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\begin{center}
\vskip 1cm{\LARGE\bf
On the Total Positivity of \\
\vskip .1in
Delannoy-Like Triangles
}
\vskip 1cm
\large
Lili Mu\\
School of Mathematics \\
Liaoning Normal University\\
Dalian, 116029\\
PR China\\
\href{mailto:lly-mu@hotmail.com }{\tt lly-mu@hotmail.com} \\
\ \\
Sai-nan Zheng\\
School of Mathematical Sciences \\
Dalian University of Technology \\
Dalian, 116024\\
PR China\\
\href{mailto: zhengsainandlut@hotmail.com}{\tt zhengsainandlut@hotmail.com}\\
\end{center}

\vskip .2in

\begin{abstract}
Define an infinite lower triangular matrix $D(e,h)=[d_{n,k}]_{n,k\geq 0}$ by the recurrence
$$d_{0,0}=d_{1,0}=d_{1,1}=1,\ d_{n,k}=d_{n-1,k-1}+ed_{n-1,k}+hd_{n-2,k-1},$$
where $e$, $h$ are both nonnegative and $ d_{n,k}=0$ unless $n\geq k \geq 0$. We call $D(e,h)$ the {\it Delannoy-like triangle}.
The entries $d_{n,k}$ count lattice paths from $(0,0)$ to $(n-k,k)$
using the steps $(0,1)$, $(1,0)$ and $(1,1)$ with assigned weights $1$, $e$,
and $h$.
 Some well-known combinatorial triangles are such matrices, including the Pascal triangle
 $D(1,0)$, the Fibonacci triangle $D(0,1)$, and the Delannoy triangle $D(1,1)$. Futhermore, the
 Schr\"oder triangle and Catalan triangle also arise as inverses of Delannoy-like triangles. Here we investigate the total positivity of Delannoy-like triangles. In addition, we show that each row and diagonal of Delannoy-like triangles are all PF sequences.
\end{abstract}


\section{Introduction}\label{Intro}

\hspace*{\parindent}
The Delannoy number $d(n,k)$ is defined as the number of lattice paths from $(0,0)$ to $(n,k)$
with steps $(0,1)$, $(1,0)$ and $(1,1)$.
Banderier and Schwer \cite{BS05} gave historical background on
Delannoy numbers. There is a link between Legendre
polynomials and Delannoy numbers \cite{Good58, Law52}.
The Delannoy triangle $D=[t_{n,k}]_{n,k\geq 0}$ is an infinite lower
triangular matrix defined by $t_{n,k}=d(n-k,k)$. 
Its entries satisfy the recurrence
$$t_{0,0}=1,\ t_{n,k}=t_{n-1,k-1}+t_{n-1,k}+t_{n-2,k-1},$$
where $t_{n,k}=0$ unless $n\geq k \geq 0$. The first few entries of $D$ are as follows:
$$D=\left[
    \begin{array}{ccccccc}
      1 &  & &  & &  & \\
      1 & 1 &  & &  & & \\
      1 & 3 & 1 & &  &  & \\
      1 & 5 & 5 & 1 &  & & \\
      1 & 7 & 13 & 7 & 1 &  & \\
      1 & 9 & 25 & 25 & 9 & 1 & \\
    \vdots & \vdots &\vdots & \vdots & \vdots & \vdots & \ddots\\
    \end{array}\right].$$
An immediate calculation shows that the row sums of Delannoy triangle form
the Pell sequence \cite{B75}.
The unsigned inverse of $D$ is the Schr\"oder triangle (\seqnum{A132372}).
The central coefficients $t_{2n,n}$ are called the central Delannoy numbers 
(\seqnum{A001850}) and have appeared in several
problems, such as the alignments between DNA sequences \cite{TCN03}.

