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\title{On total positivity of Catalan-Stieltjes matrices}

\author{Qiongqiong Pan and Jiang Zeng\\
\small Univ Lyon, Universit\'e Claude Bernard Lyon 1\\[-0.8ex]
\small CNRS UMR 5208, Institut Camille Jordan \\[-0.8ex]
\small 43 blvd. du 11 novembre 1918\\[-0.8ex]
\small F-69622 Villeurbanne cedex, France\\
\small\texttt{qpan@math.univ-lyon1.fr, zeng@math.univ-lyon1.fr}}
\date{\dateline{Jul 1, 2016}{Nov 5, 2016}{Nov 25, 2016}\\
\small Mathematics Subject Classifications: 05A18; 05A15, 05A30}


%\date{\today}
\begin{document}
\maketitle

\begin{abstract}
Recently Chen-Liang-Wang~(Linear Algebra Appl. {\bf 471} (2015) 383--393) 
present some sufficient conditions for the 
total positivity of Catalan-Stieltjes matrices. Our aim is to provide a combinatorial interpretation of their
sufficient conditions.  More precisely,  for any 
Catalan-Stieltjes matrix $A$ we construct a digraph with  a weight, which is   positive
under their sufficient conditions,  
such that  every minor of $A$  is equal to the sum of  the weights of
 families of nonintersecting paths of the digraph. 
 We have  also an  analogous  result for the minors of a
Hankel matrix associated to the first column of a
Catalan-Stieltjes matrix.
\end{abstract}
%\keywords{Totally positive matrix; Catalan-Stieltjes matrix, Hankel matrix}

%\tableofcontents

\section{Introduction}
The study of totally positive matrices appears in various areas  
such as orthogonal polynomials, combinatorics, algebraic geometry, stochastic processes, game theory, differential equations, representation theory, Brownian motion, electrical networks, and chemistry; see \cite{Karlin-Coincidence, Brenti95, Brenti96, Pinkus-1,Fallat-Johnson11}. 
In this paper we shall consider the total positivity of 
some special lower triangular matrices.  Recall that an infinite real matrix $M$ is said to 
be  \emph{totally positive} (TP) if  every 
 minor of $M$ is \emph{nonnegative}. 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Def}
An  infinite lower triangular matrix of real numbers $A^{\gamma, \sigma, \tau} :=(a_{n,k})_{n,k\geq 0}$ is called  a \emph{Catalan-Stieltjes matrix } if there are three sequences of positive numbers
$\gamma:=(r_{k})_{k\geq 0}$, $\sigma:=(s_{k})_{k\geq 0}$ and  $ \tau:=(t_{k})_{k\geq 1}$
 such that
\begin{align}\label{C-S}
a_{n,0}& = s_{0}a_{n-1,0} + t_{1}a_{n-1,1};\nonumber\\
 a_{n,k} &= r_{k-1}a_{n-1,k-1} + s_{k}a_{n-1,k} + t_{k+1}a_{n-1,k+1} \quad  (k\ge 1,\;n\geq 1),
\end{align}
where  $a_{0,0} = 1$ and $a_{n,k} = 0$ unless $n\geq k \geq 0$.
\end{Def}
Although a matrix defined by \eqref{C-S} is 
called  a \emph{Catalan matrix} in \cite[p. 291]{Aigner-course},  we prefer to call it a Catalan-Stieltjes matrix 
because, 
when $r_k=1$,  Stieltjes~\cite{Stieltjes89} first 
introduced  such a matrix  in his step-by-step method to 
expand a continued J-fraction into a power series;  see \cite[Section 53]{Wall73} and \cite{Flajolet80}.  
Indeed,  the matrix \eqref{C-S} implies
 the following continued fraction expansion of the ordinary generating function 
 of the first column of  $A^{\gamma, \sigma, \tau}$:
\begin{align}\label{CF}
\sum_{n=0}^{\infty}a_{n,0}z^n =
 \cfrac{1}{1-s_0z-\cfrac{\lambda_1z^2}{1-s_1z-\cfrac{\lambda_2z^2}{1-s_2z-\cdots}}},
\end{align}
where $\lambda_{k+1}=r_kt_{k+1}$ for $k\geq 0$.
The sequence $(a_{n,0})$ is usually called a \emph{moment sequence} 
in the theory of orthogonal polynomials (see \cite{Vi83}).
 Conversely,  starting from \eqref{CF} with nonnegative $s_k$ and $\lambda_{k+1}$ ($k\geq 0$),
 and  any factorisation $r_kt_{k+1}$ of 
 $\lambda_{k+1}$ such that $r_k, t_{k+1}\geq 0$ ($k\geq 0$), we can recover 
  the moment sequence  $(a_{n,0})$ using matrix\eqref{C-S}. 

Recently Chen-Liang-Wang~\cite{CLW15} proved  some sufficient conditions for the total positivity of Catalan-Stieltjes matrices. 
  At the end of their paper they asked for  a combinatorial interpretation of their results.
  The aim of this paper is to present such a combinatorial  interpretation using a classical lemma  of 
Lindstr\"om~\cite{Linds73}.
As in \cite{Brenti95, GV85, FZ00}, our strategy  is   to first interpret the matrix $A$ as
 a path matrix of some planar network  within two  
  boundary vertex sets,  and then  apply 
 Lindstr\"{o}m's lemma~\cite{Linds73} to write every minor of $A$ as a sum of positive 
 weights of families of nonintersecting paths.   
 We first recall some basic definitions of this methodology.
Let $G=(V,E)$ be an infinite acyclic digraph, where  $V$ is the  vertex set and $E$ the edge set. 
 If  $S:=(A_i)_{i\geq 0}$ and $T:=(B_i)_{i\geq 0}$  are  two  sequences of vertices in $G$, we say that
 the triple $(G, S, T)$ is a \emph{network}.
We  assume that there is a   weight function 
$w: E\to {\mathbb R}$ and  define the  {\em associated path matrix} $M = (m_{i,j})_{i,j\geq 0} $ by
$$
m_{i,j} = \sum_{\gamma: A_i\to B_j}w(\gamma), 
$$
where the sum is over all the  paths 
$\gamma$ from $A_i$ to $B_j$ and the {\em weight} of a path is 
 the product of its edge weights.  By convention we define $m_{i,j}=1$ if $A_i=B_j$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[t]
\centering
\begin{tikzpicture}%[scale=0.8]
\draw[->] (0,0)--(-9,0);
\draw[->] (0,0)--(0,4);
\coordinate [label=left:$x$] (x) at (-9,0);
\coordinate [label=right:$y$] (y) at (0.25,4.2);
\coordinate [label=right:$B_1$] (s1) at (0,1);
\coordinate [label=right:$B_2$] (s2) at (0,2);
\coordinate [label=right:$B_3$] (s3) at (0,3);
\node[below] at (0,0) {$A_0$};
\node[below] at (-1,0) {$A_1$};
\node[below] at (-2,0) {$A_2$};
\node[below] at (-3,0) {$A_3$};
\node[below] at (-4,0) {$A_4$};
\node[below] at (-5,0) {$A_5$};
\node[below] at (-6,0) {$A_6$};
\node[below] at (-7,0) {$A_7$};
\node[below] at (-8,0) {$A_8$};

\coordinate [label=right:$B_0$] (0) at (0,0);
\draw[dashed](0,1)--(-8,1);
\draw[dashed](0,2)--(-8,2);
\draw[dashed](0,3)--(-8,3);
%\draw[dashed](0,4)--(8,4);
\draw[dashed](-1,0)--(-1,3);
\draw[dashed](-2,0)--(-2,3);
\draw[dashed](-3,0)--(-3,3);
\draw[dashed](-4,0)--(-4,3);
\draw[dashed](-5,0)--(-5,3);
\draw[dashed](-6,0)--(-6,3);
\draw[dashed](-7,0)--(-7,3);
\draw[dashed](-8,0)--(-8,3);
%%%%
\draw[thick,->] (-8,0)--(-7,1);
\draw[thick,->] (-7,1)--(-6,1);
\draw[thick,->] (-6,1)--(-5,2);
\draw[thick,->] (-5,2)--(-4,3);
\draw[thick,->] (-4,3)--(-3,3);
\draw[thick,->] (-3,3)--(-2,2);
\draw[thick,->] (-2,2)--(-1,3);
\draw[thick,->] (-1,3)--(0,2);
%%%%%%%%%
\draw[red, thick,->] (-6,0)--(-5,1);
\draw[red, thick,->] (-5,1)--(-4,1);
\draw[red, thick,->] (-4,1)--(-3,2);
\draw[red, thick,->] (-3,2)--(-2,3);
\draw[red, thick,->] (-2,3)--(-1,2);
\draw[red, thick,->] (-1,2)--(0,1);
\end{tikzpicture}
\caption{Two nonintersecting  paths from $(8,0)$ to $(0,2)$ and from
$(6,0)$ to $(0,1)$ in the  Motzkin network $({\mathcal M}, (A_i), (B_i))$, where only the edges in the paths are depicted.}
 \label{motzkin-path}
\end{figure}


Let $I=(i_1,\ldots, i_n)$ and $J=(j_1, \ldots, j_n)$ be two positive increasing 
integer sequences. The $I$, $J$ minor of a  matrix $M$ is defined by 
$ \det M_{I,J}$, where $M_{I,J}$ is  the submatrix of $M$ corresponding to row set $I$ and column set $J$. 
Let  $A_I= \{ A_i: i\in I\}$, $B_J = \{B_j: j\in J\}$ be  two $n$-sets of 
vertices of  $\G$, which need not be disjoint.  For any permutation $\sigma\in \S_n$ denote by 
$N(\G; A_I, B_{\sigma (J)})$  the set of 
$n$-tuples $(p_1, \ldots, p_n)$ where  $p_i$ is a path from $A_{i_k}$ to $B_{j_{\sigma(k)}}$ such that any  two paths in the tuple
are vertex-disjoint. 
 The weight of $P$ is defined by 
$w(P) =\prod_{i=1}^{n}w(p_i)$.   The following result is due to  Lindstr\"{o}m's~\cite{Linds73}. See 
also\cite{GV85,Aigner-course}.
\begin{lemma}[Lindstr\"{o}m]\label{LGV} We have 
 \be\label{GV-motzkin}
\det M_{I,J}
= \sum_{\sigma\in \S_n}\sign(\sigma)
\sum_{P\in N(\G; A_I, B_{\sigma (J)})} w(P).
\ee
\end{lemma}

