<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

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

\usepackage{amsthm}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}

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

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

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

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

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

\newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}}

\begin{document}

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

\begin{center}
\vskip 1cm{\LARGE\bf A Recursive Relation for Weighted Motzkin Sequences}
\vskip 1cm
\large
Wen-jin Woan\\
Department of Mathematics\\
Howard University\\
Washington, D.C. 20059\\
USA \\
\href{mailto:wwoan@howard.edu}{\tt wwoan@howard.edu} \\
\end{center}

\vskip .2 in


\begin{abstract}
We consider those lattice paths that use the steps {\it Up\/},
{\it Level\/}, and {\it Down\/} with assigned weights $w$, $u$, and $v$.
In probability theory, the
total weight is 1. In combinatorics, we regard weight as the number of
colors and normalize by setting $w=1$. The lattice paths generate Motzkin
sequences. Here we give a combinatorial proof of a three-term recursion
for a weighted Motzkin sequence and we find the radius of convergence.
\end{abstract}

%\newtheorem{theorem}{Theorem}[section]
%\newtheorem{proposition}{Proposition}[section]
%\newtheorem{corollary}{Corollary}[section]
%\newtheorem{lemma}{Lemma}[section] 




%\maketitle

\section{Introduction}

We consider those lattice paths in the Cartesian plane starting
from $(0,0)$ that use the steps $U$, $L$, and $D$, where
$U=(1,1)$, an up-step; $L=(1,0)$, a level-step; and $D=(1,-1)$, a
down-step. Let $c$ and $d$ be positive integers, and color the $L$
steps with $d$ colors and the $D$ steps with $c$ colors. Let $
A(n,k)$ be the set of all colored paths ending at the point
$(n,k)$, and let $M(n,k)$ be the set of lattice paths in $A(n,k)$
that never go below the $x$-axis. Let $a_{n,k}=\left|
A(n,k)\right|$, $m_{n,k}=\left| M(n,k)\right|$, and $m_n=\left|
M(n,0)\right|$. The number $m_n$ is called the $ (1,d,c)$-Motzkin
number. Let $B(n,k)$ denote the set of lattice paths in $ A(n,k)$
that never return to the $x$-axis and let $b_{n,k}=\left|
B(n,k)\right|$. Note that
$a_{n,k}=a_{n-1,k-1}+da_{n-1,k}+ca_{n-1,k+1}$. Here we give a
combinatorial proof of the following three-term recursion for the
$(1,d,c)$-Motzkin sequence:
\[
(n+2)m_n=d(2n+1)m_{n-1}+(4c-d^2)(n-1)m_{n-2}.
\]


If $\frac{\sqrt{c}}2\leq d$, then
$\displaystyle
\lim_{n\to\infty }\frac{m_n}{m_{n-1}}=k=d+2\sqrt{c}$.


\begin{example}\label{example1} The first few terms of the $(1,3,2)$-Motzkin numbers are $
m_0=1,3,11,45,197,903,\ldots$,  $k=3+2\sqrt{2}$. This sequence is
the little Schroeder number sequence and is Sloane's sequence \seqnum{A001003}.
Some entries of the matrices
$(a_{n,k}), \;(m_{n,k})$ and $(b_{n,k})$ are as follows;
\[
(a_{n,k})=\left[
\begin{tabular}{cccccccccc}
$n/k$ &amp; $-4$ &amp; $-3$ &amp; $-2$ &amp; $-1$ &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ \\
$0$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $1$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $0$ \\
$1$ &amp; $0$ &amp; $0$ &amp; $0$ &amp; $2$ &amp; $3$ &amp; $1$ &amp; $0$ &amp; $0$ &amp; $0$ \\
$2$ &amp; $0$ &amp; $0$ &amp; $4$ &amp; $12$ &amp; $13$ &amp; $6$ &amp; $1$ &amp; $0$ &amp; $0$ \\
$3$ &amp; $0$ &amp; $8$ &amp; $36$ &amp; $66$ &amp; $63$ &amp; $33$ &amp; $9$ &amp; $1$ &amp; $0$ \\
$4$ &amp; $16$ &amp; $96$ &amp; $248$ &amp; $360$ &amp; $321$ &amp; $180$ &amp; $62$ &amp; $12$ &amp; $1$%
\end{tabular}
\right] ,
\]