In this paper, we study the infinite lower triangular matrix
$D(e,h)=[d_{n,k}]_{n,k\geq 0}$ defined by the recurrence
\begin{equation} \label{delannoy}
d_{0,0}=d_{1,0}=d_{1,1}=1,\ d_{n,k}=d_{n-1,k-1}+ed_{n-1,k}+hd_{n-2,k-1},
\end{equation}
where $e$, $h$ are both nonnegative and $ d_{n,k}=0$ unless $n\geq k \geq 0$. We call $D(e,h)$ the {\it Delannoy-like triangle}.
The first few rows of a Delannoy-like triangle are as follows:
$$D(e,h)=\left[
    \begin{array}{cccccc}
      1 &  & &  &   \\
      1 & 1 &  & &    \\
      e & 1+e+h & 1 & &   \\
      e^2 & e^2+2e+eh+h & 1+2e+2h & 1 &  \\
    \vdots & \vdots &\vdots &  & \ddots \\
    \end{array}\right].$$
The special cases are Delannoy triangle $D(1,1)$,
the Pascal triangle (\seqnum{A007318}) $D(1,0)$,
and the Fibonacci triangle (\seqnum{A026729}) $D(0,1)$.

On the other hand, we let $S(e,h)$ denote the unsigned inverse of $D(e,h)$,
i.e., $S(e,h)=MD(e,h)^{-1}M$, where $M$ is the diagonal matrix with diagonal
entries alternately $1$ and $-1$.
We call $S(e,h)$ the {\it generalized Schr\"oder matrix} \cite{YZYH13}.
$$S(e,h)=\left[
    \begin{array}{cccccc}
      1 &  & &  &   \\
      1 & 1 &  & &    \\
      1+h & 1+e+h & 1 & &   \\
      1+2h+eh+2h^2 & e^2+e+3eh+2h^2+2h+1 & 1+2e+2h & 1 &  \\
    \vdots & \vdots &\vdots &  & \ddots \\
    \end{array}\right].$$
Such matrices include the Catalan triangle (\seqnum{A033184})
with $e=0$, $h=1$,
and Schr\"oder triangle (\seqnum{A132372})
with $e=1$, $h=1$.
$$S(1,1)=\left[
    \begin{array}{cccccc}
      1 &  & &  & &   \\
      1 & 1 &  & &  &  \\
      2 & 3 & 1 & &  &   \\
      6 & 10 & 5 & 1 &  &  \\
      22 & 38 & 22 & 7 & 1 &   \\
    \vdots & \vdots &\vdots & \vdots & \vdots & \vdots \\
    \end{array}\right].$$
The row sums of  $S(1,1)$ are the large
Schr\"oder numbers \cite{C74}.

An infinite matrix is called {\it totally positive} ({\it TP}, for short)
if its minors of all orders are nonnegative.
The theory of totally positive matrices is rich with
deep and highly non-trivial results,
and has been studied extensively \cite{CLW-EuJC, Pin10}.
Brenti \cite{Bre95} showed the total positivity of  the
Delannoy square (\seqnum{A008288}) $[d(n,k)]_{n,k\geq0}$ by giving
a combinatorial interpretation of its minors in terms of nonintersecting
paths in a digraph.
It is natural to conjecture that Delannoy-like triangles are totally positive.
The primary purpose of this paper is to show this is true.

\begin{theorem}
The Delannoy-like triangles
defined by \eqref{delannoy} are totally positive matrices.
\label{gf}
\end{theorem}

We give the  proof of this, our main theorem,
by showing the total positivity of the generalized Schr\"oder matrix
$S(e,h)$ in Section \ref{proof}. It turns out that the Pascal triangle,
the Fibonacci triangle, the Delannoy triangle, the Catalan triangle, and
the Schr\"oder triangle  are all TP matrices.


In Section \ref{remark},
we briefly consider the PF sequence (see Section \ref{remark} for
definitions) in Delannoy-like triangles.  With a simple proof, we
conclude that each row and diagonal of Delannoy-like triangles all form
PF sequences.

\section{Proof of Theorem \ref{gf}}\label{proof}
\hspace*{\parindent}
To prove the total positivity of Delannoy-like triangles $D(e,h)$, it suffices to show that the generalized Schr\"oder matrix $S(e,h)$ is TP, because the unsigned inverse of a TP matrix is still TP \cite[Prop. 1.6]{Pin10},
and $D(e,h)$ is also the unsigned inverse of $S(e,h)$.
Actually, since the $D(e,h)$ are Riordan arrays (we will prove it first),
so are $S(e,h)$.
This implies that $S(e,h)$ possess $A$-sequences and $Z$-sequences,
since any Riordan array can be characterized by the $A$- and $Z$-sequences uniquely \cite{HS09}.
Therefore, our idea is to construct a class of Riordan arrays with $A$-sequences and $Z$-sequences having a certain form, and make sure $S(e,h)$ are such matrices. Then $S(e,h)$ is TP if we show this class of Riordan arrays is TP (Claim \ref{az}).