The basic idea is  to find a planar graph $G$ along with two  sequences of vertices $S$ and $T$  
so that $N(G; A_I, B_{\sigma (J)})$ is empty except when $\sigma$ is identity.
As there is no general  method for constructing a simple planar network in order to prove that a given  matrix is 
 totally positive, to motivate our approach, we will start with 
 the Motzkin path description of the  matrix coefficients  $a_{n,k}$ in \eqref{C-S}.
Consider
the digraph ${\mathcal M}=(V, E)$, where 
$V=\Z\times \N$ and 
$$
E=\{(i,j)\rightarrow (i+1,j+1), \; (i,j)\rightarrow (i+1, j),\; (i,j)\rightarrow (i+1, j-1)| i, j\geq 0\}.
$$
A path in ${\mathcal M}$ is called a \emph{Motzkin path}. 
An example of two nonintersecting Motzkin  paths in ${\mathcal M}$ 
is depicted in Figure~\ref{motzkin-path}.
It is well-known and easy to verify  (see \cite{Flajolet80}) that 
the coefficient $a_{n,k}$ is equal to the sum of weights of 
Motzkin paths from 
$A_n:=(n,0)$ to  $B_k:=(0,k)$, 
where the arrows are   weighted as follows:
\begin{align*}
w((i+1,j)\rightarrow (i,j+1) )&= r_i,\;\\
w((i+1,j)\rightarrow (i, j)) &= s_j,\; \\
w((i+1,j+1)\rightarrow (i, j)) &= t_{j+1},
 \end{align*}
for $i\geq 1, j\geq 0$.  
It follows that the Catalan-Stieltjes matrix $A$ is a path matrix of the Motzkin network $({\mathcal M}, (A_i), (B_i))$.
Unfortunately, the signed expression \eqref{GV-motzkin}   
does not
manifest any  positivity for $\det A_{I,J}$ in general, except for the special  case $t_k=0$; see Proposition~\ref{mongelli}.  Actually it is easy to see  that   ${\mathcal M}$ 
is not a planar graph.

In the next section we will  suitably modify the Motzkin network along with its weight
in order to  make a planar network and recover the  total positivity  conditions  in \cite{CLW15}.
We also show how to use our path model to carry  these positivity conditions over 
to the Hankel matrix associated to
the first column of $A^{\gamma, \sigma, \tau}$. In Section~3 we sepcialize our general results
to  some well-known combinatorial matrices  as well as 
coefficientwise total positivity of their polynomial analogous.
We conclude this paper with two open problems in Section~4 and  an appendix about the computation of some Catalan-Stieltjes  matrices in Section~5.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Minors of Catalan-Stieltjes and Hankel matrices}
We consider the graph $\G=(V, E)$ where 
 \begin{itemize}
\item  the vertex set $V$ is  equal to  $V=\{(i,j),(i+\frac{1}{2},j+\frac{1}{2})\,|\, (i, j)\in \Z\times \N\}$;
\item the edge set $E$ is equal to
 \begin{align*}
 &\{(i,j)\rightarrow (i-1,j); (i,j)\rightarrow (i-\frac{1}{2}, j+\frac{1}{2});
 (i,j)\rightarrow (i-\frac{1}{2}, j-\frac{1}{2});\\ 
 &(i-\frac{1}{2}, j+\frac{1}{2})\rightarrow (i-1,j+1);(i-\frac{1}{2}, j+\frac{1}{2})\rightarrow (i-1,j)\,|\,i\in\Z, j\geq 0\}.
 \end{align*}
\end{itemize}
Comparing with the Motzkin graph ${\mathcal M}$ we see that
%adding  vertices $(i+\frac{1}{2}, j+\frac{1}{2})$ ($(i, j)\in \Z\times \N$) in Motzkin graph ${\mathcal M}$ such that
%In other words, we embed the Motzkin graph ${\mathcal M}$ in a plan  by 
 each  crossing point $(i+\frac{1}{2}, j+\frac{1}{2})$  of two edges is transformed to a vertex in $\G$
as shown below:
  \begin{figure}[h]
\centering
\begin{tikzpicture}
\draw [line width=1.5pt] (0,0)--(2,0);
\draw [line width=1.5pt] (0,0)--(2,-2);
\draw [line width=1.5pt] (0,-2)--(2,0);
\draw [line width=1.5pt] (0,-2)--(2,-2);
%%%%%%%%%%%
\draw[->,line width=1.2pt] (0,0)--(1,0);
\draw[->,line width=1.2pt] (0,-2)--(1,-2);
\draw[->,line width=1.2pt] (0,0)--(1.5,-1.5);
\draw[->,line width=1.2pt] (0,-2)--(1.5,-0.5);
%%%%%%%%%%%
\fill[color=red]  (0,0) circle (2.5pt);
\fill[color=red]  (2,0) circle (2.5pt);
\fill[color=red]  (0,-2) circle (2.5pt);
\fill[color=red]  (2,-2) circle (2.5pt);
%%%%%%%%%%%%%%%%%
\draw[->,line width=1.2pt] (2.5,-1)--(3.5,-1);
%%%%%%%%%%%%5
\draw [line width=1.5pt] (4,0)--(6,0);
\draw [line width=1.5pt] (4,0)--(6,-2);
\draw [line width=1.5pt] (4,-2)--(6,0);
\draw [line width=1.5pt] (4,-2)--(6,-2);
%%%%%%%%%%%
\draw[->,line width=1.2pt] (4,0)--(5,0);
\draw[->,line width=1.2pt] (4,-2)--(5,-2);
\draw[->,line width=1.2pt] (4,0)--(5.5,-1.5);
\draw[->,line width=1.2pt] (4,-2)--(5.5,-0.5);
\draw[->,line width=1.2pt] (4,0)--(4.5,-0.5);
\draw[->,line width=1.2pt] (4,-2)--(4.5,-1.5);
%%%%%%%%%%%
\fill[color=red]  (4,0) circle (2.5pt);
\fill[color=red]  (6,0) circle (2.5pt);
\fill[color=red]  (4,-2) circle (2.5pt);
\fill[color=red]  (6,-2) circle (2.5pt);
\fill[color=red]  (5,-1) circle (2.5pt);
\end{tikzpicture}
\caption{Planar embedding  of the Motzkin graph}
\label{plan-motzkin}
\end{figure}

\parindent=0pt
Moreover, instead of three types of arrows (or edges) in ${\mathcal M}$ there are now 
five types of arrows (or edges) in ${\mathcal G}$ that are summarized in the following diagram:
\begin{figure}[h]
\centering
\begin{tikzpicture}
\draw[line width=1.5pt] (0,0)--(2,0);
\draw[line width=1.5pt] (0,0)--(1,1);
\draw[line width=1.5pt] (0,0)--(1,-1);
\draw[line width=1.5pt] (1,1)--(2,0);
\draw[line width=1.5pt] (1,-1)--(2,0);
%%%%%%%%%%
\draw[->,line width=1.2pt] (0,0)--(1,0);
\draw[->,line width=1.2pt] (0,0)--(0.5,0.5);
\draw[->,line width=1.2pt] (0,0)--(0.5,-0.5);
\draw[->,line width=1.2pt] (1,1)--(1.5,0.5);
\draw[->,line width=1.2pt] (1,-1)--(1.5,-0.5);
%%%%%%%%%%5
\node[left] at (0,0) {$(a,b)$};
\node[right] at (2,0) {$(a-1,b)$};
\node[below] at (1,-1) {$(a-\frac{1}{2},b-\frac{1}{2})$};
\node[above] at (1,1) {$(a-\frac{1}{2},b+\frac{1}{2})$};
%%%%%%%%%%
 \fill[color=red] (0,0) circle (1.7pt);
\fill[color=red] (2,0) circle (1.7pt);
\fill[color=red] (1,1) circle (1.7pt);
\fill[color=red] (1,-1) circle (1.7pt);
\end{tikzpicture}
\caption{Five types of arrows where $(a, b)\in \Z\times \N$.}
\label{fig-five}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Catalan-Stieltjes network}
\begin{figure}
\centering
\begin{tikzpicture}%[scale=0,8]
% grid
\draw[dashed,->](0,0)--(0,5);
\node[above] at (0,5) {$y$};
\draw[->](0,0)--(-6,0);
\draw (-4,0)--(4,0);
\node[left,<-] at (-6,0) {$x$};
\draw[->] (-4,0)--(-3.5,0);
\draw[->] (-3,0)--(-2.5,0);
\draw[->] (-2,0)--(-1.5,0);
\draw[->] (-1,0)--(-0.5,0);
\draw[->] (0,0)--(0.5,0);
\draw[->] (1,0)--(1.5,0);
\draw[->] (2,0)--(2.5,0);
\draw[->] (3,0)--(3.5,0);

\draw (-4,1)--(4,1);
\draw[->] (-4,1)--(-3.5,1);
\draw[->] (-3,1)--(-2.5,1);
\draw[->] (-2,1)--(-1.5,1);
\draw[->] (-1,1)--(-0.5,1);
\draw[->] (0,1)--(0.5,1);
\draw[->] (1,1)--(1.5,1);
\draw[->] (2,1)--(2.5,1);
\draw[->] (3,1)--(3.5,1);
%%
\draw[->] (-3,1)--(-2.75,0.75);
\draw[->] (-2,1)--(-1.75,0.75);
\draw[->] (-1,1)--(-0.75,0.75);
\draw[->] (0,1)--(0.25,0.75);
\draw[->] (1,1)--(1.25,0.75);
\draw[->] (2,1)--(2.25,0.75);
\draw[->] (3,1)--(3.25,0.75);
%%
%%
\draw[->] (-3,2)--(-2.75,1.75);
\draw[->] (-2,2)--(-1.75,1.75);
\draw[->] (-1,2)--(-0.75,1.75);
\draw[->] (0,2)--(0.25,1.75);
\draw[->] (1,2)--(1.25,1.75);
\draw[->] (2,2)--(2.25,1.75);
\draw[->] (3,2)--(3.25,1.75);
\draw[->] (-4,2)--(-3.75,1.75);
%%
\draw[->] (-3,3)--(-2.75,2.75);
\draw[->] (-2,3)--(-1.75,2.75);
\draw[->] (-1,3)--(-0.75,2.75);
\draw[->] (0,3)--(0.25,2.75);
\draw[->] (1,3)--(1.25,2.75);
\draw[->] (2,3)--(2.25,2.75);
\draw[->] (3,3)--(3.25,2.75);
\draw[->] (-4,3)--(-3.75,2.75);
%%
\draw[->] (-3,4)--(-2.75,3.75);
\draw[->] (-2,4)--(-1.75,3.75);
\draw[->] (-1,4)--(-0.75,3.75);
\draw[->] (0,4)--(0.25,3.75);
\draw[->] (1,4)--(1.25,3.75);
\draw[->] (2,4)--(2.25,3.75);
\draw[->] (3,4)--(3.25,3.75);
\draw[->] (-4,4)--(-3.75,3.75);
%%%

