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

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}

\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}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}}

\DeclareMathOperator{\Res}{Res}

\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 
			Reciprocal Sum of Palindromes
		}
		\vskip 1cm
		\large
		Phakhinkon Phunphayap and 
		Prapanpong Pongsriiam\footnote{Prapanpong Pongsriiam is the corresponding author.} \\
		Department of Mathematics \\
		Faculty of Science\\
		Silpakorn University\\
		Nakhon Pathom 73000 \\
		Thailand\\
		\href{mailto:phakhinkon@gmail.com}{\tt phakhinkon@gmail.com} \\
		\href{mailto:pongsriiam_p@silpakorn.edu}{\tt pongsriiam\_p@silpakorn.edu}, 
		\href{mailto:prapanpong@gmail.com}{\tt prapanpong@gmail.com} \\
	\end{center}
	
	\vskip .2 in
	
	\begin{abstract}
		A positive integer $n$ is a $b$-adic palindrome if the representation of $n$ in base $b$ reads the same backward as forward. Let $s_b$ be the reciprocal sum of all $b$-adic palindromes. In this article, we obtain upper and lower bounds, and an asymptotic formula for $s_b$. We also show that the sequence $(s_{b})_{b\geq 2}$ is strictly increasing and log-concave.
	\end{abstract}
	
	
	
	\section{Introduction}
	
	Let $n\geq 1$ and $b\geq 2$ be integers. We call $n$ a \textit{palindrome in base $b$} (or \textit{$b$-adic palindrome}) if the $b$-adic expansion of $n=(a_k a_{k-1}\cdots a_0)_b$ with $a_k\neq 0$ has the symmetric property $a_{k-i}=a_i$ for $0\leq i \leq \left\lfloor \frac{k}{2}\right\rfloor$. As usual, if we write a number without specifying the base, then it is always in base $10$. So, for example, $9 = (1001)_2 = (100)_3$ is a palindrome in bases $2$ and $10$ but not in base $3$.
	
In recent years, there has been an increasing interest in the importance of palindromes in mathematics \cite{Adam1, Adam3, Adam4, Fis, Luca, Roy}, theoretical computer science \cite{Amb, Brl, Fic}, and theoretical physics \cite{Dam, Hof}. There are also some discussions on the reciprocal sum of palindromes on the internet but as far as we are aware, our observation has not appeared in the literature. Throughout this article, we let $b\geq 2$, $s_b$ the reciprocal sum of all $b$-adic palindromes, and $s_{b,k}$ the reciprocal sum of all $b$-adic palindromes which have $k$ digits in their $b$-adic expansions. The set of all $b$-adic palindromes is infinite but quite sparse, so it is not difficult to see that $s_b$ converges. In fact, Shallit proposed the convergence of $s_b$ as a problem in the {\it Fibonacci Quarterly} in 1980 \cite{Sha, Sha2}.
	
In this article, we obtain upper and lower bounds for $s_b$, which enable us to show that $s_{b+1} &gt; s_b$ for all $b\geq 2$ and $s_{b}^2 -s_{b-1}s_{b+1} &gt; 0$ for all $b \geq 3$. That is, the sequence $(s_b)_{b\geq 2}$ is strictly increasing and log-concave. Furthermore, we give an asymptotic formula for $s_b$ of the form $s_b = g(b) + O(h(b))$ where the implied constant can be taken to be 1 and the order of magnitude of $h(b)$ is $\frac{\log b}{b^3}$ as $b \rightarrow \infty$. Our result $s_{b+1} &gt; s_b$ for all $b\geq 2$ also implies that if $b_1&gt;b_2\geq 2$ and if we use the logarithmic measure, then we can say that the palindromes in base $b_1$ occur more often than those in base $b_2$. On the other hand, if we use the usual counting measure, then we obtain from Pongsriiam and Subwattanachai's exact formula \cite{PSu} that the number of palindromes in different bases which are less than or equal to $N$ are not generally comparable. It seems that there are races between palindromes in different bases which may be similar to races between primes in different residue classes. We will get back to this problem in the near future.
	
The reciprocal sum of an integer sequence is also of general interest in mathematics and theoretical physics as proposed by Bayless and Klyve \cite{Bay}, and by Roggero, Nardelli, and Di Noto \cite{Rog}. See also the work of Nguyen and Pomerance \cite{NP} on the reciprocal sum of the amicable numbers, the preprint of Kinlaw, Kobayashi, and Pomerance \cite{KKP} on the reciprocal sum of the positive integers $n$ satisfying $\varphi (n) = \varphi (n+1)$, and the article by Lichtman \cite{Lich} on the reciprocal sum of primitive nondeficient numbers. In addition, Banks \cite{Bank}, Cilleruelo, Luca, and Baxter \cite{Cil}, and Rajasekaran, Shallit, and Smith \cite{Ra} have recently investigated some additive properties of palindromes while Banks, Hart, and Sakata \cite{Bank2} and Banks and Shparlinski \cite{Bank3} show some of their multiplicative properties. For more information concerning palindromes, we refer the reader to the entry \seqnum{A002113} in the On-Line Encyclopedia of Integer Sequences (OEIS) \cite{Slo}.
	
\section{Results}
	
Throughout this section, $a$, $c$, $m$, $n$, $k$, $\ell$ denote
positive integers, and $x$, $y$, $z$ denote positive real numbers.
Furthermore,
\begin{itemize}
\item $\left\lfloor x\right\rfloor$ is the greatest integer less than or equal to $x$;
\item $\left\lceil x\right\rceil$ is the least integer greater than or equal to $x$;
\item $\log x$ is the natural logarithm of $x$;
\item $x_b = \sum_{m=1}^{b-1} \frac{1}{m}$;
\item $y_b = \sum_{m=b}^{b^2 - 1} \frac{1}{m}$; and 
\item $z_{b} = \sum_{m=b+1}^{b^2} \frac{1}{m}$.
\end{itemize}
Note that $x_b = s_{b,1}$ and $x_b/(b+1) = s_{b,2}$ and that
	\begin{equation*}
		z_b - y_b = \frac{1}{b^2} - \frac{1}{b} = \frac{1 - b}{b^2}.
	\end{equation*}
	