\[
(m_{n,k})=\ \left[
\begin{array}{cccccc}
n/k &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 \\
0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\
2 &amp; 11 &amp; 6 &amp; 1 &amp; 0 &amp; 0 \\
3 &amp; 45 &amp; 31 &amp; 9 &amp; 1 &amp; 0 \\
4 &amp; 197 &amp; 156 &amp; 60 &amp; 12 &amp; 1
\end{array}
\right] ,
\]

\[
(b_{n,k})=\left[
\begin{array}{ccccccc}
n/k &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\
0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
1 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
2 &amp; 0 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\
3 &amp; 0 &amp; 11 &amp; 6 &amp; 1 &amp; 0 &amp; 0 \\
4 &amp; 0 &amp; 45 &amp; 31 &amp; 9 &amp; 1 &amp; 0 \\
5 &amp; 0 &amp; 197 &amp; 156 &amp; 60 &amp; 12 &amp; 1
\end{array}
\right] .
\]
\end{example}

\begin{example}\label{example2}
The first few terms of the $(1,2,1)$-Motzkin numbers are $
m_0=1,2,5,14,42,\ldots , k=2+2\sqrt{1}=4$. This sequence is the
Catalan sequence, Sloane's sequence \seqnum{A000108}.\end{example}

\begin{example}\label{example3}
The first few terms of the $(1,1,1)$-Motzkin numbers are $
m_0=1,1,2,4,9,21,\ldots , k=1+2\sqrt{1}=3$. The sequence is the
Motzkin sequence, discussed by Woan \cite{Wo}, and is Sloane's
sequence \seqnum{A001006}.\end{example}

\section{Main Results}

 We apply the cut and paste technique to
prove the following lemma. Please refer to Dershowitz and Zaks \cite{DZ}
and Pergola and Pinzani \cite{PP}
for information about the technique.

\begin{lemma}\label{lemma1}There is a combinatorial proof for the equation $
m_n=b_{n+1,1}=db_{n,1}+cb_{n,2}=dm_{n-1}+cb_{n,2}$.\end{lemma}

\begin{proof} Let $P\in B(n+1,1).$ Remove the first step $(U)$ and note that the
remaining is in $M(n,0)$.  \end{proof}

For example, the path $P=UULDLUUUDDLD\in B(12,1)$ becomes
 $Q=ULDLUUUDDLD\in M(11)$ where $\times $ marks the origin.


\[
P=
\begin{array}{ccccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  \\
&amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  \\
&amp; \circ &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ \\
\times &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;
\end{array}
,
\]
\[
Q=
\begin{array}{cccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  \\
&amp; \circ &amp; \circ &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  \\
\times &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ
\end{array}
.
\]


\begin{theorem}\label{theorem2} There is a combinatorial proof for the equation $
(n+1)b_{n+1,1}=a_{n+1,1}$.\end{theorem}

\begin{proof} Wendel \cite{We} proved a similar result. Let 
$$S(n+1)=\{P^{*}:P\in B(n+1,1)$$
where $P^{*}$ is $P$ with one vertex marked$\}$. Then $\left| S(n+1)\right|
=(n+1)b_{n+1,1}$. Let $P^{*}\in S(n+1)$. The marked vertex partitions the
path into $P=FB,$ where $F$ is the front section and $B$ is the back
section. Then $Q=BF\in A(n+1,1)$. Note that, graphically, the point of
attachment is the rightmost lowest point of $Q$.

