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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Sufficient Conditions for the \\
\vskip .12in
Log-Balancedness of Combinatorial Sequences
}
\vskip 1cm
\large
Ting-Ting Zhang \\
Department of Mathematics\\
Shanghai University \\
Shanghai 200444 \\
P. R. China\\
\href{mailto:ztingting@i.shu.edu.cn}{\tt ztingting@i.shu.edu.cn}\\
\ \\
Feng-Zhen Zhao\\
Department of Mathematics \\
Shanghai University\\
Shanghai 200444 \\
P. R. China\\
\href{mailto:fengzhenzhao@shu.edu.cn}{\tt fengzhenzhao@shu.edu.cn}\\
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\begin{abstract}
In this paper, we give some new sufficient conditions for
log-balancedness of combinatorial sequences. In particular, we show
that the product of two log-convex sequences is log-balanced under a
mild condition. Then, we apply this result to a series of special
combinatorial sequences. In addition, we show some results by using the
definition of log-balancedness directly.
\end{abstract}

\section{Introduction}

For convenience, we first recall some concepts that will be used later on. The following definition is well known in combinatorics.

\begin{definition} \label{de1-0}
(i) For a sequence of real numbers $\{z_n\}_{n\ge 0}$, we say that $\{z_n\}_{n\geq 0}$ is {\it concave} (resp., {\it convex}) if $2z_n\ge z_{n-1}+z_{n+1}$ (resp., $2z_n\le z_{n-1}+z_{n+1})$ for all $n\ge 1$.\\
(ii) For a sequence of positive numbers $\{z_n\}_{n\geq 0}$, we say that $\{z_n\}_{n\geq 0}$ is {\it log-concave} (resp., {\it log-convex}) if $z_n^2\geq z_{n-1}z_{n+1}$ (resp., $z_n^2\leq z_{n-1}z_{n+1}$) for all $n\geq 1$.
\end{definition}

Do\v{s}li\'{c} \cite{ref1} gave the following definition.

\begin{definition} \label{de1-1}
 Let $\{z_n\}_{n\geq 0}$ be a log-convex sequence. We say that $\{z_n\}_{n\geq 0}$ is {\it log-balanced} if  $\{\frac{z_n}{n!}\}_{n\ge 0}$ is log-concave.
\end{definition}

Log-concavity and log-convexity play important roles in many subjects. For example, in combinatorics, they are not only instrumental in obtaining the growth rate of a combinatorial sequence, but also fertile sources of inequalities. See, e.g., \cite{Asai, Stan} for more applications of log-concavity and log-convexity.

For a sequence of positive numbers, it is easy to see from the arithmetic-geometric mean inequality that its concavity implies its log-concavity and its log-convexity implies its convexity. Obviously, a sequence $\{z_n\}_{n\ge 0}$ is log-convex (resp., log-concave) if and only if its quotient sequence $\{\frac{z_{n+1}}{z_n}\}_{n\geq 0}$ is nondecreasing (resp., nonincreasing). A log-balanced sequence is naturally log-convex, but its quotient sequence does not grow too fast. Moreover, a sequence $\{z_n\}_{n\ge 0}$ is log-balanced if and only if $z_n^2\le z_{n-1}z_{n+1}$ and $(n+1)z_n^2\ge n z_{n-1}z_{n+1}$ for every $n\ge 1$. Do\v{s}li\'{c} \cite{ref1} showed that many combinatorial sequences, including the Motzkin numbers, the Fine numbers, the Franel numbers of order $3$ and $4$, the Ap\'{e}ry numbers, the large Schr\"{o}der numbers, and the central Delannoy numbers, are log-balanced. Zhao \cite{ref5, ref6} proved that the sequences of the exponential numbers and the Catalan-Larcombe-French numbers are respectively log-balanced.

The main purpose of this paper is to discuss log-balancedness of some combinatorial sequences. In the next section, we present some new sufficient conditions for log-balancedness of combinatorial sequences. In particular, we provide a sufficient condition for log-balancedness of the product of two log-convex sequences. Then, based on this result, we obtain some similar results for a series of special combinatorial sequences.