	\begin{theorem}\label{rpthm0}
		We have
		\begin{equation*}
		\frac{y_b}{b} - \frac{x_b}{b^3} \leq s_{b,3} \leq \frac{y_b}{b} \quad \text{and} \quad  \frac{z_b}{b^{\left\lfloor \frac{k}{2} \right\rfloor}} \leq s_{b,k} \leq \frac{y_b}{b^{\left\lfloor \frac{k}{2} \right\rfloor}} \quad \text{for every $k \geq 4$}.
		\end{equation*}	
	\end{theorem}
	
	\begin{proof}
		We first consider the case $k = 3$. The $b$-adic palindromes which have $3$ digits are of the form $(aca)_b$ where $1\leq a \leq b-1$ and $0\leq c \leq b-1$. Since
		\begin{equation*}
		\left| \frac{1}{(aca)_b} - \frac{1}{(ac0)_b} \right| = \left| \frac{-a}{(ab^{2} + cb + a)(ab^{2} + cb)} \right| \leq \frac{a}{(ab^2)^2} = \frac{1}{ab^4},
		\end{equation*}
		we obtain
		\begin{equation}\label{rpthm0eq1}
		\frac{1}{(ac0)_b} - \frac{1}{ab^4} \leq \frac{1}{(aca)_b} \leq \frac{1}{(ac0)_b}.
		\end{equation}
		Observe that
		\begin{equation*}
		\sum_{\substack{1\leq a \leq b-1 \\ 0 \leq c \leq b-1}} \frac{1}{(ac0)_b} = \frac{1}{b} \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq c \leq b-1}} \frac{1}{ab + c} = \frac{y_b}{b} \quad \text{and} \quad \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq c \leq b-1}} \frac{1}{ab^4} = \frac{x_b}{b^3}.
		\end{equation*}
		So by summing \eqref{rpthm0eq1} over all $a = 1,2,\ldots , b-1$ and $c = 0,1,\ldots ,b-1$, we obtain the inequality $\frac{y_b}{b} - \frac{x_b}{b^3} \leq s_{b,3} \leq \frac{y_b}{b}$. For $k=4$, $1\leq a \leq b-1$, and $0 \leq c \leq b-1$, we have
		\begin{equation*}
		\frac{1}{b^2 (ab + c + 1)} = \frac{1}{(ac00)_b + b^2} \leq \frac{1}{(acca)_b} \leq \frac{1}{(ac00)_b} = \frac{1}{b^2 (ab + c)}.
		\end{equation*}
		Summing over all $a = 1, 2, \ldots, b-1$ and $c = 0, 1, \ldots, b-1$ leads to
		\begin{equation*}
		\frac{z_b}{b^2} = \frac{1}{b^2} \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq c \leq b-1}} \frac{1}{ab + c + 1} \leq s_{b,4} \leq \frac{1}{b^2} \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq c \leq b-1}} \frac{1}{ab + c} = \frac{y_b}{b^2}.
		\end{equation*}
		Let $k\geq 5$. The $b$-adic palindromes which have $k$ digits are of the form $(aa_1a_2\cdots a_{k-2}a)_{b}$ where $1\leq a\leq b-1$, $0\leq a_i \leq b-1$ for all $i\in \{1, 2, \ldots, k-2\}$ with the usual symmetric property on $a_i$. We fix $a$ and $a_1$ and count the number of palindromes in this form. There are $b$ choices for $a_2\in \{0, 1, 2, \ldots, b-1\}$ and so there is only 1 choice for $a_{k-3} = a_2$. Similarly, there are $b$ choices for $a_3$ and 1 choice for $a_{k-4}$. By continuing this counting, we see that the number of palindromes in this form (when $a$ and $a_1$ are already chosen) is equal to $b^{\left\lceil \frac{k-4}{2}\right\rceil}$. Therefore the reciprocal sum of such palindromes satisfies
		\begin{equation*}
		\sum_{a_2,\ldots,a_{k-3}}\frac{1}{(aa_1a_2\cdots a_{k-2}a)_b} \leq  \frac{b^{\left\lceil \frac{k-4}{2}\right\rceil}}{(aa_{1}0\cdots 0a_{1}a)_b} \leq \frac{b^{\left\lceil \frac{k-4}{2}\right\rceil}}{b^{k-2} (ab + a_1)} = \frac{1}{b^{\left\lfloor \frac{k}{2} \right\rfloor} (ab + a_1)},
		\end{equation*}
		where $a_2, \ldots, a_{k-3}$ run over all integers $0, 1, 2, \ldots, b-1$ with the symmetric condition of palindromes. Hence
		\begin{equation*}
		s_{b,k} = \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq a_1 \leq b-1}} \sum_{a_2,\ldots,a_{k-3}} \frac{1}{(aa_1a_2\cdots a_{k-2}a)_b} \leq   \sum_{\substack{1\leq a \leq b-1 \\ 0 \leq a_1 \leq b-1}} \frac{1}{b^{\left\lfloor \frac{k}{2} \right\rfloor} (ab + a_1)} = \frac{y_b}{b^{\left\lfloor \frac{k}{2} \right\rfloor}}.
		\end{equation*}
		Similarly, if $a$ and $a_1$ are fixed, then
		\begin{equation*}
		\sum_{a_2, \ldots, a_{k-3}} \frac{1}{(aa_1a_2\cdots a_{k-2}a)_{b}} \geq \frac{b^{\left\lceil \frac{k-4}{2}\right\rceil}}{ab^{k-1} + a_{1}b^{k-2} + b^{k-2}} = \frac{1}{b^{\left\lfloor \frac{k}{2} \right\rfloor}  (ab + a_1 + 1)}.
		\end{equation*}
		Summing the above over all $a = 1, 2, \ldots, b-1$ and $a_1 = 0, 1, \ldots, b-1$, we obtain the desired lower bound for $s_{b,k}$. This completes the proof.
	\end{proof}
	
	\begin{theorem}\label{rpthm1}
		For every $b, \ell \geq 2$, we have
		\begin{equation*}
		\left( \frac{b+2}{b+1} \right)x_{b} + \sum_{k=3}^{2\ell - 1} s_{b,k} + \frac{2z_{b}}{(b-1)b^{\ell - 1}} \leq s_{b} \leq \left( \frac{b+2}{b+1} \right)x_{b} + \sum_{k=3}^{2\ell - 1} s_{b,k} + \frac{2y_{b}}{(b-1)b^{\ell -1}}.
		\end{equation*}
		In particular,
		\begin{equation}\label{rpthm1eq1}
		\left( \frac{b+2}{b+1} \right) x_{b} + \frac{y_b}{b} - \frac{x_b}{b^3} + \frac{2z_b}{b(b-1)} \leq s_{b} \leq \left( \frac{b+2}{b+1} \right) x_{b} + \left( \frac{1}{b} + \frac{2}{b(b-1)} \right) y_{b}.
		\end{equation}
	\end{theorem}
	