Conversely, starting with
any path we may find the rightmost lowest point of $Q$ and reverse the
procedure to create a marked path $P^{*}$ in $B(n+1,1)$.\end{proof}

For example,
\[
P^{*}=
\begin{array}{ccccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  \\
&amp;  &amp; \circ &amp; \bullet &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  \\
&amp; \circ &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ \\
\times &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;
\end{array}
\in S(12),
\]
\[
\rightarrow Q=
\begin{array}{ccccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp; \circ \\
\times &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp; \circ &amp;  &amp;  \\
&amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \bullet &amp;  &amp;  &amp;
\end{array}
\in A(12,1).
\]

\begin{proposition}\label{proposition3}
The total number of $L$ steps in $M(n)$ is the same
as that in $B(n+1,1)$ and is $da_{n,1}$.\end{proposition}

\begin{proof} From the proof of Lemma \ref{lemma1}, there is a bijection between $M(n)$ and $
B(n+1,1)$. Let $P=FLB\in B(n+1,1)$ with an $L$ step. Then $Q=BF\in A(n,1)$.
Note that the attachment point is the rightmost lowest point in $Q$ since $
P\in B(n+1,1)$. This identification suggests the inverse mapping. Note that
there are $d$ colors for an $L$ step.\end{proof}

For example,

\[
P=
\begin{array}{ccccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  \\
&amp;  &amp; \bullet &amp; \bullet &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  \\
&amp; \circ &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ \\
\times &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;
\end{array}
\in B(12,1),
\]
\[
\rightarrow Q=
\begin{array}{cccccccccccc}
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp; \circ \\
\times &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp; \circ &amp;  \\
&amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \bullet &amp;  &amp;
\end{array}
\in A(11,1).
\]

\begin{proposition}\label{proposition4} There is a combinatorial proof for the equation
$$\displaystyle a_{n,0}=m_n+\frac 12(nm_n-da_{n,1})=b_{n+1,1}+\frac
12(nb_{n+1,1}-dnb_{n,1}).$$
\end{proposition}

\begin{proof} Let $T(n)=\{P^e:P\in M(n)$ where $P^e$ is $P$ with a $U$ step marked$\}$.
By Theorem \ref{theorem2} and Proposition \ref{proposition3} the number of $L$ steps among all paths in $
M(n)$ is $da_{n,1}=dnb_{n,1}$. The total number of steps among all paths in $
M(n)$ is $nm_n=nb_{n+1,1}$, hence the total number of $U$ steps among all
paths in $M(n)$ is $\frac 12(nb_{n+1,1}-dnb_{n,1})=\left| T(n)\right|$.
Let $P^e=FUB\in T(n)$ with a $U$ step marked. Then $Q=BUF\in
A(n,0)-M(n,0)$ and the initial point of $U$ in $Q$ is the rightmost lowest
point in $Q$. The inverse mapping starts with the rightmost lowest point.
Note that $\left| M(n,0)\right| =m_n=b_{n+1,1}$\end{proof}

For example,
\[
P^e=
\begin{array}{cccccccccccc}
&amp;  &amp;  &amp;  &amp; \circ &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  \\
&amp;  &amp;  &amp; \bullet &amp;  &amp; \circ &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  \\
&amp; \circ &amp; \bullet &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  \\
\times &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ
\end{array}
\in T(11),
\]
\[
Q=
\begin{array}{cccccccccccc}
&amp; \circ &amp;  &amp; \circ &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
\times &amp;  &amp; \circ &amp;  &amp;  &amp; \circ &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp; \circ \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp; \circ &amp; \circ &amp;  &amp; \bullet &amp;  &amp;  \\
&amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \bullet &amp;  &amp;  &amp;
\end{array}
\in A(11,0).
\]

\begin{proposition}\label{proposition5}
There is a combinatorial proof for the equation
\begin{eqnarray*}
a_{n,0}&amp;=&amp;a_{n-1,-1}+da_{n-1,0}+ca_{n-1,1}=2ca_{n-1,1}+da_{n-1,0}\\
&amp;=&amp;2c(n-1)b_{n-1,1}+d(b_{n,1}+\frac 12((n-1)b_{n,1}-d(n-1)b_{n-1,1}))\,.
\end{eqnarray*}
\end{proposition}

\begin{proof} The first equality represents the partition of $A(n,0)$ by the last
step ($U$, $L$ or $D$). The second equality represents the fact that $
a_{n-1,-1}=ca_{n-1,1}$, since elements in $A(n-1,-1)$ have one more $D$ step
than those in $A(n-1,1)$. And the last equality holds by Theorem \ref{theorem2} and Proposition
\ref{proposition4}.\end{proof}


