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\begin{center}
\vskip 1cm{\LARGE\bf New Sufficient Conditions for \\
\vskip .02in
Log-Balancedness, With Applications to  \\
\vskip .10in
Combinatorial Sequences}
\vskip 1cm
\large Rui-Li Liu and Feng-Zhen Zhao\\
Department of Mathematics\\
 Shanghai University \\
Shanghai 200444 \\
 P. R. China\\
\href{mailto:liuruili@i.shu.edu.cn}{\tt liuruili@i.shu.edu.cn}\\
\href{mailto:fengzhenzhao@shu.edu.cn}{\tt fengzhenzhao@shu.edu.cn}\\
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\begin{abstract}
In this paper, we mainly study the log-balancedness of combinatorial sequences. We first give some new sufficient conditions for log-balancedness of some kinds of sequences. Then we use these results to derive the log-balancedness of a number of log-convex sequences related to derangement numbers, Domb numbers, numbers of tree-like polyhexes, numbers of walks on the cubic lattice, and so on.
\end{abstract}


\section{Introduction}

A sequence of positive real numbers $\{z_n\}_{n\geq 0}$ is said to be {\it log-convex} (or {\it log-concave}) if $z_n^2\leq z_{n-1}z_{n+1}$ (or $z_n^2\geq z_{n-1}z_{n+1}$) for each $n\geq 1$. A log-convex sequence $\{z_n\}_{n\geq 0}$ is said to be {\it log-balanced} if $\{\frac{z_n}{n!}\}_{n\ge 0}$ is log-concave. See Do\v{s}li\'{c} \cite{dos05} for more details about log-balanced sequences. It is well known that $\{z_n\}_{n\geq 0}$ is log-convex (or log-concave) if and only if its quotient sequence $\{\frac{z_{n+1}}{z_n}\}_{n\ge 0}$ is nondecreasing (or nonincreasing) and a log-convex sequence $\{z_n\}_{n\geq 0}$ is log-balanced if and only if $\frac{(n+1)z_n}{z_{n-1}}\ge\frac{nz_{n+1}}{z_n}$ for each $n\ge 1$. It is clear that the quotient sequence of a log-balanced sequence does not grow too quickly.

In combinatorics, log-convexity and log-concavity are not only instrumental in obtaining the growth rate of a combinatorial sequence, but also important sources of inequalities. Log-convexity and log-concavity have applications in many fields such as quantum physics, white noise theory, probability, economics and mathematical biology. See, for instance \cite{asai, bren, dos10, dos04, dos08, mil04, pre04, stan89}. Since log-balancedness is related to log-convexity and log-concavity, it can help us to find new inequalities. Hence, the log-balancedness of various sequences deserves to be studied.

In this paper, we are interested in the log-balancedness of some combinatorial sequences. In fact, there are many log-balanced sequences in combinatorics and number theory. Do\v{s}li\'{c} \cite{dos05} presented some sufficient conditions for the log-balancedness of sequences satisfying three-term linear recurrences. As consequences, a number of sequences such as the Motzkin numbers, the Fine numbers, the Franel numbers of orders $3$ and $4$, the Ap\'{e}ry numbers, the large and little Schr\"{o}der numbers, and the central Delannoy numbers, are log-balanced (see Do\v{s}li\'{c} \cite{dos05}). Recently, Zhao \cite{zhao15} gave a sufficient condition for the log-balancedness of the product of a log-balanced sequence and a log-concave sequence and she also proved that the binomial transformation preserves the log-balancedness. Zhao \cite{zhao14, zhao15} showed that the sequences of the exponential numbers and the Catalan-Larcombe-French numbers are respectively log-balanced. Zhang and Zhao \cite{zhangtt16} gave some sufficient conditions for the log-balancedness of combinatorial sequences. In addition, for a log-balanced sequence $\{z_n\}_{n\ge 0}$, Zhang and Zhao \cite{zhangtt16} proved that $\{\sqrt{z_n}\}_{n\ge 0}$ is still log-balanced.

This paper is devoted to the study of log-balancedness of some combinatorial sequences and is organized as follows. In Section 2, we give some new sufficient conditions for log-balancedness. In Section 3, using these new results, we investigate the log-balancedness of a series of log-convex sequences.




\section{Sufficient conditions for log-balancedness}

Zhang and Zhao \cite{zhangtt16} proved that the sequence of the arithmetic square root of a log-balanced sequence is still log-balanced. For a log-convex sequence $\{z_n\}_{n\ge 0}$, here we prove that $\{\sqrt[r]{z_n}\}_{n\ge 0}$ is log-balanced under some conditions, where $r$ is a fixed positive real number.

\begin{theorem}\label{thm:1a}
Let $\{z_n\}_{n\ge 0}$ be a log-convex sequence and $r$ be a fixed positive real number. For $n\ge 0$, let $x_n=\frac{z_{n+1}}{z_n}$. If there exists a nonnegative integer $N_r$ such that
\begin{eqnarray*}
 (n+2)^rx_n-(n+1)^rx_{n+1}\ge 0, \quad  n\ge N_r,
\end{eqnarray*}
 the sequence $\{\sqrt[r]{z_n}\}_{n\ge N_r}$ is log-balanced.
\end{theorem}

\begin{proof} Since the sequence $\{z_n\}_{n\ge 0}$ is log-convex, $\{\sqrt[r]{z_n}\}_{n\ge 0}$ is also log-convex. In order to prove the log-balancedness of $\{\sqrt[r]{z_n}\}_{n\ge N_r}$, it is sufficient to show that the sequence $\{\frac{\sqrt[r]{z_n}}{n!}\}_{n\ge N_r}$ is log-concave. In fact, it is clear that $\{\frac{\sqrt[r]{z_n}}{n!}\}_{n\ge N_r}$ is log-concave if and only if $\frac{\sqrt[r]{x_n}}{n+1}\ge\frac{\sqrt[r]{x_{n+1}}}{n+2}$ for every $n\ge N_r$. 
It follows from $(n+2)^rx_n-(n+1)^rx_{n+1}\ge 0$ that $\frac{\sqrt[r]{x_n}}{n+1}\ge\frac{\sqrt[r]{x_{n+1}}}{n+2}$. Hence the sequence $\{\frac{\sqrt[r]{z_n}}{n!}\}_{n\ge N_r}$ is log-concave. Therefore, $\{\sqrt[r]{z_n}\}_{n\ge N_r}$ is log-balanced.
\end{proof}


\begin{theorem}\label{thm:2a}
Suppose that $a$ and $b$ are positive real numbers with $b&lt;a$ and $\{z_n\}_{n\ge 0}$ is a log-convex sequence. If the sequence $\{z_n^a\}_{n\ge 0}$ is log-balanced, then so is the sequence $\{z_n^b\}_{n\ge 0}$.
\end{theorem}
\begin{proof} Since the sequence $\{z_n^a\}_{n\ge 0}$ is log-balanced, we have
\begin{eqnarray*}
\frac{n}{n+1}z_{n-1}^az_{n+1}^a\le z_n^{2a}\le z_{n-1}^az_{n+1}^a.
\end{eqnarray*}
Then we derive
\begin{eqnarray*}
\bigg(\frac{n}{n+1}\bigg)^{\frac{1}{a}}z_{n-1}z_{n+1}\le z_n^2\le z_{n-1}z_{n+1},
\end{eqnarray*}
\begin{eqnarray*}
\bigg(\frac{n}{n+1}\bigg)^{\frac{b}{a}}z_{n-1}^bz_{n+1}^b\le z_n^{2b}\le z_{n-1}^bz_{n+1}^b.
\end{eqnarray*}
Since $0&lt;\frac{b}{a}&lt;1$ and $0&lt;\frac{n}{n+1}&lt;1$, we have
$(\frac{n}{n+1})^{\frac{b}{a}}\geq \frac{n}{n+1}$ and hence
\begin{eqnarray*}
\frac{n}{n+1}z_{n-1}^bz_{n+1}^b\le z_n^{2b}\le z_{n-1}^bz_{n+1}^b.
\end{eqnarray*}
It follows from the definition of log-balancedness that the sequence $\{z^b_n\}_{n\ge 0}$ is log-balanced.
\end{proof}

In Theorem \ref{thm:2a}, if the condition ``$b&lt;a$" is replaced by ``$b&gt;a$", the conclusion is not valid in general. For example, the sequence $\{nn!\}_{n\ge 2}$ is log-balanced, but $\{(nn!)^2\}_{n\ge 2}$ is not log-balanced.