%%%
\draw (-4,2)--(4,2);
\draw[->] (-4,2)--(-3.75,2.25);
\draw[->] (-4,3)--(-3.75,3.25);
\draw[->] (-4,2)--(-3.5,2);
\draw[->] (-3,2)--(-2.5,2);
\draw[->] (-2,2)--(-1.5,2);
\draw[->] (-1,2)--(-0.5,2);
\draw[->] (0,2)--(0.5,2);
\draw[->] (1,2)--(1.5,2);
\draw[->] (2,2)--(2.5,2);
\draw[->] (3,2)--(3.5,2);
%%%%
\draw (-4,3)--(4,3);
\draw[->] (-4,3)--(-3.5,3);
\draw[->] (-3,3)--(-2.5,3);
\draw[->] (-2,3)--(-1.5,3);
\draw[->] (-1,3)--(-0.5,3);
\draw[->] (0,3)--(0.5,3);
\draw[->] (1,3)--(1.5,3);
\draw[->] (2,3)--(2.5,3);
\draw[->] (3,3)--(3.5,3);

\draw[->] (-4,4)--(-3.5,4);
\draw[->] (-3,4)--(-2.5,4);
\draw[->] (-2,4)--(-1.5,4);
\draw[->] (-1,4)--(-0.5,4);
\draw[->] (0,4)--(0.5,4);
\draw[->] (1,4)--(1.5,4);
\draw[->] (2,4)--(2.5,4);
\draw[->] (3,4)--(3.5,4);

\draw(-4,4)--(4,4);%
\draw(-4,0)--(-3,1);%
\draw(-3.5,1.5)--(-3, 2);
\draw[->] (-4,0)--(-3.75, 0.25);
\draw[->] (-4,0)--(-3.2,0.8);
\draw[->](-4,1)--(-3.75, 1.25);
\draw[->](-4,1)--(-3.75, 0.75);
\draw[->] (-4,1)--(-3.2,1.8);
\draw (-4,2)--(-3,3);
\draw[->] (-4,2)--(-3.2,2.8);
\draw (-4,3)--(-3,4);
\draw[->] (-4,3)--(-3.2,3.8);
\draw (-4,1)--(-3,0);
\draw[->] (-4,1)--(-3.2,0.2);
\draw (-4,2)--(-3,1);
\draw[->] (-4,2)--(-3.2,1.2);
\draw (-4,3)--(-3,2);
\draw[->] (-4,3)--(-3.2,2.2);
\draw (-4,4)--(-3,3);
\draw[->] (-4,4)--(-3.2,3.2);

\draw[->](-3,0)--(-2,1);
\draw[->](-3,0)--(-2,1);
\draw[->] (-3,0)--(-2.75,0.25);
\draw (-3,1)--(-2,2);
\draw[->] (-3,1)--(-2.2,1.8);
\draw (-3,2)--(-2,3);
\draw[->] (-3,2)--(-2.2,2.8);
\draw (-3,3)--(-2,4);
\draw[->] (-3,3)--(-2.2,3.8);
\draw (-3,1)--(-2,0);
\draw[->] (-3,1)--(-2.2,0.2);
%%%
\draw[->] (-3,2)--(-2.75,2.25);
\draw[->] (-2,2)--(-1.75, 2.25);
\draw[->] (-1,2)--(-0.75,2.25);
\draw[->] (-0,2)--(0.25,2.25);
\draw[->] (1,2)--(1.25,2.25);
\draw[->] (2,2)--(2.25,2.25);
\draw[->] (3,2)--(3.25,2.25);
%%
%%%
\draw[->] (-3,3)--(-2.75,3.25);
\draw[->] (-2,3)--(-1.75, 3.25);
\draw[->] (-1,3)--(-0.75,3.25);
\draw[->] (-0,3)--(0.25,3.25);
\draw[->] (1,3)--(1.25,3.25);
\draw[->] (2,3)--(2.25,3.25);
\draw[->] (3,3)--(3.25,3.25);
%%
\draw (-3,2)--(-2,1);
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\draw (-3,3)--(-2,2);
\draw[->] (-3,3)--(-2.2,2.2);
\draw (-3,4)--(-2,3);
\draw[->] (-3,4)--(-2.2,3.2);
%%%
\draw[->] (-3,1)--(-2.75,1.25);
\draw[->] (-2,1)--(-1.75, 1.25);
\draw[->] (-1,1)--(-0.75,1.25);
\draw[->] (-0,1)--(0.25,1.25);
\draw[->] (1,1)--(1.25,1.25);
\draw[->] (2,1)--(2.25,1.25);
\draw[->] (3,1)--(3.25,1.25);
%%
\draw (-2,0)--(-1,1);
\draw[->] (-2,0)--(-1.75,0.25);
\draw[->] (-2,0)--(-1.2,0.8);
\draw (-2,1)--(-1,2);
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\draw (-2,4)--(-1,3);
\draw[->] (-2,4)--(-1.2,3.2);

\draw (-1,0)--(0,1);
\draw[->] (-1,0)--(-0.75,0.25);
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\draw[->] (-1,3)--(-0.2,2.2);
\draw (-1,4)--(0,3);
\draw[->] (-1,4)--(-0.2,3.2);
%%%


\draw[->] (0,0)--(0.25,0.25);
\draw[->] (1,0)--(1.25,0.25);
\draw[->] (2,0)--(2.25,0.25);
\draw[->] (3,0)--(3.25,0.25);

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\draw[->] (0,3)--(0.8,3.8);
\draw (0,1)--(1,0);
\draw[->] (0,1)--(0.8,0.2);
\draw (0,2)--(1,1);
\draw[->] (0,2)--(0.8,1.2);
\draw (0,3)--(1,2);
\draw[->] (0,3)--(0.8,2.2);
\draw (0,4)--(1,3);
\draw[->] (0,4)--(0.8,3.2);

\draw (1,0)--(2,1);
\draw[->] (1,0)--(1.8,0.8);
\draw (1,1)--(2,2);
\draw[->] (1,1)--(1.8,1.8);
\draw (1,2)--(2,3);
\draw[->] (1,2)--(1.8,2.8);
\draw (1,3)--(2,4);
\draw[->] (1,3)--(1.8,3.8);
\draw (1,1)--(2,0);
\draw[->] (1,1)--(1.8,0.2);
\draw (1,2)--(2,1);
\draw[->] (1,2)--(1.8,1.2);
\draw (1,3)--(2,2);
\draw[->] (1,3)--(1.8,2.2);
\draw (1,4)--(2,3);
\draw[->] (1,4)--(1.8,3.2);

\draw (2,0)--(3,1);
\draw[->] (2,0)--(2.8,0.8);
\draw (2,1)--(3,2);
\draw[->] (2,1)--(2.8,1.8);
\draw (2,2)--(3,3);
\draw[->] (2,2)--(2.8,2.8);
\draw (2,3)--(3,4);
\draw[->] (2,3)--(2.8,3.8);
\draw (2,1)--(3,0);
\draw[->] (2,1)--(2.8,0.2);
\draw (2,2)--(3,1);
\draw[->] (2,2)--(2.8,1.2);
\draw (2,3)--(3,2);
\draw[->] (2,3)--(2.8,2.2);
\draw (2,4)--(3,3);
\draw[->] (2,4)--(2.8,3.2);

\draw (3,0)--(4,1);
\draw[->] (3,0)--(3.8,0.8);
\draw (3,1)--(4,2);
\draw[->] (3,1)--(3.8,1.8);
\draw (3,2)--(4,3);
\draw[->] (3,2)--(3.8,2.8);
\draw (3,3)--(4,4);
\draw[->] (3,3)--(3.8,3.8);
\draw (3,1)--(4,0);
\draw[->] (3,1)--(3.8,0.2);
\draw (3,2)--(4,1);
\draw[->] (3,2)--(3.8,1.2);
\draw (3,3)--(4,2);
\draw[->] (3,3)--(3.8,2.2);
\draw (3,4)--(4,3);
\draw[->] (3,4)--(3.8,3.2);

\node[below] at (-4,0) {$A_4$};
\node[below] at (-3,0) {$A_3$};
\node[below] at (-2,0) {$A_2$};
\node[below] at (-1,0) {$A_1$};
\node[below] at (0,0) {$A_0\;C_0$};
%\node[below] at (.5,0) {$B_0 $};
\node[right] at (0,4) {$B_4$};
\node[right] at (0,3) {$B_3$};
\node[right] at (0,2) {$B_2$};
\node[right] at (0,1) {$B_1$};
\node[right] at (0,0.25) {$B_0$};
%%%%%%%%%%%%%%%%
\node[below] at (4,0) {$C_4$};
\node[below] at (3,0) {$C_3$};
\node[below] at (2,0) {$C_2$};
\node[below] at (1,0) {$C_1$};

\fill (-4,0) circle (1.5pt);
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\fill (3.6,4.4) circle (0.5pt);
\fill (3.2,4.6) circle (0.5pt);
%\coordinate [label=left:FIGURE 6] (m1) at (0.5,-1.5);
\end{tikzpicture}
\caption{Catalan-Stieltjes network $(\G, (A_i),\; (B_i))$ and
Hankel network $(\G, (A_i),\; (C_i))$}
\label{fighankel}
\end{figure}

