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\begin{center}
\vskip 1cm{\LARGE\bf Continued Fractions and \\
\vskip .1in
Transformations of Integer Sequences} \vskip 1cm\large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
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\begin{abstract} We show how various transformations of integer
sequences, normally realized by Riordan or generalized Riordan arrays,
can be translated into continued fraction form. We also examine the
Deleham number triangle construction using bi-variate continued
fractions, giving examples from the field of associahedra.
\end{abstract}

\section{Introduction}

Many classical sequences have generating functions with well known
representations as continued fractions.  Many other important sequences
arise from applying transformations to such sequences with known
continued fraction representations. Thus if we can represent the result
of the transformation in continued fraction form, we can infer the
continued fraction representation of the new sequence.

The transformations that we shall discuss in this note will all be
described by (ordinary) Riordan arrays, or generalized Riordan arrays.
Thus we shall devote the next section to an overview of the Riordan
group.

Sequences will be referred to by their $Annnnnn$ number, as found in
the On-Line Encyclopedia of Integer Sequences \cite{SL1, SL2}.

The reader is referred to \cite{Wall} for a general reference on continued fractions.
\begin{example} \textbf{The Catalan numbers.}
The Catalan numbers \seqnum{A000108}
$$c_n=\frac{1}{n+1}\binom{2n}{n}$$ have generating function
$$C(x)=\frac{1-\sqrt{1-4x}}{2x},$$ which can be represented as, for instance,
$$C(x)=\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}},$$ or as
$$C(x)=\cfrac{1}{1-x-
\cfrac{x^2}{1-2x-
\cfrac{x^2}{1-2x-
\cfrac{x^2}{1-\cdots}}}}.$$
\end{example}
We shall use this notation of $C(x)$ for the generating function of the Catalan numbers and $c_n$ for the 
$n$-th Catalan number throughout. 

\noindent Similarly, the g.f. of the central binomial coefficients $\binom{2n}{n}$ \seqnum{A000984},
$\frac{1}{\sqrt{1-4x}}$, may be represented as
$$\frac{1}{\sqrt{1-4x}}=\cfrac{1}{1-
\cfrac{2x}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}}}$$ or as
$$\frac{1}{\sqrt{1-4x}}=\cfrac{1}{1-2x-
\cfrac{2x^2}{1-2x-
\cfrac{x^2}{1-2x-
\cfrac{x^2}{1-\cdots}}}}.$$

\noindent In the sequel we will have occasion to use the Iverson bracket notation \cite{Concrete},
defined by $[\mathcal{P}]=1$ if the proposition $\mathcal{P}$
is true, and
$[\mathcal{P}]=0$ if $\mathcal{P}$ is false. For instance,
$\delta_{ij}=[i=j]$, while $\delta_n=[n=0]$.
\newline\newline
Note also that if we have a sequence $a_0,a_1,a_2, \ldots$ then the \emph{aeration} of this sequence is the sequence 
$a_0,0,a_1,0,a_2,0,a_3,0,\ldots$ with interpolated zeros. If $a_n$ has generating function $g(x)$, then the aerated 
sequence has generating function $g(x^2)$.
\section{Riordan group}
The \emph{Riordan group} \cite{SGWW, Spru}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$ \cite{Spru}. The associated matrix is the matrix whose
$j$-th column is generated by $g(x)f(x)^j$ (the first column being
indexed by 0). Thus the $i$-th element of the $j$-th column is given by
$$T_{i,j}=[x^i]g(x)f(x)^j$$ where the operator $[x^n]$ \cite{Merlini_MC} extracts the coefficient of $x^n$ from the power 
series that it is applied
to. The matrix corresponding to the pair $g, f$ is
denoted by $(g, f)$ or $\cal{R}$$(g,f)$. 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 $f$.

\noindent A Riordan array of the form $(g(x),x)$, where $g(x)=\sum_{k=0}^n a_k x^k$ is the
generating function of the sequence $a_n$, is called the
\emph{sequence array} of the sequence $a_n$. Its general term is
$$T_{n,k}=[x^n]g(x)x^k=[x^{n-k}]g(x)=a_{n-k}.$$ Arrays of this form constitute a subgroup of the Riordan group, called 
the
\emph{Appell} group \cite{Appell}.
\newline\newline If $\mathbf{M}$ is the matrix $(g,f)$, and
$\mathbf{a}=(a_0,a_1,\ldots)^\top$ is an integer sequence with ordinary
generating function $\cal{A}$ $(x)$, then the sequence
$\mathbf{M}\mathbf{a}$ has ordinary generating function
$g(x)$$\cal{A}$$(f(x))$. This follows since if $\mathbf{M}=(T_{n,k})_{n,k\ge 0}$, we have
\begin{eqnarray*}
\sum_{k=0}^n T_{n,k}a_k&amp;=&amp;\sum_{k=0}^{\infty} [x^n]g(x)f(x)^k a_k\\
&amp;=&amp;[x^n]g(x)\sum_{k=0}^{\infty}f(x)^k a_k\\
&amp;=&amp;[x^n]g(x)\mathrm{\cal{A}}(f(x)).\end{eqnarray*}
\noindent The (infinite) matrix $(g,f)$ can thus be considered to act on the ring of
integer sequences $\mathbf{Z}^\mathbf{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbf{Z}[[x]]$ by
$$(g,f):\cal{A}(\mathnormal{x}) \longrightarrow \mathnormal{(g,f)}\cdot
\cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\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 $\binom{n}{k}$. More generally, $\mathbf{B}^r$ is the
element $(\frac{1}{1-r x},\frac{x}{1-rx})$ of the Riordan group,
with general term $\binom{n}{k}r^{n-k}$. It is easy to show that the
inverse $\mathbf{B}^{-r}$ of $\mathbf{B}^r$ is given by
$(\frac{1}{1+rx},\frac{x}{1+rx})$.
\end{example}

If $f_1=0$ we call the matrix a ``generalized'' Riordan array. Such a matrix is not invertible.
\section{The Binomial transform $b_n=\sum_{k=0}^n \binom{n}{k}r^{n-k}a_k$.}




