\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \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{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{mathrsfs} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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}\\ \end{center} \def\DD{\mathcal{D}} \def\SS{\mathcal{S}} \vskip .2 in \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 $ba$", 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*} && nz^2_n-(n+1)z_{n-1}z_{n+1}\le 0,\\ && z_n\le\sqrt{\frac{n+1}{n}z_{n-1}z_{n+1}}<\frac{n+1}{n}\sqrt{z_{n-1}z_{n+1}},\\ && (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} $n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ \\ \hline $d_n$ & $1$ & $0$ & $1$ & $2$ & $9$ & $44$ & $265$ & $1854$ & $14833$ \\ \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 $\lambda_n>\frac{(n+1)^3}{2(n+2)^2}$. Hence, $f$ is increasing on $[\lambda_n,+\infty)$. Note that \begin{eqnarray*} f(\lambda_n)&=&(n+2)^2\lambda_{n}^2-(n+1)^3\lambda_n-(n+1)^3 \\ &=&\frac{2n^3+3n^2-2n-2}{4} > 0 \quad (n\geq 1). \end{eqnarray*} Then we have $f(x_n)>0$ 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 $r>2$, 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} $n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ \\ \hline $h_n$ & $1$ & $1$ & $3$ & $10$ & $36$ & $137$ & $543$ & $2219$ \\ \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}&\geq& \frac{2k+5}{2(k+3)}-\frac{5k}{(k+3)\lambda_k} \\ &=&\frac{16k+15}{2(k+3)(10k+3)} \geq 0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\mu_{k+1}&\leq&\frac{k^2-3k-9}{(k+2)(k+3)}-\frac{5k}{(k+3)\mu_k} \\ &=&\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 $\lambda_n>\frac{3(n+1)(2n+3)}{2(n+2)(n+3)}$ and \begin{eqnarray*} f(\lambda_n)&=&\frac{84n^2-41n-27}{4(n+2)} > 0 \quad (n\ge 1). \end{eqnarray*} Then we have $(n+2)x_n-(n+1)x_{n+1}>0$ 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ \\ \hline $w_n$ & $1$ & $4$ & $17$ & $76$ & $354$ & $1704$ & $8421$ \\ \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}&\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)\lambda_k}-\frac{6k+19}{k+5}\\ &=&\frac{8k^2+17k+72}{(k+4)(k+5)(6k+13)} >0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\mu_{k+1}&<&\frac{4(2k+5)}{k+4}-\frac{12(k+1)}{(k+4)\mu_k}-\frac{6(k+4)}{k+5}\\ &=&-\frac{2k^2+18k+28}{(k+3)(k+4)(k+5)} <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}&\ge&(n+2)\lambda_n-(n+1)\mu_{n+1}\\ &=&\frac{n^2+7n+34}{(n+4)(n+5)} >0 \quad (n\ge 0). \end{eqnarray*} Then $\{w_n\}_{n\ge 0}$ is log-balanced. For $r>1$, it follows from Theorem \ref{thm:2a} that $\{\sqrt[r]{w_n}\}_{n\ge 0}$ is log-balanced. For $\frac{5}{6}0$, we have \begin{eqnarray*} &&\quad r\ln(n+2)-r\ln(n+1)+\ln(6n^2+43n+65)-\ln(6n^2+48n+96)\\ &&>\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 $(6r-5)n^2+(43r-41)n+65r-62>0$ for $n\ge N_r$. Then the sequence $\{\sqrt[r]{w_n}\}_{n\ge N_r}$ is log-balanced for $\frac{5}{6}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 $(n+2)^5x_n-(n+1)^5x_{n+1}>0$ 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 $r>5$, 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} & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ \\ \hline $t_n$ & $1$ & $1$ & $3$ & $14$ & $85$ & $626$ & $5387$ \\ \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}&=&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}&\ge&k+\frac{5}{4}-\sqrt{k+1}-\frac{k^2}{\lambda_k}\\ &=&\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) \\ &>&\frac{1}{\lambda_k}\bigg(\sqrt{k}+k-\sqrt{k(k+1)}-\frac{5}{16}\bigg)\\ &>&\frac{k+1-\sqrt{k(k+1)}}{\lambda_k} >0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\mu_{k+1}&\leq&k-\sqrt{k+2}-\frac{k^2}{\mu_k} \\ &=&\frac{k+k\sqrt{k+1}-(k+1)\sqrt{k+2}-\sqrt{(k+1)(k+2)}}{k+1+\sqrt{k+1}} \\ &\leq& -\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}&\ge&(n+2)^3\bigg(n+\sqrt{n}-\frac{1}{4}\bigg)-(n+1)^3(n+2+\sqrt{n+2})\\ &=&\bigg(\frac{3}{4}-\frac{2}{\sqrt{n}+\sqrt{n+2}}\bigg)n^3+\frac{3n^2-4n-8}{2}\\ &&+3n ( 2(n+2)\sqrt{n}-(n+1)\sqrt{n+2} ) +8\sqrt{n}-\sqrt{n+2} \\ &>&0 \quad (n\ge 3). \end{eqnarray*} On the other hand, we can verify that $(n+2)^3x_n-(n+1)^3x_{n+1}>0$ for $1\le n\le 2$. Hence the sequence $\{\sqrt[3]{t_n}\}_{n\ge 1}$ is log-balanced. For $r>3$, 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ \\ \hline $P_n$ & $1$ & $0$ & $1$ & $6$ & $90$ & $2040$ & $67950$ \\ \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_k0 \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}<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}&=&(n+2)^3x_n-(n+1)^4(n+2)-\frac{(n+1)^5(n+2)}{2x_n}\\ &\ge&(n+2)^3\lambda_n-(n+1)^4(n+2)-\frac{(n+1)^5(n+2)}{2\lambda_n}\\ &=&\frac{(n+1)(n+2)(n^3-n^2-5n-1)}{2n} >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 $r>3$, 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ \\ \hline $G_n$ & $1$ & $1$ & $2$ & $5$ & $17$ & $73$ & $388$ \\ \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}}<0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\lambda_{k+1}&=&1-{k+1\choose 2}\frac{1}{x_k x_{k-1}}\\ &\ge&\frac{2k(k-1)-k(k+1)}{2x_k x_{k-1}} >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}&=&(n+2)^2x_n-(n+1)^2(n+2)+\frac{(n+1)^2{n+1\choose 2}}{x_n x_{n-1}}\\ &\ge&(n+2)^2\lambda_n-(n+1)^2(n+2)+\frac{(n+1)^2{n+1\choose 2}}{\lambda_{n+1}\lambda_n}\\ &=&\frac{n^2-3}{2} >0 \quad (n\ge 2). \end{eqnarray*} On the other hand, we note that $(k+2)^2x_k-(k+1)^2x_{k+1}>0$ for $k=0, 1$. Then $(n+2)^2x_n-(n+1)^2x_{n+1}>0$ 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 $r>2$, 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$\\ \hline $g_n$ & $1$ & $0$ & $0$ & $1$ & $3$ & $12$ & $70$ & $465$ & $3507$\\ \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}}>0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\lambda_{k+2}&=&\frac{k(k+1)}{2x_k x_{k-1}}-1\\ &\le&-\frac{k(k-3)}{2x_k x_{k-1}} <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}&=&(n+2)^2x_n-(n+1)^3-\frac{(n+1)^2{n+1\choose 2}}{x_n x_{n-1}}\\ &\ge&(n+2)^2\lambda_n-(n+1)^3-\frac{(n+1)^2{n+1\choose 2}}{\lambda_n \lambda_{n-1}}\\ &=&\frac{n^3-3n^2-7n+1}{2(n-1)} >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 $r>2$, 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} & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ \\ \hline $T_n$ & $1$ & $2$ & $7$ & $33$ & $192$ & $1320$ \\ \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}&=&k-\frac{(k+1)(k-1)}{x_k}\\ &\ge&k-\frac{(k+1)(k-1)}{k} >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 $\lambda_n>\sigma_n$. Hence $f$ is increasing on $[\lambda_n,+\infty)$. We note that \begin{eqnarray*} f(\lambda_n)&=&(n^2+4n+4)\lambda_n^2-(2n^3+5n^2+4n+1)\lambda_n+n^4+2n^3-2n-1 \\ &=&n^3-3n-1 \\ &>& 0 \quad (n\geq 2). \end{eqnarray*} Then we have $f(x_n)>0$ 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 $r>2$, 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ \\ \hline $D_n$ & $1$ & $4$ & $28$ & $256$ & $2716$ & $31504$ & $387136$ & $4951552$ \\ \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_k0 \quad (k\ge 3) \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\mu_{k+1}&\le&\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}\\ &=&-\frac{2(3k^3+7k^2+4k+2)}{k(k+2)^3} <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}&\ge&\frac{16[(n-1)(n+2)^3-(n+1)^4]}{(n+1)(n+2)}\\ &=&\frac{16(n^3-8n-9)}{(n+1)(n+2)}\\ &>&0 \quad (n\ge 4). \end{eqnarray*} On the other hand, we observe that $(n+2)^2x_n-(n+1)^2x_{n+1}>0$ 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 $r>2$, 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} & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ &$8$\\ \hline $\tau_n$ & $1$ & $0$ & $1$ & $1$ & $6$ & $22$ & $130$ & $822$ & $6202$\\ \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}& \geq & 1-\frac{k+1}{2(k-1)} \\ &=&\frac{k-3}{2(k-1)} \geq 0 \end{eqnarray*} and \begin{eqnarray*} x_{k+1}-\lambda_{k+2}&\le&\frac{1}{k}-\frac{1}{2}\\ &=&-\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}&=&(n+2)^2x_n-(n+1)^3-\frac{(n+1)^3}{x_n}+\frac{n(n+1)^3}{2x_n x_{n-1}}\\ &\ge&(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}}\\ &\ge&\frac{3n^2-6n-22}{6} >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 $r>2$, 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. 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Zhao, The log-behavior of the Catalan-Larcombe-French sequence, \textit{Int. J. Number Theory.} {\bf 10} (2014), 177--182. \end{thebibliography} \bigskip \hrule \bigskip \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 \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}