\section{Main results}

Zhao \cite{ref5} gave a sufficient condition for log-balancedness of
the product of a log-balanced sequence and a log-concave sequence. Here, we consider log-balancedness of the product of two log-convex sequences.

\begin{theorem}\label{thm:1a}
Suppose that the sequences $\{x_n\}_{n\ge 0}$ and $\{y_n\}_{n\ge 0}$ are both log-convex. Let $s_n=\frac{x_{n+1}y_{n+1}}{(n+1)x_ny_n}$ for $n\ge 0$. If $\{s_n\}_{n\ge 0}$ is decreasing, then $\{x_ny_n\}_{n\ge 0}$ is log-balanced.
\end{theorem}

\begin{proof} By the log-convexity of the sequences $\{x_n\}_{n\ge 0}$ and $\{y_n\}_{n\ge 0}$, we know that $\{x_ny_n\}_{n\ge 0}$ is log-convex. Note that $\{s_n\}_{n\ge 0}$ is the quotient sequence of $\{\frac{x_ny_n}{n!}\}_{n\ge 0}$. Since  $\{s_n\}_{n\ge 0}$ is decreasing, $\{\frac{x_ny_n}{n!}\}_{n\ge 0}$ is log-concave. Hence, the sequence $\{x_ny_n\}_{n\ge 0}$ is log-balanced.
\end{proof}

Next, we apply Theorem \ref{thm:1a} to deduce log-balancedness of some combinatorial sequences.

\begin{corollary}
For the sequence $\{C_n\}_{n\ge 1}$ of the Catalan numbers, we have that $\{C_n^2\}_{n\ge 3}$ is log-balanced.
\end{corollary}

\begin{proof} Since $\{C_n\}_{n\ge 1}$ is log-convex, $\{C_n^2\}_{n\ge 1}$ is log-convex. For $n\ge 1$, let $s_n=\frac{C_{n+1}^2}{(n+1)C_n^2}$. It is well known that
$$
C_n=\frac{1}{n}{2n-2\choose n-1}, \quad n\ge 1.
$$
Then we have
\begin{eqnarray*}
s_n=\frac{4(2n-1)^2}{(n+1)^3}.
\end{eqnarray*}
It is not difficult to verify that $\{s_n\}_{n\ge 3}$ is decreasing. By Theorem \ref{thm:1a}, the sequence $\{C_n^2\}_{n\ge 3}$ is log-balanced.
\end{proof}

\begin{corollary}
For the sequence $\{M_n\}_{n\ge 0}$ of the Motzkin numbers, we have that $\{M_n^2\}_{n\ge 1}$ is log-balanced.
\end{corollary}

\begin{proof} The Motzkin numbers satisfy the recurrence
\begin{eqnarray}
(n+3)M_{n+1}=(2n+3)M_n+3nM_{n-1}, \quad M_0=M_1=1. \label{zf-2.1}
\end{eqnarray}
For $n\ge 0$, let $t_n=\frac{M_{n+1}}{M_n}$ and $s_n=\frac{t_n^2}{n+1}$. It follows from (\ref{zf-2.1}) that
\begin{eqnarray}
t_n=\frac{2n+3}{n+3}+\frac{3n}{(n+3)t_{n-1}}.  \label{zf-2.2}
\end{eqnarray}
Then we have
\begin{eqnarray*}
s_n-s_{n+1}=\frac{(n+2)t_n^2-(n+1)t_{n+1}^2}{(n+1)(n+2)}.
\end{eqnarray*}
It follows from (\ref{zf-2.2}) that
\begin{eqnarray*}
&amp;&amp;\quad (n+2)t_n^2-(n+1)t_{n+1}^2\\
&amp;&amp;=\frac{(n+2)(n+4)^2t_n^4-(n+1)(2n+5)^2t_n^2-6(n+1)^2(2n+5)t_n-9(n+1)^3}{(n+4)^2t_n^2}.
\end{eqnarray*}

