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\begin{center}
\vskip 1cm
{\LARGE{\bf Mean Value Inequalities for \\
\vskip .1in
Motzkin Numbers}}
\vskip 1cm
\large
Takashi Agoh\\
Department of Mathematics\\ 
Tokyo University of Science\\ 
Noda, Chiba 278-8510\\ 
Japan\\
\href{mailto:agoh$\_$takashi@ma.noda.tus.ac.jp}{\tt agoh$\_$takashi@ma.noda.tus.ac.jp}\\
\ \\
Horst Alzer\\
Morsbacher Stra\ss e 10\\ 
51545 Waldbr\"ol\\ 
Germany\\
\href{mailto:h.alzer@gmx.de}{\tt h.alzer@gmx.de}
\end{center}

\vskip .2in

\begin{abstract}
Following up on results of Aigner, we present some inequalities for 
Motzkin numbers $M_n$. In particular, we prove that the sequence 
$(1/M_n)_{n\geq 1}$ is strictly convex.
\end{abstract}

\section{Introduction}

The Motzkin numbers, named after the American mathematician Theodore
S. Motzkin (1908--1970), are defined by
$$
M_n=\sum_{k=0}^{\lfloor n/2 \rfloor} \frac{1}{k+1} {2k\choose k} {n\choose 2k} \quad(n=0,1,2,\ldots).
$$
They satisfy the recurrence relation
\begin{align}~\label{eq:1}
M_n=\frac{2n+1}{n+2}M_{n-1}+\frac{3(n-1)}{n+2}M_{n-2}\quad (n\geq 2).
\end{align}
The generating function is given by
$$
\frac{1}{2x^2}\Bigl( 1-x-\sqrt{1-2x-3x^2}\Bigl) =\sum_{n=0}^\infty M_n x^n.
$$
Here are the first few Motzkin numbers (see the sequence 
\seqnum{A001006} in \cite{Slo}):
$$
1, \quad 1, \quad 2, \quad 4, \quad 9, \quad 21, \quad 51, 
\quad 127, \quad 323, \quad 835.
$$
The Motzkin numbers have interesting applications in number theory and
geometry, and they play an important role in various counting problems. 
For instance, $M_n$ is  the number of paths from $(0,0)$ to $(n,0)$ 
in the integer plane $\mathbb{Z} \times \mathbb{Z}$ which never dip
below the $x$-axis and use only steps $(1,0)$, $(1,1)$ and $(1,-1)$. 
Donaghey and Shapiro \cite{DS} presented a selection of $14$
situations where  these numbers occur; also see Stanley \cite{S}.

This note was inspired by an interesting paper published by Aigner
\cite{A} in 1998. He used tools from linear algebra to prove 
the inequalities
\begin{align}
&\hspace{0.2cm} M_n < 3M_{n-1} \quad (n\geq 1),~\label{eq:2}\\[3pt]
% \end{equation}
% \begin{equation}
& M^2_{n}\leq {M_{n-1} M_{n+1}} \quad (n\geq 1),~\label{eq:3}
\end{align}
and the limit relation
$$
\lim_{n\to\infty} \frac{M_{n-1}}{M_n}=\frac{1}{3}.
$$
A combinatorial proof of \eqref{eq:3} was given by Sun and 
Wang \cite{SW}. From \eqref{eq:3} we conclude that the sequence 
$(M_n)_{n\geq 0}$ is log-convex. 

Here, we present counterparts of \eqref{eq:3}. In particular, 
we show that the sequence $(1/M_n)_{n\geq 1}$ is strictly convex. 
This result and \eqref{eq:3} lead to the double-inequality
$$
\frac{2}{1/M_{n-1}+1/M_{n+1}}<M_n\leq \sqrt{M_{n-1} M_{n+1} }\quad(n\geq 2).
$$
Therefore, we see that if $n\geq 2$, then $M_n$ separates the harmonic 
and geometric means of $M_{n-1}$ and $M_{n+1}$.