We let $(g(x),f(x))$  denote a Riordan array $R=[r_{n,k}]_{n,k\geq 0}$ \cite{SGWW91}, and $(g(x),f(x))$
is an infinite lower triangular matrix whose generating function
of the $k$th column is $g(x)f^k(x)$ for $k=0,1,2,\ldots,$
where $g(x)$  and $f(x)$  are formal power series with
$g(0)=1$ and $f(0)=0$, $f'(0)\neq 0$.
Suppose we multiply the array $(g(x), f(x))$ by a column vector
    $ (b_0,b_1, b_2, \ldots )^T  $ with    generating function
  $b(x)$.  Then we get a column vector whose  generating function
  is given by $g(x)b( f (x))$.
     If we identify a sequence  with its   generating
   function, the composition rule can be rewritten as
      \begin{equation} \label{vector}
         (g(x), f(x)) b(x) = g(x)b( f (x)).
     \end{equation}
This is called the fundamental theorem for Riordan arrays and this lead to the multiplication
rule for the Riordan arrays (see Shapiro et al.~\cite{SGWW91}):
\begin{equation}\label{multi}
\left(g(x),f(x)\right)\left(d(x),h(x)\right)=\left(g(x)d(f(x)),h(f(x))\right).
\end{equation}
The inverse of $(g(x),f(x))$ is
\begin{equation} \label{inverse}
(g(x),f(x))^{-1}=(1/g(\bar{f}(x)),\bar{f}(x)),
\end{equation}
where $\bar{f}(x)$ is the compositional inverse of $f(x)$,
such that $f(\bar{f}(x))=\bar{f}(f(x))=x$.
The bivariate generating function $R(x,y)$ of the Riordan array $R$ is given
by
$$R(x,y)=(g(x),f(x))\frac{1}{1-yx}=\frac{g(x)}{1-yf(x)},$$
following \eqref{vector}.

A Riordan array $R=[r_{n,k}]_{n,k\geq 0}$ can also be characterized
by two sequences $(a_n)_{n\geq 0}$ and $(z_n)_{n\geq 0}$ such that
$$r_{0,0}=1,\ r_{n+1,0}=\sum_{j\geq 0}z_jr_{n,j},\ r_{n+1,k+1}=\sum_{j\geq 0}a_jr_{n,k+j}$$
for $n,k \geq 0$ (see \cite{Spr94} for instance).
Call $(a_n)_{n\geq 0}$ and $(z_n)_{n\geq 0}$ the $A$- and $Z$-sequences of $R$,
respectively.
Following \cite{CLW-EuJC}, call
$$J(R)=\left[
       \begin{array}{cccccc}
           z_{0} & a_{0} &  & & \\
           z_{1} & a_{1} &a_{0}  &  & \\
           z_{2} &a_{2} & a_{1} & a_{0}&  \\
           z_{3} &a_{3} & a_{2} & a_{1} & a_{0} \\
            \vdots & \vdots &  & \cdots &  & \ddots \\
      \end{array}
      \right]$$
the {\it coefficient matrix} of the Riordan array $R$.
Let $A(x)$ and $Z(x)$ be the generating functions of $A$-sequence and $Z$-sequence respectively.
Then
\begin{equation}\label{AZ}
A(x)=\frac{x}{\bar{f}(x)};\qquad  Z(x)=\frac{1}{\bar{f}(x)}\left(1-\frac{1}{g(\bar{f}(x))}\right).
\end{equation}