%%%%%%%%%%
\begin{Def}
 Let $A_i:=(i,0)$ and  $B_i:=(0,i)$\ for  $i\geq 0$. 
 The Catalan-Stieltjes network is defined to be 
 the triple $(\G, (A_i),\; (B_i))$. See Figure~\ref{fighankel}.
\end{Def}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma} \label{keylemma}
For any of the following four weight functions on the edges of $\G$:
\begin{itemize}
\item $w((i,j)\rightarrow (i-1,j)) = s_j - t_j - r_j$ (resp. $s_j - t_{j+1} - r_{j-1}, s_j - t_{j}r_{j-1} - 1, s_j - t_{j+1}r_{j} - 1$);
\item $w((i,j)\rightarrow (i-\frac{1}{2}, j+\frac{1}{2})) = r_j$ (resp. $1, 1, r_j$);
\item $w((i,j+1)\rightarrow (i-\frac{1}{2}, j+\frac{1}{2})) = t_{j+1}$ (resp. $1, t_{j+1}, 1$);
 \item $w((i-\frac{1}{2}, j+\frac{1}{2})\rightarrow (i-1,j+1) )= 1$ (resp. $r_{j}, r_{j}, 1$);
 \item $w((i-\frac{1}{2}, j+\frac{1}{2})\rightarrow (i-1,j)) = 1$ (resp. $t_{j+1}, 1, t_{j+1}$);
 \end{itemize}
 the matrix \eqref{C-S} is a  path matrix 
 from $(A_i)$ to $(B_j)$ of the network $(\G, (A_i),\; (B_i))$, namely,
\be\label{new-interpretation}
a_{i,j} = \sum_{\gamma: A_i\to B_j}w(\gamma).
\ee
%where the sum is over all  paths from $A_i$ to $B_j$ in ${\mathcal C}$.
%the weighted path matrix associated to the boundary vertices in $\bar{G}$ is the same as the Stieltjes matrix corresponding to the three sequences: $(r_k)_{k\geq 0}$, $(s_k)_{k\geq 0}$, $(t_k)_{k\geq 1}$.
\end{lemma}

\begin{proof} Let $w_{i,j}$ be  the right-hand side of \eqref{new-interpretation}.
It
 suffices to prove that $w_{i,j}$ satisfy the recurrence \eqref{C-S}. 
 Among the four weight functions, we just prove the first one because the other cases can be verified in the same manner.
 Firstly, by definition $w_{0,0} = 1$. 
 %Now let's consider the paths from $A_{n+1}:=(n+1,0)$ to $B_{k}:=(0,k)$.  
 We can 
classify  the paths from $A_{n+1}$ to $B_{k}$  according to their intersecting points with the line $x=1$ as follows:
\begin{itemize}
\item 
All the paths from $A_{n+1}$ to $(1,k)$ plus the last step $(1,k)\to B_k$;
\item
All the paths from $A_{n+1}$ to $(1,k)$ plus the last two steps $(1,k)\to (\frac{1}{2},k-\frac{1}{2})$ and $(\frac{1}{2},k-\frac{1}{2})\to  B_k$;
\item
All the paths from $A_{n+1}$ to $(1,k)$ plus the last two 
steps $(1,k)\to (\frac{1}{2},k+\frac{1}{2})$ and $(\frac{1}{2},k+\frac{1}{2})\to B_k$;
\item
All the paths from $A_{n+1}$ to $(1,k-1)$ plus the last two steps 
$(1,k-1)\to (\frac{1}{2},k-\frac{1}{2})$ and $(\frac{1}{2},k-\frac{1}{2})\to B_k$;
\item
All the paths from $A_{n+1}$ to $(1,k+1)$ plus the last two steps 
$(1,k+1)\to (\frac{1}{2},k+\frac{1}{2})$ and $(\frac{1}{2},k+\frac{1}{2})\to (0,k)$.
\end{itemize}
It is clear that the sum of the weights of the paths from $A_{n+1}$ to $(1,k)$ 
(resp. $(1,k-1)$,  $(1,k+1)$) is $w_{n,k}$ (resp. $w_{n,k-1}$,  $w_{n,k+1}$).
So,
\begin{eqnarray*}
w_{n+1,k} & = & r_{k-1}w_{n,k-1} + (s_{k} - t_{k} - r_{k})w_{n,k} + t_{k+1}w_{n,k+1} + t_{k}w_{n,k} + r_
{k}w_{n,k}\\
 & = & r_{k-1}w_{n,k-1} + s_{k}w_{n,k} + t_{k+1}w_{n,k+1},
\end{eqnarray*}
which is the recurrence $(1.1)$.
\end{proof}

\begin{theorem}\label{new-thm}
Let $I=(i_1,\ldots, i_n)$ and $J=(j_1, \ldots, j_n)$ be two positive increasing 
integer sequences. For any of  the four weight functions in Lemma~\ref{keylemma}, 
we have 
\be\label{GV-new}
\det A_{I,J}=\sum_{P\in N(\G; A_I, B_J)} w(P).
\ee
\end{theorem}
\begin{proof}
By Lemma 2.2 and Lindstr\"{o}m's lemma, we can write 
$\det A_{I,J}$ as a double sum as \eqref{GV-motzkin} except that the nonintersection 
condition forces the path $p_k$ to go from $A_{i_k}$ to $B_{j_k}$ for all $k=1,\ldots, n$, namely $\sigma\in \S_n$ must be identity. 
\end{proof}

From Theorem~\ref{new-thm}  we derive immediately the main results of  
Chen-Liang-Wang~\cite[Theorems 2.10 and 2.11, Corollary 2.12]{CLW15}.
%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{corollary}\label{wang1}
If the three sequences of nonnegative numbers 
$(r_{k})_{k\geq 0}, (s_{k})_{k\geq 0}$ and  $(t_{k})_{k\geq 1}$ 
 satisfy one of the following conditions:
\renewcommand{\labelenumi}{$($\roman{enumi}$)$}
\begin{enumerate}
\item 
$s_{0}\geq r_{0}$ and $s_{k}\geq r_{k} + t_{k}$ for $k\geq 1$;
\item
$s_{0}\geq t_{1}$ and $s_{k}\geq t_{k+1} + r_{k-1}$ for $k\geq 1$;
\item
$s_{0}\geq 1$ and $s_{k}\geq r_{k-1}\cdot t_{k} + 1$ for $k\geq 1$;
\item
$s_{0}\geq r_{0}\cdot t_{1}$ and $s_{k}\geq r_{k}\cdot t_{k+1} + 1$ for $k\geq 1$;
\end{enumerate}
then the Catalan-Stieltjes matrix $A$ defined by  $(1.1)$  is totally positive.
\end{corollary}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{remark} The above conditions  (i)-(iii) are exactly  Theorems 2.10 and 2.11
of Chen-Liang-Wang~\cite{CLW15},  
while the special 
 $r_k=1$ case of  (iv) is  Corollary~2.12 of \cite{CLW15}.
\end{remark}

We will give several examples of Catalan-Stieltjes networks 
in Section~3.  For the reader's convenience,  in
Figure~\ref{figC-S} we present 
a rotated version of the Catalan-Stieltjes network in
Figure~\ref{fighankel}.
 %%%%%%%%%%%%%



\begin{figure}[t]
\centering
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\draw[<-] (2,2.5)--(2,2);
\draw[<-] (3,2.5)--(3,2);
\draw[<-] (4,2.5)--(4,2);
\draw[<-] (5,2.5)--(5,2);
\draw[<-] (6,2.5)--(6,2);
\draw[<-] (7,2.5)--(7,2);
\draw[<-] (8,2.5)--(8,2);
\draw[<-] (3,3.5)--(3,3);
\draw[<-] (4,3.5)--(4,3);
\draw[<-] (5,3.5)--(5,3);
\draw[<-] (6,3.5)--(6,3);
\draw[<-] (7,3.5)--(7,3);
\draw[<-] (4,4.5)--(4,4);
\draw[<-] (5,4.5)--(5,4);
\draw[<-] (6,4.5)--(6,4);


\fill (0,-0.7) circle (1.2pt);
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\fill (10,-0.7) circle (1.2pt);

\fill (0,-0.9) circle (1.2pt);
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\fill (10,-0.9) circle (1.2pt);

\fill (0,-1.1) circle (1.2pt);
\fill (1,-1.1) circle (1.2pt);
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\fill (3,-1.1) circle (1.2pt);
\fill (4,-1.1) circle (1.2pt);
\fill (5,-1.1) circle (1.2pt);
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\fill (7,-1.1) circle (1.2pt);
\fill (8,-1.1) circle (1.2pt);
\fill (9,-1.1) circle (1.2pt);
\fill (10,-1.1) circle (1.2pt);
\fill (-0.2,0) circle (1.2pt);
\fill (-0.4,0) circle (1.2pt);
\fill (-0.6,0) circle (1.2pt);
\fill (10.2,0) circle (1.2pt);
\fill (10.4,0) circle (1.2pt);
\fill (10.6,0) circle (1.2pt);

\draw (0,-0.5)--(0,0);
\draw (1,-0.5)--(1,0);
\draw (2,-0.5)--(2,0);
\draw (3,-0.5)--(3,0);
\draw (4,-0.5)--(4,0);
\draw (5,-0.5)--(5,0);
\draw (6,-0.5)--(6,0);
\draw (7,-0.5)--(7,0);
\draw (8,-0.5)--(8,0);
\draw (9,-0.5)--(9,0);
\draw (10,-0.5)--(10,0);

\coordinate [label=right:$B_5$] (b5) at (10,1);
\coordinate [label=right:$B_4$] (b4) at (9,2);
\coordinate [label=right:$B_3$] (b3) at (8,3);
\coordinate [label=right:$B_2$] (b2) at (7,4);
\coordinate [label=right:$B_1$] (b1) at (6,5);
\coordinate [label=left:$A_5$] (a5) at (0,1);
\coordinate [label=left:$A_4$] (a4) at (1,2);
\coordinate [label=left:$A_3$] (a3) at (2,3);
\coordinate [label=left:$A_2$] (a2) at (3,4);
\coordinate [label=left:$A_1$] (a1) at (4,5);

\draw (5,5)--(5,6);
\draw[->] (5,6)--(-2,-1);
\node[below] at (-2,-1) {$x$};
\draw[dashed,->] (5,6)--(12,-1);
\node[below] at (12,-1) {$y$};
\fill (5,6) circle (2pt);
\draw[<-] (5,5.5)--(5,5);
\coordinate [label=left:$A_0$] (a0b0) at (5,6);
\coordinate [label=right:$B_0$] (a0b0) at (5,6);
\draw (9,0)--(10,1);
\draw[<-] (9.5,0.5)--(9,0);
%\coordinate [label=left:level 5] (l5) at (6,-3);

\draw (9,2)--(7,0);
%\coordinate [label=left:level 4] (l4) at (4,-3);
\draw[<-] (7.5,0.5)--(7,0);
\draw[<-] (8.5,1.5)--(8,1);

\draw (8,3)--(5,0);
%\coordinate [label=left:level 3] (l3) at (2,-3);
\draw[<-] (5.5,0.5)--(5,0);
\draw[<-] (6.5,1.5)--(6,1);
\draw[<-] (7.5,2.5)--(7,2);

\draw (7,4)--(3,0);
%\coordinate [label=left:level 2] (l2) at (1,-2);
\draw[<-] (3.5,0.5)--(3,0);
\draw[<-] (4.5,1.5)--(4,1);
\draw[<-] (5.5,2.5)--(5,2);
\draw[<-] (6.5,3.5)--(6,3);

\draw (6,5)--(1,0);
%\coordinate [label=left:level 1] (l1) at (0,-1);
\draw[<-] (1.5,0.5)--(1,0);
\draw[<-] (2.5,1.5)--(2,1);
\draw[<-] (3.5,2.5)--(3,2);
\draw[<-] (4.5,3.5)--(4,3);
\draw[<-] (5.5,4.5)--(5,4);

\draw (5,6)--(-1,0);
%\coordinate [label=left:level 0] (l0) at (-1,0);
\draw[<-] (4.5,5.5)--(4,5);
\draw[<-] (3.5,4.5)--(3,4);
\draw[<-] (2.5,3.5)--(2,3);
\draw[<-] (1.5,2.5)--(1,2);
\draw[<-] (0.5,1.5)--(0,1);
\draw[<-] (-0.5,0.5)--(-1,0);
\draw[red,dashed] (4,5)--(10,-1);
\end{tikzpicture}
\vspace{-0.5cm}
\caption{Catalan-Stieltjes network $(\G, (A_i),\; (B_i))$}
\label{figC-S}
\end{figure}
%%%%%%%%%%%%%%%
\subsection{Hankel network}
A  sequence $\alpha = (a_n)_{n\geq 0}$ of real numbers  is \emph{Hankel-totally positive} if 
 the associated Hankel matrix $H:=H(\alpha)=(a_{i+j})_{i,j\geq 0}$ is totally positive.
 For brevity, we use  \emph{H-TP} to denote Hankel-totally positive or Hankel-totally positivity in what follows.
 It is known (see \cite[Theorem 4.4]{Pinkus-1}) that the H-TP condition on $\alpha$ is equivalent to sat that 
$\alpha$ is  a {\em Stieltjes moment sequence}, i.e., there 
 is  an  integral representation of the form
 $$
a_n = \int_0^{+\infty} x^n d\mu(x),\quad \textrm{for}\quad n\geq 0,
$$
where $\mu$ is a non-negative measure $\mu$ on $[0,+\infty)$. 