	\begin{proof}
		For simplicity, we write $x$, $y$, $z$ instead of $x_b$, $y_b$, $z_b$, respectively. We consider $s_{b,k}$ for each $k$ as follows. Obviously $s_{b,1} = 1 + \frac{1}{2} + \cdots + \frac{1}{b-1} = x$. For $k=2$, $s_{b,k}$ is 
		\begin{equation*}
		\sum_{a=1}^{b-1} \frac{1}{(aa)_b} = \sum_{a=1}^{b-1} \frac{1}{a(b+1)} = \frac{x}{b+1}.
		\end{equation*}
		By writing $s_b = x + \frac{x}{b+1} + \sum_{k=3}^{2\ell -1} s_{b,k}  + \sum_{k=2\ell}^\infty s_{b,k}$ and applying Theorem \ref{rpthm0}, we obtain
		\begin{equation*}
		s_b \leq \frac{b+2}{b+1}x + \sum_{k=3}^{2\ell -1} s_{b,k} +  \sum_{k=2\ell}^\infty \frac{y}{b^{\left\lfloor \frac{k}{2} \right\rfloor}} = \frac{b+2}{b+1}x  + \sum_{k=3}^{2\ell -1} s_{b,k} + \frac{2y}{(b-1)b^{\ell - 1}}.
		\end{equation*}
		Similarly,
		\begin{equation*}
		s_b \geq \frac{b+2}{b+1}x + \sum_{k=3}^{2\ell - 1} s_{b,k} + \sum_{k=2\ell}^\infty \frac{z}{b^{\left\lfloor  \frac{k}{2}\right\rfloor}} = \left( \frac{b+2}{b+1} \right)x + \sum_{k=3}^{2\ell - 1} s_{b,k} + \frac{2z}{(b-1)b^{\ell -1}}.		
		\end{equation*}
		This proves the first part of this theorem. The second part follows from Theorem \ref{rpthm0} and the substitution $\ell =2$ in the first part.
	\end{proof}
	
	\begin{theorem}\label{rpthm2}
		The sequence $(s_b)_{b \geq 2}$ is strictly increasing.
	\end{theorem}
	
	\begin{proof}
		We first verify that $s_{b+1} &gt; s_b$ for $2 \leq b \leq 16$. Myers gives the decimal expansion of $s_2$ in the entry \seqnum{A244162} in
		the OEIS \cite{Slo}. Myers also describe the algorithm in his calculation, which can be found in the web page \cite{Mye}. So we know that $s_2 &lt; 2.3787957$. Alternatively, substituting $\ell = 3$ and $b = 2, 3$ in Theorem \ref{rpthm1} and running the computation in a computer, we obtain $2.32137259 \leq s_2 \leq 2.44637260$ and $2.60503980 \leq s_3 \leq 2.62973117$, which implies that $s_2 &lt; s_3$. Similarly, we apply \eqref{rpthm1eq1} to obtain upper and lower bounds for $s_b$ and we see that $s_b &lt; s_{b+1}$ for $3 \leq b \leq 16$. So we assume throughout that $b\geq 16$. We first observe that
		\begin{equation*}
		y_b = z_b + \frac{1}{b} - \frac{1}{b^2} \quad \text{and} \quad \frac{y_b}{b} - \frac{z_b}{b} = \frac{b-1}{b^3} \geq \frac{x_b}{b^3}.
		\end{equation*}
		Therefore $\frac{y_b}{b} - \frac{x_b}{b^3} \geq \frac{z_b}{b}$ for all $b \geq 2$. So the term $\frac{y_b}{b} - \frac{x_b}{b^3}$ in \eqref{rpthm1eq1} can be replaced by $\frac{z_b}{b}$. Therefore we obtain by \eqref{rpthm1eq1} that $s_{b+1} - s_b$ is larger than
		\begin{equation}\label{rpthm2eq1}
		\left( \frac{b+3}{b+2} \right)x_{b+1} - \left( \frac{b+2}{b+1} \right)x_b + \left( \frac{1}{b+1} + \frac{2}{(b+1)b} \right)z_{b+1} - \left( \frac{1}{b} + \frac{2}{b(b-1)} \right)y_b.
		\end{equation}
		In addition, $z_{b+1} - z_b$ is equal to
		\begin{equation*}
		- \frac{1}{b+1} + \sum_{m = b^2 + 1}^{b^2 + 2b + 1} \frac{1}{m} \geq - \frac{1}{b+1} + \frac{2b + 1}{b^2 + 2b + 1} = \frac{b}{(b+1)^2} &gt; 0.
		\end{equation*}
		Since $x_{b+1} = x_b + \frac{1}{b}$ and $z_{b+1} &gt; z_b = y_b - \frac{1}{b} + \frac{1}{b^2}$, we obtain from \eqref{rpthm2eq1} that $s_{b+1} - s_b$ is larger than or equal to
		\begin{align*}
		&amp; \left( \frac{b+3}{b+2} \right)\left( \frac{1}{b} \right) + x_{b}\left( \frac{b+3}{b+2} - \frac{b+2}{b+1} \right) + y_{b} \left( \frac{1}{b+1} + \frac{2}{b(b+1)} - \frac{1}{b} - \frac{2}{b(b-1)} \right)\\
		&amp;\phantom{\geq} + \left( \frac{1}{b^2} - \frac{1}{b} \right) \left( \frac{1}{b+1} + \frac{2}{b(b+1)} \right)\\
		&amp;= \frac{1}{b} + \frac{1}{b(b+2)} - \frac{x_b}{(b+1)(b+2)} - \frac{(b+3)y_b}{b(b-1)(b+1)} - \frac{(b-1)(b+2)}{b^3 (b+1)}.
		\end{align*}
		Recall that if $a$ and $b$ are integers, $a &lt; b$, and $f$ is monotone on $[a,b]$, then 
		\begin{equation}\label{rpthm2eq2}
		\min\{ f(a),f(b) \} \leq \sum_{n=a}^b f(n) - \int_{a}^b f(t) dt \leq \max\{f(a),f(b)\}.
		\end{equation}
		From \eqref{rpthm2eq2}, we obtain
		\begin{align*}
		x_b &amp;= \sum_{m=1}^{b-1} \frac{1}{m} \leq 1 + \log (b-1) \leq \frac{3}{2}\log b,\\
		y_b &amp;= - \frac{1}{b-1} + \sum_{m=b-1}^{b^2 - 1} \frac{1}{m} \leq \log (b+1) \leq \frac{5}{4}\log b.
		\end{align*}
		In addition, it is straightforward to verify that
		\begin{equation*}
		\frac{1}{b(b+2)} - \frac{(b-1)(b+2)}{b^3 (b+1)} &gt; - \frac{1}{b^2}.
		\end{equation*}
		Therefore $s_{b+1} - s_b$ is larger than
		\begin{align}\label{rpthm2eq3}
		\frac{1}{b} - \frac{3\log b}{2(b+1)(b+2)} - \frac{5(b+3)\log b}{4b(b-1)(b+1)} - \frac{1}{b^2} &gt; \frac{1}{b} - \frac{3\log b}{2b^2} - \frac{3\log b}{2b^2} - \frac{1}{b^2} = \frac{1}{b} - \frac{1}{b^2} - \frac{3\log b}{b^2}.
		\end{align}
		Observe that the function $x\mapsto \frac{\log x}{x}$ is decreasing on $[3,\infty)$. Since $b\geq 16$, we obtain
		\begin{equation*}
		\frac{3\log b}{b} \leq \frac{3\log 16}{16} &lt;\frac{7}{10} \quad \text{and} \quad \frac{1}{b} &lt; \frac{1}{10}.
		\end{equation*}
		Hence we obtain from \eqref{rpthm2eq3} that
		\begin{equation*}
		s_{b+1} - s_b &gt; \frac{1}{b} - \frac{1}{b^2} - \frac{3\log b}{b^2} &gt; \frac{1}{b} - \frac{1}{10b} - \frac{7}{10b} = \frac{1}{5b} &gt; 0.
		\end{equation*}
		This completes the proof.
	\end{proof}
	