For any real number $x$, we let
$$
f(x)=(n+2)(n+4)^2x^4-(n+1)(2n+5)^2x^2-6(n+1)^2(2n+5)x-9(n+1)^3.
$$
Then we have
\begin{eqnarray*}
f^{\prime}(x)=4(n+2)(n+4)^2x^3-2(n+1)(2n+5)^2x-6(n+1)^2(2n+5)
\end{eqnarray*}
and
\begin{eqnarray*}
f^{\prime\prime}(x)=12(n+2)(n+4)^2x^2-2(n+1)(2n+5)^2.
\end{eqnarray*}
Since $f^{\prime\prime}(x)&gt;0$ when $x\ge 1$, we know that $f^{\prime}$ is increasing over $[1,\infty)$. Do\v{s}li\'{c} and Veljan \cite{ref2} showed that
\begin{eqnarray*}
t_n\ge q_n,
\end{eqnarray*}
where $q_n=\frac{6(n+1)}{2n+5}$. Since
\begin{eqnarray*}
f^{\prime}(q_n)=\frac{18(n+1)^2[48(n+1)(n+2)(n+4)^2-(2n+5)^4]}{(2n+5)^3}&gt;0,
\end{eqnarray*}
the function $f$ is increasing over $[q_n, \infty)$. Note that
\begin{eqnarray*}
f(q_n)=\frac{81(n+1)^3(16n^3+72n^2+24n-113)}{(2n+5)^4}&gt;0.
\end{eqnarray*}

By the definition of $f$, we have
$$(n+2)t_n^2-(n+1)t_{n+1}^2=\frac{f(t_n)}{(n+4)^2t_n^2}&gt;0$$ for each $n$.
This means that $\{s_n\}_{n\ge 0}$ is decreasing. On the other hand, $\{M_n\}_{n\ge 1}$ is log-balanced.
It follows from Theorem \ref{thm:1a} that the sequence $\{M_n^2\}_{n\ge 1}$ is log-balanced.
\end{proof}


Denote by $A_n$ the number of directed animals of size $n$ (see \cite[Exercise 6.46]{ref4}), which satisfies the recurrence
\begin{eqnarray}
(n+1)A_{n+1}=2(n+1)A_n+3(n-1)A_{n-1}   \label{zf-2.3}
\end{eqnarray}
with $A_0=1$, $A_1=1$, and $A_2=2$.

\begin{corollary}
Both $\{A_n^2\}_{n\ge 2}$ and $\{\frac{A_n}{n}\}_{n\ge 2}$ are log-balanced.
\end{corollary}

\begin{proof} It is clear that the sequence $\{\frac{1}{n}\}_{n\ge 1}$ is log-convex. Liu and Wang \cite{ref3} proved that the sequence $\{A_n\}_{n\ge 0}$ is log-convex. For $n\ge 0$, let $t_n=\frac{A_{n+1}}{A_n}$ and $s_n=\frac{t_n^2}{n+1}$. By (\ref{zf-2.3}), we have
\begin{eqnarray}
t_n=2+\frac{3(n-1)}{(n+1)t_{n-1}}, \quad n\ge 1.  \label{zf-2.4}
\end{eqnarray}
It follows from (\ref{zf-2.4}) that
\begin{eqnarray*}
&amp;&amp;\quad (n+2)t_n^2-(n+1)t_{n+1}^2\\
&amp;&amp;=\frac{(n+2)^3t_n^4-4(n+1)(n+2)^2t_n^2-12n(n+1)(n+2)t_n-9n^2(n+1)}{(n+2)^2t_n^2}
\end{eqnarray*}
and
\begin{eqnarray*}
&amp;&amp;\quad n(n+2)^2t_n-(n+1)^3t_{n+1}\\
&amp;&amp;=\frac{n(n+2)^3t_n^2-2(n+2)(n+1)^3t_n-3n(n+1)^3}{(n+2)t_n}.
\end{eqnarray*}
For any real number $x$, let
\begin{eqnarray*}
f(x)&amp;=&amp;(n+2)^3x^4-4(n+1)(n+2)^2x^2-12n(n+1)(n+2)x-9n^2(n+1),\\
g(x)&amp;=&amp;n(n+2)^3x^2-2(n+2)(n+1)^3x-3n(n+1)^3.
\end{eqnarray*}
Then we have
\begin{eqnarray*}
f^{\prime}(x)&amp;=&amp;4(n+2)^3x^3-8(n+1)(n+2)^2x-12n(n+1)(n+2),\\
f^{\prime\prime}(x)&amp;=&amp;12(n+2)^3x^2-8(n+1)(n+2)^2,\\
g^{\prime}(x)&amp;=&amp;2n(n+2)^3x-2(n+2)(n+1)^3.
\end{eqnarray*}
It is obvious that $f^{\prime\prime}(x)&gt;0$ when $x\ge 1$ and hence $f^{\prime}$ is increasing over $[1, +\infty)$. Noting that $f^{\prime}(2)&gt;0$, we have $f^{\prime}(x)&gt;0$ when $x\ge 2$.