A sequence $\alpha=(a_n)_{n\geq 0}$  is said to be  generated by 
a Catalan-Stieltjes matrix \eqref{C-S}
if it coincides with its first column, namely, $a_n=a_{n,0}$ for all $n\in \N$.
In \cite{LMW16} Liang-Mu-Wang gave some sufficient conditions on the H-TP  of a sequence generated by 
a Catalan-Stieltjes matrix.  Actually, for such a sequence $\alpha$, 
we can derive from Lemma~\ref{keylemma} a lattice path interpretation for each 
minor of the associated Hankel matrix $H(\alpha)$, which implies 
sufficient conditions on the  H-TP of  $\alpha$. 
 Recall  that a sequence $\alpha=(a_n)_{n\geq 0}$ is {\em strongly log-convex} 
    if $a_{n}a_{m+1}\geq a_{m}a_{n+1}$ for all $m\geq n\geq 0$.
Clearly  the H-TP of $\alpha$ implies that it is 
strongly log-convex. 

%%%%%%%%%%%%%%%%%%%%%%%%
\begin{Def}
 Let $A_i:=(i,0)$ and  $C_i:=(-i, 0)$\ for  $i\geq 0$. 
 The Hankel network is defined to be 
 the triple ${\mathcal H}:=(\G, (A_i),\; (C_i))$. See Figure~\ref{fighankel}.
\end{Def}

\begin{theorem} 
 Let $\alpha=(a_n)_{n\geq 0}$ be a sequence generated by 
a Catalan-Stieltjes matrix \eqref{C-S} and $H=(a_{i+j})$ the associated Hankel matrix.
Then, for any    two positive increasing 
integer sequences $I=(i_1,\ldots, i_n)$ and $J=(j_1, \ldots, j_n)$   we have 
$$
\det H_{I,J}=\sum_{P\in N(\G; A_I, C_J)} w(P),
$$
where $w$ is  any of  the four weight functions in Lemma~\ref{keylemma}. 
\end{theorem}
\begin{proof} 
By Lemma~2.2, the coefficient $a_{i+j,0}$ is the sum of weights of  paths from 
$A_i$ to $C_j$ in Hankel network ${\mathcal H}$; see Figure~\ref{fighankel}.
In other words, the matrix $H$ is the path matrix of ${\mathcal H}$ from $(A_i)$ to $(C_j)$.
It is clear that  the only possible permutation in Lindstr\"{o}m's lemma is identity, so each minor of $H$ reduces to  the sum of positive weights of   
 nonintersecting paths families from $(A_i)_{i\in I}$ to $(C_j)_{j\in J}$ in ${\mathcal H}$.
 \end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%
\begin{corollary}\label{Hankel-minor}
The sequence  $(a_{n,0})_{n\geq 0}$ of \eqref{C-S}
is Hankel totally positive  if the three sequences $(r_k), (s_k)$ and 
 $(t_{k+1})$ satisfy one of the four conditions (i)--(iv) of  Corollary~\ref{wang1}.
%\renewcommand{\labelenumi}{$($\roman{enumi}$)$}
\end{corollary}
%\begin{proof}
%One can also derive Corollary~\ref{Hankel-minor} directly from Corollary~\ref{wang1}.
%Indeed, it is well-known\cite[p. 291-292]{Aigner-course}  that for the Catalan-Stieltjes matrix $A$ in \eqref{C-S}
%one has the following fundamental  formula:
%$$
%a_{m+n,0}=\sum_{k\geq 0} a_{m,k}a_{n,k}T_k,
%$$
%where $T_0=1$ and $T_k=t_1\ldots t_k$ for $k\geq 1$. This is equivalent to the matrix identity
%\begin{align}\label{addition}
%H=ATA^t,
%\end{align}
%where $A^t$ is the transpose of $A$ and $T$ the diagonal matrix $\diag(T_0, T_1,\ldots)$. 
%Therefore, by Cauchy-Binet formula, 
%the Hankel matrix $H$ is totally positive if $A$ is totally positive and $t_k\geq  0$. 
%Note that the above formula is equivalent to the Stieltjes-Rogers addition formula of the formal power 
%series 
%$\sum_{n\geq 0} a_{n,0} x^n/n!$ (see \cite[p. 295-296]{GJ83}.
%
%\end{proof}
\begin{remark}
\begin{enumerate}
\item  The first condition (i) of Corollary~\ref{wang1} is Corollary~2.4 of \cite{LMW16}.
 \item If $(a_{n,0})$ is the  first column of 
  a Catalan-Stieltjes matrix \eqref{C-S}, then it is also generated by the Catalan-Stieltjes matrix 
$(\tilde a_{n,k})$ defined by 
\begin{align}\label{C-S-tilde}
 \tilde a_{n,k} = t_{k}\tilde a_{n-1,k-1} + s_{k}\tilde a_{n-1,k} + r_{k}\tilde a_{n-1,k+1} \quad  (k\ge 0,\;n\geq 1),
\end{align}
where  $\tilde a_{0,0} = 1$ and $\tilde a_{n,k} = 0$ unless $n\geq k \geq 0$. Applying the condition (i) (resp. (iii))
of Corollary~\ref{wang1} to  the matrix~\eqref{C-S-tilde} 
we get the condition (ii) (resp. (iv)) of Corollary~\ref{wang1}, so we need only to verify conditions (i) and (iii)
of Corollary~\ref{Hankel-minor}.
 \end{enumerate}

\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Applications to some combinatorial matrices}
It is known \cite{Zhu13, CLW15} that instead of the total order of 
real numbers we can consider the  \emph{partial order} of the 
 commutative ring ${\mathbb R}[x]$ of polynomials with real coefficients as follows:
 a polynomial in $\mathcal{R}[x]$ is \emph{coefficientwise nonnegative} if it has nonnegative coefficients
 and  $p(x) \succcurlyeq q(x)$  if $p(x)-q(x)$ is  coefficientwise nonnegative. Thus  we can 
 generalize the previous  notions to coefficentwise log-convexity and coefficentwise-Hankel total positivity. For example, a  sequence  in $\mathcal{R}[x]$ is called \emph{coefficientwise-Hankel totally positive} if the associated Hankel matrix is coefficientwise totally positive. Clearly the totally positive 
 results in the previous sections can be restated in terms of 
 coefficientwise totally positive sequence. 
 In what follows  we consider some special cases of  Catalan-Stieltjes 
network and Hankel network  in connection with some classical  combinatorial sequences.
One souce of such examples can be found in Viennot's Lecture Note~\cite{Vi83} because 
 almost all the moment sequences of classical orthogonal 
polynomials have interesting combinatorial interpretations.  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Stirling network}
%%%%%%%%%%%%%%%
\begin{figure}[t]
\centering
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\draw[dashed,->] (0,0)--(0,4);
\node[left] at (-7,0) {$x$};
\node[above] at (0,4) {$y$};
\node[below] at (0,0) {$A_0$};
%\node[above] at (0,0) {$B_0$};
\node[below] at (-1,0) {$A_1$};
\node[below] at (-2,0) {$A_2$};
\node[below] at (-3,0) {$A_3$};
\node[below] at (-4,0) {$A_4$};
\node[right] at (0,1) {$B_1$};
\node[right] at (0,2) {$B_2$};
\node[right] at (0,3) {$B_3$};
\node[right] at (0,0) {$B_0$};
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%\draw (-4,4)--(0,4);

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\fill (-4.2,2.8) circle (0.5pt);
\fill (-4.4,2.6) circle (0.5pt);
\fill (-4.6,2.4) circle (0.5pt);
\end{tikzpicture}
\caption{Stirling network}
\label{stirling}
\end{figure}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


If $t_k=0$ for all $k\geq 1$,  the recurrence \eqref{C-S} reduces to 
\be\label{eq-stirling}
a_{n,k} = r_{k-1}a_{n-1,k-1} + s_k a_{n-1,k}\quad  (k\geq 0,\;n\geq 1).
\ee
Since the Stirling numbers of second kind $S(n,k)$ satisfy \eqref{eq-stirling} with
$r_k=1$ and $t_k=k$ we call the corresponding graph \emph{Stirling network}.
%As  $t_k=0$ for all $k\geq 1$,  
On the other hand, 
as there is no \emph{down} step in the 
 Motzkin paths,  the corresponding network (see Figure~\ref{motzkin-path}) reduces  to Figure~\ref{stirling}.
  
\begin{proposition}\label{mongelli}  Let $I=(i_1,\ldots, i_n)$ and $J=(j_1, \ldots, j_n)$ be two positive increasing  integer sequences. 
The minors of the  matrix $A=(a_{i,j})$ satisfying  \eqref{eq-stirling}
has the following combinatorial interpretation
\be\label{GV-stirling}
\det A_{I,J}
= \sum_{P\in N(\G; A_I, B_J)} w(P).
\ee
In particular $A$ is coefficientwise totally positive if $r_k$ and $s_k$ are polynomials in $x$ with nonnegative 
coefficients for all $k\geq 0$.
\end{proposition}
%%%%%%%%%%%%%%%%%
\begin{remark} 
Mongelli~\cite[Theorem 5]{Pietro-Total} 
gave the above combinatorial interpretation in the special case 
$r_k=1$ and $s_k=k(z+1)$. Generalizing the positivity part of Mongelli's result
Zhu~\cite{Bao}
 proved the above total positivity  result in the 
 special case where $r_k$ and $s_k$ are quadratic polynomials of $k$. 
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Narayana network of type A}
The \emph{Narayana polynomials} $N_n(x)$ (see \cite{Vi83, Zhu13}) are defined  by
$$
N_n(x) = \sum_{k=1}^n\frac{1}{n}{n\choose k-1}{n\choose k}x^k.
$$
It is known (see Appendix) that $(N_n(x))$ is the moment sequence 
generated by  the Catalan-Stieltjes matrix $A^{\gamma, \sigma, \tau}$ (see \eqref{C-S}) with
$$
\gamma=(0, 1, 1, \ldots), \quad \sigma=(x, 1+x, 1+x, \ldots),\quad \tau=(x, x, \ldots).
$$