Sulanke \cite{S} proved the following result for the Motzkin sequence.

\begin{theorem}\label{theorem6} $(n+2)m_n=(2dn+d)m_{n-1}+(4c-d^2)(n-1)m_{n-2}.$
\end{theorem}

\begin{proof} By Propositions \ref{proposition4} and \ref{proposition5}
\begin{eqnarray*}
b_{n+1,1}+\frac
12(nb_{n+1,1}-dnb_{n,1})&amp;=&amp;2c(n-1)b_{n-1,1}+db_{n,1}+(\frac
12((n-1)b_{n,1}-d(n-1)b_{n-1,1}))\,.
\end{eqnarray*}
By Lemma \ref{lemma1}
\begin{eqnarray*}
m_n+\frac 12(nm_n-dnm_{n-1}).&amp;=&amp;2c(n-1)m_{n-2}+dm_{n-1}+d(\frac
12((n-1)m_{n-1}-d(n-1)m_{n-2}))\,.
\end{eqnarray*}
Equivalently
\[
(n+2)m_n=(2dn+d)m_{n-1}+(4c-d^2)(n-1)m_{n-2}\,.
\]
\end{proof}



\begin{theorem}\label{theorem7} $\displaystyle \lim_{n\to \infty}
\frac{m_n}{m_{n-1}}=k=d+2\sqrt{c}$. \end{theorem}

\begin{proof} By Theorem \ref{theorem6}, let
\[
s_n:=\frac{m_n}{m_{n-1}}=\frac{d(2n+1)}{n+2}+(\frac{(4c-d^2)(n-1)}{
n+2})\frac{m_{n-2}}{m_{n-1}}\,,
\]
and let
\begin{eqnarray*}
a_n&amp;:=&amp;\frac{d(2n+1)}{n+2}=2d-\frac{3d}{n+2}\\
b_n&amp;:=&amp;\frac{
(4c-d^2)(n-1)}{n+2}=\ (4c-d^2)(1-\frac 3{n+2}).
\end{eqnarray*}
Then $s_n=a_{n+}\frac{b_n}{s_{n-1}}$.


If the sequence $s_n=a_{n+}\frac{b_n}{s_{n-1}}$ has limit $k$, then $
k^2=2dk+(4c-d^2)$ and $k=\frac{2d+\sqrt{4d^2+4(4c-d^2)}}2=d+2\sqrt{c}$.

\medskip\noindent
{\bf Case 1}. $4c-d^2\geq 0$.

If $s_{n-1}\leq k,$ then
\begin{eqnarray*}
s_n &amp;=&amp;a_{n+}\frac{b_n}{s_{n-1}}\geq 2d-\frac{3d}{n+2}+\frac{(4c-d^2)\frac{
n-1}{n+2}}k=k-\frac{3dk+3(4c-d^2)}{k(n+2)} \\
\ \ \ &amp;=&amp;k-\frac{3k^2-3dk}{k(n+2)}=k-\frac{3(k-d)}{n+2}
\end{eqnarray*}
and
\begin{eqnarray*}
s_{n+1}&amp;=&amp;a_{n+1}+\frac{b_{n+1}}{s_n}\leq 2d-\frac{3d}{n+3}+\frac{
(4c-d^2)(\frac n{n+3})}{k-\frac{3(k-d)}{n+2}}\\
&amp;=&amp;2d-\frac{3d}{n+3}+\frac{\left( 4c-d^2\right) }k(1-\frac{3\ dn+9d-3k}{%
(n+3)(kn-k+3d)})\\
&amp;=&amp;2d+\frac{\left( 4c-d^2\right) }k-\frac{3d}{n+3}-
\frac{\left( 4c-d^2\right) }k\frac{3\ dn+9d-3k}{(n+3)(kn-k+3d)}\\
&amp;=&amp;k-\frac{3(2dn(k-d)-(k-d)(k-3d))}{(n+3)(kn-k+3d)}\\
&amp;=&amp;k-\frac{%
6(k-d)(d(n+1)-\sqrt{c})}{(n+3)(kn-k+3d)}.
\end{eqnarray*}

Note that $s_1=\frac d1$ and $s_2=\frac{d^2+c}d=d+\frac cd$.
If $\frac{\sqrt{c}}
2\leq d$, then $s_1,s_2\leq k.$
By induction on both odd and even $n$ we
have
\[k-\frac{3(k-d)}{n+3}\leq s_{n+1}\leq k-\frac{6(k-d)(d(n+1)-\sqrt{c})}{
(n+3)(kn-k+3d)}\leq k\,.\]

and $\displaystyle \lim_{n\to \infty }\frac{m_n}{m_{n-1}}=k$.