	Recall that if we write $f(b) = g(b) + O^\ast (h(b))$, then it means that $f(b) = g(b) + O(h(b))$ and the implied constant can be taken to be $1$. In addition, $f(b) = g(b) + \Omega_+ (h(b))$ means $\limsup_{b \rightarrow \infty} \frac{f(b) - g(b)}{h(b)} &gt; 0$. From this point on, we use \eqref{rpthm2eq2} without further reference.
	
	\begin{theorem}\label{rpthm3}
		Uniformly for $b \geq 2$,
		\begin{equation*}
		s_b = \left( \frac{b+2}{b+1} \right)x_b + \left( \frac{1}{b} + \frac{2}{b^2} \right)y_b + O^\ast \left( \frac{5\log b}{b^3} \right).
		\end{equation*}
		This estimate is sharp in the sense that $O^\ast \left( \frac{5\log b}{b^3} \right)$ can be replaced by $\Omega_+ \left( \frac{\log b}{b^3} \right)$. 
	\end{theorem}
	
	\begin{proof}
		Let $g(b) = \left( \frac{b+2}{b+1} \right)x_b + \left( \frac{1}{b} + \frac{2}{b^2} \right)y_b$ be the main term above. Since $y_b \leq \log (b+1)$, we obtain by \eqref{rpthm1eq1} that
		\begin{equation*}
		s_{b} - g(b) \leq \frac{2y_b}{b^2 (b-1)} \leq \frac{2\log (b+1)}{b^2 (b-1)}.
		\end{equation*} 
		If $b = 2$, then it is easy to check that $\frac{2y_b}{b^2 (b-1)} = \frac{5}{12} &lt; \frac{5\log b}{b^3}$. If $b \geq 3$, then we assert that $\frac{2 \log (b+1)}{b^2 (b-1)} \leq \frac{5\log b}{b^3}$. To verify this assertion, we observe that it is equivalent to $\left( \frac{5}{2}\log b \right)\left( \frac{b-1}{b} \right) \geq \log (b+1)$. Since $b \geq 3$, we obtain
		\begin{equation*}
		\left( \frac{5}{2}\log b \right)\left( \frac{b-1}{b} \right) \geq \frac{5}{3} \log b \geq \log 2 + \log b = \log (2b) \geq \log (b+1), \ \text{as desired.}
		\end{equation*} So in any case,
		\begin{equation}\label{rpthm3eq1}
		s_b - g(b) \leq \frac{5 \log b}{b^3}.
		\end{equation}
		We also obtain by \eqref{rpthm1eq1} that $s_b - g(b)$ is larger than or equal to 
		\begin{equation}\label{rpthm3eq2}
		\frac{2z_b}{b(b-1)} - \frac{2y_b}{b^2} - \frac{x_b}{b^3} = \frac{2z_b}{b(b-1)} - \frac{2\left( z_b + \frac{1}{b} - \frac{1}{b^2} \right)}{b^2} - \frac{x_b}{b^3} = \frac{2z_b}{b^2 (b-1)} - \frac{x_b}{b^3} - \frac{2}{b^3} + \frac{2}{b^4}.
		\end{equation} 
		We have
		\begin{align}
		x_b &amp;= \sum_{m=1}^{b-1} \frac{1}{m} \leq 1 + \log (b-1) \leq 2\log b, \label{rpthm3eq3} \\
		z_b &amp;= \sum_{m=b}^{b^2} \frac{1}{m} - \frac{1}{b} \geq \int_{b}^{b^2} \frac{1}{t} dt + \frac{1}{b^2} - \frac{1}{b} \geq \log b - \frac{1}{b} \geq \frac{\log b}{4}. \label{rpthm3eq4}
		\end{align}
		Therefore \eqref{rpthm3eq2} implies that
		\begin{equation}\label{rpthm3eq5}
		s_b - g(b) \geq \frac{\log b}{2b^{2}(b-1)} - \frac{2\log b}{b^3} - \frac{2}{b^3} \geq \frac{\log b}{2b^3} - \frac{2\log b}{b^3} - \frac{3\log b}{b^3} &gt; - \frac{5\log b}{b^3}.
		\end{equation}
		By \eqref{rpthm3eq1} and \eqref{rpthm3eq5}, we obtain $|s_b - g(b)| \leq \frac{5\log b}{b^3}$. This proves the first part of this theorem. For the $\Omega_+$ result, we only need to observe that as $b \rightarrow \infty$, \eqref{rpthm3eq2} and the inequalities $x_b\leq 1 + \log (b-1)$ and $z_b \geq \log b - \frac{1}{b}$ given in \eqref{rpthm3eq3} and \eqref{rpthm3eq4} imply that
		\begin{equation*}
		s_b - g(b) \geq \frac{2\left( \log b - \frac{1}{b} \right)}{b^2 (b-1)} - \frac{1 + \log (b-1)}{b^3} - \frac{2}{b^3} + \frac{2}{b^4} &gt; \frac{3\log b}{2b^{2}(b-1)} - \frac{11\log b}{10b^3} - \frac{2}{b^3} + \frac{2}{b^4},
		\end{equation*} 
		so
		\begin{equation*} 
		\limsup_{b \rightarrow \infty} \frac{s_b - g(b)}{\left( \frac{\log b}{b^3} \right)} \geq \frac{3}{2} - \frac{11}{10} &gt; 0.
		\end{equation*}
		This completes the proof.
	\end{proof}
	