Liu and Wang \cite{ref3} showed that $$t_n\ge\mu_n,$$
where $\mu_n=\frac{6n}{2n+1}$. Since $f^{\prime}(\mu_n)&gt;0$, $f$ is increasing over $[\mu_n, \infty)$. It is evident that $g^{\prime}(x)&gt;0$ for $x\ge 1$ and hence the function $g$ is also increasing over $[1, \infty)$.

Note that
\begin{eqnarray*}
f(\mu_n)&amp;=&amp;\frac{9n^2}{(2n+1)^4}\bigg[144n^2(n+2)^3-16(n+1)(n+2)^2(2n+1)^2\\
&amp;&amp;-8(n+1)(n+2)(2n+1)^3-(n+1)92n+1)^4\bigg]\\
&amp;=&amp;\frac{9n^2}{(2n+1)^4}\bigg(144n^4+370n^2-72n^2-513n-81\bigg)
\end{eqnarray*}
and
\begin{eqnarray*}
g(\mu_n)&amp;=&amp;\frac{3n}{(2n+1)^2}\bigg[12n^3(n+2)^3-4(n+2)(2n+1)(n+1)^3-(2n+1)^2(n+1)^3\bigg]\\
&amp;=&amp;\frac{3n}{(2n+1)^2}\bigg(12n^4+27n^3-15n^2-51n-9\bigg).
\end{eqnarray*}
Clearly, $f(\mu_n)&gt;0$ and $g(\mu_n)&gt;0$ for $n\ge 2$. This implies that
\begin{eqnarray*}
(n+2)t_n^2-(n+1)t_{n+1}^2&gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
n(n+2)^2t_n-(n+1)^3t_{n+1}&gt;0
\end{eqnarray*}
for $n\ge 2$. Then $\{s_n\}_{n\ge 2}$ and $\{\frac{nt_n}{(n+1)^2}\}_{n\ge 2}$ are both decreasing. It follows from Theorem \ref{thm:1a} that the sequences $\{A_n^2\}_{n\ge 2}$ and $\{\frac{A_n}{n}\}_{n\ge 2}$ are both log-balanced.
\end{proof}

\begin{corollary}
For the sequence $\{B_n\}_{n\ge 0}$ of the Fine numbers, we have that $\{\frac{B_n}{n}\}_{n\ge 2}$ is log-balanced.
\end{corollary}

\begin{proof} The Fine numbers satisfy the recurrence
\begin{eqnarray}
B_{n+1}=\frac{7n+2}{2(n+2)}B_n+\frac{2n+1}{n+2}B_{n-1}, \quad B_0=1,\ \ B_1=0. \label{zf-2.5}
\end{eqnarray}
For $n\ge 2$, let $t_n=\frac{B_{n+1}}{B_n}$. Do\v{s}li\'{c} \cite{ref1} showed that the sequence $\{B_n\}_{n\ge 2}$ is log-balanced. We next prove that $\{\frac{nt_n}{(n+1)^2}\}_{n\ge 2}$ is decreasing.

By (\ref{zf-2.5}), we have
\begin{eqnarray*}
t_n=\frac{7n+2}{2(n+2)}+\frac{2n+1}{(n+2)t_{n-1}}.  \label{zf-2.6}
\end{eqnarray*}
Then we have
\begin{eqnarray*}
&amp;&amp;\quad n(n+2)^2t_n-(n+1)^3t_{n+1}\\
&amp;&amp;=\frac{2n(n+3)(n+2)^2t_n^2-(7n+9)(n+1)^3t_n-2(n+1)^3(2n+3)}{2(n+3)t_n}.
\end{eqnarray*}
For any real number $x$, let
$$
f(x)=2n(n+3)(n+2)^2x^2-(7n+9)(n+1)^3x-2(n+1)^3(2n+3).
$$
Then we obtain
\begin{eqnarray*}
f^{\prime}(x)=4n(n+3)(n+2)^2x-(7n+9)(n+1)^3.
\end{eqnarray*}
It is obvious that $f^{\prime}(x)&gt;0$ for $x\ge 3$. Then $f$ is increasing over $[3, \infty)$.