\medskip\noindent
{\bf Case 2}. $4c-d^2&lt;0$.

Inductively, assuming that $s_{n-1}\leq k$ and $\{s_i\}$ is nondecreasing
up to $n-1$. Then

\begin{eqnarray*}
s_n &amp;=&amp;a_{n+}\frac{b_n}{s_{n-1}}\leq 2d-\frac{3d}{n+2}+\frac{(4c-d^2)\frac{
n-1}{n+2}}k=k-\frac{3dk+3(4c-d^2)}{k(n+2)} \\
&amp;=&amp;k-\frac{3k^2-3dk}{k(n+2)}=k-\frac{3(k-d)}{n+2}&lt;k
\end{eqnarray*}


and

\begin{eqnarray*}
s_n-s_{n-1}&amp;=&amp;(\frac{d(2n+1)}{n+2}+\frac{(4c-d^2)(n-1)}{s_{n-1}(n+2)})-\hbox{}\\
&amp;&amp;(\frac{
d(2n-1)}{n+1}+\frac{(4c-d^2)(n-2)}{s_{n-2}(n+1)}) \\
&amp;=&amp;d(2-\frac 3{n+2})+\frac{(4c-d^2)}{s_{n-1}}(1-\frac 3{n+2})-\hbox{}\\
&amp;&amp;(d(2-\frac
3{n+1})+\frac{(4c-d^2)}{s_{n-2}}(1-\frac 3{n+1}))\\
&amp;\geq&amp; -\frac{3d}{n+2}+\frac{3d}{n+1}+\frac{(4c-d^2)}{s_{n-1}}(-\frac
3{n+2}+\frac 3{n+1})\\
&amp;=&amp;\frac{3d}{(n+2)(n+1)}+\frac{3(4c-d^2)}{s_{n-1}(n+2)(n+1)}\\
&amp;=&amp;\frac{3ds_{n-1}+3(4c-d^2)}{s_{n-1}(n+2)(n+1)}\geq
\frac{12c}{s_{n-1}(n+2)(n+1)}&gt;0\,.
\end{eqnarray*}

By induction $\{s_i\}$ is a bounded nondecreasing sequence and
\[\lim_{n\to \infty }\frac{m_n}{m_{n-1}}=k\,.\]
\end{proof}

\begin{thebibliography}{9}

\bibitem{DZ}  N. Dershowitz and S. Zaks, The cycle lemma and
some applications, {\it European
J. Combin.\/} {\bf 11} (1990), 34--40.

\bibitem{PP} E. Pergola and R. Pinzani, A combinatorial interpretation of the area of
Schroeder paths, {\it Electron. J. Combin.\/} {\bf 6} (1999), Research paper 40, 13 pp.

\bibitem{S} R.  Sulanke, Moments of generalized Motzkin paths, {\it J. Integer Seq.\/}
{\bf 3} (2000), Article 00.1.1.

\bibitem{We} J. Wendel, Left-continuous random walk and the Lagrange expansion,
{\it Amer. Math. Monthly} {\bf 82} (1975), 494--499.

\bibitem{Wo} W. Woan, A Combinatorial proof of a recursive relation of the
Motzkin sequence by Lattice Paths {\it Fibonacci Quart.\/} {\bf 40}
(2002), 3--8.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 05A19.

\noindent \emph{Keywords:} Motzkin paths, combinatorial identity.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000108},
\seqnum{A001003}, and
\seqnum{A001006}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 9 2005;
revised version received February 24 2005.  
Published in {\it Journal of Integer Sequences},
February 28 2005.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                


\end{document}
</pre></body></html>