	Recall that by applying Euler-Maclaurin summation formula, we get
	\begin{equation}\label{rpthm3eq6}
	\sum_{m\leq n} \frac{1}{m} = \log n +\gamma + \frac{1}{2n} - \frac{1}{12n^2} + \frac{\theta_n}{60n^4}, 
	\end{equation}
	where $\gamma$ is Euler's constant and $\theta_n \in [0,1]$. The calculation of \eqref{rpthm3eq6} can be found in Tenenbaum \cite[p.\ 6]{Ten}. From this, we obtain another form of Theorem \ref{rpthm3} as follows.
	
	\begin{theorem}\label{rpthm4}
		Uniformly for $b \geq 2$, 
		\begin{equation}\label{rpthm4eq1}
		s_b = \log b + \gamma + \left( \frac{1}{b} + \frac{1}{b+1} \right)\log b + \frac{\gamma}{b+1} - \frac{1}{2b} + \frac{2\log b}{b^2} - \frac{1}{12b(b+1)} + O^\ast\left( \frac{6\log b}{b^3} \right). 
		\end{equation}
		This estimate is sharp in the sense that $O^\ast \left( \frac{6\log b}{b^3} \right)$ is also $\Omega_+ \left( \frac{\log b}{b^3} \right)$.
	\end{theorem}
	
	\begin{proof}
		By \eqref{rpthm3eq6}, we have
		\begin{align*}
		x_b &amp;= \sum_{m \leq b} \frac{1}{m} - \frac{1}{b} = \log b + \gamma - \frac{1}{2b} - \frac{1}{12b^2} + \frac{\theta_b}{60b^4},\\
		z_b &amp;= \sum_{m \leq b^2} \frac{1}{m} - \sum_{m \leq b} \frac{1}{m} = \left( \log b^2 + \gamma + \frac{1}{2b^2}  - \frac{1}{12b^4} + \frac{\theta_{b^2}}{60b^8} \right) - \left( \log b + \gamma + \frac{1}{2b} - \frac{1}{12b^2} + \frac{\theta_b}{60b^4} \right) \\
		&amp;= \log b - \frac{1}{2b} + \frac{7}{12b^2} - \frac{5 + \theta_b}{60b^4} + \frac{\theta_{b^2}}{60b^8}, \\
		y_b &amp;= z_b + \frac{1}{b} - \frac{1}{b^2} = \log b + \frac{1}{2b} - \frac{5}{12b^2} - \frac{5 + \theta_b}{60b^4} + \frac{\theta_{b^2}}{60b^8}.
		\end{align*}
		Writing $\frac{b+2}{b+1} = 1 + \frac{1}{b+1}$ and substituting $x_b$ and $y_b$ in Theorem \ref{rpthm3}, we obtain
		\begin{equation}\label{rpthm4eq2}
		s_b = h(b) + h_1 (b) + O^\ast\left( \frac{5\log b}{b^3} \right)
		\end{equation}
		where $h(b)$ is the main term given in \eqref{rpthm4eq1} and
		\begin{equation*}
		h_1 (b) = \frac{11b^2 - 3b - 10}{12b^4 (b+1)} + \frac{\theta_b}{60b^4} + \frac{\theta_b}{60b^4 (b+1)} + \frac{\theta_{b^2}}{60b^9} + \frac{\theta_{b^2}}{30b^{10}} - \frac{5 + \theta_b}{60b^5} - \frac{5 + \theta_b}{30b^6}.
		\end{equation*}
		It is not difficult to see that $h_1 (b) \geq 0$ and 
		\begin{equation}\label{rpthm4eq3}
		h_1 (b) \leq \frac{11b^2}{12b^4 (b+1)} + \frac{1}{60b^4} + \frac{1}{60b^4 (b+1)} \leq \frac{11\log b}{12b^3} + \frac{\log b}{60b^3} + \frac{\log b}{60b^3} \leq \frac{\log b}{b^3}.
		\end{equation}
		Therefore \eqref{rpthm4eq2} implies that $s_b = h(b) + O^\ast \left( \frac{6\log b}{b^3} \right)$, which is the same as \eqref{rpthm4eq1}. In addition, by the first inequality given in \eqref{rpthm4eq3}, we see that $h_1 (b) \ll \frac{1}{b^3}$. Since $O^\ast\left( \frac{5\log b}{b^3} \right)$ in \eqref{rpthm4eq2} is $\Omega_+ \left( \frac{\log b}{b^3} \right)$ and $h_1 (b) \ll \frac{1}{b^3}$, $h_1 (b) + O^* \left( \frac{5\log b}{b^3} \right)$ in \eqref{rpthm4eq2} is $\Omega_+ \left( \frac{\log b}{b^3} \right)$. This completes the proof.
	\end{proof}
	
	\begin{corollary}\label{rpcor4}
		The sequence $(s_b)_{b\geq 2}$ diverges to $+\infty$ and the sequence $(s_b - s_{b-1})_{b\geq 3}$ converges to zero as $b \rightarrow \infty$.
	\end{corollary}
	
	\begin{proof}
		The first assertion follows immediately from Theorem \ref{rpthm4}. Recall that $\log (b-1) = \log b + O\left(\frac{1}{b}\right)$. So we obtain by Theorem \ref{rpthm4} that as $b \rightarrow \infty$,
		\begin{equation*}
		0 &lt; s_b - s_{b-1} = \log b - \log (b-1) + O\left( \frac{\log b}{b} \right) \ll \frac{\log b}{b},
		\end{equation*}
		which implies our assertion.
	\end{proof}
	