We introduce the following notation. Let $a$ and $b$ be positive real
numbers. The weighted harmonic and geometric means of $a$ and $b$ are
defined by
$$
H_t(a,b)=\frac{1}{t/a+(1-t)/b} \quad\mbox{and} 
\quad  G_t(a,b)=a^t b^{1-t} \quad (0\leq t\leq 1),
$$
respectively. For $t=1/2$ we obtain the unweighted harmonic and 
geometric means of $a$ and $b$ as follows:
$$
H(a,b)=\frac{2ab}{a+b}\quad\mbox{and}\quad G(a,b)=\sqrt{ab}.
$$
Moreover, let
$$
F_t(a,b)=tG(a,b)+(1-t)H(a,b) \quad (t\in\mathbb{R}).
$$
We remark that there is a connection between $F_t(a,b)$ and the
so-called Heron means, 
$$
K_t(a,b)=tA(a,b)+(1-t)G(a,b),
$$
where $A(a,b)=(a+b)/2$ denotes the arithmetic mean of $a$ and $b$. We have
$$
A(a,b)F_t(a,b)=G(a,b) K_t(a,b).
$$

\vspace{0.4cm}
\section{Inequalities}

First, we offer lower and upper bounds for $M_n$ in terms of weighted
harmonic and geometric means of $M_{n-1}$ and $M_{n+1}$.
\begin{theorem}~\label{thm:1}
Let $v,w \in (0,1)$. The inequalities
\begin{equation}~\label{eq:4}
H_{v}(M_{n-1}, M_{n+1}) \leq M_n \leq G_{w}(M_{n-1}, M_{n+1}) 
\end{equation}
hold for all $n\geq 2$ if and only if $v\geq 5/14=0.35714\ldots$ 
and $w\leq 1/2$.
\end{theorem}

\begin{proof}
We assume that \eqref{eq:4} is valid for all $n\geq 2$. Then,
$$
H_{v}(M_{2}, M_{4}) \leq M_3 \quad\mbox{and} \quad  M_2\leq G_{w}(M_{1}, M_{3}).
$$
This leads to
$$
\frac{1}{v/2+(1-v)/9}\leq 4 \quad\mbox{and} \quad 2\leq 4^{1-w}.
$$
It follows that  $v\geq 5/14$ and $w\leq 1/2$.

\vspace{0.2cm}
\noindent 
Next, let $w\leq 1/2$ and $\lambda_n=M_{n-1}/M_n$. From \eqref{eq:3} 
we obtain $\lambda_n \leq \lambda_2=1/2$ for $n\geq 2$. Thus, 
\begin{align*}
G_{w}(M_{n-1}, M_{n+1})&=(\lambda_n \lambda_{n+1})^w M_{n+1}\\ 
&\geq (\lambda_n \lambda_{n+1})^{1/2} M_{n+1}\\ 
&=G_{1/2}(M_{n-1}, M_{n+1}) \geq M_n.
\end{align*}
The left-hand side of (4) is equivalent to
$$
Q_n\leq v,
$$
where
$$
Q_n=\frac{M_{n-1}(M_{n+1}-M_n)}{M_n(M_{n+1}-M_{n-1})}
=\frac{\lambda_n(1-\lambda_{n+1})}{1-\lambda_n\lambda_{n+1}}.
$$
It remains to show that for $n\geq 2$,
\begin{align}~\label{eq:5}
Q_n\leq \frac{5}{14}.
\end{align}
We have $Q_2=1/3$ and $Q_3=5/14$. Let $n\geq 4$. Inequality
\eqref{eq:5} can be written as
$$
0\leq \frac{5}{\lambda_n}+9\lambda_{n+1}-14.
$$
Applying \eqref{eq:1} gives
\begin{align}~\label{eq:6}
\frac{1}{\lambda_n}=\frac{2n+1+3(n-1)\lambda_{n-1}}{n+2} \quad\mbox{and}
\quad \lambda_{n+1}=\frac{n+3}{2n+3+3n\lambda_n}.
\end{align}
From \eqref{eq:3} we conclude that $\lambda_n\leq \lambda_{n-1}$, 
so that we get
\begin{align}~\label{eq:7}
\lambda_{n+1}\geq \frac{n+3}{2n+3+3n\lambda_{n-1}}.
\end{align}
Using \eqref{eq:6} and \eqref{eq:7} yields
\begin{eqnarray}\nonumber
\frac{5}{\lambda_n}+9\lambda_{n+1}-14 & \geq & \frac{5(2n+1+3(n-1)\lambda_{n-1})}{n+2}
+\frac{9(n+3)}{2n+3+3n\lambda_{n-1}}-14 \\ \nonumber
&=& \frac{S_n}{(n+2)(2n+3+3n\lambda_{n-1})}, \nonumber
\end{eqnarray}
where
$$
S_n=n^2-13n-15+9(2n^2-6n-5)\lambda_{n-1}+45(n^2-n)\lambda^2_{n-1}.
$$
From \eqref{eq:2} we obtain $\lambda_{n-1}>1/3$. Thus,
$$
S_n>n^2-13n-15+9(2n^2-6n-5)\frac{1}{3}+45(n^2-n)\frac{1}{9}=6(2n^2-6n-5)>0.
$$
This implies that \eqref{eq:5} holds. The proof of Theorem 1 is now
complete. 
\end{proof}

