which implies that
$$K(n-k,n-k,n+1,q)=-K(n-k-1,n-k-1,n,q)+qK(n-k-2,n-k-2,n,q).$$
\end{example}

\begin{example} We now consider the expression
$(-1)^{n-k}K(n-k,n-k,k,q)$. We have
\begin{eqnarray*}(-1)^{n-k}K(n-k,n-k,k,q)&=&(-1)^{n-k}\sum_{j=0}^{n-k}(-1)^j
\binom{k}{j}\binom{n-k-k}{n-k-j}(q-1)^{n-k-j}\\
&=&\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-2k}{n-k-j}(1-q)^{n-k-j}.\end{eqnarray*}

This is the $(n,k)$-th element $T(n,k;q)$ of the Riordan array
$$\left(\frac{1}{1+(q-1)x},x(1+qx)\right).$$ Other expressions for
$T(n,k;q)$ include \begin{eqnarray*}
T(n,k;q)&=&\sum_{j=0}^{n-k}\binom{k}{n-k-j}(1-q)^j q^{n-k-j}\\
&=&\sum_{j=0}^{n-k}\sum_{i=0}^k
\binom{k}{i}\binom{k-i}{n-k-i-j}(-1)^j (q-1)^{i+j},\end{eqnarray*}
hence these provide alternative expressions for
$(-1)^{n-k}K(n-k,n-k,k,q)$.

We note that for $q=1$, we obtain the Riordan array $(1,x(1+x))$
whose inverse is the array $(1,xc(x))$. The row sums of $(1,x(1+x))$
are $F(n+1)$, thus giving us
$$\sum_{k=0}^n (-1)^{n-k}K(n-k,n-k,k,1)=F(n+1).$$ Similarly, we find
$$\sum_{k=0}^n (-1)^{n-k}K(n-k,n-k,k,0)=n+1.$$
$\sum_{k=0}^n (-1)^{n-k}K(n-k,n-k,k,-1)$ is the sequence $ 1,3,6,
11,21,42,\ldots$ \seqnum{A024495} with generating function
$\frac{1}{(1-x)^3-x^3}$.

These matrices have the interesting property that $T(2n,n;q)=1$.
This is so since
\begin{eqnarray*} T(2n,n;q)&=&\sum_{j=0}^{2n-n}
\binom{n}{2n-n-j}(1-q)^jq^{2n-n-j}\\
&=&\sum_{j=0}^n \binom{n}{n-j}(1-q)^j q^{n-j}\\
&=&\sum_{j=0}^n \binom{n}{j} (1-q)^j q^{n-j}\\
&=&(1-q+q)^n = 1.\end{eqnarray*} Thus we have $$K(n,n,n,q)=(-1)^n.$$
The $A$-sequence for these arrays has generating function
$$A(x)=\frac{1+\sqrt{1+4qx}}{2}$$ and thus we have
$$a_0=1, \qquad a_n=(-1)^{n-1}q^n C_{n-1}, \qquad n>0.$$
With these values we therefore have
\begin{eqnarray*}(-1)^{n-k}K(n-k,n-k,k+1,q)&=&(-1)^{n-k}K(n-k,n-k,k,q)\\
& &+a_1(-1)^{n-k-1}K(n-k-1,n-k-1,k+1,q)+\ldots\end{eqnarray*}
\end{example}

\begin{example} We next look at the expression $(-1)^{n-k}K(n,n-k,k,q)$. We have
\begin{eqnarray*}(-1)^{n-k}K(n,n-k,k,q)&=&(-1)^{n-k}\sum_{j=0}^{n-k}(-1)^j\binom{k}{j}\binom{n-k}{n-k-j}(q-1)^{n-k-j}\\
&=&\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{n-k-j}(1-q)^{n-k-j}.\end{eqnarray*}
This is the general term $T(n,k;q)$ of the Riordan array
$$\left(\frac{1}{1+(q-1)x}, \frac{x(1+qx)}{1+(q-1)x}\right).$$
Expressing $T(n,k;q)$ differently allows us to write
$$\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{n-k-j}(1-q)^{n-k-j}=\sum_{j=0}^k\binom{n}{j}\binom{n-j}{n-k-j}q^j(1-q)^{n-k-j}.$$
The central coefficients of these arrays, $T(2n,n;q)$, have
generating function $e^{(2-q)x}I_0(2\sqrt{1-q}x)$ and represent the
$n$-th terms in the expansion of $(1+(2-q)x+(1-q)x^2)^n$.