	Recall that a sequence $(a_n)_{n\geq 0}$ is said to be log-concave if $a_n ^2 - a_{n-1}a_{n+1} &gt; 0$ for every $n\geq 1$ and is said to be log-convex if $a_n ^2 - a_{n-1}a_{n+1} &lt; 0$ for every $n\geq 1$. For a survey article concerning the log-concavity and log-convexity of sequences, we refer the reader to Stanley \cite{Stan}. See also Pongsriiam \cite{Pong} for some combinatorial sequences which are log-concave or log-convex, and some open problems concerning the log-properties of a certain sequence.
	
	\begin{theorem}\label{rpthm6}
		The sequence $(s_b)_{b \geq 2}$ is log-concave.
	\end{theorem}
	
	\begin{proof}
		For each $b=2, 3, \ldots, 15$, we use Theorem \ref{rpthm1} with $\ell = 4$ to get an upper bound $C_b$ and a lower bound $D_b$ for $s_b$. In addition, for each $b\geq 13$, let $U_b$ and $L_b$ be the upper and lower bounds of $s_b$ given in \eqref{rpthm1eq1}, respectively. Then 
		\begin{equation*}
		s_{b}^2 - s_{b-1}s_{b+1} &gt; D_{b}^2 - C_{b-1}C_{b+1} \text{ for $3\leq b \leq 14$ and } s_{b}^2 - s_{b-1}s_{b+1} &gt; L_{b}^2 - U_{b-1}U_{b+1} \text{ for $b\geq 14$.}
		\end{equation*}
		We use MATLAB to check that $D_{b}^2 - C_{b-1}C_{b+1} &gt; 0$ for $3\leq b \leq 14$ and $L_{b}^2 - U_{b-1}U_{b+1} &gt; 0$ for $14 \leq b \leq 1500$. So $s_{b}^2 - s_{b-1}s_{b+1} &gt;0$ for $3\leq b \leq 1500$. So we assume throughout that $b &gt; 1500$. Then
		\begin{align}\label{rpthm5eq1}
		U_{b-1}U_{b+1} &amp;= \frac{(b+1)(b+3)}{b(b+2)}x_{b-1}x_{b+1} + \frac{b+2}{(b-2)(b-1)(b+1)}y_{b-1}y_{b+1} \nonumber\\
		&amp;\phantom{=} + \frac{b(b+3)}{(b-2)(b-1)(b+2)}x_{b+1}y_{b-1} + \frac{b+2}{b^2}x_{b-1}y_{b+1} \nonumber\\
		&amp;= A_1 + A_2 + A_3 + A_4, \text{ say.}
		\end{align}
		Since 
		\begin{align*}
		z_b &amp;= y_b + \frac{1}{b^2} - \frac{1}{b}, \quad b^4 + 2b^3 - b - 1 \geq b^4 + 2b^3 - 2b - 1 = (b-1)(b+1)^3, \; \text{and} \\
		L_{b} &amp;= \frac{b^4 + 2b^3 - b - 1}{b^3 (b+1)}x_b + \frac{y_b}{b} + \frac{2z_b}{b(b-1)},
		\end{align*}
		we obtain
		\begin{equation*}
		L_b = \frac{b^4 + 2b^3 - b - 1}{b^3(b+1)}x_b + \frac{b+1}{b(b-1)}y_b - \frac{2}{b^3} \geq \frac{(b-1)(b+1)^2}{b^3}x_b + \frac{b+1}{b(b-1)}y_b - \frac{2}{b^3}.
		\end{equation*}
		Therefore $L_{b}^2$ is larger than or equal to
		\begin{align}\label{rpthm5eq2}
		&amp;\frac{(b-1)^2 (b+1)^4}{b^6}x_{b}^2 + \frac{(b+1)^2}{b^2 (b-1)^2}y_{b}^2 + \frac{4}{b^6} + \frac{2(b+1)^3}{b^4}x_{b}y_{b} - \frac{4(b-1)(b+1)^2}{b^6}x_{b} - \frac{4(b+1)}{b^4 (b-1)}y_b \nonumber\\
		&amp;\geq \frac{(b-1)^2 (b+1)^4}{b^6}x_{b}^2 + \frac{(b+1)^2}{b^2 (b-1)^2}y_{b}^2 + \frac{2(b+1)^3}{b^4}x_{b}y_{b} - \frac{4(b-1)(b+1)^2}{b^6}x_{b} - \frac{4(b+1)}{b^4 (b-1)}y_b \nonumber\\
		&amp;= B_1 + B_2 + B_3 - B_4 - B_5, \text{ say.}
		\end{align}
		In addition, we see that 
		\begin{align*}
		y_{b-1} &amp;= y_b + \frac{1}{b-1} - \sum_{m=(b-1)^2}^{b^2 - 1} \frac{1}{m} \leq y_b + \frac{1}{b-1} - \frac{2b - 1}{b^2 - 1} \leq y_b - \frac{b-2}{b^2},\\
		y_{b+1} &amp;= y_b - \frac{1}{b} + \sum_{m=b^2}^{b^2 + 2b} \frac{1}{m} \leq y_{b} - \frac{1}{b} + \frac{2b+1}{b^2} = y_b + \frac{b+1}{b^2},\\
		x_{b-1} &amp;= x_{b} - \frac{1}{b-1}, \quad \text{and} \quad x_{b+1} = x_b + \frac{1}{b}.
		\end{align*}
		From these, we obtain the following inequalities:
		\begin{align*}
		A_1 &amp;= \frac{(b+1)(b+3)}{b(b+2)}x_{b}^2 - \frac{(b+1)(b+3)}{b^2 (b-1)(b+2)}x_b - \frac{(b+1)(b+3)}{b^2 (b-1)(b+2)} \\
		&amp;\leq \frac{(b+1)(b+3)}{b(b+2)}x_{b}^2 - \frac{(b+1)(b+3)}{b^2 (b-1)(b+2)}x_b - \frac{1}{b^2}, \\
		A_2 &amp;\leq \frac{b+2}{(b-2)(b-1)(b+1)}y_{b}^2 + \frac{3(b+2)}{b^2 (b-2)(b-1)(b+1)}y_b - \frac{b+2}{b^4 (b-1)}, \\ 
		A_3 &amp;\leq \frac{b(b+3)}{(b-2)(b-1)(b+2)}x_{b}y_{b} + \frac{b+3}{(b-2)(b-1)(b+2)}y_b - \frac{b+3}{b(b-1)(b+2)}x_b - \frac{b+3}{b^2 (b-1)(b+2)}\\
		&amp;\leq \frac{b(b+3)}{(b-2)(b-1)(b+2)}x_{b}y_{b} + \frac{b+3}{(b-2)(b-1)(b+2)}y_b - \frac{b+3}{b(b-1)(b+2)}x_b, \\
		A_4 &amp;\leq \frac{b+2}{b^2}x_{b}y_{b} - \frac{b+2}{b^2 (b-1)}y_b + \frac{(b+1)(b+2)}{b^4}x_b - \frac{(b+1)(b+2)}{b^4 (b-1)} \\
		&amp;\leq \frac{b+2}{b^2}x_{b}y_{b} - \frac{b+2}{b^2 (b-1)}y_b + \frac{(b+1)(b+2)}{b^4}x_b.
		\end{align*}
		Since $b &gt; 1500$, it is not difficult to verify that 
		\begin{align*}
		B_1 - B_4 - A_1 &amp;\geq - \frac{(b+1)(6b^3 + 3b^2 - 3b - 2)}{b^6 (b+2)}x_{b}^2 + \frac{b^6 - 5b^4 + 8b^3 + 16b^2 -4b -8}{b^6 (b-1)(b+2)}x_b + \frac{1}{b^2}\\
		&amp;\geq - \frac{7}{b^3}x_{b}^2 + \frac{1}{b^2}, \\
		B_2 - B_5 - A_2 &amp; \geq - \frac{b^2 + 5b + 2}{b^2 (b-2)(b-1)^2 (b+1)}y_{b}^2 - \frac{7b^3 + 6b^2 - 12b - 8}{b^4 (b-2)(b-1)(b+1)}y_b + \frac{b+2}{b^4 (b-1)} \\
		&amp;\geq - \frac{1}{b^3}y_{b}^2 - \frac{1}{b^3}y_b,\\
		B_3 - A_3 - A_4 &amp;\geq - \frac{2(b^4 + 8b^3 + 5b^2 - 8b - 4)}{b^4 (b-2)(b-1)(b+2)}x_{b}y_{b} - \frac{b^2 + 4b + 8}{b^2 (b-2)(b-1)(b+2)}y_b \\
		&amp;\phantom{\geq} - \frac{b^3 + 3b^2 - 4b - 4}{b^4 (b-1)(b+2)}x_b \\
		&amp;\geq -\frac{4}{b^3}x_{b}y_{b} - \frac{3}{b^3}y_b - \frac{3}{b^3}x_b.
		\end{align*}
		From \eqref{rpthm5eq1}, \eqref{rpthm5eq2}, and the above inequalities, we obtain
		\begin{equation*}
		L_{b}^2 - U_{b-1}U_{b+1} \geq \frac{1}{b^2} - \frac{7}{b^3}x_{b}^2 - \frac{1}{b^3}y_{b}^2 - \frac{4}{b^3}x_{b}y_{b} - \frac{4}{b^3}y_b - \frac{3}{b^3}x_b.
		\end{equation*}
		Since $x_b \leq \frac{4}{3}\log b$ and $y_b \leq \frac{5}{4}\log b$, 
		\begin{equation*}
		L_{b}^2 - U_{b-1}U_{b+1} \geq \frac{1}{b^2} - \frac{2977}{144}\frac{(\log b)^2}{b^3} - \frac{9\log b}{b^3} \geq \frac{b-23(\log b)^2}{b^3}.
		\end{equation*}
		Observe that the function $x \mapsto x - 23 (\log x)^2$ is strictly increasing on $[300,\infty)$. Since $b \geq 1500$,
		\begin{equation*}
		b-23(\log b)^2 \geq 1500 - 23(\log 1500)^2 &gt; 0.
		\end{equation*}
		Therefore $s_{b}^2 - s_{b-1}s_{b+1} \geq L_{b}^2 - U_{b-1}U_{b+1} &gt; 0$. Hence $(s_b)_{b\geq 2}$ is log-concave, as desired.
	\end{proof}
	