Liu and Wang \cite{ref3} proved that
\begin{eqnarray*}
t_n\ge \lambda_n,
\end{eqnarray*}
where $\lambda_n=\frac{2(2n+5)}{n+4}$.
Since
\begin{eqnarray*}
f(\lambda_n)&amp;=&amp;\frac{1}{(n+4)^2}[8n(n+3)(n+2)^2(2n+5)^2\\
&amp;&amp;-(7n+9)(2n+5)(n+4)(n+1)^3-2(2n+3)(n+4)^2(n+1)^3]\\
&amp;=&amp;\frac{14n^6+185n^5+968n^4+458n^3+4314n^2+1023n-596}{(n+4)^2}\\
&amp;&gt;&amp;0,
\end{eqnarray*}
we have
\begin{eqnarray*}
2n(n+3)(n+2)^2t_n^2-(7n+9)(n+1)^3t_n-2(n+1)^3(2n+3)&gt;0.
\end{eqnarray*}
Then $n(n+2)^2t_n-(n+1)^3t_{n+1}&gt;0$, and $\{\frac{nt_n}{(n+1)^2}\}_{n\ge 2}$ is decreasing. It follows from Theorem \ref{thm:1a} that the sequence $\{\frac{B_n}{n}\}_{n\ge 2}$ is log-balanced.
\end{proof}

\begin{theorem}\label{thm:2a}
For a given sequence $\{z_n\}_{n\ge 0}$, if it is log-balanced, then $\{\sqrt{z_{n}}\}_{n\ge 0}$ is also log-balanced.
\end{theorem}

\begin{proof}
Suppose that $\{z_{n}\}_{n\geq0}$ is log-balanced, that is,
\begin{eqnarray*}
z_n^2\le z_{n-1}z_{n+1},\quad (n+1)z_n^2\ge n z_{n-1}z_{n+1}, \quad n\ge 1.
\end{eqnarray*}
For $n\ge 1$, we immediately derive
\begin{eqnarray*}
z_n\le \sqrt{z_{n-1}z_{n+1}}
\end{eqnarray*}
and
\begin{eqnarray*}
z_n \ge \sqrt{\frac{n}{n+1}z_{n-1}z_{n+1}}&gt;\frac{n}{n+1}\sqrt{z_{n-1}z_{n+1}}.
\end{eqnarray*}
This means that the sequence $\{\sqrt{z_n}\}_{n\ge 0}$ is log-convex and the sequence $\{\frac{\sqrt{z_n}}{n!}\}_{n\ge 0}$ is log-concave. As a result, the sequence $\{\sqrt{z_{n}}\}_{n\ge 0}$ is log-balanced.
\end{proof}

\begin{theorem}\label{thm:3a} Suppose that the sequences $\{x_n\}_{n\ge 0}$ and $\{y_n\}_{n\ge 0}$ are both log-convex.
If both $\{\frac{x_n}{n!}\}_{n\ge 0}$ and $\{\frac{y_n}{n!}\}_{n\ge 0}$ are concave, then
$\{x_n+y_n\}_{n\ge 0}$ is log-balanced.
\end{theorem}

\begin{proof} Since $\{x_n\}_{n\ge 0}$ and $\{y_n\}_{n\ge 0}$ are both log-convex, the sequence $\{x_n+y_n\}_{n\ge 0}$ is log-convex. We next prove that $\{\frac{x_n+y_n}{n!}\}_{n\ge 0}$ is log-concave.