\noindent A common transformation on integer sequences is the so-called ``binomial transform'', which maps the sequence
with general term $a_n$ to the sequence with general term $b_n$ defined by
$$b_n = \sum_{k=0}^n \binom{n}{k} a_k.$$ More generally, for $r \in \mathbb{Z}$, we can define the ``$r$-th binomial
transform'' of $a_n$ to be the sequence with general term
$$b_n = \sum_{k=0}^n r^{n-k}\binom{n}{k} a_k.$$ The theory of Riordan arrays now tells us that this transformation can be
represented by the matrix
$$\left(\frac{1}{1-rx},\frac{x}{1-rx}\right).$$ We recall that if $g(x)$ is the g.f. of the sequence $a_n$, then the g.f.
of the sequence $b_n$ will be given by
$$ \left(\frac{1}{1-rx},\frac{x}{1-rx}\right)\cdot g(x)=\frac{1}{1-rx}g\left(\frac{x}{1-rx}\right).$$ Applying this to 
the
continued fraction representations above, we obtain the following expressions for the generating function of the $r$-th
binomial transform of the Catalan numbers. Firstly
\begin{eqnarray*}
\left(\frac{1}{1-rx},\frac{x}{1-rx}\right)\cdot C(x)&amp;=&amp;\frac{1}{1-rx}
\cfrac{1}{1-
\cfrac{\frac{x}{1-rx}}{1-
\cfrac{\frac{x}{1-rx}}{1-
\cfrac{\frac{x}{1-rx}}{1-\cdots}}}}\\&amp;=&amp;
\cfrac{1}{1-rx-
\cfrac{x}{1-
\cfrac{\frac{x}{1-rx}}{1-
\cfrac{\frac{x}{1-rx}}{1-\cdots}}}}\\
&amp;=&amp;\cfrac{1}{1-rx-
\cfrac{x}{1-
\cfrac{x}{1-rx-
\cfrac{x}{1-
\cfrac{x}{1-rx-
\cfrac{x}{1-\cdots}}}}}}\end{eqnarray*} and secondly,
\begin{eqnarray*}
\left(\frac{1}{1-rx},\frac{x}{1-rx}\right)\cdot C(x)&amp;=&amp;\frac{1}{1-rx}
\cfrac{1}{1-\frac{x}{1-rx}-
\cfrac{\frac{x^2}{(1-rx)^2}}{1-2\frac{x}{1-rx}-
\cfrac{\frac{x^2}{(1-rx)^2}}{1-2\frac{x}{1-rx}-
\cfrac{\frac{x^2}{(1-rx)^2}}{1-\cdots}}}} \\&amp;=&amp;
\cfrac{1}{1-rx-x-
\cfrac{\frac{x^2}{1-rx}}{1-2\frac{x}{1-rx}-
\cfrac{\frac{x^2}{(1-rx)^2}}{1-2\frac{x}{1-rx}-
\cfrac{\frac{x^2}{(1-rx)^2}}{1-\cdots}}}} \\
&amp;=&amp;\cfrac{1}{1-rx-x-
\cfrac{x^2}{1-rx-2x-
\cfrac{x^2}{1-rx-2x-
\cfrac{x^2}{1-\cdots}}}}.\end{eqnarray*}

\noindent Generalizing this example in an obvious way, we obtain the following two propositions.

\begin{proposition} Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the $r$-th binomial transform of $a_n$ has generating function given by
$$\cfrac{1}{1-rx-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-rx-
\cfrac{\alpha_3 x}{1-
\cfrac{\alpha_4 x}{1-rx-
\cfrac{\alpha_5 x}{1-\cdots}}}}}}.$$
\end{proposition}

\begin{proposition} Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the $r$-th binomial transform of $a_n$ has generating function given by
$$\cfrac{1}{1-rx-\alpha_1 x-
\cfrac{\beta_1 x^2}{1-rx-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-rx-\alpha_3 x-
\cfrac{\beta_3 x^2}{1-rx-\cdots}}}}.$$
\end{proposition}

\noindent We note that the last expression may be written as
$$\cfrac{1}{1-(\alpha_1+r) x-
\cfrac{\beta_1 x^2}{1-(\alpha_2+r) x-
\cfrac{\beta_2 x^2}{1-(\alpha_3+r) x-
\cfrac{\beta_3 x^2}{1-rx-\cdots}}}}.$$
Thus the form of the continued fraction in this case does not change: only the coefficient of $x$ in each case is
incremented (or decremented). It can be shown that through the mechanism of series reversion, this translates the fact
that
\begin{equation*}\begin{split}\left(\frac{1}{1-\alpha x-\beta x^2},\frac{x}{1-\alpha x- \beta
x^2}\right)\cdot\left(\frac{1}{1-rx},\frac{x}{1-rx}\right)=\\\left(\frac{1}{1-(\alpha+r) x-\beta
x^2},\frac{x}{1-(\alpha+r)
x- \beta x^2}\right).\end{split}\end{equation*}
\begin{example} \textbf{The central trinomial coefficients.} The central binomial coefficients $\binom{2n}{n}$ have g.f.
expressible as
$$\cfrac{1}{1-
\cfrac{2x}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}}}.$$ Thus the central trinomial coefficients \seqnum{A002426}, which are the (first) inverse
binomial
transform of $\binom{2n}{n}$, have g.f expressible as
$$\cfrac{1}{1+x-
\cfrac{2x}{1-
\cfrac{x}{1+x-
\cfrac{x}{1-
\cfrac{x}{1+x-
\cfrac{x}{1-\cdots}}}}}}.$$
\end{example}

\begin{example} \textbf{The Motzkin numbers.} The Motzkin numbers $M_n$ \seqnum{A001006} are given by the binomial
transform of the ``aerated'' Catalan numbers $1,0,1,0,2,0,5,\ldots$, which have g.f. $C(x^2)$. Now
$$C(x^2)=\cfrac{1}{1-\cfrac{x^2}{1-\cfrac{x^2}{1-\cdots}}}$$ and so the Motzkin numbers have g.f. given by
$$M(x)=\cfrac{1}{1-x-\cfrac{x^2}{1-x-\cfrac{x^2}{1-x-\cfrac{x^2}{1-\cdots}}}}.$$
\end{example}

\begin{example} \textbf{The central trinomial coefficients (revisited).}
The central trinomial coefficients may also be expressed as the binomial transform of the aerated central binomial
coefficients. These latter have g.f. expressible as
$$\frac{1}{\sqrt{1-4x^2}}=\cfrac{1}{1-\cfrac{2x^2}{1-\cfrac{x^2}{1-\cfrac{x^2}{1-\cdots}}}}$$ and thus the central
trinomial coefficients have g.f. expressible as
$$\cfrac{1}{1-x-\cfrac{2x^2}{1-x-\cfrac{x^2}{1-x-\cfrac{x^2}{1-\cdots}}}}.$$ \end{example}

\section{The transformation $b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{n-2k}a_k$.}

The matrix that has general term
$$\left[k \le {\lfloor \frac{n}{2} \rfloor}\right] \binom{n-k}{k}r^{n-2k}$$ can be represented by the generalized Riordan
array
$$\left(\frac{1}{1-rx},\frac{x^2}{1-rx}\right).$$ This follows since the general term of
$$\left(\frac{1}{1-rx},\frac{x^2}{1-rx}\right)$$ is given by
\begin{eqnarray*}
[x^n]\left(\frac{1}{1-rx}\right)\left(\frac{x^{2k}}{(1-rx)^k}\right)&amp;=&amp;[x^{n-2k}](1-rx)^{-k-1}\\
&amp;=&amp;[x^{n-2k}]\sum_{j=0}^{\infty}\binom{-k-1}{j}(-1)^j r^j x^j\\
&amp;=&amp;[x^{n-2k}]\sum_{j=0}^{\infty} \binom{n+k}{j} r^j x^j\\
&amp;=&amp;[n-2k\ge 0]\binom{n-k}{n-2k}r^{n-2k}.\end{eqnarray*} We thus have
\begin{proposition} Let $a_n$ be a sequence with generating function expressible in the form
$$g(x)=\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence $b_n$ with general term
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{n-2k}a_k$$ has g.f. expressible as
$$\cfrac{1}{1-rx-
\cfrac{\alpha_1 x^2}{1-
\cfrac{\alpha_2 x^2}{1-rx-
\cfrac{\alpha_3 x^2}{1-
\cfrac{\alpha_4 x^2}{1-\cdots}}}}}.$$
\end{proposition}
\begin{proof} The g.f. of the transformed sequence is given by
\begin{align*}
\left(\frac{1}{1-rx},\frac{x^2}{1-rx}\right)\cdot g(x)&amp;=\frac{1}{1-x}\cfrac{1}{1-
\cfrac{\alpha_1 \frac{x^2}{1-rx}}{1-
\cfrac{\alpha_2 \frac{x^2}{1-rx}}{1-\cdots}}}\\
&amp;=\cfrac{1}{1-rx-
\cfrac{\alpha_1 x^2}{1-
\cfrac{\alpha_2 \frac{x^2}{1-rx}}{1-\cdots}}}\\
&amp;=\cfrac{1}{1-rx
\cfrac{\alpha_1 x^2}{1-
\cfrac{\alpha_2 x^2}{1-rx-
\cfrac{\alpha_3 x^2}{1-
\cfrac{\alpha_4 x^2}{1-rx-\cdots}}}}}.\end{align*}
\end{proof}