Our second result yields sharp bounds for $M_n$ in terms of
$F_t(M_{n-1}, M_{n+1})$. The following companion to \eqref{eq:4} is
valid.

\begin{theorem}~\label{thm:2}
Let $\alpha, \beta \in \mathbb{R}$. The inequalities
\begin{align}~\label{eq:8}
F_{\alpha}(M_{n-1}, M_{n+1}) \leq M_n \leq F_{\beta}(M_{n-1}, M_{n+1}) 
\end{align}
hold for all $n\geq 2$ if and only if
$$
\alpha\leq \frac{8}{3(11\sqrt{2}-12)}=0.74983\ldots 
\quad{and} \quad \beta\geq 1.
$$
\end{theorem}

\begin{proof}
First, we assume that \eqref{eq:8} is valid for all $n\geq 2$. 
Then we have  
$$
F_{\alpha}(M_2,M_4)\leq M_3 \quad\mbox{and} \quad M_2\leq F_{\beta}(M_1,M_3),
$$
and hence, 
$$
\sqrt{18}\alpha+\frac{36}{11}(1-\alpha)\leq 4\quad\mbox{and}
\quad 2\leq 2\beta+\frac{8}{5}(1-\beta).
$$
Therefore, it follows that 
$$
\alpha\leq \frac{8}{3(11\sqrt{2}-12)} \quad\mbox{and} \quad 1 \leq \beta.
$$
Applying \eqref{eq:3} and the fact that $F_t(a,b)$ is increasing with
respect to $t$ we obtain for $\beta\geq 1$ and $n\geq 2$, 
$$
 M_n \leq F_{1}(M_{n-1}, M_{n+1})  \leq F_{\beta}(M_{n-1}, M_{n+1}).
$$
Let
$$
\alpha_0=  \frac{8}{3(11\sqrt{2}-12)}.
$$
We have to show that for $n\geq 2$,
$$
 F_{\alpha_0}(M_{n-1}, M_{n+1})  \leq M_{n},
$$
or, equivalently,
$$
\alpha_0\leq R_n,
$$
where
$$
R_n=\frac{M_n -H(M_{n-1},M_{n+1})}{G(M_{n-1},M_{n+1})-H(M_{n-1},M_{n+1})}.
$$
By direct computation we find
$$
R_2=1, \quad R_3=\alpha_0, \quad R_4=0.932\ldots, 
\quad R_5=0.930\ldots, 
\quad  R_6=0.958\ldots,
$$
$$
 R_7=0.966\ldots, \quad R_8=0.974\ldots, \quad R_9=0.979\ldots, \quad R_{10}=0.983\ldots.
$$
Let $n\geq 11$. It suffices to prove that $R_n>3/4$, or, equivalently,
\begin{align}~\label{eq:9}
3 G(M_{n-1}, M_{n+1})  + H(M_{n-1}, M_{n+1})  <4 M_{n}.
\end{align}
As before, let  $\lambda_n=M_{n-1}/M_n$. Then, \eqref{eq:9} can be
written as 
$$
0<U_n-V_n,
$$
where
$$
U_n=4-2\Bigl(\frac{1}{\lambda_n}+\lambda_{n+1}\Bigr)^{-1} 
\quad\mbox{and}\quad V_n=3\sqrt{\lambda_n \frac{1}{\lambda_{n+1}}}.
$$
Using \eqref{eq:6} gives
\begin{align*}
U_n&=4-2\Bigl(\frac{1}{\lambda_n}+\frac{n+3}{2n+3+3n\lambda_n}\Bigr)^{-1}\\
&=\frac{2\bigl(  4n+6+3(n+1)\lambda_n+3n\lambda_n(1-\lambda_n)\bigr)}{2n+3+(4n+3)\lambda_n}
\end{align*}
and
$$
V_n=3\sqrt{\lambda_n \frac{2n+3+3n\lambda_n}{n+3}}.
$$
Now we have
\begin{align}~\label{eq:10}
\begin{split}
&(U^2_n-V^2_n)(n+3)\bigl(2n+3+(4n+3)\lambda_n\bigr)^2\\
&=P_1(\lambda_n) n^3+P_2(\lambda_n) n^2
+P_3(\lambda_n) n+P_4(\lambda_n)
\end{split}
\end{align}
with
\begin{align*}
P_1(t)&=-396t^4-864t^3-348t^2+120t+64,\\
P_2(t)&=-540t^4-2340t^3-1260t^2+636t+384,\\
P_3(t)&=-243t^4-1512t^3-1503t^2+810t+720,\\
P_4(t)&=-243t^3-378t^2+189t+432.
\end{align*}