Since these three sequences satisfy all the conditions in Corollary~\ref{wang1}, the matrix 
$A^{\gamma, \sigma, \tau}$ is TP and the sequence
$(N_n(x))$ is coefficientwise-H-TP. When $x=1$ the matrix
 reduces to
the Catalan triangle of Aigner~\cite{Aigner-Catalan}:
$$
C = (C_{n,k}) =
\left(\begin{array}{cccccc}
1 \\
1 & 1 \\
2 & 3 & 1 \\
5 & 9 & 5 & 1 \\
14 & 28 & 20 & 7 & 1\\
\vdots &\vdots &\vdots & \vdots&\vdots & \ddots
\end{array}\right),
$$
where $C_{n+1,0} = C_{n,0} + C_{n,1}, \; C_{n+1,k} = C_{n,k-1} + 2C_{n,k} + C_{n,k+1}$.
The corresponding \emph{Narayana network of type A} is depicted in Figure~\ref{figCA}.
As the weight of arrows
$(i,j)\rightarrow (i-1,j)$  is 
$s_j-t_j-r_j = 0$ for all $i\geq 1$ and $j\geq 0$, 
so there is no such arrows  in Figure~\ref{figCA}.
For example,  if we choose $I = \{2,3\}$ and $J = \{0,1\}$, then:
$$
\det
\left(\begin{array}{cc}
 2 & 3  \\
 5 & 9 
\end{array}\right) = 3
$$
and the three  pairs of nonintersecting paths from  $\{A_2, A_3\}$ to
$\{B_0,B_1\}$ are
 drawn as  red,  green and  blue pairs of paths in Figure~\ref{figCA}.

\begin{figure}[ht]
\centering
\scalebox{.8}{
\begin{tikzpicture}%[scale=0,8]
\draw[step=1] (0,0) grid (10,1);
\draw[step=1] (1,2) grid (9,1);
\draw[step=1] (2,3) grid (8,1);
\draw[step=1] (3,4) grid (7,1);
\draw[step=1] (4,5) grid (6,1);
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\draw[->] (6,3)--(6.5,3);
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\draw[->] (5,4)--(5.5,4);
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\draw[->] (4,5)--(4.5,5);
\draw[->] (5,5)--(5.5,5);
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\draw[<-] (1,0.5)--(1,0);
\draw[<-] (2,0.5)--(2,0);
\draw[<-] (3,0.5)--(3,0);
\draw[<-] (4,0.5)--(4,0);
\draw[<-] (5,0.5)--(5,0);
\draw[<-] (6,0.5)--(6,0);
\draw[<-] (7,0.5)--(7,0);
\draw[<-] (8,0.5)--(8,0);
\draw[<-] (9,0.5)--(9,0);
\draw[<-] (10,0.5)--(10,0);
\draw[<-] (1,1.5)--(1,1);
\draw[<-] (2,1.5)--(2,1);
\draw[<-] (3,1.5)--(3,1);
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\draw[<-] (7,1.5)--(7,1);
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\draw[<-] (9,1.5)--(9,1);
\draw[<-] (2,2.5)--(2,2);
\draw[<-] (3,2.5)--(3,2);
\draw[<-] (4,2.5)--(4,2);
\draw[<-] (5,2.5)--(5,2);
\draw[<-] (6,2.5)--(6,2);
\draw[<-] (7,2.5)--(7,2);
\draw[<-] (8,2.5)--(8,2);
\draw[<-] (3,3.5)--(3,3);
\draw[<-] (4,3.5)--(4,3);
\draw[<-] (5,3.5)--(5,3);
\draw[<-] (6,3.5)--(6,3);
\draw[<-] (7,3.5)--(7,3);
\draw[<-] (4,4.5)--(4,4);
\draw[<-] (5,4.5)--(5,4);
\draw[<-] (6,4.5)--(6,4);


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\fill (8,-0.7) circle (1.2pt);
\fill (9,-0.7) circle (1.2pt);
\fill (10,-0.7) circle (1.2pt);

\fill (0,-0.9) circle (1.2pt);
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\fill (4,-0.9) circle (1.2pt);
\fill (5,-0.9) circle (1.2pt);
\fill (6,-0.9) circle (1.2pt);
\fill (7,-0.9) circle (1.2pt);
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\fill (9,-0.9) circle (1.2pt);
\fill (10,-0.9) circle (1.2pt);

\fill (0,-1.1) circle (1.2pt);
\fill (1,-1.1) circle (1.2pt);
\fill (2,-1.1) circle (1.2pt);
\fill (3,-1.1) circle (1.2pt);
\fill (4,-1.1) circle (1.2pt);
\fill (5,-1.1) circle (1.2pt);
\fill (6,-1.1) circle (1.2pt);
\fill (7,-1.1) circle (1.2pt);
\fill (8,-1.1) circle (1.2pt);
\fill (9,-1.1) circle (1.2pt);
\fill (10,-1.1) circle (1.2pt);
\fill (-0.2,0) circle (1.2pt);
\fill (-0.4,0) circle (1.2pt);
\fill (-0.6,0) circle (1.2pt);
\fill (10.2,0) circle (1.2pt);
\fill (10.4,0) circle (1.2pt);
\fill (10.6,0) circle (1.2pt);

\draw (0,-0.5)--(0,0);
\draw (1,-0.5)--(1,0);
\draw (2,-0.5)--(2,0);
\draw (3,-0.5)--(3,0);
\draw (4,-0.5)--(4,0);
\draw (5,-0.5)--(5,0);
\draw (6,-0.5)--(6,0);
\draw (7,-0.5)--(7,0);
\draw (8,-0.5)--(8,0);
\draw (9,-0.5)--(9,0);
\draw (10,-0.5)--(10,0);

\coordinate [label=right:$B_5$] (t5) at (10,1);
\coordinate [label=right:$B_4$] (t4) at (9,2);
\coordinate [label=right:$B_3$] (t3) at (8,3);
\coordinate [label=right:$B_2$] (t2) at (7,4);
\coordinate [label=right:$B_1$] (t1) at (6,5);
\coordinate [label=left:$A_5$] (s5) at (0,1);
\coordinate [label=left:$A_4$] (s4) at (1,2);
\coordinate [label=left:$A_3$] (s3) at (2,3);
\coordinate [label=left:$A_2$] (s2) at (3,4);
\coordinate [label=left:$A_1$] (s1) at (4,5);

\draw (5,5)--(5,6);
\fill (5,6) circle (2pt);
\draw[<-] (5,5.5)--(5,5);
\coordinate [label=left:$A_0$] (s0t0) at (5,6);
\coordinate [label=right:$B_0$] (s0t0) at (5,6);
%\coordinate [label=left:FIGURE 2\quad Each edge has weight 1] (m1) at (7,-2);

\draw[red,line width=1pt] (3,4)--(4,4);
\draw[green,line width=1pt] (3,3.95)--(4,3.95);
\draw[red,line width=1pt] (4,4)--(4,5);
\draw[green,line width=1pt] (4.05,4)--(4.05,5);
\draw[red,line width=1pt] (4,5)--(5,5);
\draw[green,line width=1pt] (4,4.95)--(5,4.95);
\draw[red,line width=1pt] (5,5)--(5,6);
\draw[green,line width=1pt] (5.05,5)--(5.05,6);
\draw[red,line width=1pt] (2,3)--(5,3);
\draw[green,line width=1pt] (2,2.95)--(6,2.95);
\draw[red,line width=1pt] (5,3)--(5,4);
\draw[red,line width=1pt] (5,4)--(6,4);
\draw[red,line width=1pt] (6,4)--(6,5);
\draw[green,line width=1pt] (6.05,3)--(6.05,4);
\draw[green,line width=1pt] (6.05,4)--(6.05,5);
\draw[blue,line width=1pt] (3,3.9)--(4,3.9);
\draw[blue,line width=1pt] (4,4)--(5,4);
\draw[blue,line width=1pt] (5,4)--(5,5);
\draw[blue,line width=1pt] (5,5)--(5.1,5);
\draw[blue,line width=1pt] (5.1,5)--(5.1,6);
\draw[blue,line width=1pt] (6.1,3)--(6.1,4);
\draw[blue,line width=1pt] (6.1,4)--(6.1,5);
\draw[blue,line width=1pt] (2,2.9)--(6,2.9);
\draw[blue,line width=2pt] (6,2.9)--(6.1,2.9);
\draw[blue,line width=2pt] (6.1,2.9)--(6.1,3);
\end{tikzpicture}}
\caption[fig5]{Narayana network of type A}
\label{figCA}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Catalan-Shapiro network}
Shapiro~\cite{Shapiro-triangle} proved that the ballot numbers 
$B_{n,k}=\frac{k}{n}{2n\choose n+k}$ ($n,k\geq 1$) satisfy  the recurrence
$$
B_{n+1,k} = B_{n,k-1} + 2B_{n,k} + B_{n,k+1}.
$$ 
Thus the sequence $(B_{n,1})_{n\geq 1})$ of Catalan numbers is 
the moment sequence generated by 
the  Catalan-Stieltjes matrix $\bigl(B_{n+1,k+1}\bigr)_{n,k\geq 0}$:
$$
B=\bigl(B_{n+1,k+1}\bigr)_{n,k\geq 0} = 
\left(\begin{array}{cccccc}
1 \\
2 & 1 \\
5 & 4 & 1 \\
14 & 14 & 6 & 1 \\
42 & 48 & 27 & 8 & 1 \\
\vdots &\vdots & \vdots&\vdots &\vdots & \ddots
\end{array}\right).
$$
Clearly  all the four  conditions of  Corollary~\ref{wang1} are satisfied, so the matrix $B$ is TP and 
the sequence $(B_{n,1})_{n\geq 1})$ is H-TP.   Note that the total positivity of $B$ 
 was first proved in \cite{WW15}. The corresponding \emph{Catalan-Shapiro network} is depicted  in  Figure~\ref{figCS}, where
the edge $(i,j)\rightarrow (i-1,j)$ has weight 1  if $j = 0$, and 0 otherwise. 