In the next section, we will use the results of Theorems \ref{thm:1a}--\ref{thm:2a} to derive log-balancedness of a series of sequences.
\begin{theorem}\label{thm:3a}
Let $\{z_n\}_{n\ge 0}$ be a log-concave sequence. If the sequence $\{n!z_n\}_{n\ge 0}$ is log-balanced, then so is the sequence $\{n!\sqrt{z_n}\}_{n\ge 0}$.
\end{theorem}
\begin{proof} Since the sequence $\{z_n\}_{n\ge 0}$ is log-concave, then so is the sequence $\{\sqrt{z_n}\}_{n\ge 0}$. In order to prove the log-balancedness of $\{n!\sqrt{z_n}\}_{n\ge 0}$, we only need to show that $\{n!\sqrt{z_n}\}_{n\ge 0}$ is log-convex. Since $\{n!z_n\}_{n\ge 0}$ is log-balanced, we get
\begin{eqnarray*}
&amp;&amp; nz^2_n-(n+1)z_{n-1}z_{n+1}\le 0,\\
&amp;&amp; z_n\le\sqrt{\frac{n+1}{n}z_{n-1}z_{n+1}}&lt;\frac{n+1}{n}\sqrt{z_{n-1}z_{n+1}},\\
&amp;&amp; (n!\sqrt{z_n})^2\le(n-1)!(n+1)!\sqrt{z_{n-1}z_{n+1}}.
\end{eqnarray*}
Hence $\{n!\sqrt{z_n}\}_{n\ge 0}$ is log-convex.
\end{proof}

\section{Log-balancedness of some
sequences}

In this section, we discuss the log-balancedness of a number of log-convex sequences involving many combinatorial numbers.

\subsection{The derangement numbers}

The derangement numbers $d_n$ (sequence \seqnum{A000166} in the OEIS)
count the number of permutations of $n$ elements with no fixed points.
The sequence $\{d_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
d_{n+1}=n(d_n+d_{n-1}), \quad n\ge 1, \label{zf-3.1}
\end{eqnarray}
with $d_0=1$, $d_1=0$, $d_2=1$, $d_3=2$ and $d_4=9$; see Table \ref{tab:table1} for some information about it. In particular, Liu and Wang \cite{liu07} proved that $\{d_n\}_{n\geq2}$ is log-convex.
\begin{table} [!htbp]
\begin{center}
\begin{tabular}{c|ccccccccc}
$n$ &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ &amp; $8$ \\
   \hline
    $d_n$ &amp; $1$ &amp; $0$ &amp; $1$ &amp; $2$  &amp; $9$ &amp; $44$ &amp; $265$ &amp; $1854$ &amp; $14833$ \\
\end{tabular}
\end{center}
\caption {Some initial values of $\{d_n\}_{n\ge 0}$} \label{tab:table1}
\end{table}
\begin{theorem} \label{thm:4a}
For $r\ge 2$, the sequence $\{\sqrt[r]{d_n}\}_{n\ge 3}$ is log-balanced.
\end{theorem}
\begin{proof} We first prove that the sequence $\{\sqrt{d_n}\}_{n\ge 3}$ is log-balanced.

For $n\geq0$, let $x_n=\frac{d_{n+1}}{d_n}$. We prove by induction that
\begin{eqnarray*}
\lambda_n\leq x_n\leq \lambda_{n+1},\quad n\ge 3,
\end{eqnarray*}
where $\lambda_n=\frac{2n+1}{2}$. It follows from (\ref{zf-3.1}) that
\begin{eqnarray}
x_n=n+\frac{n}{x_{n-1}},\quad n\ge 3. \label{zf-3.2}
\end{eqnarray}
It is clear that $\lambda_3\leq x_3\leq \lambda_4$.
Assume that $\lambda_k\leq x_k\leq \lambda_{k+1}$ for $k\geq 3$.
By applying (\ref{zf-3.2}), we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}=\frac{k+1}{x_{k}}-\frac{1}{2} \quad {\rm and} \quad x_{k+1}-\lambda_{k+2}=\frac{k+1}{x_{k}}-\frac{3}{2}.
\end{eqnarray*}
Due to $\frac{1}{\lambda_{k+1}}\leq \frac{1}{x_k}\leq \frac{1}{\lambda_k}$ ($k\geq3$), we have
$$
x_{k+1}-\lambda_{k+1}\geq 0 \quad {\rm and} \quad x_{k+1}-\lambda_{k+2}\leq 0.
$$
Then we derive $\lambda_n\leq x_n\leq \lambda_{n+1}$ for $n\geq 3$.

By means of (\ref{zf-3.2}), we obtain
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}=\frac{(n+2)^2x_n^2-(n+1)^3x_n-(n+1)^3}{x_n}.
\end{eqnarray*}
For any $x\in (-\infty,+\infty)$, define a function
$$
f(x)=(n+2)^2x^2-(n+1)^3x-(n+1)^3.
$$
Then we have
$$
f'(x)=2(n+2)^2x-(n+1)^3.
$$
Since $f'(x)\geq0$ for $x\geq\frac{(n+1)^3}{2(n+2)^2}$, $f$ is increasing on $[\frac{(n+1)^3}{2(n+2)^2},+\infty)$.
We can verify that $\lambda_n&gt;\frac{(n+1)^3}{2(n+2)^2}$. Hence, $f$ is increasing on $[\lambda_n,+\infty)$.
Note that
\begin{eqnarray*}
f(\lambda_n)&amp;=&amp;(n+2)^2\lambda_{n}^2-(n+1)^3\lambda_n-(n+1)^3 \\
&amp;=&amp;\frac{2n^3+3n^2-2n-2}{4} 
&gt; 0 \quad (n\geq 1).
\end{eqnarray*}
Then we have $f(x_n)&gt;0$ for $n\geq3$. This implies that $(n+2)^2x_n-(n+1)^2x_{n+1}\geq0$ for $n\geq3$. It follows from Theorem \ref{thm:1a} that the sequence $\{\sqrt{d_n}\}_{n\geq3}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{d_n}\}_{n\ge 3}$ is log-balanced.
\end{proof}

\subsection{Numbers counting tree-like polyhexes}

Let $h_n$ denote the number of tree-like polyhexes with $n+1$ hexagons
(Harary and Read \cite{hara70}); it is sequence
\seqnum{A002212} in the OEIS.  It is well known that $h_n$ is equal
to the number of lattice paths, from $(0, 0)$ to $(2n, 0)$ with steps
$(1, 1)$, $(1, -1)$ and $(2, 0)$, never falling below the $x$-axis and
with no peaks at odd level. The sequence $\{h_n\}_{n\ge 0}$ satisfies
the recurrence
\begin{eqnarray}
(n+1)h_n=3(2n-1)h_{n-1}-5(n-2)h_{n-2}, \quad n\ge 2, \label{zf-3.3}
\end{eqnarray}
with $h_0=h_1=1$, $h_2=3$ and $h_3=10$; see Table \ref{tab:table2} for some information about it. In particular, Liu and Wang \cite{liu07} showed that the sequence $\{h_n\}_{n\geq0}$ is log-convex.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|cccccccc}
$n$ &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ \\
  \hline
  $h_n$ &amp; $1$ &amp; $1$ &amp; $3$ &amp; $10$  &amp; $36$ &amp; $137$ &amp; $543$ &amp; $2219$ \\
\end{tabular}
\end{center}
 \caption {Some initial values of $\{h_n\}_{n\ge 0}$} \label{tab:table2}
\end{table}

\begin{theorem} \label{thm:5a}
For $r\ge 1$, the sequence $\{\sqrt[r]{h_n}\}_{n\ge 1}$ is log-balanced.
\end{theorem}
\begin{proof} In order to prove that $\{\sqrt[r]{h_n}\}_{n\ge 1}$ is log-balanced for $r\ge 1$, we only need to show that $\{h_n\}_{n\ge 1}$ is log-balanced by Theorem \ref{thm:2a}.