The $A$-sequence for this family of arrays has generating function
$$A(x)=\frac{1+(1-q)x+\sqrt{1+2x(1+q)+(q-1)^2x^2}}{2}.$$ Expanding this as $a_0,a_1,a_2,\ldots$ we thus
obtain
$$(-1)^{n-k}K(n+1,n-k,k+1,q)=a_0(-1)^{n-k}K(n,n-k,k,q)+a_1(-1)^{n-k-1}K(n,n-k,k+1,q)\ldots$$
\end{example}
\begin{example} The expression $K(n,n-k,N,q)$ is the general term of the Riordan array
$$\left(\frac{(1-qx)^N}{1-(q-1)x},\frac{x}{1-(q-1)x}\right).$$
This implies that
$$\sum_{j=0}^{n-k}\binom{N}{j}\binom{n-N}{n-k-j}(-1)^j(q-1)^{n-k-j}=\sum_{j=0}^{n-k}\binom{N}{j}\binom{n-j}{n-k-j}(-1)^jq^j(q-1)^{n-k-j}.$$
The $A$-sequence for this family of arrays is given by $1+(q-1)x$. Thus we obtain
$$K(n+1,n-k,N,q)=K(n,n-k,N,q)+(q-1)K(n,n-k-1,N,q).$$
\end{example}
\begin{example}
In this example, we indicate that summing over one of the parameters can still lead to a Riordan array. Thus the
expression
$$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,n,q)$$ is equivalent to the general term of the Riordan array
$$\left(\frac{1}{1-2x},\frac{x}{1-qx}\right)$$ while the expression
$$\sum_{i=0}^{n-k}K(n-k,i,n,q)$$ is equivalent to the general term of the Riordan array
$$\left(1, \frac{x}{1+qx}\right).$$
Thus
\begin{eqnarray*}\sum_{i=0}^{n-k}(-1)^i K(n-k,i,n,q)&=&\sum_{i=0}^{n-k}\sum_{j=0}^i\binom{n}{j}\binom{k+i-j-1}{i-j}(q-1)^{i-j}\\
&=&\sum_{j=0}^{n-k}\binom{j+k-1}{j}2^{n-k-j}q^j\end{eqnarray*}
and
$$\sum_{i=0}^{n-k}K(n-k,i,n,q)=\binom{n-1}{n-k}(-q)^{n-k}.$$
The $A$-sequence for this example is given by $1+qx$, and so for example we have
$$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,n+1,q)=\sum_{i=0}^{n-k}(-1)^i K(n-k,i,n,q)+q\sum_{i=0}^{n-k-1}(-1)^i K(n-k-1,i,n,q).$$
\end{example}
\begin{example} The Riordan arrays encountered so far have all been of an elementary nature. The next example indicates that this is not always so.
We make the simple change of $2n$ for $n$ in the third parameter in the previous example. We then find that
$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,2n,q)$ is the general term of the Riordan array
$$\left(\frac{1-2x-q(2-q)x^2}{1+qx},\frac{x}{(1+qx)^2}\right)^{-1}.$$
For instance, $\sum_{i=0}^{n-k}(-1)^i K(n-k,i,2n,1)$ represents the general term of the Riordan array
$$\left(\frac{1}{2}\left(\frac{1}{1-4x}+\frac{1}{\sqrt{1-4x}}\right),\frac{1-2x-\sqrt{1-4x}}{2x}\right)$$ while
$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,2n,2)$ represents the general term of
$$\left(\frac{1}{\sqrt{1-8x}},\frac{1-4x-\sqrt{1-8x}}{2x}\right).$$
The $A$-sequence for the first array above is $(1+x)^2$, so that we obtain
\begin{eqnarray*}\sum_{i=0}^{n-k}(-1)^iK(n-k,i,2(n+1),1)&=&\sum_{i=0}^{n-k}(-1)^iK(n-k,i,2n,1)\\
& &+2\sum_{i=0}^{n-k-1}(-1)^iK(n-k-1,i,2n,1)\\& &+\sum_{i=0}^{n-k-2}(-1)^iK(n-k-2,i,2n,1)\end{eqnarray*}
while that of the second array is $(1+2x)^2$ and so
\begin{eqnarray*}\sum_{i=0}^{n-k}(-1)^iK(n-k,i,2(n+1),2)&=&\sum_{i=0}^{n-k}(-1)^iK(n-k,i,2n,2)\\
& &+4\sum_{i=0}^{n-k-1}(-1)^iK(n-k-1,i,2n,2)\\& &+4\sum_{i=0}^{n-k-2}(-1)^iK(n-k-2,i,2n,2).\end{eqnarray*}
\end{example}
We summarize these examples in the following table.
\begin{center}
\newpage
\textbf{Table 1. \ Summary of Riordan arrays }
\begin{tabular}{|c|c|c|}\hline
Krawtchouk expression & Riordan array & g.f. for A-sequence\\ \hline
$K(k,n-k,r,q)$ & $\left(\left(\frac{1-x}{1+(q-1)x}\right)^r,x(1+(q-1)x)\right)$ & $\frac{1+\sqrt{1+4(q-1)x}}{2}$\\ \hline
$(-1)^{n-k}K(k,n-k,r,q)$ & $\left(\left(\frac{1+x}{1-(q-1)x}\right)^r,x(1-(q-1)x)\right)$ & $\frac{1+\sqrt{1-4(q-1)x}}{2}$\\ \hline
$(-1)^k K(n,k,k,q)$ & $\left(\frac{1}{1-x},\frac{x(1-qx)}{1-x}\right)$ & $\frac{1+x+\sqrt{1+2x(1-2q)+x^2}}{2}$\\ \hline
$(-1)^{n-k}K(n-k,n-k,k,q)$ & $\left(\frac{1}{1-x},\frac{x}{1-qx}\right)$ & $1+qx$ \\ \hline
$(-1)^{n-k}K(n-k,n-k,k,q)$ & $\left(\frac{1}{1+(q-1)x},x(1+qx)\right)$ & $\frac{1+\sqrt{1+4x}}{2}$ \\ \hline
$(-1)^{n-k}K(n,n-k,k,q)$ & $\left(\frac{1}{1+(q-1)x},\frac{x(1+qx)}{1+(q-1)x}\right)$ & $\frac{1+(1-q)x+\sqrt{1+2x(1+q)+(q-1)^2x^2}}{2}$\\ \hline
$K(n,n-k,N,q)$ & $ \left(\frac{(1-qx)^N}{1-(q-1)x}, \frac{x}{1-(q-1)x}\right)$ & $1+(q-1)x$ \\ \hline
$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,n,q)$ & $\left(\frac{1}{1-2x},\frac{x}{1-qx}\right)$ & $1+qx$ \\ \hline
$\sum_{i=0}^{n-k}K(n-k,i,n,q)$ & $\left(1,\frac{x}{1+qx}\right)$ & $1-qx$ \\ \hline
$\sum_{i=0}^{n-k}(-1)^i K(n-k,i,2n,q)$ & $\left(\frac{1-2x-q(2-q)x^2}{1+qx},\frac{x}{(1+qx)^2}\right)^{-1}$ & $(1+qx)^2$ \\ \hline
\end{tabular}\end{center}

