\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} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \DeclareMathOperator{\lcm}{lcm} \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 On the Lcm-Sum Function} \vskip 1cm \large Soichi Ikeda and Kaneaki Matsuoka\\ Graduate School of Mathematics\\ Nagoya University\\ Furocho Chikusaku Nagoya 464-8602\\ Japan\\ \href{mailto:m10004u@math.nagoya-u.ac.jp}{\tt m10004u@math.nagoya-u.ac.jp}\\ \href{mailto:m10041v@math.nagoya-u.ac.jp}{\tt m10041v@math.nagoya-u.ac.jp}\\ \end{center} \vskip .2 in \begin{abstract} We consider a generalization of the lcm-sum function, and we give two kinds of asymptotic formulas for the sum of that function. Our results include a generalization of Bordell\`{e}s's results and a refinement of the error estimate of Alladi's result. We prove these results by the method similar to those of Bordell\`{e}s. \end{abstract} \section{Introduction} Pillai \cite{pillai} first defined the gcd-sum function \[ g(n) = \sum_{j=1}^n \gcd(j,n) \] and studied this function. The function $g(n)$ was defined again by Broughan \cite{broughan1}. Broughan considered \[ G_{\alpha}(x) = \sum_{n \le x} n^{- \alpha} g(n) \] for $\alpha \in \mathbb{R}$ and $x \ge 1$, and obtained some asymptotic formulas for $G_{\alpha}(x)$. The function $G_{\alpha}(x)$ was studied by some authors (see, for example, \cite{broughan2, tanigawa}). Some generalizations of the function $g(n)$ was considered (see, for example, \cite{bordelles, toth}). On the other hand, the lcm-sum function \[ l(n) := \sum_{j=1}^n \lcm(j,n) \] was considered by some authors. Alladi \cite{alladi} studied the sum \[ \sum_{j=1}^n (\lcm(j,n))^r \qquad (r \in \mathbb{R}, \, r \ge 1) \] and obtained \begin{equation} \label{res_alladi} \sum_{n \le x} \sum_{j=1}^n (\lcm(j,n))^r = \frac{\zeta(r+2)}{2(r+1)^2 \zeta(2)} x^{2r+2} + O(x^{2r+1+ \epsilon}). \end{equation} We define the functions \begin{gather} L_a(n) := \sum_{j=1}^n (\lcm(j,n))^a \notag \\ T_a(x) := \sum_{n \le x} L_a(n) \notag \end{gather} for $a \in \mathbb{Z}$ and $x \ge 1$. Bordell\`{e}s studied the sums $T_1(x)$ and $T_{-1}(x)$ and obtained \begin{gather} l(n) = \frac{1}{2}((\mathrm{Id}^2 \cdot (\varphi + \tau_0)) \ast \mathrm{Id})(n), \notag \\ \sum_{n \le x} \sum_{j=1}^n \lcm(j,n) = \frac{\zeta(3)}{8 \zeta(2)} x^4 + O(x^3 (\log x)^{2/3} (\log \log x)^{4/3}) \quad (x > e), \notag \\ \sum_{n \le x} \sum_{j=1}^n \frac{1}{\lcm(j,n)} = \frac{(\log x)^3}{6 \zeta(2)} + \frac{(\log x)^2}{2 \zeta(2)} \biggl( \gamma + \log \biggl( \frac{\mathcal{A}^{12}}{2 \pi} \biggr) \biggr) + O( \log x), \notag \end{gather} where $\mathrm{Id}^a(n) = n^a$ ($a \in \mathbb{Z}$), \[ \tau_0(n) = \begin{cases} 1, & \text{if $n = 1$;} \\ 0, & \text{otherwise}, \end{cases} \] $F*G$ is the usual Dirichlet convolution product, and $\mathcal{A}$ is the Glaisher-Kinkelin constant \cite[p.