\begin{figure}[ht]
\centering
\scalebox{.8}{
\begin{tikzpicture}%[scale=0,8]
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\draw[step=1] (1,2) grid (9,1);
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\draw[<-] (5,2.5)--(5,2);
\draw[<-] (6,2.5)--(6,2);
\draw[<-] (7,2.5)--(7,2);
\draw[<-] (8,2.5)--(8,2);
\draw[<-] (3,3.5)--(3,3);
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\draw[<-] (6,4.5)--(6,4);


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\fill (9,-0.7) circle (1.2pt);
\fill (10,-0.7) circle (1.2pt);

\fill (0,-0.9) circle (1.2pt);
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\fill (9,-0.9) circle (1.2pt);
\fill (10,-0.9) circle (1.2pt);

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\fill (1,-1.1) circle (1.2pt);
\fill (2,-1.1) circle (1.2pt);
\fill (3,-1.1) circle (1.2pt);
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\fill (10.2,0) circle (1.2pt);
\fill (10.4,0) circle (1.2pt);
\fill (10.6,0) circle (1.2pt);

\draw (0,-0.5)--(0,0);
\draw (1,-0.5)--(1,0);
\draw (2,-0.5)--(2,0);
\draw (3,-0.5)--(3,0);
\draw (4,-0.5)--(4,0);
\draw (5,-0.5)--(5,0);
\draw (6,-0.5)--(6,0);
\draw (7,-0.5)--(7,0);
\draw (8,-0.5)--(8,0);
\draw (9,-0.5)--(9,0);
\draw (10,-0.5)--(10,0);

\coordinate [label=right:$B_5$] (t5) at (10,1);
\coordinate [label=right:$B_4$] (t4) at (9,2);
\coordinate [label=right:$B_3$] (t3) at (8,3);
\coordinate [label=right:$B_2$] (t2) at (7,4);
\coordinate [label=right:$B_1$] (t1) at (6,5);
\coordinate [label=left:$A_5$] (s5) at (0,1);
\coordinate [label=left:$A_4$] (s4) at (1,2);
\coordinate [label=left:$A_3$] (s3) at (2,3);
\coordinate [label=left:$A_2$] (s2) at (3,4);
\coordinate [label=left:$A_1$] ($A_1$) at (4,5);

\draw (-1,0)--(5,6);
\draw[->] (-1,0)--(-0.5,0.5);
\draw[->] (0,1)--(0.5,1.5);
\draw[->] (1,2)--(1.5,2.5);
\draw[->] (2,3)--(2.5,3.5);
\draw[->] (3,4)--(3.5,4.5);
\draw[->] (4,5)--(4.5,5.5);
\draw (5,5)--(5,6);
\fill (5,6) circle (2pt);
\draw[<-] (5,5.5)--(5,5);
\coordinate [label=right:$B_0$] (s0t0) at (5,6);
\coordinate [label=left:$A_0$] (s0t0) at (5,6);
%\coordinate [label=left:FIGURE 3 \quad Each edge has weight 1] (m1) at (7,-2);
\end{tikzpicture}}
\caption[fig6]{Catalan-Shapiro network}
\label{figCS}
\end{figure}


\subsection{Bell network}
The  \emph{Bell polynomials} are defined by
$$
B_n(x) = \sum_{k=0}^nS(n,k)x^k.
$$
 It is known \cite{Vi83, Flajolet80} that 
$B_n(x)$'s are generated by   the  Catalan-Stieltjes matrix $(a_{n,k})$: 
\begin{align}\label{charlier}
a_{n,k} = xa_{n-1,k-1} + (k+x)a_{n-1,k} + (k+1)a_{n-1,k+1},
\end{align}
where  $a_{n,0}=B_n(x)$ for $n\geq 0$. Since recurrence \eqref{charlier}  satisfies just the second condition (ii) of Corollary~\ref{Hankel-minor}, 
the sequence $(B_{n}(x))$ is coefficientwise-H-TP. 
When $x=1$ it reduces to 
the Bell triangle~\cite{Aigner-characterization} 
$$
X = (X_{n,k}) = 
\left(\begin{array}{cccccc}
1 \\
1 & 1 \\
2 & 3 & 1 \\
5 & 10 & 6 & 1 \\
15 & 37 & 31 & 10 & 1 \\
\vdots &\vdots & \vdots&\vdots &\vdots & \ddots 
\end{array}\right),
$$
where
$X_{n+1,k} = X_{n,k-1} + (k+1)X_{n,k} + (k+1)X_{n,k+1}$. It
 satisfies the first and the third conditions in Corollary~\ref{wang1}.
The corresponding \emph{Bell network} is depicted
in Figure~\ref{figBell}.

\begin{figure}[t]
\centering
  \scalebox{.8}{
\begin{tikzpicture}%[scale=0,8]
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\draw[step=1] (1,2) grid (9,1);
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\draw (4,-0.5)--(4,0);
\draw (5,-0.5)--(5,0);
\draw (6,-0.5)--(6,0);
\draw (7,-0.5)--(7,0);
\draw (8,-0.5)--(8,0);
\draw (9,-0.5)--(9,0);
\draw (10,-0.5)--(10,0);

\coordinate [label=right:$B_5$] (t5) at (10,1);
\coordinate [label=right:$B_4$] (t4) at (9,2);
\coordinate [label=right:$B_3$] (t3) at (8,3);
\coordinate [label=right:$B_2$] (t2) at (7,4);
\coordinate [label=right:$B_1$] (t1) at (6,5);
\coordinate [label=left:$A_5$] (s5) at (0,1);
\coordinate [label=left:$A_4$] (s4) at (1,2);
\coordinate [label=left:$A_3$] (s3) at (2,3);
\coordinate [label=left:$A_2$] (s2) at (3,4);
\coordinate [label=left:$A_1$] (s1) at (4,5);
\coordinate [label=left:$A_0$] (s0) at (5,6);
\coordinate [label=right:$B_0$] (t0) at (5,6);
\draw (5,5)--(5,6);
\fill (5,6) circle (2pt);
\draw[<-] (5,5.5)--(5,5);

\coordinate [label=right:$5$] (5) at (9,0.5);
\coordinate [label=right:$4$] (4) at (7,0.5);
\coordinate [label=right:$3$] (3) at (5,0.5);
\coordinate [label=right:$2$] (2) at (3,0.5);
\coordinate [label=right:$4$] (41) at (8,1.5);
\coordinate [label=right:$3$] (31) at (6,1.5);
\coordinate [label=right:$2$] (21) at (4,1.5);
\coordinate [label=right:$3$] (32) at (7,2.5);
\coordinate [label=right:$2$] (22) at (5,2.5);
\coordinate [label=right:$2$] (23) at (6,3.5);
%\coordinate [label=left:FIGURE 4 \quad The edge  not be labeled has weight 1] (m1) at (9,-2);
\end{tikzpicture}}
\caption{Bell network}
\label{figBell}
\end{figure}

\subsection{Restricted hexagonal network}
A \emph{hex tree} is an ordred tree of which each vertex has updegree 0, 1, or 2, 
and an edge from a vertex of updegree 1 is either left, median, or right. 
The so-called \emph{restricted hexagonal number}  (see \cite{LMW16}) 
$h_n$  is also
the number of hex trees with $n$ edges (see \cite[A002212]{A002212}). 
The numbers $h_n$  have the explicit formula
$h_n=\sum_{j=0}^nC_{j+1} {n\choose j}$ where $C_j=\frac{1}{j+1}{2j\choose j}$ 
are  the Catalan numbers.
 The sequence $(h_n)$ is generated by 
the Catalan-Stieltjes  matrix (see Appendix):
$$
H = (h_{n,k}) = 
\left(\begin{array}{cccccc}
1 \\
3 & 1 \\
10 & 6 & 1 \\
36 & 29 & 9 & 1 \\
137 & 132 & 57 & 12 & 1 \\
\vdots &\vdots &\vdots &\vdots &\vdots & \ddots 
\end{array}\right),
$$
where
$h_{n+1,k} = h_{n,k-1} + 3h_{n,k} + h_{n,k+1}$.  It is easy to see that
 all the four conditons of Corollary~\ref{wang1} are satisfied. Thus the matrix $H$ is TP and 
 the sequence $(h_n)$ is  H-TP.
The corresponding \emph{restricted hexagonal network} is depicted in
Figure~\ref{fig8}.

%%%%%%%%%%%%%%%%%%%%%%%%%%
In a recent paper~\cite{H-R} Kim and Stanley studied a related polynomial sequence
$\bigl(p_n(x)\bigr)$ where $p_n(x)=\sum_{j=0}^n\frac{1}{j+1}{2j\choose j} {n\choose j}x^{n-j}$ for $n\geq 0$.
In particular they proved that 
this sequence is  the moment sequence associated to the Catalan-Stieltjes matrix $(p_{n,k}(x))$:
\begin{align*}
p_{n+1,0}(x) & =  (x+1)p_{n,0}(x) + p_{n,1}(x),\\
p_{n+1,k+1}(x) & =  p_{n,k}(x) + (x+2)p_{n,k+1} + p_{n,k+2}\quad (k\geq 1).
\end{align*}
where  $p_{n,0}=p_n(x)$ for $n\geq 0$. Clearly all the conditions of ~\eqref{Hankel-minor} are satisfied, so
the sequence $(p_n(x))$ is coefficientwise-H-TP.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}[t]
\centering
\scalebox{.8}{
\begin{tikzpicture}%[scale=0,8]
\draw[step=1] (0,0) grid (10,1);
\draw[step=1] (1,2) grid (9,1);
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\draw[<-] (-0.5,0.5)--(-1,0);
\end{tikzpicture}}
\caption[fig5]{Restricted Hexagonal network}
\label{fig8}
\end{figure}