\begin{example} \textbf{Number of Motzkin paths of length $n$ with no level steps at odd level}.
The sequence \seqnum{A090344} with general term
$$a_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}c_k$$ has generating function
given by
$$\cfrac{1}{1-x
\cfrac{x^2}{1-
\cfrac{x^2}{1-x-
\cfrac{x^2}{1-
\cfrac{x^2}{1-\cdots}}}}}.$$
\end{example}
\begin{proposition}
Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the g.f. of the sequence
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{n-2k}a_k$$ is given by
$$\cfrac{1}{1-rx-\alpha_1 x^2 -
\cfrac{\beta_1 x^4}{1-rx-\alpha_2 x^2-
\cfrac{\beta_2 x^4}{1-\cdots}}}.$$
\end{proposition}
\begin{proof}
We have
\begin{align*}
\left(\frac{1}{1-rx},\frac{x^2}{1-rx}\right)\cdot g(x)&amp;=
\frac{1}{1-rx}\cfrac{1}{1-\alpha_1 \frac{x^2}{1-rx} -
\cfrac{\beta_1 \frac{x^4}{(1-rx)^2}}{1-\alpha_2 \frac{x^2}{1-rx}-
\cfrac{\beta_2 \frac{x^4}{(1-rx)^2}}{1-\cdots}}}\\
&amp;=\cfrac{1}{1-rx-\alpha_1 x^2-
\cfrac{\beta_1 \frac{x^4}{1-rx}}{1-\alpha_2 \frac{x^2}{1-rx}-
\cfrac{\beta_2 \frac{x^4}{(1-rx)^2}}{1-\cdots}}}\\
&amp;=\cfrac{1}{1-rx-\alpha_1 x^2 -
\cfrac{\beta_1 x^4}{1-rx-\alpha_2 x^2-
\cfrac{\beta_2 x^4}{1-\cdots}}}.\end{align*}
\end{proof}
\begin{example} \textbf{Number of ordered trees with $n$ edges and having no branches of length $1$.}
This is given by \seqnum{A026418}, which begins $1,0,1,1, 2, 3, 6, 11, 22,\ldots$ Now the sequence which begins
$1,1,2,3,6,\ldots$ has general term
$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^{n-2k}M_k$$ and therefore has g.f. given by

$$\cfrac{1}{1-x- x^2 -
\cfrac{ x^4}{1-x- x^2-
\cfrac{x^4}{1-\cdots}}}.$$
\end{example}

\section{The transformation $b_n=\sum_{k=0}^n \binom{n+k}{2k}a_k$.}

The transformation which maps the sequence with general term $a_n$ to the sequence with general term
$$b_n=\sum_{k=0}^n \binom{n+k}{2k}a_k$$ can be represented by the Riordan array
$$\left(\frac{1}{1-x},\frac{x}{(1-x)^2}\right).$$ We then have the following proposition.
\begin{proposition} Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence with general term $b_n$ given by
$$b_n=\sum_{k=0}^n \binom{n+k}{2k}a_k$$ has g.f. expressible as
$$\left(\frac{1}{1-x},\frac{x}{(1-x)^2}\right)\cdot g(x)=\cfrac{1}{1-x-
\cfrac{\alpha_1 x}{1-x-
\cfrac{\alpha_2 x}{1-x-\cdots}}}.$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
\left(\frac{1}{1-x},\frac{x}{(1-x)^2}\right)\cdot g(x)&amp;=&amp;
\frac{1}{1-x}\cfrac{1}{1-
\cfrac{\alpha_1\frac{x}{(1-x)^2}}{1-
\cfrac{\alpha_2\frac{x}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1 \frac{x}{1-x}}{1-
\cfrac{\alpha_2\frac{x}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;\cfrac{1}{1-x-
\cfrac{\alpha_1 x}{1-x-
\cfrac{\alpha_2 x}{1-x-\cdots}}}.\end{eqnarray*}

\end{proof}

\begin{example} \textbf{The large Schr\"oder numbers.} The large Schr\"oder numbers $S_n$ \seqnum{A006318} can be defined
by
$$S_n = \sum_{k=0}^n \binom{n+k}{2k} c_k.$$ Thus they have g.f. expressible as
$$S(x)=\cfrac{1}{1-x-\cfrac{x}{1-x-\cfrac{x}{1-x-\cfrac{x}{1-x-\cdots}}}}.$$
\end{example}

\begin{example} \textbf{The central Delannoy numbers.} The central Delannoy numbers $d_n$ \seqnum{A001850} can be defined
by
$$d_n=\sum_{k=0}^n \binom{n+k}{2k} \binom{2k}{k}.$$ Thus they have g.f. expressible as
$$\cfrac{1}{1-x-\cfrac{2x}{1-x-\cfrac{x}{1-x-\cfrac{x}{1-x-\cdots}}}}.$$
\end{example}

\noindent More generally, we can consider the action of the Riordan array
$$\left(\frac{1}{1-rx},\frac{x}{(1-rx)^2}\right)$$ for $r \in \mathbb{Z}$. We obtain
\begin{proposition} Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence $b_n$ with g.f. given by
$$\frac{1}{1-rx}g\left(\frac{x}{(1-rx)^2}\right)$$ is the expansion of
$$\cfrac{1}{1-rx-\cfrac{\alpha_1 x}{1-rx-\cfrac{\alpha_2 x}{1-rx-\cfrac{\alpha_3 x}{1-rx-\cdots}}}}.$$ In this case,
$$b_n=\sum_{k=0}^n \binom{n+k}{2k}r^{n-k} a_k$$ is the general element of the transformed sequence.
\end{proposition}

\begin{example} \textbf{The case $r=-1$.}
\noindent This case corresponds to the Riordan array
$$\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right).$$
Now we have, for instance,
$$\sum_{k=0}^n \binom{n+k}{2k} (-1)^{n-k} c_k = 0^n=\delta_{0n},$$ where $0^n$ is the sequence $1,0,0,0,\ldots$ with g.f.
equal to $1$. Thus we obtain the identity
\begin{equation}1=\frac{1}{1+x-\cfrac{x}{1+x-\cfrac{x}{1+x-\cfrac{x}{1+x-\cdots}}}}.\end{equation}