For $n\geq0$, put $x_n=\frac{h_{n+1}}{h_n}$. We next prove by induction that
\begin{eqnarray*}
\lambda_n\leq x_n\leq \mu_n,\quad n\ge 0,
\end{eqnarray*}
where $\lambda_n=\frac{10n+3}{2n+4}$ and $\mu_n=\frac{5n+4}{n+1}$.  It follows from (\ref{zf-3.3}) that
\begin{eqnarray}
x_n=\frac{3(2n+1)}{n+2}-\frac{5(n-1)}{(n+2)x_{n-1}},\quad n\ge 1, \label{zf-3.4}
\end{eqnarray}
It is easy to find that $\lambda_0\leq x_0\leq \mu_0$. Assume that $\lambda_k\leq x_k\leq \mu_k$ for $k\geq 0$.
By using (\ref{zf-3.4}), we derive
\begin{eqnarray*}
 x_{k+1}-\lambda_{k+1}=\frac{3(2k+3)}{k+3}-\frac{10k+13}{2k+6}-\frac{5k}{(k+3)x_k}
\end{eqnarray*}
and
\begin{eqnarray*}
 x_{k+1}-\mu_{k+1}=\frac{3(2k+3)}{k+3}-\frac{5k+9}{k+2}-\frac{5k}{(k+3)x_k}.
\end{eqnarray*}
Since $\frac{1}{\mu_k}\leq \frac{1}{x_k}\leq \frac{1}{\lambda_k}$ ($k\geq0$), we have
\begin{eqnarray*}
 x_{k+1}-\lambda_{k+1}&amp;\geq&amp; \frac{2k+5}{2(k+3)}-\frac{5k}{(k+3)\lambda_k} \\
&amp;=&amp;\frac{16k+15}{2(k+3)(10k+3)} 
\geq 0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\mu_{k+1}&amp;\leq&amp;\frac{k^2-3k-9}{(k+2)(k+3)}-\frac{5k}{(k+3)\mu_k} \\
&amp;=&amp;\frac{-26k^2-67k-36}{(k+2)(k+3)(5k+4)} 
\leq 0.
\end{eqnarray*}
Then we have $\lambda_n\leq x_n\leq \mu_n$ for $n\geq 0$.

By means of (\ref{zf-3.4}), we obtain
\begin{eqnarray*}
(n+2)x_n-(n+1)x_{n+1}=\frac{(n+2)(n+3)x_n^2-3(n+1)(2n+3)x_n+5n(n+1)}{(n+3)x_n}.
\end{eqnarray*}
For any $x\in (-\infty,+\infty)$, define a function
$$
f(x)=(n+2)(n+3)x^2-3(n+1)(2n+3)x+5n(n+1).
$$
It is clear that
\begin{eqnarray*}
(n+2)x_n-(n+1)x_{n+1}=\frac{f(x_n)}{(n+3)x_n}.
\end{eqnarray*}
We note that $f$ is increasing on $[\frac{3(n+1)(2n+3)}{2(n+2)(n+3)}, +\infty]$. We find that $\lambda_n&gt;\frac{3(n+1)(2n+3)}{2(n+2)(n+3)}$ and
\begin{eqnarray*}
f(\lambda_n)&amp;=&amp;\frac{84n^2-41n-27}{4(n+2)} &gt; 0 \quad (n\ge 1).
\end{eqnarray*}
Then we have $(n+2)x_n-(n+1)x_{n+1}&gt;0$ for $n\ge 1$. Hence $\{h_n\}_{n\ge 1}$ is log-balanced.
\end{proof}

\subsection{Numbers counting walks on the cubic lattice}

Consider the sequence $\{w_n\}_{n\ge 0}$ counting the number of walks on
the cubic lattice with $n$ steps, starting and finishing on the $xy$ plane and never going below it (Guy \cite{guy}); it is sequence
\seqnum{A005572} in the OEIS.  The sequence $\{w_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
(n+2)w_n=4(2n+1)w_{n-1}-12(n-1)w_{n-2}, \quad n\ge 2, \label{zf-3.5}
\end{eqnarray}
where $w_0=1$, $w_1=4$ and $w_2=17$; see Table \ref{tab:table3} for some information about it. In particular, Liu and Wang \cite{liu07} showed that $\{w_n\}_{n\ge 0}$ is log-convex.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|ccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$  \\
  \hline
  $w_n$ &amp; $1$ &amp; $4$ &amp; $17$ &amp; $76$  &amp; $354$ &amp; $1704$ &amp; $8421$  \\
\end{tabular}
\end{center}
 \caption {Some initial values of $\{w_n\}_{n\ge 0}$} \label{tab:table3}
\end{table}

\begin{theorem} \label{thm:6a} Let $r$ be a positive real number.
For $r\ge 1$, the sequence $\{\sqrt[r]{w_n}\}_{n\ge 0}$ is log-balanced. For $\frac{5}{6}&lt;r&lt;1$, there exists a positive integer $N_r$ such that $\{\sqrt[r]{w_n}\}_{n\ge N_r}$ is log-balanced.
\end{theorem}
\begin{proof} For $n\geq0$, let $x_n=\frac{w_{n+1}}{w_n}$. Now we prove by induction that
\begin{eqnarray}
\lambda_n\leq x_n\leq \mu_n,\quad n\ge 0, \label{zf-3.6}
\end{eqnarray}
where $\lambda_n=\frac{6n+13}{n+4}$ and $\mu_n=\frac{6(n+3)}{n+4}$. It follows from (\ref{zf-3.5}) that
\begin{eqnarray}
x_n=\frac{4(2n+3)}{n+3}-\frac{12n}{(n+3)x_{n-1}},\quad n\ge 1, \label{zf-3.7}
\end{eqnarray}
 We observe that $\lambda_k\leq x_k\leq \mu_k$ for $k=0, 1, 2$.
Assume that $\lambda_k\leq x_k\leq \mu_k$ for $k\geq 2$.
By using (\ref{zf-3.7}), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}=\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)x_k}-\frac{6k+19}{k+5}
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\mu_{k+1}=\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)x_k}-\frac{6(k+4)}{k+5}.
\end{eqnarray*}
Since $\frac{1}{\mu_k}\le\frac{1}{x_k}\le\frac{1}{\lambda_k}$, we derive
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;&gt;&amp;\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)\lambda_k}-\frac{6k+19}{k+5}\\
&amp;=&amp;\frac{8k^2+17k+72}{(k+4)(k+5)(6k+13)} &gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\mu_{k+1}&amp;&lt;&amp;\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)\mu_k}-\frac{6(k+4)}{k+5}\\
&amp;=&amp;-\frac{2k^2+18k+28}{(k+3)(k+4)(k+5)}
&lt;0.
\end{eqnarray*}
Hence we have $\lambda_n\le x_n\le \mu_n$ for $n\ge 0$.

By applying (\ref{zf-3.6}), we obtain
\begin{eqnarray*}
(n+2)x_n-(n+1)x_{n+1}&amp;\ge&amp;(n+2)\lambda_n-(n+1)\mu_{n+1}\\
&amp;=&amp;\frac{n^2+7n+34}{(n+4)(n+5)} &gt;0 \quad (n\ge 0).
\end{eqnarray*}
Then $\{w_n\}_{n\ge 0}$ is log-balanced. For $r&gt;1$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{w_n}\}_{n\ge 0}$ is log-balanced.