	\begin{remark}
		We have uploaded the numerical data on the computation of $s_b$ in the second author's ResearchGate account \cite{Pong2} which are freely downloadable by everyone.
	\end{remark}
	
	\section{Acknowledgments}
	
	We are grateful to the referee for pointing out References \cite{KKP, Lich, NP} and for his/her suggestions which improve the presentation of this article. We also thank the Editor-in-Chief for letting us know about the problem in the {\it Fibonacci Quarterly} \cite{Sha, Sha2} and the algorithm of Myers \cite{Mye}. Furthermore, we would like to thank Wannarut Rungrottheera for her support. Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). Prapanpong Pongsriiam received financial support jointly from the Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040. Prapanpong Pongsriiam is the corresponding author.
	
	\begin{thebibliography}{99}
		\bibitem{Adam1} B. Adamczewski and J.-P. Allouche, Reversal and palindromes in continued fractions, \emph{Theoret. Comput. Sci.} \textbf{380} (2007), 220--237.
		
		\bibitem{Adam3} B. Adamczewski and Y. Bugeaud, Palindromic continued fractions, \emph{Ann. Inst. Fourier} \textbf{57} (2007), 1557--1574.
		
		\bibitem{Adam4} B. Adamczewski and Y. Bugeaud, Transcendence measure for continued fractions involving repetitive or symmetric patterns, \emph{J. Eur. Math. Soc.} \textbf{12} (2010), 883--914.
		