\noindent Similarly the identity
$$\sum_{k=0}^n \binom{n+k}{2k} (-1)^{n-k} \binom{2k}{k}=1$$ yields the identity
\begin{equation}\frac{1}{1-x}=\frac{1}{1+x-\cfrac{2x}{1+x-\cfrac{x}{1+x-\cfrac{x}{1+x-\cdots}}}}.\end{equation}
\end{example}

\begin{proposition}
Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the generating function of the sequence
$$b_n=\sum_{k=0}^n \binom{n+k}{2k}r^{n-k}a_k,$$ obtained by applying the transformation represented by
$$\left(\frac{1}{1-rx},\frac{x}{(1-rx)^2}\right)$$ to $a_n$ has g.f. given by
$$\cfrac{1-rx}{(1-rx)^2-\alpha_1 x-
\cfrac{\beta_1 x^2}{(1-rx)^2-\alpha_2 x -
\cfrac{\beta_2 x^2}{(1-rx)^2-\cdots}}}.$$
\end{proposition}
\begin{proof} The result follows by considering the expression $$\frac{1-rx}{(1-rx)^2}\cfrac{1}{1-\alpha_1
\frac{x}{(1-rx)^2}-\cfrac{\beta_1 \frac{x^2}{(1-rx)^4}}{1-\alpha_2 \frac{x}{(1-rx)^2}-
\cfrac{\beta_2\frac{x^2}{(1-rx)^4}}{1-\cdots}}}.$$
\end{proof}

\begin{example} \textbf{Large Schr\"oder numbers (revisited).} Since the large Schr\"oder numbers are given by
$$S_n=\sum_{k=0}^n \binom{n+k}{2k}c_k$$ we obtain the following expression for their generating function\,:

$$S(x)=\cfrac{1-x}{(1-x)^2-x-\cfrac{x^2}{(1-x)^2-2x-\cfrac{x^2}{(1-x)^2-2x-\cfrac{x^2}{1-\cdots}}}}$$
or
$$S(x)=\cfrac{1-x}{1-3x+x^2-\cfrac{x^2}{1-4x+x^2-\cfrac{x^2}{1-4x+x^2-\cfrac{x^2}{1-\cdots}}}}.$$
\end{example}

Note that the above example shows that the partial sums of the large Schr\"oder numbers, which have g.f. given by
$\frac{1}{1-x}S(x)$, can be obtained from the expansion of
\begin{equation}\label{sLS}\frac{1}{1-x}S(x)=\cfrac{1}{(1-x)^2-x-\cfrac{x^2}{(1-x)^2-2x-\cfrac{x^2}{(1-x)^2-2x-\cfrac{x^2}{1-\cdots}}}}.\end{equation}
This can be put in a more general context by observing that the general term of the matrix
$$\left(\frac{1}{(1-rx)^2},\frac{x}{(1-rx)^2}\right)$$ is given by
$$\binom{n+k+1}{n-k}r^{n-k}=\binom{n+k+1}{2k+1}r^{n-k}.$$ Thus we have the proposition
\begin{proposition}
Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the generating function of the sequence
$$b_n=\sum_{k=0}^n \binom{n+k+1}{2k+1}r^{n-k}a_k,$$ obtained by applying the transformation represented by
$$\left(\frac{1}{(1-rx)^2},\frac{x}{(1-rx)^2}\right)$$ to $a_n$ has g.f. given by
$$\cfrac{1}{(1-rx)^2-\alpha_1 x-
\cfrac{\beta_1 x^2}{(1-rx)^2-\alpha_2 x -
\cfrac{\beta_2 x^2}{(1-rx)^2-\cdots}}}.$$
\end{proposition}

\noindent For instance, the g.f. of the partial sums of the large Schr\"oder numbers, which have general term
$$\sum_{k=0}^n \binom{n+k+1}{2k+1}c_k,$$ can be represented as above in Eq. (\ref{sLS}). This sequence is
\seqnum{A086616}.

In a similar fashion, we have the result.
\begin{proposition}
Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the generating function of the sequence
$$b_n=0^n+\sum_{k=0}^{n-1} \binom{n+k+1}{2k+1}r^{n-k-1}a_{k+1},$$ obtained by applying the transformation represented by
$$\left(1,\frac{x}{(1-rx)^2}\right)$$ to $a_n$ has g.f. given by
$$\cfrac{(1-rx)^2}{(1-rx)^2-\alpha_1 x-
\cfrac{\beta_1 x^2}{(1-rx)^2-\alpha_2 x -
\cfrac{\beta_2 x^2}{(1-rx)^2-\cdots}}}.$$
\end{proposition}
\begin{example} \textbf{Royal paths in a lattice} \seqnum{A006319}. This is the sequence $1, 1, 4, 16, 68, 304, 1412, 
\ldots$, which also gives the number of peaks at level $1$ in all Schr\"oder paths of semi-length $n$, ($n\ge 1$). It has
general element
$$0^n+\sum_{k=0}^{n-1} \binom{n+k+1}{2k+1}C_{k+1},$$ and g.f.
$$\cfrac{(1-x)^2}{(1-x)^2-x-
\cfrac{x^2}{(1-x)^2-2x-
\cfrac{x^2}{(1-x)^2-2x-
\cfrac{x^2}{(1-x)^2-2x-\cdots}}}}.$$
\end{example}

\noindent We note similarly that the sequence with g.f. given by
$$\cfrac{(1-x)^2}{(1-x)^2-x-
\cfrac{x^2}{(1-x)^2-x-
\cfrac{x^2}{(1-x)^2-x-
\cfrac{x^2}{(1-x)^2-x-\cdots}}}}$$ is the transformation of the Motzkin numbers $M_n$ with general element
$$0^n+\sum_{k=0}^{n-1} \binom{n+k+1}{2k+1}M_{k+1}.$$

Finally, the sequence with g.f.
$$\cfrac{(1-x)^2}{(1-x)^2-2x-
\cfrac{x^2}{(1-x)^2-2x-
\cfrac{x^2}{(1-x)^2-2x-
\cfrac{x^2}{(1-x)^2-2x-\cdots}}}}$$ is the transformation of the shifted Catalan numbers $c_{n+1}$ with general term
$$0^n+\sum_{k=0}^{n-1} \binom{n+k+1}{2k+1}c_{k+2}.$$

\section{The transformation $b_n=\sum_{k=0}^n \binom{n}{2k} a_k$.}

The matrix with general element $\binom{n}{2k}$ is the generalized Riordan array
$$\left(\frac{1}{1-x},\frac{x^2}{(1-x)^2}\right).$$
\begin{proposition} Let $a_n$ be a sequence with generating function expressible in the form
$$g(x)=\cfrac{1}{1-\cfrac{\alpha_1 x}{1-\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence with general term
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k}a_k$$ has g.f. expressible in the form
$$\cfrac{1}{1-x-\cfrac{\alpha_1 x^2}{1-x-\cfrac{\alpha_2 x^2}{1-x-\cdots}}}.$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}\frac{1}{1-x}g\left(\frac{x^2}{(1-x)^2}\right)&amp;=&amp;
\frac{1}{1-x}\cfrac{1}{1-\cfrac{\alpha_1\frac{x^2}{(1-x)^2}}{1-
\cfrac{\alpha_2 \frac{x^2}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;
\cfrac{1}{1-x-\cfrac{\alpha_1 \frac{x^2}{1-x}}{1-
\cfrac{\alpha_2 \frac{x^2}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;\cfrac{1}{1-x-\cfrac{\alpha_1 x^2}{1-x-\cfrac{\alpha_2 x^2}{1-x-\cdots}}}.\end{eqnarray*}
\end{proof}
\noindent In similar fashion, we can show that for $a_n$ as above, the sequence
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k}r^{n-2k}a_k$$
has g.f. expressible in continued fraction form as
$$\cfrac{1}{1-rx-\cfrac{\alpha_1 x^2}{1-rx-\cfrac{\alpha_2 x^2}{1-rx-\cdots}}}.$$
\noindent We also have the following\,:
\begin{proposition}
Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-\alpha_1 x -
\cfrac{\beta_1 x^2}{1-\alpha_2 x-
\cfrac{\beta_2 x^2}{1-\cdots}}}.$$ Then the generating function of the sequence
$$b_n=\sum_{k=0}^n \binom{n}{2k}a_k$$ is expressible in the form
$$\cfrac{1-x}{(1-x)^2-\alpha_1 x^2-
\cfrac{\beta_1 x^4}{(1-x)^2-\alpha_2 x^2-
\cfrac{\beta_2 x^4}{(1-x)^2-\alpha_3 x^2-\cdots}}}.$$
\end{proposition}

\begin{example} \textbf{The Motzkin numbers $M_n$.} We have
$$M_n = \sum_{k=0}^n \binom{n}{2k} c_k.$$ Hence the Motzkin numbers have g.f. expressible as
\begin{eqnarray*}M(x)&amp;=&amp;
\cfrac{1-x}{(1-x)^2-x^2-
\cfrac{x^4}{(1-x)^2-2x^2-
\cfrac{x^4}{(1-x)^2-2x^2-\cdots}}}\\
&amp;=&amp;\cfrac{1-x}{1-2x-
\cfrac{x^4}{1-2x-x^2-
\cfrac{x^4}{1-2x-x^2-
\cfrac{x^4}{1-\cdots}}}}.\end{eqnarray*} A simple consequence of this is the fact that the
partial sums of the Motzkin numbers, \seqnum{A086615}, have generating function expressible as
$$\cfrac{1}{1-2x-
\cfrac{x^4}{1-2x-x^2-
\cfrac{x^4}{1-2x-x^2-
\cfrac{x^4}{1-\cdots}}}}.$$
\end{example}