For $\frac{5}{6}&lt;r&lt;1$, by using (\ref{zf-3.6}), we get
\begin{eqnarray*}
(n+2)^rx_n-(n+1)^rx_{n+1}\ge\frac{(n+2)^r(n+5)(6n+13)-6(n+1)^r(n+4)^2}{(n+4)(n+5)}.
\end{eqnarray*}
It is obvious that $(n+2)^r(n+5)(6n+13)\ge6(n+1)^r(n+4)^2$ if and only if
\begin{eqnarray*}
r\ln(n+2)-r\ln(n+1)+\ln(6n^2+43n+65)-\ln(6n^2+48n+96)\ge 0.
\end{eqnarray*}
We note that
\begin{eqnarray*}
&amp;&amp;\quad r\ln(n+2)-r\ln(n+1)+\ln(6n^2+43n+65)-\ln(6n^2+48n+96)\\
&amp;&amp;=r\ln\bigg(1+\frac{1}{n+1}\bigg)-\ln\bigg(1+\frac{5n+31}{6n^2+43n+65}\bigg).
\end{eqnarray*}
Due to $\frac{x}{1+x}&lt;\ln(1+x)&lt;x$ for $x&gt;0$, we have
\begin{eqnarray*}
&amp;&amp;\quad r\ln(n+2)-r\ln(n+1)+\ln(6n^2+43n+65)-\ln(6n^2+48n+96)\\
&amp;&amp;&gt;\frac{(6r-5)n^2+(43r-41)n+65r-62}{(n+2)(6n^2+43n+65)}.
\end{eqnarray*}
Since
\begin{eqnarray*}
\lim_{n\to+\infty}[(6r-5)n^2+(43r-41)n+65r-62]=+\infty,
\end{eqnarray*}
there exists a positive integer $N_r$ such that $(6r-5)n^2+(43r-41)n+65r-62&gt;0$ for $n\ge N_r$. Then the sequence $\{\sqrt[r]{w_n}\}_{n\ge N_r}$ is log-balanced for $\frac{5}{6}&lt;r&lt;1$.
\end{proof}

\subsection{Numbers counting a class of arrays}

For an integer $r\ge 0$, let $Q(n, r)$ denote the number of arrays (or matrices) of integers $a_{i, j}\ge 0$ ($1\le i, j\le n$) such that
$$
\sum_{i=1}^na_{i, j}=\sum_{j=1}^na_{i, j}=r
$$
holds for all $i$ and $j$. Consider the sequence $\{A_n\}_{n\ge 0}$, where $A_n=Q(n, 2)$. The sequence $\{A_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
A_{n+1}=(n+1)^2A_n-n{n+1\choose 2}A_{n-1}, \quad n\ge 1,  \label{zf-3.8}
\end{eqnarray}
where $A_0=A_1=1$ and $A_2=3$;
see Table \ref{tab:table4} for some information about it.  It is sequence
\seqnum{A000681} in the OEIS.  In particular, Zhao \cite{zhao15J} proved that the sequence $\{A_n\}_{n\ge 1}$ is log-convex (it is clear that $\{A_n\}_{n\ge 0}$ is also log-convex). See \cite{comt74} for more properties of $\{A_n\}_{n\ge 0}$.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|ccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ \\
  \hline
  $A_n$ &amp; $1$ &amp; $1$ &amp; $3$ &amp; $21$ &amp; $282$ &amp; $6210$ &amp; $202410$ \\
\end{tabular}
\end{center}
 \caption {Some initial values of $\{A_n\}_{n\ge 0}$} \label{tab:table4}
\end{table}

\begin{theorem} \label{thm:7a}
For $r\ge 5$, the sequence $\{\sqrt[r]{A_n}\}_{n\ge 0}$ is log-balanced.
\end{theorem}

\begin{proof} For $n\geq 0$, set $x_n=\frac{A_{n+1}}{A_n}$. We next prove by induction that
\begin{eqnarray}
\lambda_n\leq x_n\leq \lambda_{n+1},\quad n\ge 0,\label{zf-3.9}
\end{eqnarray}
where $\lambda_n=n^2$. It follows from (\ref{zf-3.8}) that
\begin{eqnarray}
x_n=(n+1)^2-\frac{n^2(n+1)}{2x_{n-1}},\quad n\ge 1, \label{zf-3.10}
\end{eqnarray}
It is clear that $\lambda_k \leq x_k \leq \lambda_{k+1}$ for $k=0, 1, 2$.
Assume that $\lambda_k\leq x_k\leq \lambda_{k+1}$ for $k\geq 2$.
By applying (\ref{zf-3.10}), we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;=&amp;2k+3-\frac{(k+1)^2(k+2)}{2x_k}, \\
x_{k+1}-\lambda_{k+2}&amp;=&amp;-\frac{(k+1)^2(k+2)}{2x_k}.
\end{eqnarray*}
Due to $\frac{1}{\lambda_{k+1}}\leq \frac{1}{x_{k}}\leq \frac{1}{\lambda_k}$ ($k\geq2$), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp; \geq &amp; 2k+3-\frac{(k+1)^2(k+2)}{2\lambda_k} \\
  &amp;=&amp;\frac{3k^3+2k^2-5k-2}{2k^2}
\geq  0 \quad (k\geq 2)
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}&amp;\leq&amp;-\frac{(k+1)^2(k+2)}{2\lambda_{k+1}} \\
&amp;=&amp;-\frac{k+2}{2} \leq 0.
\end{eqnarray*}
Then we derive $\lambda_n\leq x_n\leq \lambda_{n+1}$ for $n\geq 0$.

By using (\ref{zf-3.9}), we have
\begin{eqnarray*}
(n+2)^5x_n-(n+1)^5x_{n+1}&amp;\ge&amp;(n+2)^5\lambda_n-(n+1)^5\lambda_{n+2}\\
&amp;=&amp;(n+2)^2(n^4+2n^3-2n^2-5n-1)
&gt;0 \quad (n\ge 2).
\end{eqnarray*}
On the other hand, we note that $(k+2)^5x_k-(k+1)^5x_{k+1}$ for $k=0, 1$. Then $(n+2)^5x_n-(n+1)^5x_{n+1}&gt;0$ holds for $n\ge 0$. It follows from Theorem \ref{thm:1a} that the sequence $\{\sqrt[5]{A_n}\}_{n\ge 0}$ is log-balanced. For $r&gt;5$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{A_n}\}_{n\ge 0}$ is log-balanced.
\end{proof}

\subsection{Numbers satisfying a three-term recurrence}

Let $t_n$ counting the number of integer sequences $(f_j, \ldots, f_2, f_1, 1, 1, g_1, g_2, \ldots, g_k)$ with $j+k+2=n$ in which every $f_i$ is the sum of one or more contiguous terms immediately to its right, and $g_i$ is likewise the sum of one or more contiguous terms immediately to its left; see Odlyzko \cite{Odly95}. Fishburn et al. \cite{fish89} proved that the sequence $\{t_n\}_{n\ge 1}$ satisfies the recurrence
\begin{eqnarray}
t_{n+1}=2nt_n-(n-1)^2t_{n-1}, \quad n\ge 2, \label{zf-3.11}
\end{eqnarray}
where $t_1=t_2=1$ and $t_3=3$; see Table \ref{tab:table5} for some information about it. It is sequence \seqnum{A005189} in the OEIS.  Zhao \cite{zhao15J} showed that the sequence $\{t_n\}_{n\ge 1}$ is log-convex.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|ccccccc}
{\em n} &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ \\
  \hline
  $t_n$ &amp; $1$ &amp; $1$ &amp; $3$ &amp; $14$ &amp; $85$ &amp; $626$ &amp; $5387$ \\
\end{tabular}
\end{center}
\caption {Some initial values of $\{t_n\}_{n\ge 0}$} \label{tab:table5}
\end{table}