\vspace{0.2cm}
\noindent
Since $P''_j(t)<0$ ($j=1,2,3,4$) for $t\geq 0$, we conclude that 
the functions $P_1$, $P_2$, $P_3$, and $P_4$ are concave on 
$[0,\infty)$.
It follows that, for $t\in [0,  0.38]$,
$$
P_j(t)\geq m_j \quad(j=1,2,3,4),
$$
where
$$
m_j=
\min\{P_j(0), P_j(0.38) \}.
$$
We have
$$
m_1=3.68\ldots, \quad m_2=304.07\ldots, \quad m_3=720, \quad m_4=432.
$$
Thus, it follows that $P_j(t)>0$ $(j=1,2,3,4)$. 
Since $(\lambda_n)_{n\geq 1}$ is decreasing, we obtain 
$\lambda_n\leq\lambda_{11}=0.377\ldots$. From \eqref{eq:10} 
we conclude that $U^2_n-V^2_n>0$, so that $U_n+V_n>0$ leads to 
$U_n-V_n>0$. This completes the proof of Theorem~\ref{thm:2}.
\end{proof}

\begin{remark}
\leavevmode
\begin{itemize}
\item[(i)]  Computer calculations reveal that for 
$n=4,5,\ldots,500$ the lower bound given in \eqref{eq:8} with 
$\alpha=8/(33\sqrt{2}-36)$ is greater than the lower bound 
in \eqref{eq:4} with $v=5/14$.

\item[(ii)] \ Aigner's inequality \eqref{eq:3} can be written as
\begin{align}~\label{eq:11}
\frac{M_n - M_{n-1}}{M_{n+1}-M_n}\leq \frac{M_n}{M_{n+1}}.
\end{align}
\end{itemize}
\end{remark}

It is possible to show that the following counterpart of \eqref{eq:11}
holds. 
\begin{theorem}~\label{thm:3}
For all $n\geq 2$ we have
$$
c_0  \frac{M_n}{M_{n+1}}
\leq  \frac{M_n-M_{n-1}}{M_{n+1}-M_n}\leq
c_1 \frac{M_n}{M_{n+1}}
$$
with the best possible constant factors $c_0=9/10$ and $c_1=1$.
\end{theorem}

This result can be proved by using the same method as in the proof 
of Theorem~\ref{thm:2}. So we omit the details.

\section{Acknowledgments} 
We thank the referee for helpful comments.

\begin{thebibliography}{9}

\bibitem{A}
M. Aigner, Motzkin numbers, \textit{European J. Combin.}  
{\bf{19}} (1998), 663--675.

\bibitem{DS}
R. Donaghey and L.~W. Shapiro, Motzkin numbers, 
\textit{J. Combin. Theory Ser. A} {\bf{23}} (1977), 291--301.

\bibitem{Slo}
N.~J.~A.~Sloane et al., The On-Line Encyclopedia of Integer Sequences, 
2020. Available at \url{https://oeis.org}.

\bibitem{S}
R.~P. Stanley, \textit{Enumerative Combinatorics}, Vol.~2, 
Cambridge University Press, 1999.

\bibitem{SW}
H. Sun and Y. Wang, A combinatorial proof of the log-convexity 
of Catalan-like numbers, \textit{J. Integer Sequences} {\bf{17}} (2014), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL23/Frontczak/front16.html}{Article 14.5.2}.

\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 05A20, 11B83.

\noindent \emph{Keywords: }
Motzkin number, inequality, mean value.

\bigskip
\hrule
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\noindent (Concerned with sequences 
\seqnum{A001006} and
\seqnum{A114473}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 1 2020; 
revised version received  May 27 2021.
Published in {\it Journal of Integer Sequences}, June 2 2021.

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\noindent
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