\section{The transformation $b_n=\sum_{k=0}^n \binom{n-k}{2k}a_k$.}

We consider the generalized Riordan array
$$\left(\frac{1}{1-x},\frac{x^3}{(1-x)^2}\right).$$
The general term of this matrix is given by
\begin{eqnarray*}
T_{n,k}&amp;=&amp;[x^n]\frac{x^{3k}}{(1-x)^{2k+1}}\\
&amp;=&amp;[x^{n-3k}]\sum_{i=0}\binom{-2k-1}{i}(-1)^i x^i \\
&amp;=&amp;[x^{n-3k}]\sum_{i=0}\binom{2k+1+i-1}{i}x^i \\
&amp;=&amp;\binom{2k+n-3k}{n-3k}\\
&amp;=&amp;\binom{n-k}{2k}.\end{eqnarray*}

\noindent We then have
\begin{proposition} Let $a_n$ be a sequence with generating function $g(x)$ expressible in the form
$$g(x)=\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence
$$b_n=\sum_{k=0}^n \binom{n-k}{2k} a_k$$ has g.f. expressible as
$$\cfrac{1}{1-x-
\cfrac{\alpha_1 x^3}{1-x-
\cfrac{\alpha_2 x^3}{1-x-\cdots}}}.$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
\frac{1}{1-x}g\left(\frac{x^3}{(1-x)^2}\right)&amp;=&amp;\frac{1}{1-x}
\cfrac{1}{1-
\cfrac{\alpha_1\frac{x^3}{(1-x)^2}}{1-
\cfrac{\alpha_2\frac{x^3}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1\frac{x^3}{1-x}}{1-
\cfrac{\alpha_2\frac{x^3}{(1-x)^2}}{1-\cdots}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1 x^3}{1-x-
\cfrac{\alpha_2 x^3}{1-x-\cdots}}}.
\end{eqnarray*}
\end{proof}

\begin{example} \textbf{Generalized Catalan numbers.}
We take $a_n=C_n$ and form the sequence
$$b_n=\sum_{k=0}^n \binom{n-k}{2k}c_k,$$ with g.f.
$$\frac{1}{1-x}C\left(\frac{x^3}{(1-x)^2}\right).$$
Then we can express this g.f. as
$$\cfrac{1}{1-x-
\cfrac{x^3}{1-x-
\cfrac{x^3}{1-x-
\cfrac{x^3}{1-\cdots}}}}.$$
These are the generalized Catalan numbers \seqnum{A023431}.

\end{example}

\noindent Similarly, the sequence with g.f.
$$\cfrac{1}{1-x-
\cfrac{2x^3}{1-x-
\cfrac{x^3}{1-x-
\cfrac{x^3}{1-\cdots}}}}$$ has general term
$$\sum_{k=0}^n \binom{n-k}{2k}\binom{2k}{k}.$$ This is \seqnum{A098479}.

In general, we can show that the sequence with g.f. given by
$$\cfrac{1}{1-x-
\cfrac{rx^3}{1-x-
\cfrac{x^3}{1-x-
\cfrac{x^3}{1-\cdots}}}}$$ is the image of the power sequence $r^n$ by the product of Riordan arrays
$$\left(\frac{1}{1-x},\frac{x^3}{(1-x)^2}\right)\cdot (1, xC(x)).$$ Thus the g.f. of the image sequence is given by
$$\frac{1}{1-x}\frac{1}{1-r\frac{x^3}{1-x^3}C\left(\frac{x^3}{1-x^3}\right)}.$$

\section{The transformation $b_n=\sum_{k=0}^n \binom{n+k}{3k}a_k$.}
The general term of the generalized Riordan array
$$\left(\frac{1}{1-x},\frac{x^2}{(1-x)^3}\right)$$ is given by
$$T_{n,k}=\binom{n+k}{3k}.$$ Thus if the sequence $a_n$ has g.f. expressible in the form
$$g(x)=
\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-
\cfrac{\alpha_3 x}{1-\cdots}}}}$$ then the sequence
$$b_n=\sum_{k=0}^n \binom{n+k}{3k}a_k$$ will have g.f.
$$\frac{1}{1-x}g\left(\frac{x^2}{(1-x)^3}\right),$$ which is equal to
\begin{eqnarray*}
&amp;&amp;\frac{1}{1-x}\cfrac{1}{1-
\cfrac{\alpha_1 \frac{x^2}{(1-x)^3}}{1-
\cfrac{\alpha_2 \frac{x^2}{(1-x)^3}}{1-
\cfrac{\alpha_3 \frac{x^2}{(1-x)^3}}{1-\cdots}}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1 \frac{x^2}{(1-x)^2}}{1-
\cfrac{\alpha_2 \frac{x^2}{(1-x)^3}}{1-
\cfrac{\alpha_3 \frac{x^2}{(1-x)^3}}{1-\cdots}}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1 x^2}{(1-x)^2-
\cfrac{\alpha_2 \frac{x^2}{1-x}}{1-
\cfrac{\alpha_3 \frac{x^2}{(1-x)^3}}{1-\cdots}}}}\\
&amp;=&amp;
\cfrac{1}{1-x-
\cfrac{\alpha_1 x^2}{(1-x)^2-
\cfrac{\alpha_2 x^2}{1-x-
\cfrac{\alpha_3 \frac{x^2}{(1-x)^2}}{1-\cdots}}}}.\end{eqnarray*}
Thus we arrive at the
\begin{proposition} In the circumstances above, the sequence
$$b_n=\sum_{k=0}^n \binom{n+k}{3k}a_k$$ has g.f. expressible in the form
$$\cfrac{1}{1-x-
\cfrac{\alpha_1 x^2}{(1-x)^2-
\cfrac{\alpha_2 x^2}{1-x-
\cfrac{\alpha_3 x^2}{(1-x)^2-\cdots}}}}.$$
\end{proposition}