It is well known that $\{x_n\}_{n\ge 0}$ is concave if and only if its difference sequence $\{x_{n+1}-x_n\}_{n\ge 0}$ is decreasing. Therefore, by the concavity of $\{\frac{x_n}{n!}\}_{n\ge 0}$ and $\{\frac{y_n}{n!}\}_{n\ge 0}$, the sequence $\{\frac{x_{n+1}+y_{n+1}}{n!}-\frac{x_n+y_n}{n!}\}_{n\ge 0}$ is decreasing. Then the sequence
$\{\frac{x_n+y_n}{n!}\}_{n\ge 0}$ is concave and it is also log-concave. Hence, the sequence
$\{x_n+y_n\}_{n\ge 0}$ is log-balanced.
\end{proof}

It follows from Theorem \ref{thm:3a} that the sequence $\{n!+(n+1)!\}_{n\ge 0}$ is log-balanced.

In the rest of this section, we devote to discuss the log-balancedness of some sequences by means of Definition \ref{de1-1} directly. Our first example is to consider some sequences related to harmonic numbers. Let $H_n$ denote the $n^{th}$ harmonic number. Then we have the following result.

\begin{proposition}
Both $\{\frac{H_n}{n}\}_{n\ge 1}$ and $\{\frac{H_n}{n^2}\}_{n\ge 1}$ are log-balanced.
\end{proposition}

\begin{proof} In order to prove the log-balancedness of $\{\frac{H_n}{n}\}_{n\ge 1}$, it is sufficient to show that $\{\frac{H_n}{n}\}_{n\geq1} $ is log-convex and the sequence $\{\frac{H_n}{nn!}\}_{n\geq1}$ is log-concave. In fact, for $n\geq2$, we have
\begin{eqnarray*}
\quad \frac{H_n^2}{n^2}-\frac{H_{n-1}H_{n+1}}{n^2-1}
&amp;=&amp;\frac{1}{n^2(n^2-1)}\bigg[(n^2-1)H_n^2-n^2\bigg(H_n-\frac{1}{n}\bigg)\bigg(H_n+\frac{1}{n+1}\bigg)\bigg]\\
&amp;=&amp;-\frac{n(H_n^2-H_n-1)+H_n^2}{n^2(n+1)^2(n-1)}.
\end{eqnarray*}
Note that
\begin{eqnarray*}
2(H_2^2-H_2-1)+H_2^2&gt;0, \quad  H_n&gt;H_2&gt;2 \quad(n\ge 3).
\end{eqnarray*}
Now we prove that $n(H_n^2-H_n-1)+H_n^2&gt;0$ for $n\ge 3$.
For any real number $x$, let
$$
f(x)=x^2-x-1.
$$
It is clear that $f^{\prime}(x)=2x-1&gt;0$ for $x\ge 2$. Then $f$ is increasing over $[2, \infty)$ and $f(H_n)&gt;f(H_{3})=\frac{19}{36}&gt;0$ for $n\ge 3$.
Hence, the sequence $\{\frac{H_n}{n}\}_{n\geq1}$ is log-convex. On the other hand, for $n\ge 2$, we have
\begin{eqnarray*}
&amp;&amp;\quad\frac{H_n^2}{n n!}-\frac{H_{n-1}H_{n+1}}{(n^2-1)(n-1)!(n+1)(n+1)!}\\
&amp;&amp;=\frac{1}{n^{2}(n^{2}-1)n!(n+1)!}\bigg[(n^{2}-n-1)H_n^2+n^{2}\bigg(\frac{H_n}{n+1}+\frac{1}{n+1}\bigg)\bigg]\\
&amp;&amp;&gt;0.
\end{eqnarray*}
Hence the sequence $\{\frac{H_n}{nn!}\}_{n\geq1}$ is log-concave.
It follows from  Definition \ref{de1-1} that the sequence $\{\frac{H_n}{n}\}_{n\ge 1}$ is log-balanced.