Now we start by proving that Delannoy-like triangles are Riordan arrays,
and deduce the generating functions
of $A$-sequence and $Z$-sequence of $S(e,h)$.
Let $B(x,y)$ denote
the bivariate generating function of $D(e,h)$.
By \eqref{delannoy}, we have
\begin{align*}
B(x,y)&=\sum_{n=0}^ {\infty}\sum_{k=0}^{\infty}d_{n,k}x^ny^k\\
&=1+x+xy+\sum_{n=2}^ {\infty}\sum_{k=0}^{\infty}d_{n,k}x^ny^k\\
&=1+x+xy+\sum_{n=2}^ {\infty}\sum_{k=0}^{\infty}(d_{n-1,k-1}+ed_{n-1,k}+hd_{n-2,k-1})x^ny^k\\
&=1+x+xy+xy(B(x,y)-1)+ex(B(x,y)-1)+hx^2yB(x,y).
\end{align*}
It follows that
\begin{align*}
B(x,y)&=\frac{1+x-ex}{1-ex-xy-hx^2y}=\frac{1+x-ex}{1-ex}\frac{1}{1-y\frac{x+hx^2}{1-ex}}.
\end{align*}
Thus,
$$D(e,h)=(d(x),h(x))=\left(\frac{1+x-ex}{1-ex},\frac{x+hx^2}{1-ex}\right).$$
Then $$(S(e,h))^{-1}=(1,-x)D(e,h)(1,-x)=
(d(-x),-h(-x))=\left(\frac{1-x+ex}{1+ex},\frac{x-hx^2}{1+ex}\right),$$
since $S(e,h)=(1,-x)D^{-1}(e,h)(1,-x)$ and \eqref{multi}.
From \eqref{inverse} and \eqref{AZ} ,
we have
$${A}^{\ast}(x)=\frac{x}{-h(-x)}=\frac{1+ex}{1-hx};\quad
{Z}^{\ast}(x)=\frac{1}{-h(-x)}\left(1-d(-x)\right)=\frac{1}{1-hx},$$
where ${A}^{\ast}(x)$ and ${Z}^{\ast}(x)$ are the generating functions of $A$-sequence and $Z$-sequence of $S(e,h)$, respectively.

Next we construct a class of Riordan arrays by expanding the
$A$-sequence of $S(e,h)$ and have the following result.

\begin{lemma}\label{az}
Let $A(x)$ and $Z(x)$ be the generating functions of $A$-sequence and $Z$-sequence of
Riordan array $R$.
Suppose that $Z(x)=\frac{1+ux}{1-hx}$ and $A(x)=Z(x)(1+ex)$, where
$e,h,u\geq 0$.
Then $R$ is TP.
\end{lemma}

\begin{proof}
The first few rows of the coefficient matrix of $R$ are as follows:
$$J(R)=
\left[
\begin{array}{cccccc}
    1   & 1       &        &     \\
    u+h   & e+h+u     &1       &   \\
    (u+h)h & ue+eh+uh+h^2   & e+h+u    & 1    \\
    (u+h)h^2 &(ue+eh+uh+h^2)h & ue+eh+uh+h^2 & e+h+u  \\
    \vdots & \vdots &  & \cdots &  & \ddots\\
\end{array}
\right].$$
A Riordan array is totally positive if its coefficient matrix is TP \cite{CLW-EuJC}.
That means our next task is to show the total positivity of $J(R)$.
Let
$$H=
\left[
\begin{array}{ccccc}
     1     &      &     & \\
     h     &1     &     & \\
     h^2   & h    & 1   &  \\
     h^3   & h^2  & h   & 1 \\
      \vdots &  & \cdots &  & \ddots \\
\end{array}
\right],
\quad
Q=
\left[
\begin{array}{cccccc}
   1  & 1 &   &     & \\
   u  & e+u &1  &     & \\
      & eu &e+u  & 1   &  \\
      &   & eu & e+u  & 1 \\
       &  &   &    & \ddots & \ddots \\
\end{array}
\right].$$
Then $J(R)$ is obtained from $Q$ by adding $h$ times each row
to the succeeding row.
It follows that  $J(R)=HQ$.
Note that the product of two totally positive matrices is TP.
Hence, it suffices to show that $H$ and $Q$ are TP.
The coefficient matrix of $H$ is
$$J(H)=
\left[
\begin{array}{ccccc}
   h  & 1 &   &     \\
     &  0& 1 &     \\
      &  &  0&1      \\
       &  &   & &  \ddots  \\
\end{array}
\right].$$
Thus $H$ is TP since $J(H)$ is TP.
Note that
$$Q=
\left[
\begin{array}{cccccc}
   1  &  &   &     & \\
   u & 1 &  &     & \\
      &  u&1  &    &  \\
      &   & u& 1  &  \\
       &  &   &  \ddots  & \ddots &  \\
\end{array}
\right]
\left[
\begin{array}{cccccc}
   1  & 1 &   &     & \\
      & e &1  &     & \\
      &  &e  & 1   &  \\
      &   &  & e  & 1 \\
       &  &   &    & \ddots & \ddots \\
\end{array}
\right].$$
Clearly, $Q$ is TP if $e,u\geq 0$.
\end{proof}
Therefore,  $S(e,h)$ is TP by the case of $u=0$ in Lemma \ref{az},
and so $D(e,h)$ is, too.