\begin{example} \textbf{Number of Dyck paths of semi-length $n$ with no $UUDD$}. We let $a_n=c_n$. Then the sequence with 
general term
$$b_n=\sum_{k=0}^n \binom{n+k}{3k} c_k$$ has g.f. $\frac{1}{1-x}C\left(\frac{x^2}{(1-x)^3}\right)$ expressible as
$$\cfrac{1}{1-x-
\cfrac{x^2}{(1-x)^2-
\cfrac{x^2}{1-x-
\cfrac{x^2}{(1-x)^2-\cdots}}}}.$$
This is \seqnum{A086581} (number of Dyck paths of semi-length $n$ with no $UUDD$).
\end{example}
\section{Bi-variate continued fractions and number triangles}
We have seen that
$$
\left(\frac{1}{1-rx},\frac{x}{1-rx}\right)\cdot C(x)=
\cfrac{1}{1-rx-
\cfrac{x}{1-
\cfrac{x}{1-rx-
\cfrac{x}{1-
\cfrac{x}{1-rx-
\cfrac{x}{1-\cdots}}}}}}.$$
Now treating $r$ as an independent variable (and writing it as $y$), we consider the bi-variate expression
\begin{equation}\label{Biv}g(x,y)=\cfrac{1}{1-xy-
\cfrac{x}{1-
\cfrac{x}{1-xy-
\cfrac{x}{1-
\cfrac{x}{1-xy-
\cfrac{x}{1-\cdots}}}}}}.\end{equation}
This is the bi-variate generating function of the number triangle with general term
$$[k\le n] \binom{n}{k}c_{n-k},$$ which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 2
&amp; 1 &amp; 0 &amp; 0 &amp;
0 &amp; \ldots \\ 5 &amp; 6 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 14 &amp; 20 &amp; 12
&amp; 4 &amp; 1 &amp; 0 &amp; \ldots \\42 &amp; 70 &amp; 50 &amp; 20 &amp; 5 &amp; 1
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} This is \seqnum{A124644}.
The image of the sequence $r^n$ by this matrix is given by
$$b_n=\sum_{k=0}^n \binom{n}{k}c_{n-k}r^k=\sum_{k=0} \binom{n}{k}c_k r^{n-k},$$
using $\binom{n}{k}=\binom{n}{n-k}$. Thus applying the matrix with bi-variate generating
function given by Eq. (\ref{Biv}) to the sequence $r^n$ is equivalent to calculating the
$r$-th binomial transformation of the Catalan numbers $C_n$.

This example may be generalized in many ways.
\section{The transformation $b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n-k}{k}r^{n-2k}a_{n-2k}$.}
Setting $y=x$ in Eq. (\ref{Biv}) of the last section gives us the generating function of the diagonal sums of the matrix.
Thus the
sequence with general term
$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}c_{n-2k}$$ has generating function expressible as
$$
\cfrac{1}{1-x^2-
\cfrac{x}{1-
\cfrac{x}{1-x^2-
\cfrac{x}{1-\cdots}}}}.$$

\noindent This is \seqnum{A105864}. By the construction above, it is the result of
applying the Riordan array
$$\left(\frac{1}{1-x^2},\frac{x}{1-x^2}\right)$$ to the Catalan numbers. In fact, we have the following proposition\,:
\begin{proposition}
Let $a_n$ be a sequence with generating function expressible as
$$\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\cdots}}}.$$ Then the sequence $b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n-k}{k}a_{n-2k}$
which results from applying the Riordan array
$$\left(\frac{1}{1-rx^2},\frac{x}{1-rx^2}\right)$$ to $a_n$ will have g.f. of the form
$$\frac{1}{1-rx^2-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-rx^2-
\cfrac{\alpha_3 x}{1-
\cfrac{\alpha_4 x}{1-rx^2-
\cfrac{\alpha_5 x}{1-\cdots}}}}}}.$$
\end{proposition}
\begin{example}\textbf{A transform of the large Schr\"oder numbers}.
The large Schr\"oder numbers have g.f. expressible as
$$\cfrac{1}{1-
\cfrac{2x}{1-
\cfrac{x}{1-
\cfrac{2x}{1-
\cfrac{x}{1-\cdots}}}}}.$$ Thus the sequence with general term
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}S_{n-2k}$$ has g.f. given by
$$\cfrac{1}{1-x^2-
\cfrac{2x}{1-
\cfrac{x}{1-x^2-
\cfrac{2x}{1-
\cfrac{x}{1-x^2-\cdots}}}}}.$$ This is equal to
$$\frac{1-x-x^2-\sqrt{1-6x-x^2+6x^3+x^4}}{2x(1-x^2)}.$$
\end{example}

\noindent The special case $r=-1$ which corresponds to the Riordan array $\left(\frac{1}{1+x^2},\frac{x}{1+x^2}\right)$
corresponds to the so-called ``Chebyshev transform''. Thus the transform of $c_k$ for this matrix has
general term
$$b_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k}c_{n-2k},$$ reminiscent of the formula for the
Chebyshev polynomials of the second kind. This sequence is \seqnum{A101499}. By the above, it has generating function

$$
\cfrac{1}{1+x^2-
\cfrac{x}{1-
\cfrac{x}{1+x^2-
\cfrac{x}{1-
\cfrac{x}{1+x^2-\cdots}}}}}.$$
\section{The Deleham construction}
The Deleham construction is a means of using bi-variate continued fraction generating functions, based on two base
sequences, to construct number triangles. Many triangles of classical importance may be so constructed.
Numerous examples are to be found in
\cite{SL1}.
For the purposes of this note, we can define the \emph{Deleham construction} as follows.
Given two sequences $r_n$ and $s_n$,we use the notation
$$ r  \,\Delta\, s = [r_0,r_1,r_2,\ldots]  \,\Delta\, [s_0,s_1,s_2,\ldots]$$ to denote the
number triangle whose bi-variate generating function is given by
$$\cfrac{1}{1-
\cfrac{(r_0x+s_0 xy)}{1-
\cfrac{(r_1x+s_1 xy)}{1-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}}.$$ We furthermore define
$$ r \, \Delta^{(1)} \, s = [r_0,r_1,r_2,\ldots]  \,\Delta^{(1)}\,  [s_0,s_1,s_2,\ldots]$$ to denote
the number triangle whose bi-variate generating function is given by
$$\cfrac{1}{1-(r_0x+s_0 xy)-
\cfrac{(r_1x+s_1 xy)}{1-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}.$$
\noindent See \seqnum{A084938} for the original definition.

\begin{example} \textbf{The Narayana triangles.}
Three common versions of the Narayana triangle can be expressed as follows\,:
$$[1,0,1,0,1,0,1,\ldots]  \,\Delta \,
[0,1,0,1,0,1,\ldots]$$ which is \seqnum{A131198},
$$[0,1,0,1,0,1,\ldots]  \,\Delta\,
[1,0,1,0,1,0,1,\ldots],$$ which is the reverse of that matrix, and
$$[0,1,0,1,0,1,\ldots]  \,\Delta^{(1)} \,
[1,0,1,0,1,0,1,\ldots]$$ which is \seqnum{A090181}.
\end{example}