Now we consider the sequence $\{\frac{H_n}{n^2}\}_{n\ge 1}$. Since both $\{\frac{H_n}{n}\}_{n\ge 1}$ and $\{\frac{1}{n}\}_{n\ge 1}$ are log-convex, $\{\frac{H_n}{n^2}\}_{n\ge 1}$ is log-convex.
On the other hand, for $n\ge 2$, we get
\begin{eqnarray*}
&amp;&amp;\quad \bigg(\frac{H_n}{n^2n!}\bigg)^2-\frac{H_{n-1}H_{n+1}}{(n-1)^2(n-1)!(n+1)^2(n+1)!}\\
&amp;&amp;=\frac{1}{n^2(n-1)^2(n+1)^2n!(n+1)!}\bigg[(n^4-2n^3-2n^2+3n+1)H_n^2\\
&amp;&amp;\quad+n^4\bigg(\frac{H_n}{n+1}+\frac{1}{n+1}\bigg)\bigg].
\end{eqnarray*}
For $n=2$,
$$
\bigg(\frac{H_n}{n^2n!}\bigg)^2-\frac{H_{n-1}H_{n+1}}{(n-1)^2(n-1)!(n+1)^2(n+1)!}=\frac{25}{20736}.
$$
We find that $n^{4}-2n^{3}-2n^{2}+3n+1&gt;0$ for $n\ge 3$. Thus the sequence $\{\frac{H_n}{n^2n!}\}_{n\geq1}$ is log-concave.
It follows from Definition \ref{de1-1} that the sequence $\{\frac{H_n}{n^2}\}_{n\ge 1}$ is log-balanced.
\end{proof}

Our second example is to consider some sequences related to the Fibonacci (Lucas) sequence. The Binet form of the Fibonacci sequence $\{F_n\}_{n\ge 0}$ and the Lucas sequence $\{L_n\}_{n\ge 0}$ are
\begin{eqnarray*}
F_n=\frac{\alpha^n-(-1)^n\alpha^{-n}}{\sqrt 5}, \quad L_n=\alpha^n+(-1)^n\alpha^{-n},
\end{eqnarray*}
where $\alpha=\frac{1+\sqrt 5}{2}$.
It is well known that log-convexity and log-concavity of $\{F_n\}_{n\ge 0}$ and $\{L_n\}_{n\ge 0}$ depend on the parity of $n$. In fact, by using the definition of log-convexity, we can easily prove that both
$\{F_{2n+1}\}_{n\ge 0}$ and $\{L_{2n}\}_{n\ge 2}$ are log-convex. Now we discuss the log-balancedness of some sequences related to $F_n$ and $L_n$. We first give a lemma.

\begin{lemma} \label{lem:4a}
For $n\ge 1$, we have
\begin{eqnarray}
F_{2n+1}\ge 2n \label{zf-2.7}
\end{eqnarray}
and
\begin{eqnarray}
L_{2n}\ge 3n. \label{zf-2.8}
\end{eqnarray}
\end{lemma}

\begin{proof}
It is well known that $\{F_n\}_{n\ge 0}$ and $\{L_n\}_{n\ge 0}$ satisfy the recurrence relation
\begin{eqnarray}
W_{n+1}=W_n+W_{n-1}, \quad n\ge 1. \label{zf-2.9}
\end{eqnarray}
We can prove (\ref{zf-2.7})--(\ref{zf-2.8}) by induction. We only give a proof of \eqref{zf-2.7} and (\ref{zf-2.8}) can be shown in a similar way. In fact, it is clear that $F_{2n+1}\ge 2n$ for $1\le n\le 5$. Assume that $F_{2n+1}\ge 2n$ for $n\ge 5$. By (\ref{zf-2.9}), we have
\begin{eqnarray*}
F_{2n+3}=F_{2n+1}+F_{2n+2}.
\end{eqnarray*}
Then we have $F_{2n+3}\ge F_{2n+1}+2\ge2n+2$.
By mathematical induction, (\ref{zf-2.7}) holds for each $n\ge 1$.
\end{proof}

\begin{proposition}
The sequences $\{\frac{F_{2n+1}}{n}\}_{n\ge 1}$ and $\{\frac{L_{2n}}{n}\}_{n\ge 2}$ are log-balanced.
\end{proposition}

\begin{proof} Because $\{F_{2n+1}\}_{n\ge 0}$, $\{L_{2n}\}_{n\ge 2}$ and $\{\frac{1}{n}\}_{n\ge 1}$ are log-convex, the sequences $\{\frac{F_{2n+1}}{n}\}_{n\ge 1}$ and $\{\frac{L_{2n}}{n}\}_{n\ge 2}$ are log-convex.
Next we show that $\{\frac{F_{2n+1}}{nn!}\}_{n\ge 1}$ and $\{\frac{L_{2n}}{nn!}\}_{n\ge 2}$ are log-concave.