\begin{theorem} \label{thm:8a}
For $r\ge 3$, the sequence $\{\sqrt[r]{t_n}\}_{n\ge 3}$ is log-balanced.
\end{theorem}
\begin{proof} For $n\geq 0$, put $x_n=\frac{t_{n+1}}{t_n}$. We first prove by induction that
\begin{eqnarray}
\lambda_k\leq x_k\leq \mu_k,\quad k\ge 3, \label{zf-3.12}
\end{eqnarray}
where $\lambda_k=k+\sqrt{k}-\frac{1}{4}$ and $\mu_k=k+1+\sqrt{k+1}$. It follows from (\ref{zf-3.11}) that
\begin{eqnarray}
x_n=2n-\frac{(n-1)^2}{x_{n-1}},\quad n\ge 2, \label{zf-3.13}
\end{eqnarray}
It is clear that $\lambda_k \leq x_k \leq \mu_k$ for $k=3, 4$. Assume that $\lambda_k\leq x_k\leq \mu_k$ for $k\geq 4$.
By using (\ref{zf-3.13}), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}=k-\sqrt{k+1}+\frac{5}{4}-\frac{k^2}{x_k} \quad {\rm and} \quad x_{k+1}-\mu_{k+1}&amp;=&amp;k-\sqrt{k+2}-\frac{k^2}{x_k}.
\end{eqnarray*}
Since $\frac{1}{\mu_k}\leq \frac{1}{x_k}\leq \frac{1}{\lambda_k}$ for $k\geq4$, we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;\ge&amp;k+\frac{5}{4}-\sqrt{k+1}-\frac{k^2}{\lambda_k}\\
&amp;=&amp;\frac{1}{\lambda_k}\bigg(k\sqrt{k}+k+\frac{5\sqrt{k}}{4}+\frac{\sqrt{k+1}}{4}-\frac{5}{16}-k\sqrt{k+1}-\sqrt{k(k+1)}\bigg)      \\
&amp;&gt;&amp;\frac{1}{\lambda_k}\bigg(\sqrt{k}+k-\sqrt{k(k+1)}-\frac{5}{16}\bigg)\\
&amp;&gt;&amp;\frac{k+1-\sqrt{k(k+1)}}{\lambda_k}
&gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\mu_{k+1}&amp;\leq&amp;k-\sqrt{k+2}-\frac{k^2}{\mu_k} \\
&amp;=&amp;\frac{k+k\sqrt{k+1}-(k+1)\sqrt{k+2}-\sqrt{(k+1)(k+2)}}{k+1+\sqrt{k+1}} \\
&amp;\leq&amp; -\frac{\sqrt{k+2}}{k+1+\sqrt{k+1}}
\leq 0.
\end{eqnarray*}
Then we derive $\lambda_k\leq x_k\leq \mu_k$ for $k\geq 3$.

It follows from (\ref{zf-3.12}) that
\begin{eqnarray*}
(n+2)^3x_n-(n+1)^3x_{n+1}&amp;\ge&amp;(n+2)^3\bigg(n+\sqrt{n}-\frac{1}{4}\bigg)-(n+1)^3(n+2+\sqrt{n+2})\\
&amp;=&amp;\bigg(\frac{3}{4}-\frac{2}{\sqrt{n}+\sqrt{n+2}}\bigg)n^3+\frac{3n^2-4n-8}{2}\\
&amp;&amp;+3n ( 2(n+2)\sqrt{n}-(n+1)\sqrt{n+2} ) +8\sqrt{n}-\sqrt{n+2}  \\
&amp;&gt;&amp;0 \quad (n\ge 3).
\end{eqnarray*}
On the other hand, we can verify that $(n+2)^3x_n-(n+1)^3x_{n+1}&gt;0$ for $1\le n\le 2$. Hence the sequence $\{\sqrt[3]{t_n}\}_{n\ge 1}$ is log-balanced. For $r&gt;3$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{t_n}\}_{n\ge 1}$ is log-balanced.
\end{proof}

\subsection{Numbers counting bipermutations}

For a given nonnegative integer $k$, a relation $\Re$ is called a $k$-permutation of $[n]=\{1, 2, \ldots, n\}$ if all vertical sections and all horizontal sections have $k$ elements. The $k$-permutation $\Re$ is called a bipermutation when $k=2$. Let $P(n, k)$ denote the number of these relations. Let $P_n=P(n, 2)$. The sequence $\{P_n\}$ satisfies the recurrence
\begin{eqnarray}
P_{n+1}={n+1\choose 2}(2P_n+nP_{n-1}), \quad n\ge1, \label{zf-3.14}
\end{eqnarray}
where $P_0=1$ and $P_1=0$; see Table \ref{tab:table6} for some information about it.  It is sequence~\seqnum{A001499} in the OEIS.
In particular, Zhao \cite{zhao15J} showed that the sequence $\{P_n\}_{n\ge 2}$ is log-convex.
\begin{table}[!htbp]

\begin{center}
\begin{tabular}{c|ccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$  \\
  \hline
  $P_n$ &amp; $1$ &amp; $0$ &amp; $1$ &amp; $6$ &amp; $90$ &amp; $2040$ &amp; $67950$  \\
\end{tabular}
\end{center}
\caption {Some initial values of $\{P_n\}_{n\ge 0}$} \label{tab:table6}
\end{table}

\begin{theorem} \label{thm:9a}
For $r\ge 3$, the sequence $\{\sqrt[r]{P_n}\}_{n\ge 3}$ is log-balanced.
\end{theorem}
\begin{proof} For $n\ge 2$, put $x_n=\frac{P_{n+1}}{P_n}$. It follows from (\ref{zf-3.14}) that
\begin{eqnarray}
x_k=k(k+1)+k{k+1\choose 2}\frac{1}{x_{k-1}}, \quad k\ge 3. \label{zf-3.15}
\end{eqnarray}
 We prove that
\begin{eqnarray}
\lambda_n\le x_n\le\lambda_{n+1}, \quad n\ge 2,  \label{zf-3.16}
\end{eqnarray}
where $\lambda_n=n(n+1)$. It is evident that
$\lambda_k&lt;x_k&lt;\lambda_{k+1}$ for $k=2, 3$. Assume that $\lambda_k\leq x_k\leq\lambda_{k+1}$ for $k\ge 3$. By applying (\ref{zf-3.15}), we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}=(k+1){k+2\choose 2}\frac{1}{x_k}&gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}=-2(k+2)+\frac{(k+1)^2(k+2)}{2x_k}.
\end{eqnarray*}
Since $\frac{1}{x_k}\le\frac{1}{\lambda_k}$, we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}\le-\frac{3k^2+5k-2}{2k}&lt;0.
\end{eqnarray*}
Then we have $\lambda_n\le x_n\le\lambda_{n+1}$ for $n\ge 2$.

It follows from (\ref{zf-3.15}) and (\ref{zf-3.16}) that
\begin{eqnarray*}
(n+2)^3x_n-(n+1)^3x_{n+1}&amp;=&amp;(n+2)^3x_n-(n+1)^4(n+2)-\frac{(n+1)^5(n+2)}{2x_n}\\
&amp;\ge&amp;(n+2)^3\lambda_n-(n+1)^4(n+2)-\frac{(n+1)^5(n+2)}{2\lambda_n}\\
&amp;=&amp;\frac{(n+1)(n+2)(n^3-n^2-5n-1)}{2n}
&gt;0 \quad (n\ge 3).
\end{eqnarray*}
We have from Theorem \ref{thm:1a} that the sequence $\{\sqrt[3]{P_n}\}_{n\ge 3}$ is log-balanced. For $r&gt;3$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{P_n}\}_{n\ge 3}$ is log-balanced.
\end{proof}