\subsection{Eulerian polynomials}
The \emph{Eulerian polynomials} $(A_n(x))$ can be defined by
$$
\sum_{k\geq 0} (k+1)^nx^k=\frac{A_n(x)}{(1-x)^{n+1}}.
$$
They have many combinatorial interpretations and can be generated by the following Catalan-Stieltjes matrix~\cite{Vi83, Flajolet80}:
$$
a_{n,k} = ka_{n-1,k-1} + (k(x + 1)+1)a_{n-1,k} + (k+1)xa_{n-1,k+1},
$$
where  $a_{n,0}=A_n(x)$ for $n\geq 0$.
By the condition (i) of Corollary~\ref{Hankel-minor}, 
the sequence $(A_n(x))$ is coefficientwise-H-TP.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Rising factorials}
The rising factorial $\mu_n:=(x)_n$ is defined by
$$
\mu_0=1, \quad \mu_n=x(x+1)\cdots (x+n-1) \quad \textrm{for}\quad n\geq 1.
$$
They can be generated by the 
Catalan-Stieltjes matrix (see \cite{Vi83})
\begin{align}\label{second-CS}
a_{n,k} = ka_{n-1,k-1} + (x+2k)a_{n-1,k} + (x-1+k)a_{n-1,k+1}
\end{align}
where $a_{n,0} = \mu_n$.
Since recurrence \eqref{second-CS} satisfies only the second point of Corollary~\ref{Hankel-minor}, the sequence $(\mu_n)$ is coefficientwise-H-TP in ${\mathbb R}[x]$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Schr\"{o}der polynomials}
The \emph{Schr\"{o}der polynomials} $r_n(x)$ (see \cite{BSS93, Zhu13}) are defined by $r_n(x)=N_n(x+1)$, i.e., 
$$
r_n(x) = \sum_{k=0}^n{n+k\choose n-k}\frac{1}{k+1}{2k\choose k}x^{n-k}
$$
and generated by  the Catalan-Stieltjes matrix~(see Appendix):
\begin{align*}
a_{n,0} & =  (x+1)a_{n-1,0} + (x+1)a_{n-1,1}\\
a_{n,k} & =  a_{n-1,k-1} + (x+2)a_{n-1,k} + (x+1)a_{n-1,k+1}\quad (k\geq 1).
\end{align*}
Since the recurrence satisfies only the condition (i)  of Corollary~\ref{Hankel-minor}, so
 the sequence $(r_n(x))$ is coefficientwise-H-TP.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Central Delannoy polynomials}
The \emph{central Delannoy numbers}~\cite{Sagan97, Zhu13} are defined by 
\begin{align}\label{delannoy}
D_n(x) = \sum_{k=0}^n{n+k\choose n-k}{2k\choose k}x^{k}
\end{align}
and generated by  the Catalan-Stieltjes matrix~(see Appendix):
\begin{align*}
a_{n,0} & =   (2x+1)a_{n-1,0} +(x+1)a_{n-1,1},\\
a_{n,1} & =  2xa_{n-1,0} + (2x+1)a_{n-1,1} +(x+1)a_{n-1,2},\\
a_{n,k} & =  xa_{n-1,k-1} + (2x+1)a_{n-1,k} + (x+1)a_{n-1,k+1}\quad  (k\geq 2).
\end{align*}
Since the recurrence satisfies only the condition (i) of Corollary~\ref{Hankel-minor}, so
the sequence $(D_n(x))$ is coefficientwise-H-TP.

%%%%%%%%%%%%%%%%%%%%%%
\subsection{Narayana polynomials of type B}
The {\em Narayana polynomials of type B}~\cite{Zhu13} are defined by 
$$
W_n(x) = \sum_{k\geq 0}{n\choose k}^2
x^k$$ 
and generated by  the  Catalan-Stieltjes matrix~(see Appendix)
\begin{align*}
a_{n,0} & =  2 a_{n-1,0} +xa_{n-1,1},\\
%a_{n,1} & = a_{n-1,0} + (x+1)a_{n-1,1} +(x+1)a_{n-1,2},\\
a_{n,k} & =  a_{n-1,k-1} + (x+1)a_{n-1,k} + xa_{n-1,k+1}\quad  (k\geq 1).
\end{align*}
%where $r_k=1$,  $s_k = 1+x$ for $k\geq 0$ and $t_1 = 2x$, $t_k = x$ for $k\geq 2$. 
%We note that  $(W_n(x))$ is also generated by 
%the  Catalan-Stieltjes matrix \eqref{C-S} with 
%where
%$r_0=2, r_k=1$ ($k\geq 1$),  $s_k = 1+x$ for $k\geq 0$ and $t_k = x$ for $k\geq 1$. 
Since the recurrence satisfies only the condition (i) of Corollary~\ref{Hankel-minor}, so the sequence  $(W_n(x))$ 
is Hankel totally positive if $x\geq 1$.  

%%%%%%%%%%%%%%%%
\section{Two open problems}
Sokal~\cite{Sokal} and Wang-Zhu~\cite{WZ16} actually independently proved
 that  the polynomial sequence $(W_n(x))_{n\geq 0}$ is  coefficientwise-Hankel totally positive. 
Can one  find a planar network proof of this result~?
A toy example of Lindstr\"om-Gessel-Viennot's methodology is a lattice path model for 
the  total positivity of 
the Pascal matrix $P:=\bigl({n\choose k}\bigr)_{n,k\geq 0}$.  As 
the Hadamard product of two totally positive matrices is not totally positive in general (see \cite{Fallat-Johnson07}),   
 we speculate that  the  Hadamard product  $P\circ P=\bigl({n\choose k}^2\bigr)_{n,k\geq 0}$ is 
 totally positive and have checked this until $n=9$.  
\begin{conj}
The matrix 
$$
\biggl({n\choose k}^2\biggr)_{n,k\geq 0}=
\begin{pmatrix}
1&0&0&0&0&0&\cdots\\
1&1&0&0&0&0&\cdots\\
1&4&1&0&0&0&\cdots\\
1&9&9&1&0&0&\cdots\\
1&16&36&16&1&0&\cdots\\
1&25&100&100&25&1&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix}
$$
 is totally positive.
\end{conj}
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%
\section{Appendix}
For the reader's convenience, we indicate a quick path  to  the last four 
 Catalan-Stieltjes matrices of Section~4. 
Our derivation of the Catalan-Stieltjes matrices relies  on
 the correspondance  between \eqref{C-S} and \eqref{CF}.
Introduce  two formal power series $F$ and $V$ given by
\begin{align}
F:=F(z,a,b) :&=\frac{1-az-\sqrt{1-2az+(a^2-4b)z^2}}{2bz^2},\label{F}\\
V:=V(z,a,b):&=\frac{1}{\sqrt{1-2az+(a^2-4b)z^2}}.\label{V-def}
\end{align}
As $F(1-az-bz^2F)=1$ we have
\begin{align}\label{F-CF}
F&=\frac{1}{1-az-bz^2F}
=\frac{1}{1-az-\displaystyle
\frac{bz^2}{1-az-\displaystyle
\frac{bz^2}{1-az-\displaystyle
\frac{bz^2}{\ddots}}}}.
\end{align}
From $1-az-2bz^2F=1/V$ we derive that
\begin{align}\label{V-CF}
V=\frac{1}{1-az-2bz^2F}
=\frac{1}{1-az-\displaystyle
\frac{2bz^2}{1-az-\displaystyle
\frac{bz^2}{1-az-\displaystyle
\frac{bz^2}{\ddots}}}}.
\end{align}
\begin{enumerate}
\item{}(\emph{Nayanara polynomials of type A})
Let $G(z,x)=1+\sum_{n\geq 1} N_n(x)z^n$ be the generating function of Narayana polynomials $N_n(x)$. 
It is known \cite{Aigner-course} that
$$
G(z,x)=\frac{1-(x-1)z-\sqrt{1-2(1+x)z+(1-x)^2z^2}}{2z}.
$$
Setting $a=x+1$ and $b=x$ in \eqref{F}  
we see that $G(z,x)=1+xzF(z,x+1, x)$. Note that
$$
1+xzF(z,x+1, x)=\frac{1}{1-xz-xz^2F(z, x+1,x)}.
$$
We derive then from \eqref{F-CF} the continued fraction expansion of $G(z,x)$
with $r_k=1$ ($k\geq 0$), $s_0=x$, $s_k=x+1$ and $t_k=x$ for $k\geq 1$.
%%%%%%%%%%%%
\item{}(\emph{the restricted hexagonal numbers}) 
The numbers $h_n$ has 
the generating function
$H(z)=\sum_{n\geq 0}h_n z^n$ is given by 
\begin{align}\label{hex}
H(z)=\frac{1-3z-\sqrt{1-6z+5z^2}}{2z^2}=1+3z+10z^2+36z^3+137z^4+\cdots;
\end{align}
see \cite[A002212]{A002212}. 
Hence $H(z)=F(z, 3,1)$. We derive from 
  \eqref{F-CF} the continued fraction \eqref{CF} with 
  $s_k=3$ and $r_k=t_{k+1}=1$ for $k\geq 0$.  
%%%%%%%%%%%%%%%
\item{}(\emph{Schr\"oder polynomials})
Since the Schr\"oder polynomials  $d_n(x)$ are shifted Narayana 
polynomials  $d_n(x)=N_n(x+1)$ (see \cite{BSS93}),
from the  previous result we derive immediately the 
corresponding Catalan-Stieltjes matrix with
 $$
r_k=1\; (k\geq 0), \quad s_0=x+1,\; s_k=x+2,\; t_k=x+1 \quad \text{for}\quad k\geq 1.
$$
\item{}(\emph{Central Delannoy polynomials})
The Legendre polynomials $P_n(x)$ are defined by
$$
P_n(x)=\sum_{j=0}^n {n+j\choose j}{n\choose j} \left(\frac{x-1}{2}\right)^j
$$
and have the generating function 
$$
\sum_{n=0}^\infty P_n(x)z^n=\frac{1}{\sqrt{1-2xz+z^2}}.
$$
It follows from \eqref{delannoy} that 
$D_n(x)=P_n(2x+1)$. Thus
$$
\sum_{n\geq 0}D_n(x)z^n=\frac{1}{\sqrt{1-2(2x+1)z+z^2}}.
$$
This is the special case of \eqref{V-def} with $a=2x+1$ and $b=x^2+x$. 
We derive then from \eqref{V-CF} the continued fraction expansion with 
$r_k=1$ ($k\geq 0$), $s_k=2x+1$, $t_1=2x(x+1)$ and $t_k=x(x+1)$ for $k\geq 1$.
\item{}(\emph{Narayana polynomials of type B})
Using another formula for Legendre polynomials
$$
P_n(x)
=\left(\frac{x-1}{2}\right)^n\sum_{k=0}^n {n\choose k}^2\left(\frac{x+1}{x-1}\right)^k,
$$
we see that the polynomials $W_n(x)$ are related to $P_n(x)$ by
$$
W_n(x)=\sum_{k=0}^n{n\choose k}^2 x^k=(x-1)^nP_n\left(\frac{x+1}{x-1}\right).
$$
It follows from \eqref{V-CF} that 
$$
\sum_{n=0}^\infty W_n(x)z^n=\frac{1}{\sqrt{1-2(1+x)z+(x-1)^2z^2}}.
$$
This is \eqref{V-def} with $a=x+1$ and $b=x$.
 We derive from \eqref{V-CF} the continued fraction expansion \eqref{CF}
with %$s_k = 1+x$ for $k\geq 0$ and $\lambda_1 = 2x$, $\lambda_k = x$ for $k\geq 2$. So we can choose 
$r_0=2, r_k=1$ ($k\geq 1$),  $s_k = 1+x$ for $k\geq 0$ and $t_k = x$ for $k\geq 1$.
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%
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\end{document}