\section{Acknowledgements}
The author would like to express his appreciation to an anonymous
reviewer, whose careful reading of the manuscript has led to
significant clarifications.

\begin{thebibliography}{13}

\bibitem{Tri} P. Barry, \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html}{On integer-sequence-based constructions of 
generalized Pascal triangles}, \emph{J. Integer Sequences}, {\bf 9}
(2006), Article 06.2.4.

\bibitem{Lah} P. Barry,
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL10/Barry2/barry401.html}{Some observations on the Lah and Laguerre transforms of integer sequences},
\emph{J. Integer Sequences},
{\bf 10} (2007), Article 07.4.6.

\bibitem{PasTri} P. Barry,
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL10/Barry/barry202.html}{On a family of generalized Pascal triangles
defined by exponential Riordan arrays}, \emph{J. Integer Sequences}, {\bf 10}
(2007), Article 07.3.5.

\bibitem{Cameron} P. J. Cameron and J. H. van Lint, \emph{Designs,
Graphs, Codes and their Links}, Cambridge University Press, 2000

\bibitem{Chihara} T. S. Chihara, \emph{An Introduction to Orthogonal
Polynomials}, Gordon and Breach, New York, 1978.

\bibitem{ProdMat} E. Deutsch, L. Ferrari, and S. Rinaldi, Production 
matrices and Riordan arrays, 
\href{http://arxiv.org/abs/math/0702638v1}{\tt http://arxiv.org/abs/math/0702638v1}, February 22 2007.