\subsection{Numbers satisfying a four-term recurrence (``minus" case)}

Let $G_n$ stand for the number of graphs on the vertex set $[n]=\{1, 2, \ldots, n\}$, whose every component is a cycle, and put $G_0=1$. The sequence $\{G_n\}$ satisfies the recurrence
\begin{eqnarray}
G_{n+1}=(n+1)G_n-{n\choose 2}G_{n-2}, \quad n\ge 2, \label{zf-3.17}
\end{eqnarray}
where $G_1=1$, $G_2=2$, and $G_3=5$; see Table \ref{tab:table7} for some information about it.  It is sequence \seqnum{A002135} in the OEIS.
This example is Exercise 5.22 of Stanley \cite{stan99},
and one can find its combinatorial proof in Stanley \cite[p.~121]{stan99}.
In addition, Do\v{s}li\'{c} \cite{dos08} showed that the sequence $\{G_n\}_{n\ge 0}$ is log-convex.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|ccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$  \\
  \hline
  $G_n$ &amp; $1$ &amp; $1$ &amp; $2$ &amp; $5$ &amp; $17$ &amp; $73$ &amp; $388$  \\
\end{tabular}
\end{center}
\caption {Some initial values of $\{G_n\}_{n\ge 0}$} \label{tab:table7}
\end{table}

\begin{theorem} \label{thm:10a}
For $r\ge 2$, the sequence $\{\sqrt[r]{G_n}\}_{n\ge 0}$ is log-balanced.
\end{theorem}

\begin{proof} For $n\ge 0$, let $x_n=\frac{G_{n+1}}{G_n}$. We next prove by induction that
\begin{eqnarray}
\lambda_n\le x_n\le\lambda_{n+1}, \quad n\ge 0, \label{zf-3.18}
\end{eqnarray}
where $\lambda_n=n$. It follows from (\ref{zf-3.17}) that
\begin{eqnarray}
x_n=n+1-{n\choose 2}\frac{1}{x_{n-1}x_{n-2}}, \quad n\ge 2. \label{zf-3.19}
\end{eqnarray}
Firstly, we have $\lambda_k\le x_k\le\lambda_{k+1}$ for $0\le k\le 4$. Assume that $\lambda_k\le x_k\le\lambda_{k+1}$ for $k\ge 4$. By using (\ref{zf-3.19}), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}=-{k+1\choose 2}\frac{1}{x_k x_{k-1}}&lt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;=&amp;1-{k+1\choose 2}\frac{1}{x_k x_{k-1}}\\
&amp;\ge&amp;\frac{2k(k-1)-k(k+1)}{2x_k x_{k-1}}
&gt;0.
\end{eqnarray*}
Then we derive $\lambda_n\le x_n\le\lambda_{n+1}$ for $n\ge 0$.

It follows from (\ref{zf-3.17}) and (\ref{zf-3.18}) that
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}&amp;=&amp;(n+2)^2x_n-(n+1)^2(n+2)+\frac{(n+1)^2{n+1\choose 2}}{x_n x_{n-1}}\\
&amp;\ge&amp;(n+2)^2\lambda_n-(n+1)^2(n+2)+\frac{(n+1)^2{n+1\choose 2}}{\lambda_{n+1}\lambda_n}\\
&amp;=&amp;\frac{n^2-3}{2}
&gt;0 \quad (n\ge 2).
\end{eqnarray*}
On the other hand, we note that $(k+2)^2x_k-(k+1)^2x_{k+1}&gt;0$ for $k=0, 1$. Then $(n+2)^2x_n-(n+1)^2x_{n+1}&gt;0$ holds for $n\ge 0$. We have from Theorem \ref{thm:1a} that the sequence $\{\sqrt{G_n}\}_{n\ge 0}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{G_n}\}_{n\ge 0}$ is log-balanced.
\end{proof}

\subsection{Numbers satisfying a four-term recurrence (``plus" case)}

Let be given a set of $\Delta$ of $n$ straight lines in the plane, $\delta_1, \delta_2, \ldots, \delta_n$, lying in general position (no two among them are parallel, and no three among are concurrent). Let $P$ be the set of their points of intersection, $|P|={n\choose 2}$. We call any set of $n$ points from $P$ such that any three different points are not collinear, a {\it cloud}. Let $\mathscr{G}(\Delta)$ stand for the set of clouds of $\Delta$ and $g_n=|\mathscr{G}(\Delta)|$. The sequence $\{g_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
g_{n+1}=n g_n+{n\choose 2}g_{n-2}, \quad n\ge 2, \label{zf-3.20}
\end{eqnarray}
where $g_0=1$, $g_1=g_2=0$, $g_3=1$, $g_4=3$ and $g_5=12$; see Table \ref{tab:table8} for some information about it.  It is sequence
\seqnum{A001205} in the OEIS.
In particular, Zhao \cite{zhao15J} proved that the sequence $\{g_n\}_{n\ge 3}$ is log-convex. For more properties of $\{g_n\}_{n\ge 0}$, see Comtet \cite{comt74}.
\begin{table}[!htbp]
\begin{center}
\begin{tabular}{c|ccccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ &amp; $8$\\
  \hline
  $g_n$ &amp; $1$ &amp; $0$ &amp; $0$ &amp; $1$  &amp; $3$ &amp; $12$ &amp; $70$ &amp; $465$ &amp; $3507$\\
\end{tabular}
\end{center}
 \caption {Some initial values of $\{g_n\}_{n\ge 0}$} \label{tab:table8}
\end{table}

\begin{theorem} \label{thm:11a}
For $r\ge 2$, the sequence $\{\sqrt[r]{g_n}\}_{n\ge 5}$ is log-balanced.
\end{theorem}

\begin{proof} For $n\ge 3$, let $x_n=\frac{g_{n+1}}{g_n}$. It follows from (\ref{zf-3.20}) that
\begin{eqnarray}
x_n=n+{n\choose 2}\frac{1}{x_{n-1}x_{n-2}}, \quad n\ge 5. \label{zf-3.21}
\end{eqnarray}
Now we show that
\begin{eqnarray}
\lambda_n\le x_n\le\lambda_{n+1}, \quad n\ge 3,  \label{zf-3.22}
\end{eqnarray}
where $\lambda_n=n$. We can verify that $\lambda_k\le x_k\le\lambda_{k+1}$ for $3\le k\le 5$. Assume that $\lambda_k\le x_k\le\lambda_{k+1}$ for $k\ge 5$. By applying (\ref{zf-3.21}), we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}={k+1\choose 2}\frac{1}{x_k x_{k-1}}&gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}&amp;=&amp;\frac{k(k+1)}{2x_k x_{k-1}}-1\\
&amp;\le&amp;-\frac{k(k-3)}{2x_k x_{k-1}}
&lt;0.
\end{eqnarray*}
Then we have $\lambda_n\le x_n\le\lambda_{n+1}$ for $n\ge 3$.

It follows from (\ref{zf-3.21}) and (\ref{zf-3.22}) that
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}&amp;=&amp;(n+2)^2x_n-(n+1)^3-\frac{(n+1)^2{n+1\choose 2}}{x_n x_{n-1}}\\
&amp;\ge&amp;(n+2)^2\lambda_n-(n+1)^3-\frac{(n+1)^2{n+1\choose 2}}{\lambda_n \lambda_{n-1}}\\
&amp;=&amp;\frac{n^3-3n^2-7n+1}{2(n-1)}
&gt;0 \quad (n\ge 5).
\end{eqnarray*}
We have from Theorem \ref{thm:1a} that the sequence $\{\sqrt{g_n}\}_{n\ge 5}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{g_n}\}_{n\ge 5}$ is log-balanced.
\end{proof}

\subsection{Numbers counting permutation with ordered orbits}

Consider the sequence $\{T_n\}_{n\ge 2}$ defined by
\begin{eqnarray}
T_{n+1}=(n-1)T_n+\frac{n!}{2}, \quad n\ge 2, \label{zf-3.23}
\end{eqnarray}
where $T_2=1$; see Table \ref{tab:table9} for some information about it. The value of $T_n$ is related to the number of permutations with ordered orbits. In particular, Zhao \cite{zhao15J} proved that the sequence $\{T_n\}_{n\ge 2}$ is log-convex.  It is sequence \seqnum{A006595} in the OEIS.
For more properties of $\{T_n\}_{n\ge 2}$, see Comtet \cite{comt74}.
\begin{table}[!htbp]

\begin{center}
\begin{tabular}{c|cccccc}
{\em n} &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ \\
  \hline
  $T_n$ &amp; $1$ &amp; $2$ &amp; $7$ &amp; $33$ &amp; $192$ &amp; $1320$ \\
\end{tabular}
\end{center}
\caption {Some initial values of $\{T_n\}_{n\ge 0}$} \label{tab:table9}