\noindent We have the following result:
\begin{theorem}
The first column of the Deleham array
$$[r_0,r_1,r_2,r_3,\ldots] \, \Delta \,
[s_0,s_1,s_2,s_3,\ldots]$$ has g.f.
$$\cfrac{1}{1-
\cfrac{r_0x}{1-
\cfrac{r_1x}{1-
\cfrac{r_2x}{1-\cdots}}}}.$$
The row sums of the array have
g.f.
$$\cfrac{1}{1-
\cfrac{(r_0+s_0)x}{1-
\cfrac{(r_1+s_1)x}{1-
\cfrac{(r_2+s_2)x}{1-\cdots}}}}.$$ The
diagonal sums of the array have g.f.
$$\cfrac{1}{1-
\cfrac{(r_0x+s_0x^2)}{1-
\cfrac{(r_1x+s_1x^2)}{1-
\cfrac{(r_2x+s_2x^2)}{1-\cdots}}}}.$$
The product of the array with $\mathbf{B}$ has generating
function
$$\cfrac{1}{1-
\cfrac{((r_0+s_0)x+s_0 xy)}{1-
\cfrac{((r_1+s_1)x+s_1 xy)}{1-
\cfrac{((r_2+s_2)x+s_2 xy)}{1-\cdots}}}}=\cfrac{1}{1-
\cfrac{r_0 x+s_0x(1+y)}{1-
\cfrac{r_1 x+s_1x(1+y)}{1-
\cfrac{r_2 x+s_2x(1+y)}{1-\cdots}}}},$$ and is thus the Deleham array
$$(r+s)\, \Delta \, s.$$
The product of $\mathbf{B}$ and the array has generating function
$$\cfrac{1}{1-x-
\cfrac{(r_0x+s_0 xy)}{1-
\cfrac{(r_1x+s_1 xy)}{1-x-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}}.$$
\end{theorem}
\begin{proof}
The g.f. of the first column is obtained by setting $y=0$ in the bivariate g.f. Similarly, the
g.f. of the row sums is obtained by setting $y=1$, while that of the diagonal sums is found by setting
$y=x$. \newline\newline
The product of the array $r \,\Delta\,s$ and $\mathbf{B}$ has g.f. given by
$$(1,x,x^2,\ldots)\, (r \,\Delta\,s)\cdot\mathbf{B}\, (1,y,y^2,\ldots)^\top.$$ But this is
$$(1,x,x^2,\ldots)\, (r \,\Delta\,s)\, (1,1+y,(1+y)^2,\ldots)^\top$$ which by assumption is
$$\cfrac{1}{1-
\cfrac{r_0 x+s_0x(1+y)}{1-
\cfrac{r_1 x+s_1x(1+y)}{1-
\cfrac{r_2 x+s_2x(1+y)}{1-\cdots}}}}.$$
The g.f. of the binomial transform of the array (that is, of the product of $\mathbf{B}$ and $\mathbf{r}\,\Delta\, 
\mathbf{s}$) will be given by
$$\frac{1}{1-x}\cfrac{1}{1-
\cfrac{(r_0+s_0y)\frac{x}{1-x}}{1-
\cfrac{(r_1+s_1y)\frac{x}{1-x}}{1-
\cfrac{(r_2+s_2y)\frac{x}{1-x}}{1-\cdots}}}},$$ which simplifies to
$$\cfrac{1}{1-x-
\cfrac{(r_0x+s_0 xy)}{1-
\cfrac{(r_1x+s_1 xy)}{1-x-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}}.$$
\end{proof}
\begin{example} \textbf{\seqnum{A088874}}. The number triangle
$$[0,2,6,12,20,30,\ldots]\,\Delta \,[1,2,3,4,5,6,\ldots]$$ with g.f.
$$\cfrac{1}{1-
\cfrac{ xy}{1-
\cfrac{(2x+2 xy)}{1-
\cfrac{(6x+3 xy)}{1-
\cfrac{(10x+4xy)}{1-\cdots}}}}}$$ is studied in \cite{Franssens}.
\end{example}
\noindent
The Deleham construction leads to many interesting triangular
arrays of numbers. The field of \emph{associahedra}
\cite{Burgiel, Chapoton, Fomin_Reading, Postnikov} is rich in
such triangles,
including the Narayana triangle. We finish with some examples from
this
area. An associahedron is a special type of polytope.  The $f$-vector is a vector $(f_{-1}, f_0, \ldots , f_{n-1})$ where 
$f_i$ denotes the number
of $i$-dimensional faces. The unique ``($-1$)-dimensional" face is the empty face. The
$h$-vector $(h_0, h_1, \ldots, h_n)$ is determined from the $f$-vector by a process equivalent to that described below. In 
the sequel, $A_n$ and $B_n$ refer to standard root systems connected to rotation groups \cite{Fomin_Reading}.
\begin{example} \textbf{The coefficient array for the $f$-vector for $B_n$.} The triangle with general term
$$\binom{n}{k}\binom{n+k}{k}=\binom{n+k}{2k}\binom{2k}{k},$$ which is \seqnum{A063007}, is the coefficient array for the 
$f$-vector of $B_n$ \cite{Chapoton}. In other words, the $f$-vector for $B_n$ is given by $\sum_{k=0}^n 
\binom{n}{k}\binom{n+k}{k}x^k$. By our previous results, the bi-variate generating function of this triangle is given by
$$\cfrac{1}{1-x-
\cfrac{2xy}{1-x-
\cfrac{xy}{1-x-
\cfrac{xy}{1-x-
\cfrac{xy}{1-x-\cdots}}}}}.$$
This may also be expressed as
$$\cfrac{1}{1-x-
\cfrac{2xy}{1-
\cfrac{x+xy}{1-
\cfrac{xy}{1-
\cfrac{x+xy}{1-\cdots}}}}},$$ which
is
$$[1,0,1,0,1,0,\ldots]\, \Delta^{(1)}\,[0,2,1,1,1,1,\ldots].$$
\end{example}

\begin{example} \textbf{The $h$-vector array for $B_n$.}
Reversing the above array to get the array
$$[0,2,1,1,1,1,\ldots]\, \Delta^{(1)}\,[1,0,1,0,1,0,\ldots],$$ we see that this array, which has general term 
$\binom{n}{k}\binom{2n-k}{n}$, has generating function
$$\cfrac{1}{1-xy-
\cfrac{2x}{1-
\cfrac{x+xy}{1-
\cfrac{x}{1-
\cfrac{x+xy}{1-
\cfrac{x}{1-\cdots}}}}}},$$ or equivalently,
$$\cfrac{1}{1-xy-
\cfrac{2x}{1-xy-
\cfrac{x}{1-xy-
\cfrac{x}{1-xy-
\cfrac{x}{1-xy-\cdots}}}}}.$$

\noindent Taking the product of this matrix with $\mathbf{B}^{-1}$ we obtain the matrix with general term 
$\binom{n}{k}^2$. This is the $h$-vector array for $B_n$. Its generating function is thus expressible as
$$\cfrac{1}{1-x(y-1)-
\cfrac{2x}{1-
\cfrac{x+x(y-1)}{1-
\cfrac{x}{1-
\cfrac{x+x(y-1)}{1-
\cfrac{x}{1-\cdots}}}}}},$$ or
$$\cfrac{1}{1-xy+x-
\cfrac{2x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}}}.$$ \noindent It may also be expressed as
$$\cfrac{1}{1-xy+x-
\cfrac{2x}{1-xy+x-
\cfrac{x}{1-xy+x-
\cfrac{x}{1-xy+x-
\cfrac{x}{1-xy+x-\cdots}}}}}.$$
This is thus the Deleham array
$$[-1,2,0,1,0,1,\ldots]\, \Delta^{(1)}\,[1,0,1,0,1,0,\ldots].$$
\end{example}