For $n\ge 2$, we obtain
\begin{eqnarray*}
\bigg(\frac{F_{2n+1}}{nn!}\bigg)^2-\frac{F_{2n-1}F_{2n+3}}{(n^2-1)(n-1)!(n+1)!}&amp;=&amp;\frac{(n+1)(n^2-1)F_{2n+1}^2-n^3F_{2n-1}F_{2n+3}}{n^2(n^2-1)n!(n+1)!}\\
&amp;=&amp;\frac{n^3(F_{2n+1}^2-F_{2n-1}F_{2n+3})+(n^2-n-1)F_{2n+1}^2}{n^2(n^2-1)n!(n+1)!}
\end{eqnarray*}
and
\begin{eqnarray*}
\bigg(\frac{L_{2n}}{n n!}\bigg)^2-\frac{L_{2n-2}L_{2n+2}}{(n^2-1)(n-1)!(n+1)!}&amp;=&amp;\frac{(n+1)(n^2-1)L_{2n}^2-n^3L_{2n-2}L_{2n+2}}{n^2(n^2-1)n!(n+1)!}\\
&amp;=&amp;\frac{n^3(L_{2n}^2-L_{2n-2}L_{2n+2})+(n^2-n-1)L_{2n}^2}{n^2(n^2-1)n!(n+1)!}.
\end{eqnarray*}
By means of the equalities
$$
F_{2n+1}^2-F_{2n-1}F_{2n+3}=-1 \quad {\rm and} \quad L_{2n}^2-L_{2n-2}L_{2n+2}=-5,
$$
we have
\begin{eqnarray*}
\bigg(\frac{F_{2n+1}}{nn!}\bigg)^2-\frac{F_{2n-1}F_{2n+3}}{(n^2-1)(n-1)!(n+1)!}&amp;=&amp;\frac{-n^3+(n^2-n-1)F_{2n+1}^2}{n^2(n^2-1)n!(n+1)!},\\
\bigg(\frac{L_{2n}}{n n!}\bigg)^2-\frac{L_{2n-2}L_{2n+2}}{(n^2-1)(n-1)!(n+1)!}&amp;=&amp;\frac{-5n^3+(n^2-n-1)L_{2n}^2}{n^2(n^2-1)n!(n+1)!}.
\end{eqnarray*}
For $n\ge 2$, put
$$
R(n)=-n^3+(n^2-n-1)F_{2n+1}^2 \quad {\rm and} \quad S(n)=-5n^3+(n^2-n-1)L_{2n}^2.
$$
It follows from Lemma {\ref{lem:4a}} that
\begin{eqnarray*}
R(n)\ge n^2(4n^2-5n-4), \quad S(n)\ge n^2(9n^2-14n-9).
\end{eqnarray*}
Note that
\begin{eqnarray*}
R(n)&gt;0 \ (n\ge 2), \quad
S(n)&gt;0 \ (n\ge 3).
\end{eqnarray*}
This implies that $\{\frac{F_{2n+1}}{nn!}\}_{n\ge 1}$ and $\{\frac{L_{2n}}{nn!}\}_{n\ge 2}$ are both log-concave. By Definition \ref{de1-1}, $\{\frac{F_{2n+1}}{n}\}_{n\ge 1}$ and $\{\frac{L_{2n}}{n}\}_{n\ge 2}$ are both log-balanced. This completes the proof.
\end{proof}

\section{Conclusions}

We have derived some new sufficient conditions for log-balancedness of combinatorial sequences. We have further applied these results to show log-balancedness of some special sequences. One future work is to study log-balancedness of the partial sums for log-balanced sequences.

\section{Acknowledgment}

This work was supported in part by a grant of the First-Class Discipline of Universities in Shanghai and the Innovation Fund of Shanghai University. The authors would like to thank an anonymous referee for his or her helpful comments and suggestions.



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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:  Primary
05A20; Secondary 11B37, 11B39.

\noindent \emph{Keywords: } log-convexity, log-concavity, log-balancedness.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000957},
\seqnum{A001006}, and
\seqnum{A005773}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 9 2016;
revised versions received  May 11 2016; June 22 2016; July 7 2016.
Published in {\it Journal of Integer Sequences}, August 29 2016.

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