\end{table}

\begin{theorem} \label{thm:12a}
For $r\ge 2$, the sequence $\{\sqrt[r]{T_n}\}_{n\ge 2}$ is log-balanced.
\end{theorem}
\begin{proof} For $n\ge 2$, let $x_n=\frac{T_{n+1}}{T_n}$. It is easy to verify that
\begin{eqnarray}
T_{n+1}=(2n-1)T_n-(n-2)n T_{n-1}, \quad n\ge 3. \label{zf-3.24}
\end{eqnarray}
It follows from (\ref{zf-3.24}) that
\begin{eqnarray}
x_n=2n-1-\frac{(n-2)n}{x_{n-1}}, \quad n\ge 3.  \label{zf-3.25}
\end{eqnarray}
Now we prove by induction that
\begin{eqnarray*}
\lambda_n\le x_n\le\lambda_{n+1}, \quad n\ge 2,
\end{eqnarray*}
where $\lambda_n=n$. It is not difficult to verify that $\lambda_k\le x_k\le\lambda_{k+1}$ for $2\le k\le 4$. Assume that $\lambda_k\le x_k\le\lambda_{k+1}$ for $k\ge 4$. Using (\ref{zf-3.25}), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;=&amp;k-\frac{(k+1)(k-1)}{x_k}\\
&amp;\ge&amp;k-\frac{(k+1)(k-1)}{k}
&gt;0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}\le k-1-\frac{(k+1)(k-1)}{\lambda_{k+1}}=0.
\end{eqnarray*}
Then we derive that $\lambda_n\le x_n\le\lambda_{n+1}$ for $n\ge 2$.
By means of (\ref{zf-3.26}), we obtain
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}=\frac{(n^2+4n+4)x_n^2-(2n^3+5n^2+4n+1)x_n+n^4+2n^3-2n-1}{x_n}
\end{eqnarray*}
For any $x\in (-\infty,+\infty)$, define a function
$$
f(x)=(n^2+4n+4)x^2-(2n^3+5n^2+4n+1)x+n^4+2n^3-2n-1.
$$
We can prove that $f$ is increasing on $[\sigma_n, +\infty)$, where $\sigma_n=\frac{2n^3+5n^2+4n+1}{2(n^2+4n+4)}$, $f$ is increasing on $[\sigma_n, +\infty)$.
We can verify that $\lambda_n&gt;\sigma_n$. Hence $f$ is increasing on $[\lambda_n,+\infty)$.
We note that
\begin{eqnarray*}
f(\lambda_n)&amp;=&amp;(n^2+4n+4)\lambda_n^2-(2n^3+5n^2+4n+1)\lambda_n+n^4+2n^3-2n-1 \\
&amp;=&amp;n^3-3n-1 \\
&amp;&gt;&amp; 0 \quad (n\geq 2).
\end{eqnarray*}
Then we have $f(x_n)&gt;0$ for $n\geq2$. This implies that $(n+2)^2x_n-(n+1)^2x_{n+1}\geq0$ for $n\geq2$. It follows from Theorem \ref{thm:1a} that the sequence $\{\sqrt{T_n}\}_{n\geq2}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{T_n}\}_{n\ge 2}$ is log-balanced.
\end{proof}

\subsection{The Domb numbers}

Let $\{D_n\}_{n\ge 0}$ be the sequence of the Domb numbers. The value of $D_n$ is the number of $2n$-step polygons on the diamond lattice. The sequence $\{D_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
n^3D_n=2(2n-1)(5n^2-5n+2)D_{n-1}-64(n-1)^3D_{n-2}, \quad n\ge 2, \label{zf-3.26}
\end{eqnarray}
where $D_0=1$ and $D_1=4$; see Table \ref{tab:table10} for some information about it. It is sequence \seqnum{A002895} in the OEIS.
In particular, Wang and Zhu \cite{wang14} proved that the sequence $\{D_n\}_{n\ge 0}$ is log-convex.
\begin{table}[!htbp]

\begin{center}
\begin{tabular}{c|cccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ \\
  \hline
  $D_n$ &amp; $1$ &amp; $4$ &amp; $28$ &amp; $256$ &amp; $2716$ &amp; $31504$ &amp; $387136$ &amp; $4951552$ \\
\end{tabular}
\end{center}
 \caption {Some initial values of $\{D_n\}_{n\ge 0}$} \label{tab:table10}
\end{table}

\begin{theorem} \label{thm:13a}
For $r\ge 2$, the sequence $\{\sqrt[r]{D_n}\}_{n\ge 1}$ is log-balanced.
\end{theorem}

\begin{proof} For $n\ge 0$, let $x_n=\frac{D_{n+1}}{D_n}$. It follows from (\ref{zf-3.26}) that
\begin{eqnarray}
x_n=\frac{2(2n+1)(5n^2+5n+2)}{(n+1)^3}-\frac{64n^3}{(n+1)^3x_{n-1}}, \quad n\ge 1.  \label{zf-3.27}
\end{eqnarray}
We first show that
\begin{eqnarray}
\lambda_n\le x_n\le\mu_n,  \quad n\ge 1, \label{zf-3.28}
\end{eqnarray}
where $\lambda_n=\frac{16(n-1)}{n+1}$ and $\mu_n=\frac{16n}{n+1}$. It is obvious that $\lambda_k&lt;x_k&lt;\mu_k$ for $1\le k\le3$. Assume that $\lambda_k\le x_k\le\mu_k$ for $k\ge 3$. By means of (\ref{zf-3.27}), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp;\ge&amp;\frac{2(2k+3)(5k^2+15k+12)}{(k+2)^3}-\frac{16k}{k+2}-\frac{64(k+1)^3}{(k+2)^3\lambda_k}\\
&amp;=&amp;\frac{2(3k^3+12k^2-9k-38)}{(k-1)(k+2)^3}
&gt;0 \quad (k\ge 3)
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\mu_{k+1}&amp;\le&amp;\frac{2(2k+3)(5k^2+15k+12)}{(k+2)^3}-\frac{16(k+1)}{k+2}-\frac{64(k+1)^3}{(k+2)^3\mu_k}\\
&amp;=&amp;-\frac{2(3k^3+7k^2+4k+2)}{k(k+2)^3}
&lt;0.
\end{eqnarray*}
Then we have $\lambda_n\le x_n\le\mu_n$ for $n\ge 1$.

It follows from (\ref{zf-3.28}) that
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}&amp;\ge&amp;\frac{16[(n-1)(n+2)^3-(n+1)^4]}{(n+1)(n+2)}\\
&amp;=&amp;\frac{16(n^3-8n-9)}{(n+1)(n+2)}\\
&amp;&gt;&amp;0 \quad (n\ge 4).
\end{eqnarray*}
On the other hand, we observe that $(n+2)^2x_n-(n+1)^2x_{n+1}&gt;0$ for $0\le n\le 3$. We have from Theorem \ref{thm:1a} that the sequence $\{\sqrt{D_n}\}_{n\ge 0}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{D_n}\}_{n\ge 0}$ is log-balanced.
\end{proof}