\begin{example} \textbf{The $f$- and $h$-vectors for $A_n$}.
The triangle with general term
$$\frac{1}{k+1}\binom{n}{k}\binom{n+k+2}{k}$$ is given by
$$[1,0,1,0,1,\ldots] \quad \Delta^{(1)} \quad
[1,1,1,1,\ldots].$$ This is the coefficient array for the
$f$-vector for $A_n$ \cite{Burgiel, Chapoton}. We recall that
$$[1,0,1,0,1,\ldots]\, \Delta \,
[1,1,1,1,\ldots]$$ has generating function
$$\cfrac{1}{1-
\cfrac{x+xy}{1-
\cfrac{xy}{1-
\cfrac{x+xy}{1-\cdots}}}},$$ and thus the triangle $$[1,0,1,0,1,\ldots] \,
\Delta^{(1)} \,
[1,1,1,1,\ldots]$$ has generating function
$$\cfrac{1}{1-(x+xy)-
\cfrac{xy}{1-
\cfrac{x+xy}{1-
\cfrac{xy}{1-\cdots}}}}.$$ This is the array \seqnum{A033282}
that begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \ldots \\1 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 5 &amp; 5 &amp;
0 &amp; 0
&amp; 0 &amp; \ldots \\ 1 &amp; 9 &amp; 21 &amp; 14 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 14 &amp;
56 &amp; 84 &amp; 42 &amp; 0 &amp; \ldots \\1 &amp; 20  &amp; 120 &amp; 1300 &amp; 330 &amp; 132
&amp;\ldots\\ \vdots &amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
Reversing this triangle to get the triangle with general element
$$[k\le n] \frac{1}{n-k+1}\binom{n}{k}\binom{2n-k+2}{n-k},$$ and then forming the product of this matrix and 
$\mathbf{B}^{-1}$, we obtain the coefficient array of the $h$-vector for $A_n$ \cite{Fomin_Reading}. This turns out to be 
the version of the triangle of Narayana numbers which begins \begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 
0
&amp;
0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots
\\ 1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp;
\ldots \\ 1 &amp; 10 &amp; 20 &amp; 10 &amp; 1 &amp; 0 &amp; \ldots \\1 &amp; 15 &amp; 50 &amp; 50 &amp;
15
&amp; 1 &amp;\ldots\\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath} and has
generating function
$$\cfrac{1}{1-x-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}}.$$
\end{example}
\begin{example} \textbf{The Narayana numbers \seqnum{A090181}.} In this example, we start with the number array with 
general element
$$\binom{n+k}{2k}c_k=\frac{1}{k+1}\binom{n}{k}\binom{n+k}{k}.$$ This is \seqnum{A088617}, whose terms count the number of 
Schr\"oder paths from $(0,0)$ to $(2n,0)$ with $k$ up-steps.
It has generating function
$$\cfrac{1}{1-x-
\cfrac{xy}{1-x-
\cfrac{xy}{1-x-
\cfrac{xy}{1-x-
\cfrac{xy}{1-x-\cdots}}}}}.$$
This may also be expressed as
$$\cfrac{1}{1-x-
\cfrac{xy}{1-
\cfrac{x+xy}{1-
\cfrac{xy}{1-
\cfrac{x+xy}{1-\cdots}}}}},$$ which
is
$$[1,0,1,0,1,0,\ldots]\, \Delta^{(1)}\,[0,1,1,1,1,1,\ldots].$$
Reversing the above array to get the array
$$[0,1,1,1,1,1,\ldots]\, \Delta^{(1)}\,[1,0,1,0,1,0,\ldots],$$ we see that this array, which has general term $$[k\le 
n]\binom{2n-k}{k}c_{n-k}=[k \le n]\frac{1}{n-k+1}\binom{n}{k}\binom{2n-k}{n-k},$$ has generating function
$$\cfrac{1}{1-xy-
\cfrac{x}{1-
\cfrac{x+xy}{1-
\cfrac{x}{1-
\cfrac{x+xy}{1-
\cfrac{x}{1-\cdots}}}}}},$$ or equivalently,
$$\cfrac{1}{1-xy-
\cfrac{x}{1-xy-
\cfrac{x}{1-xy-
\cfrac{x}{1-xy-
\cfrac{x}{1-xy-\cdots}}}}}.$$
Taking the product of this matrix with $\mathbf{B}^{-1}$ we obtain the matrix of the Narayana numbers that begins 
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \ldots \\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 1 &amp;
0 &amp; 0
&amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 6 &amp;
6
&amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 1  &amp; 10 &amp; 20 &amp; 10 &amp; 1 &amp;\ldots\\ \vdots
&amp;
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath} which is the
Narayana
triangle \seqnum{A090181}. Its generating function is thus expressible as
$$\cfrac{1}{1-x(y-1)-
\cfrac{x}{1-
\cfrac{x+x(y-1)}{1-
\cfrac{x}{1-
\cfrac{x+x(y-1)}{1-
\cfrac{x}{1-\cdots}}}}}},$$ or
$$\cfrac{1}{1-xy+x-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}}}.$$ \noindent It may also be expressed as
$$\cfrac{1}{1-xy+x-
\cfrac{x}{1-xy+x-
\cfrac{x}{1-xy+x-
\cfrac{x}{1-xy+x-
\cfrac{x}{1-xy+x-\cdots}}}}}.$$
This is thus the Deleham array
$$[-1,1,0,1,0,1,\ldots]\, \Delta^{(1)}\,[1,0,1,0,1,0,\ldots].$$
In fact, it is also expressible as the Deleham array
$$[0,1,0,1,0,1,\ldots]\, \Delta\,[1,0,1,0,1,0,\ldots]$$ with generating function
$$\cfrac{1}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}}.$$
\end{example}

\section{Acknowledgements} The author is happy to acknowledge the input of an anonymous referee, whose careful reading of 
the original text and cogent comments have hopefully led to a clearer paper.

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\bibitem{Chapoton} F. Chapoton, Enumerative properties of generalized associahedra, \emph{S\'eminaire Lotharingien de
    Combinatoire}, \textbf{51} (2004), Article B51b.

\bibitem{Fomin_Reading} S. Fomin, N. Reading, Root systems and generalized associahedra, Preprint,
    \href{http://arxiv.org/abs/math/0505518}{http://arxiv.org/abs/math/0505518}.

\bibitem{Franssens} G.~R. Franssens, On a number pyramid related to the
binomial, Deleham, Eulerian, MacMahon and
Stirling number triangles, \emph{J. Integer Sequences}, {\bf 9} (2006)
    Article 06.4.1, published electronically at
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Franssens/franssens13.html}{\texttt{http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Franssens/franssens13.html}}

\bibitem{Concrete} I. Graham, D. E. Knuth \&amp; O. Patashnik,
    \emph{Concrete Mathematics}, Addison-Wesley, 
    1995.

\bibitem{Appell} C. Kimberling, Matrix transformations of integer sequences,
\emph{J. Integer Sequences}, {\bf 6} (2003) Article 03.3.3.

\bibitem{Merlini_MC} D. Merlini, R. Sprugnoli and M.~C.
    Verri, The method of coefficients, \emph{Amer. Math. Monthly}, \textbf{114} (2007), 40--57.

\bibitem{Postnikov} A. Postnikov, V. Reiner, and  L. Williams,
    Faces of generalized permutohedra, \emph{Documenta Math.}
    \textbf{13} (2008) 207--273.

\bibitem{SGWW} L.~W.~Shapiro, S. Getu, W-J. Woan and L~.C.~Woodson,
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\bibitem {SL1} N.~J.~A.~Sloane, {\em \htmladdnormallink {The On-Line Encyclopedia of Integer
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\end{thebibliography}

\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary  15A36, 30B70, 15A04.

\noindent \emph{Keywords:} Integer sequence, Riordan array, continued fraction, generating function, linear
transformation, Deleham construction.

\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences
\seqnum{A000108},
\seqnum{A000984},
\seqnum{A001850},
\seqnum{A006318},
\seqnum{A006319},
\seqnum{A023431},
\seqnum{A026418},
\seqnum{A063007},
\seqnum{A084938},
\seqnum{A086581},
\seqnum{A086615},
\seqnum{A088617},
\seqnum{A090181},
\seqnum{A090344},
\seqnum{A098479},
\seqnum{A101499},
\seqnum{A105864},
\seqnum{A124644}, and
\seqnum{A131198}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 7 2009;
revised version received  November 2 2009.
Published in {\it Journal of Integer Sequences}, November 4 2009.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                

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