\bibitem{DeutschShap} E. Deutsch, L. Shapiro, Exponential
Riordan Arrays, Lecture Notes, Nankai University, 2004, available
electronically at
\href{http://www.combinatorics.net/ppt2004/Louis%20W.%20Shapiro/shapiro.htm}{\tt http://www.combinatorics.net/ppt2004/Louis\%20W.\%20Shapiro/shapiro.htm}

\bibitem{Feinsilver1} P. Feinsilver, J. Kocik, Krawtchouk polynomials
and Krawtchouk matrices, in R. Baeza-Yates, J. Glaz, H. Gzyl, J. Husler, and
J. L. Palacios, eds.,  \emph{Recent Advances in Applied Probability},
Springer, 2005, pp.\ 115--141.

\bibitem{Feinsilver2} 
P. Feinsilver, J. Kocik, Krawtchouk matrices from classical and
quantum random walks, \href{http://arxiv.org/abs/quant-ph/0702173v1}{\tt http://arxiv.org/abs/quant-ph/0702173v1}, February 16 2007.

\bibitem{Gautschi} W. Gautschi, \emph{Orthogonal Polynomials:
Computation and Approximation}, Clarendon Press, Oxford, 2003.

\bibitem{MS} F. J. MacWilliams, N. J. A.~Sloane, \textit{The Theory of
Error-Correcting Codes}, North Holland, Amsterdam, 2003.

\bibitem{Alter} D. Merlini, D. G. Rogers, R. Sprugnoli, and M. C. Verri,
On some alternative characterizations of Riordan arrays,
\emph{Canadian J. Mathematics}, \textbf{49} (2) (1997), 301--320.

\bibitem{Riordan} J. Riordan, \emph{An Introduction to Combinatorial Analysis}, Dover, 2002.

\bibitem{Riordan2} J. Riordan, \emph{Combinatorial Identities}, John
Wiley \& Sons, 1968.

\bibitem{Roman} S. Roman, \emph{The Umbral Calculus}, Dover,
2005.

\bibitem{Szego} G. Szeg\"\o, \emph{Orthogonal Polynomials}, 4th ed.,
American Mathematical Society, 1975,
pp.\ 35--37.

\bibitem{SGWW} L. W. Shapiro, S. Getu, W-J. Woan and L.C. Woodson,
The Riordan Group, \emph{Discr. Appl. Math.} \textbf{34} (1991)
229--239.

\bibitem{Sloane} N. J. A.~Sloane, \emph{The
On-Line Encyclopedia of Integer Sequences}. Published electronically
at \href{http://www.research.att.com/~njas/sequences/}{\tt http://www.research.att.com/$\sim$njas/sequences/}, 2008.

\bibitem{SpruSums} R. Sprugnoli, Riordan arrays and combinatorial sums,
\emph{Discrete Math.} \textbf{132} (1994), 267--290.

\bibitem{SpruId} R. Sprugnoli, \emph{Riordan Array Proofs of Identities
in Gould's Book}. Published electronically at
\href{http://www.dsi.unifi.it/~resp/GouldBK.pdf}{\tt http://www.dsi.unifi.it/$\sim$resp/GouldBK.pdf}, 2007.

\bibitem{Vilenkin1} N. Ja. Vilenkin and A. U. Klimyk, \emph{Representation of Lie Groups and Special Functions}, Vol.\ 1, Kluwer Academic Publishers, 1991.

\bibitem{Vilenkin2} N. Ja. Vilenkin and A. U. Klimyk, \emph{Representation of Lie Groups and Special Functions}, Vol.\ 2, Kluwer Academic Publishers, 1992.

\bibitem{W_K} E. W. Weisstein,
\href{http://mathworld.wolfram.com/KrawtchoukPolynomial.html/}{\tt http://mathworld.wolfram.com/KrawtchoukPolynomial.html/}, 2007.


\end{thebibliography}

\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}: Primary
33C45; Secondary 11B83,11C20.

\noindent \emph{Keywords:} Krawtchouk polynomials, orthogonal
polynomials, Riordan arrays, integer sequences.


\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences \seqnum{A000108}, \seqnum{A001850},
\seqnum{A008288}, \seqnum{A024495}, \seqnum{A081578},
\seqnum{A098593}, \seqnum{A106566}, \seqnum{A132364}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 5 2007;
revised version received May 8 2008. 
Published in {\it Journal of Integer Sequences}, June 3 2008.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