\section{Remarks}\label{remark}
\hspace*{\parindent}
There are various total positivity properties of Riordan arrays. For example,
the Pascal triangle $P$ is a totally positive matrix and each row of $P$ is a
P\'{o}lya frequency sequence.
We now consider similar properties of Delannoy-like triangles.

An infinite sequence $(a_n)_{n\geq 0}$ is called
a {\it P\'{o}lya frequency sequence} (or shortly, a {\it PF} sequence) if its Toeplitz matrix
$[a_{i-j}]_{i,j\geq 0}$
is TP.
P\'{o}lya frequency sequences
often arise in combinatorics  \cite{Bre89}.
 A fundamental representation theorem of Schoenberg and Edrei
states that a sequence $(a_n)_{n\geq 0}$ of real numbers
is PF if and only if its generating function has the form
$$\sum_{n\geq 0}a_nz^n=\frac{\prod_{j\geq 1}(1+\alpha_jz)}{\prod_{j\geq 1}(1-\beta_jz)}e^{\gamma z}$$
in some open disk center at the origin, where $\alpha_j, \beta_j, \gamma \geq 0$
and $\sum_{j\geq 1}(\alpha_j+\beta_j)<+\infty$
(see Karlin \cite[p.\ 412]{Kar68}, for instance).
Aissen, Schoenberg and Whitney showed that
a finite sequence of nonnegative numbers is PF if and only if
its generating function has only real zeros \cite[p.\ 399]{Kar68}.


We let ${r_n(x)}$ denote the $n$th row generating function
$\sum_{k\geq0}d_{n,k}x^{k}$ of a Delannoy-like triangle $D(e,h)$. Then
${r_n(x)}$ satisfies the recurrence
$$r_0(x)=1,\ r_1(x)=e+x,\ r_n(x)=(e+x)r_{n-1}+hxr_{n-2}(x).$$ 
It is easy to check that $r_n(x)$ has only real zeros \cite[Theorem
2.1]{LW07}.  Hence each row of a Delannoy-like triangle is a PF
sequence.

Moreover,
let ${m_n(x)}$ denote
the $n$th diagonal generating function $\sum_{k\geq0}d_{n+k,k}x^{k}$ of $D(e,h)$. Then ${m_n(x)}$ satisfies the recurrence
$$m_n(x)=\frac{e+hx}{1-x}m_{n-1}(x),$$
where $m_0(x)=1/(1-x)$, and
we have 
$$m_n(x)=\frac{(e+hx)^{n-1}}{(1-x)^n}.$$
This means that each diagonal of
Delannoy-like triangles is also a PF sequence.

Thus we have the following result:
\begin{theorem}
Each row and diagonal of Delannoy-like triangles form PF sequences.
\end{theorem}
In addition,
the generating function of  the $n$th column  of $D(e,h)$ is
$$g(x)f^{n}(x)=\frac{(1+x-ex)(x+hx^2)^n}{(1-ex)^{n+1}}.$$
Hence each column of a Delannoy-like triangle is a PF sequence if $e=0,1$.
It follows that each column of the Delannoy triangle,
the Pascal triangle and the Fibonacci triangle
is a PF sequence.
However, not all lines of every Delannoy-like triangle are PF sequences.
For example, the central Delannoy numbers, the central coefficient of the Delannoy triangle, is a log-convex sequence \cite{Ben11}.

\section{Acknowledgments}
The authors would like to thank the anonymous referees for their careful reading and helpful comments 
which have hopefully led to a clearer paper. 

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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 05A20; Secondary 05B36, 15A45. \\
\noindent \emph{Keywords: } Delannoy number, Schr\"oder number, totally positive matrix.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001850},
\seqnum{A007318},
\seqnum{A008288},
\seqnum{A026729}, and
\seqnum{A132372}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent
Received June 29 2016;
revised version received  December 20 2016.
Published in {\it Journal of Integer Sequences}, December 26 2016.

\bigskip
\hrule
\bigskip

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