		\bibitem{Amb} P. Ambro\v{z}, C. Frougny, Z. Mas\'{a}kov\'{a}, and E. Pelantov\'{a}, Palindromic complexity of infinite words associated with simple Parry numbers, \emph{Ann. Inst. Fourier} \textbf{56} (2006), 2131--2160.
		
		\bibitem{Bank} W. D. Banks, Every natural number is the sum of forty-nine palindromes, \emph{Integers} \textbf{16} (2016), Paper No.~A3, 1--9. 
		
		\bibitem{Bank2} W. D. Banks, D. N. Hart, and M. Sakata, Almost all palindromes are composite, \emph{Math. Res. Lett.} \textbf{11} (2004), 853--868.
		
		\bibitem{Bank3}  W. D. Banks and I. Shparlinski, Prime divisors of palindromes, \emph{Period. Math. Hungar.} \textbf{51} (2005), 1--10.
		
		\bibitem{Bay} J. Bayless and D. Klyve, Reciprocal sums as a knowledge metric: theory, computation, and perfect numbers, \emph{Amer. Math. Monthly} \textbf{120} (2013), 822--831.
		
		\bibitem{Brl} S. Brlek, S. Hamel, M. Nivat, and C. Reutenauer, On the palindromic complexity of infinite words, \emph{Internat. J. Found. Comput. Sci.} \textbf{15} (2004), 293--306.
		
		\bibitem{Cil} J. Cilleruelo, F. Luca, and L. Baxter, Every positive integer is a sum of three palindromes, \emph{Math. Comp.} \textbf{87} (2018), 3023--3055.
		
		\bibitem{Dam} D. Damanik and L. Q. Zamboni, Combinatorial properties of Arnoux-Rauzy subshifts and applications to Schr\"{o}dinger operators, \emph{Rev. Math. Phys.} \textbf{15} (2003), 745--763.
		
		\bibitem{Fic} G. Ficia and L. Q. Zamboni, On the least number of palindromes contained in an infinite word, \emph{Theoret. Comput. Sci.} \textbf{481} (2013), 1--8.
		
		\bibitem{Fis} S. Fischler, Palindromic prefixes and diophantine approximation, \emph{Monatsh. Math.} \textbf{151} (2007), 11--37.
		
		\bibitem{Hof} A. Hof, O. Knill, and B. Simon, Singular continuous spectrum for palindromic schr\"{o}dinger operators, \emph{Comm. Math. Phys.} \textbf{174} (1995), 149--159.
		
		\bibitem{KKP} P. Kinlaw, M. Kobayashi, and C. Pomerance, On the equation $\varphi (n) = \varphi (n+1)$, preprint, 2019. Available at \url{https://math.dartmouth.edu/~carlp/Updated_Euler_Function_Project.pdf}.
		
		\bibitem{Lich} J. Lichtman, The reciprocal sum of primitive nondeficient numbers, \emph{J. Number Theory}, \textbf{191} (2018), 104--118.
		
		\bibitem{Luca} F. Luca, Palindromes in various sequences, \emph{Gac. R. Soc. Mat. Esp.} \textbf{20} (2017), 49--56. 
		
		\bibitem{Mye} J. S. Myers, Palindrome reciprocal sums, \url{http://list.seqfan.eu/pipermail/seqfan/2014-June/013186.html}.
				
		\bibitem{NP} H. M. Nguyen and C. Pomerance, The reciprocal sum of the amicable numbers, \emph{Math. Comp.} \textbf{88} (2019), 1503--1526.
	
\bibitem{Pong2} P. Phunphayap and P. Pongsriiam, Estimates for the reciprocal sum of $b$-adic palindromes, preprint, 2019. Available at \url{https://www.researchgate.net/publication/336304247_Estimates_for_the_Reciprocal_Sum_of_b-adic_Palindromes}.

\bibitem{Pong} P. Pongsriiam, Local behaviors of the number of relatively prime sets, \emph{Int. J. Number Theory} \textbf{12} (2016), 1575--1593.

\bibitem{PSu} P. Pongsriiam and K. Subwattanachai, Exact formulas for the number of palindromes up to a given positive integer, \emph{Int. J. Math. Comput. Sci.} \textbf{14} (2019), 27--46.
		
\bibitem{Ra} A. Rajasekaran, J. Shallit, and T. Smith, Sums of
palindromes: an approach via automata, in \emph{35th Symposium on
Theoretical Aspects of Computer Science} (STACS 2018), Article no.~54,
pp.~54:1--54:12.

		\bibitem{Rog} P. Roggero, M. Nardelli, and F. Di Noto,
		Sum of the reciprocals of famous series: mathematical
		connections with some sectors of theoretical physics
		and string theory, preprint, 2016. Available at
		\url{http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/nardelli2017a.pdf}.

		\bibitem{Roy} D. Roy, Approximation to real numbers
		by cubic algebraic integers, II, \emph{Ann. of Math.}
		\textbf{158} (2003), 1081--1087.

		\bibitem{Sha} J. Shallit, Problem B-441, \emph{Fibonacci
		Quart.} \textbf{18} (1980), 370.

		\bibitem{Sha2} J. Shallit and H. Klauser, Sum of
		base-$b$ palindrome reciprocals, \emph{Fibonacci Quart.}
		\textbf{19} (1981), 469.

		\bibitem{Slo} N. J. A. Sloane et al., The On-Line Encyclopedia
		of Integer Sequences, \url{https://oeis.org}, 2019.

		\bibitem{Stan} R. P. Stanley, Log-concave and unimodal
		sequences in algebra, combinatorics, and geometry,
		\emph{Ann. New York Acad. Sci.} \textbf{576} (1989),
		500--535.

		\bibitem{Ten} G. Tenenbaum, \emph{Introduction to Analytic
		and Probabilistic Number Theory}, Cambridge University
		Press, 1995.

	\end{thebibliography}

	\bigskip \hrule \bigskip

	\noindent 2010 {\it Mathematics Subject Classification}: Primary
	11A63; Secondary 11N37, 11Y60, 40A25.

	\noindent \emph{Keywords: } palindrome, palindromic number,
	reciprocal, asymptotic, log-concave.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A002113} and
\seqnum{A244162}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 27 2019; 
revised versions received  October 7 2019; December 10 2019.
Published in {\it Journal of Integer Sequences}, December 27 2019.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

\end{document}


</pre></body></html>