\subsection{Numbers counting a class of $n\times n$ symmetric matrices}

Let $\tau_n$ denote the number of $n\times n$ symmetric $\mathbb{N}_{0}$-matrices with every row(and hence every column) sum equals to 2 with trace zero (i.e., all main diagonal entries are zero). The sequence $\{\tau_n\}_{n\ge 0}$ satisfies the recurrence
\begin{eqnarray}
\tau_n=(n-1)\tau_{n-1}+(n-1)\tau_{n-2}-{n-1\choose 2}\tau_{n-3}, \label{zf-3.29}
\end{eqnarray}
where $\tau_0=1,\tau_1=0,\tau_2=\tau_3=1$; see Table \ref{tab:table11} for some information about it.
It is sequence \seqnum{A002137} in the OEIS.
In particular, Do\v{s}li\'{c} \cite{dos08} showed that the sequence $\{\tau_n\}_{n\ge 6}$ is log-convex.
\begin{table}[!htbp]

\begin{center}
\begin{tabular}{c|ccccccccc}
{\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$  &amp;$8$\\
  \hline
  $\tau_n$ &amp; $1$ &amp; $0$ &amp; $1$ &amp; $1$ &amp; $6$ &amp; $22$ &amp; $130$ &amp; $822$ &amp; $6202$\\
\end{tabular}
\end{center}
\caption {Some initial values of $\{\tau_n\}_{n\ge 0}$} \label{tab:table11}
\end{table}

\begin{theorem} \label{thm:14a}
For $r\ge 2$, the sequence $\{\sqrt[r]{\tau_n}\}_{n\ge 6}$ is log-balanced.
\end{theorem}

\begin{proof} For $n\geq 2$, set $x_n=\frac{\tau_{n+1}}{\tau_n}$. We now prove by induction that
\begin{eqnarray}
\lambda_n\leq x_n\leq \lambda_{n+1},\quad n\ge 6,\label{zf-3.30}
\end{eqnarray}
where $\lambda_n=n$. It is clear that $\lambda_k \leq x_k \leq \lambda_{k+1}$ for $k=6, 7$.
Assume that $\lambda_k\leq x_k\leq \lambda_{k+1}$ for $k\geq 7$. It follows from (\ref{zf-3.29}) that
\begin{eqnarray}
x_n=n+\frac{n}{x_{n-1}}-\frac{n(n-1)}{2x_{n-2}x_{n-1}},\quad n\ge 4, \label{zf-3.31}
\end{eqnarray}
By applying (\ref{zf-3.31}), we get
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}=\frac{k+1}{x_{k}}-\frac{(k+1)k}{2x_{k-1}x_k}\quad {\rm and} \quad x_{k+1}-\lambda_{k+2}=-1+\frac{k+1}{x_k}-\frac{(k+1)k}{2x_{k-1}x_k}.
\end{eqnarray*}
Due to $\frac{1}{\lambda_{k+1}}\leq \frac{1}{x_{k}}\leq \frac{1}{\lambda_k}$ ($k\geq7$), we have
\begin{eqnarray*}
x_{k+1}-\lambda_{k+1}&amp; \geq &amp; 1-\frac{k+1}{2(k-1)} \\
  &amp;=&amp;\frac{k-3}{2(k-1)}
  \geq  0
\end{eqnarray*}
and
\begin{eqnarray*}
x_{k+1}-\lambda_{k+2}&amp;\le&amp;\frac{1}{k}-\frac{1}{2}\\
&amp;=&amp;-\frac{k-2}{2k}
\leq 0.
\end{eqnarray*}
Then we derive $\lambda_n\leq x_n\leq \lambda_{n+1}$ for $n\geq 6$.

By using (\ref{zf-3.30}) and (\ref{zf-3.31}), we have
\begin{eqnarray*}
(n+2)^2x_n-(n+1)^2x_{n+1}&amp;=&amp;(n+2)^2x_n-(n+1)^3-\frac{(n+1)^3}{x_n}+\frac{n(n+1)^3}{2x_n x_{n-1}}\\
&amp;\ge&amp;(n+2)^2\lambda_n-(n+1)^3-\frac{(n+1)^3}{\lambda_n}+\frac{n(n+1)^3}{2\lambda_n \lambda_{n+1}}\\
&amp;\ge&amp;\frac{3n^2-6n-22}{6} 
&gt;0 \quad (n\ge 6).
\end{eqnarray*}
It follows from Theorem \ref{thm:1a} that the sequence $\{\sqrt{\tau_n}\}_{n\ge 6}$ is log-balanced. For $r&gt;2$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{\tau_n}\}_{n\ge 6}$ is log-balanced.
\end{proof}

In the rest of this section, we discuss log-balancedness of some sequences by mens of Theorem \ref{thm:3a}.
\subsection{The harmonic numbers}

Let $\{H_n\}_{n\ge 1}$ be the sequence of harmonic numbers. It is well known that
\begin{eqnarray*}
H_n=\sum_{k=1}^n\frac{1}{k}, \quad n\ge 1.
\end{eqnarray*}

\begin{theorem} \label{thm:15a}
The sequence $\{n!\sqrt{H_n}\}_{n\ge 1}$ is log-balanced.
\end{theorem}
\begin{proof}
Using the definition of log-concavity, one can immediately prove that $\{H_n\}_{n\ge 1}$ is log-concave. Moreover, from Zhao \cite{zhao15}, $\{n!H_n\}_{n\ge 1}$ is log-balanced. It follows from Theorem \ref{thm:3a} that the sequence $\{n!\sqrt{H_n}\}_{n\ge 1}$ is log-balanced.
\end{proof}

\subsection{The Fibonacci and Lucas numbers}

Let $\{F_n\}_{n\ge 0}$ and $\{L_n\}_{n\ge 0}$ denote the Fibonacci and Lucas sequence, respectively. The Binet's forms of $F_n$ and $L_n$ respectively are
\begin{eqnarray*}
F_n=\frac{\alpha^n-(-1)^n\alpha^{-n}}{\sqrt 5}, \quad L_n=\alpha^n+(-1)^n\alpha^{-n}, \quad n\ge 0,
\end{eqnarray*}
where $\alpha=\frac{1+\sqrt 5}{2}$.


\begin{theorem} \label{thm:16a}
The sequences $\{n!\sqrt{F_{2n}}\}_{n\ge 1}$ and $\{n!\sqrt{L_{2n+1}}\}_{n\ge 1}$ are both log-balanced.
\end{theorem}
\begin{proof}
Using the definition of log-concavity, we can prove that the sequences $\{F_{2n}\}_{n\ge 1}$ and $\{L_{2n-1}\}_{n\ge 1}$ are both log-concave. Zhao \cite{zhao15} showed that $\{n!F_{2n}\}_{n\ge 1}$ and $\{n!L_{2n+1}\}_{n\ge 1}$ are log-balanced. It follows from Theorem \ref{thm:3a} that the sequences $\{n!\sqrt{F_{2n}}\}_{n\ge 1}$ and $\{n!\sqrt{L_{2n+1}}\}_{n\ge 1}$ are both log-balanced.
\end{proof}


\section{Conclusions}

For a log-convex sequence $\{z_n\}_{n\ge 0}$, we have shown that the arithmetic root sequence $\{\sqrt[r]{z_n}\}_{n\ge 0}$ is log-balanced under suitable conditions. We have also derived the log-balancedness of a number of log-convex sequences related to many famous combinatorial numbers. However, we cannot give the minimum value of $r$ such that $\{\sqrt[r]{z_n}\}_{n\ge 0}$ is log-balanced. We hope to solve this question in the future work. In addition, we also hope to find more functions $f$ defined in $(-\infty, +\infty)$ such that $\{f(z_n)\}_{n\ge 0}$ is log-balanced.

\section{Acknowledgment}

The authors would like to thank the anonymous referee for his (her) many helpful comments and suggestions.



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\end{thebibliography}

\bigskip
\hrule
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\noindent 2010 {\it Mathematics Subject Classification}:  Primary 05A20;
Secondary 11B37, 11B83, 11B39.

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


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000166},
\seqnum{A000681},
\seqnum{A001205},
\seqnum{A001499},
\seqnum{A002135},
\seqnum{A002137},
\seqnum{A002212},
\seqnum{A002895},
\seqnum{A005189},
\seqnum{A005572}, and
\seqnum{A006595}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 9 2017;
revised versions received  March 16 2018; June 3 2018; June 27 2018.
Published in {\it Journal of Integer Sequences}, June 29 2018.

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