\ 25]{srivastava}). Gould and Shonhiwa \cite{gould} stated that the $\log$-factors in the error term in the second formula can be removed. In this paper we study $T_a(x)$ for $a \ge 2$ and $a \le -2$. The following theorems are our main results. These results are proved by the methods similar to those of Bordell\`{e}s \cite[Section 6]{bordelles}. We write $f(x) = O(g(x))$, or equivalently $f(x) \ll g(x)$, where there is a constant $C > 0$ such that $|f(x)| \le C g(x)$ for all values of $x$ under consideration. \begin{theorem} \label{th_conv} Let $B_n$ be Bernoulli numbers defined by \[ \frac{z}{e^z - 1} = \sum_{n=0}^{\infty} B_n \frac{z^n}{n!}. \] If we define \[ \varphi_k(n) := \sum_{d \mid n} \mu(d) \Bigl( \frac{d}{n} \Bigr)^k \] and \[ M_a(n) := \biggl( \mathrm{Id}^{2a} \cdot \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr) \biggr)(n), \] then for $a \in \mathbb{Z}$ we have \[ L_a(n) = (M_a \ast \mathrm{Id}^a)(n). \] \end{theorem} \begin{theorem} \label{th_pos} Let $x > e$ and $a \in \mathbb{N}$. Then we have \[ \sum_{n \le x} L_a(n) = \frac{\zeta(a+2)}{2(a+1)^2 \zeta(2)} x^{2a+2} + O(x^{2a+1} (\log x)^{2/3} (\log \log x)^{4/3}) \quad (\text{as $x \to \infty$}), \] where the implied constant depends on $a$. \end{theorem} \begin{theorem} \label{th_neg} Let $x \ge 1$ and $k \in \mathbb{N}$ with $k \ge 2$. Then we have \begin{equation} \label{series_lkn} \sum_{n=1}^{\infty}L_{-k}(n)=\frac{\zeta(k)}{2} \biggl( 1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr) \end{equation} and \[ \sum_{n \le x}L_{-k}(n)=\frac{\zeta(k)}{2} \biggl( 1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr) - \frac{\zeta(k)x^{-k+1}\log x}{(k-1)\zeta(k+1)}+O(x^{-k+1}) \quad (\text{as $x \to \infty$}), \] where the implied constant depends on $k$. \end{theorem} We note that the function $L_a(n)$ is not multiplicative for all $a \in \mathbb{Z} \setminus \{0\}$, but we can write $L_a(n)$ ($a \ge 1$) explicitly by Dirichlet convolution. In the proof of Theorem \ref{th_pos} we use this fact. The error estimates in Theorem \ref{th_pos} are better than (\ref{res_alladi}). Since we have \[ g_r(n) := \sum_{j=1}^n (\gcd(j,n))^r > \varphi(n) \] for all $r \in \mathbb{R}$, the sum \begin{equation*} \label{sum_gcd_r} \sum_{n = 1}^{\infty} g_r(n) \end{equation*} is divergent for all $r$. Therefore the behavior of the sum $T_a(x)$ ($a \in \mathbb{Z}$ and $a \le -2$) is completely different from that of the sum \[ \sum_{n \le x} g_a(n). \] \section{Lemmas for the proof of theorems} In this section, we collect some auxiliary results and definitions. Let $B_n(x)$ be Bernoulli polynomials defined by \[ \frac{z e^{xz}}{e^z - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{z^n}{n!}. \] The following relations are well-known \cite[p.\ 59]{srivastava}. \begin{gather} B_n(x + 1) - B_n(x) = n x^{n-1}, \notag \\ B_n(x) = \sum_{k=0}^n \binom{n}{k} B_k x^{n-k}, \notag \\ B_n(0) = B_n(1) = B_n \quad (n > 1). \notag \end{gather} \begin{lemma} \label{lem_berpoly} Let $m,n \in \mathbb{N}$ and \[ S_n(m) := \sum_{l=1}^m l^n. \] Then we have \[ S_n(m) = \frac{m^{n+1}}{n+1} + \frac{1}{2} m^n + \frac{1}{n+1} \sum_{k=1}^{n-1} \binom{n+1}{k+1} B_{k+1} m^{n-k}. \] \end{lemma} \begin{proof} We have \[ \begin{split} S_n(m) &= \frac{1}{n+1} (B_{n+1}(m+1) - B_{n+1}(1)) \\ &= \frac{1}{n+1}(B_{n+1}(m) + (n+1)m^n - B_{n+1}) \\ &= \frac{1}{n+1} \biggl( \sum_{k=0}^{n+1} \binom{n+1}{k} B_k m^{n+1-k} + (n+1)m^n - B_{n+1} \biggr) \\ &= \frac{m^{n+1}}{n+1} + \frac{1}{2} m^n + \frac{1}{n+1} \sum_{k=1}^{n-1} \binom{n+1}{k+1} B_{k+1} m^{n-k}. \end{split} \] \end{proof} We use the following lemmas in the proof of Theorem \ref{th_pos}. \begin{lemma} \label{lem_phik} Let $r, k \in \mathbb{N}$ with $r > k$ and $x \ge 1$. We have \[ \sum_{n \le x} n^r \varphi_k(n) \le x^{r+1}. \] \end{lemma} \begin{proof} We have \begin{align*} \sum_{n \le x} n^r \varphi_k(n) &= \sum_{n \le x} n^{r-k} \sum_{d \mid n} \mu(d) d^k \\ &= \sum_{d \le x} \mu(d) d^k \bigl(d^{r-k} + (2d)^{r-k} + \cdots + (d \lfloor x/d \rfloor)^{r-k} \bigr) \\ &= \sum_{d \le x} \mu(d) d^r \sum_{j \le x/d} j^{r-k} \\ &\le x^{r+1}. \end{align*} \end{proof} \begin{lemma} \label{lem_nphin} Let $r \in \mathbb{N}$ and $x > e$. We have \[ \sum_{n \le x} n^r \varphi(n) = \frac{x^{r+2}}{(r+2) \zeta(2)} + O(x^{r+1} (\log x)^{2/3} (\log \log x)^{4/3}) \quad (\text{as $x \to \infty$}), \] where the implied constant depends on $r$. \end{lemma} \begin{proof} We can obtain the lemma by the estimate \cite{walfisz} \[ \sum_{n \le x} \varphi(n) = \frac{x^2}{2 \zeta(2)} + O(x (\log x)^{2/3} (\log \log x)^{4/3}) \] and the partial summation formula. \end{proof} \section{Proof of Theorem \ref{th_conv} and Theorem \ref{th_pos}} \begin{proof}[Proof of Theorem \ref{th_conv}] We have \[ \begin{split} \sum_{j=1}^n \biggl( \frac{j}{\gcd(n,j)} \biggr)^a &= \sum_{d \mid n} \frac{1}{d^a} \sum_{\substack{j=1 \\ \gcd(j,n)=d}}^n j^a \\ &= \sum_{d \mid n} \frac{1}{d^a} \sum_{\substack{k \le n/d \\ \gcd(k,n/d)=1}} (kd)^a = \sum_{d \mid n} \sum_{\substack{k \le n/d \\ \gcd(k,n/d)=1}} k^a. \end{split} \] By Lemma \ref{lem_berpoly} we have \begin{align*} \sum_{\substack{k \le N \\ \gcd(k,N)=1}} k^a &= \sum_{k \le N} k^a \sum_{d \mid \gcd(k,N)} \mu(d) = \sum_{d \mid N} d^a \mu(d) \sum_{m \le N/d} m^a \\ &= \sum_{d \mid N} d^a \mu(d) \biggl( \frac{1}{a+1} \biggl( \frac{N}{d} \biggr)^{a+1} + \frac{1}{2} \biggl( \frac{N}{d} \biggr)^a + \\ &\qquad + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \biggl( \frac{N}{d} \biggr)^{a-k} \biggr) \\ &= \frac{N^a}{a+1} \sum_{d \mid N} \mu(d) \frac{N}{d} + \frac{N^a}{2} \sum_{d \mid N} \mu(d) + \\ &\qquad + \frac{N^a}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \sum_{d \mid N} \mu(d) \biggl( \frac{d}{N} \biggr)^k \\ &= N^a \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr)(N). \end{align*} Hence we obtain \begin{align*} L_a(n) &= n^a \sum_{j=1}^n \biggl( \frac{j}{\gcd(n,j)} \biggr)^a \\ &= \sum_{d \mid n} \biggl( \frac{n}{d} \biggr)^{2a} \cdot \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr) (n/d) \cdot d^a \\ &= (M_a * \mathrm{Id}^a)(n). \end{align*} \end{proof} \begin{proof}[Proof of Theorem \ref{th_pos}] By Lemma \ref{lem_phik}, Lemma \ref{lem_nphin} and Theorem \ref{th_conv}, we have \begin{align*} \sum_{n \le x} L_a(n) &= \sum_{n \le x} (M_a \ast \mathrm{Id}^a)(n) = \sum_{d \le x} d^a \sum_{m \le x/d} M_a(m) \\ &= \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a} \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr) (m) \\ &= \frac{1}{a+1} \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a} \varphi(m) + O(x^{a+1}) + \\ &\qquad + \frac{1}{a+1} \sum_{k=1}^{a-1} \binom{a+1}{k+1} B_{k+1} \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a} \varphi_k(m) \\ &= \frac{1}{a+1} \sum_{d \le x} d^a \biggl( \frac{1}{(2a+2) \zeta(2)} \biggl( \frac{x}{d} \biggr)^{2a+2} + \\ &\qquad + O \biggl( \biggl( \frac{x}{d} \biggr)^{2a+1} (\log (x/d))^{2/3} (\log \log (x/d))^{4/3} \biggr) \biggr) + \\ &\qquad + O(x^{a+1}) + O \biggl( \sum_{d \le x} d^a (x/d)^{2a+1} \biggr) \\ &= \frac{x^{2a+2}}{(a+1)(2a+2) \zeta(2)} \sum_{d \le x} \frac{1}{d^{a+2}} + O(x^{2a+1} (\log x)^{2/3} (\log \log x)^{4/3}) + \\ &\qquad + O(x^{2a+1}). \end{align*} This implies the theorem. \end{proof} \section{Proof of Theorem \ref{th_neg}} \begin{proof}[Proof of Theorem \ref{th_neg}] Since we have \begin{align*} L_{-k}(n)&=\sum_{j=1}^{n}\frac{1}{(\lcm(n,j))^{k}}= \frac{1}{n^{k}}\sum_{j=1}^{n}\frac{(\gcd(n,j))^{k}}{j^{k}}= \frac{1}{n^{k}}\sum_{d|n}d^{k} \sum_{\substack{ j=1\\ \gcd(j,n)=d}}^{n} \frac{1}{j^{k}}\\ &=\frac{1}{n^{k}}\sum_{d|n}d^{k} \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}} \frac{1}{i^{k}d^{k}}\\ &=\frac{1}{n^{k}}\sum_{d|n} \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}}\frac{1}{i^{k}}, \end{align*} we obtain \begin{align*} \sum_{n=1}^{\infty}L_{-k}(n)&= \sum_{n=1}^{\infty}\frac{1}{n^{k}}\sum_{d|n} \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}} \frac{1}{i^{k}}=\sum_{d=1}^{\infty}\sum_{j=1}^{\infty} \frac{1}{j^{k}d^{k}}\sum_{\substack{i\leq j\\ \gcd(i,j)=1}} \frac{1}{i^{k}}\\ &=\zeta(k)\sum_{j=1}^{\infty} \frac{1}{j^{k}} \sum_{\substack{i\leq j\\ \gcd(i,j)=1}}\frac{1}{i^{k}}\\ &=\zeta(k)\sum_{n=1}^{\infty}\frac{1}{n^{k}} (\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1). \end{align*} Also we have \[ \sum_{n=1}^{\infty}\frac{1}{n^{k}} \left(\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1 \right) = 1+\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n^{k}} \left(\sum_{\substack{ \gcd(i,j)=1\\ ij=n}}1 \right) \] and the relation \cite[1.2.8]{titchmarsh} \[ \sum_{n=1}^{\infty}\frac{1}{n^{k}} \left(\sum_{\substack{ \gcd(i,j)=1\\ ij=n}}1 \right)= \frac{\zeta(k)^{2}}{\zeta(2k)}. \] Therefore we obtain \[ \sum_{n=1}^{\infty} \frac{1}{n^{k}} \left(\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1 \right)= 1+\frac{1}{2} \biggl( \frac{\zeta(k)^{2}}{\zeta(2k)}-1 \biggr)= \frac{1}{2} \biggl(1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr). \] This implies (\ref{series_lkn}). By the relation \[ \sum_{n \le x} L_{-k}(n) = \frac{\zeta(k)}{2} \biggl( 1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr) - \sum_{n > x} L_{-k} (n), \] the remaining task is to estimate the sum $\sum_{n > x} L_{-k} (n)$. We have \begin{align*} \sum_{n>x}L_{-k}(n)&=\sum_{n>x}\frac{1}{n^{k}}\sum_{d|n} \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}} \frac{1}{i^{k}}=\sum_{d=1}^{\infty} \sum_{h>\frac{x}{d}}\frac{1}{(hd)^{k}} \sum_{\substack{j\leq h\\ \gcd(j,h)=1}}\frac{1}{j^{k}}\\ &=\sum_{d=1}^{\infty}\sum_{h>\frac{x}{d}}\frac{1}{(hd)^{k}} \sum_{j\leq h}\frac{1}{j^{k}}\sum_{\delta| \gcd(j,h)}\mu(\delta)\\ &=\sum_{d=1}^{\infty}\sum_{h>\frac{x}{d}}\frac{1}{(hd)^{k}} \sum_{\delta|h}\sum_{m\leq\frac{h}{\delta}} \frac{\mu(\delta)}{m^{k}\delta^{k}}\\ &=\sum_{d=1}^{\infty}\frac{1}{d^{k}}\sum_{\delta=1}^{\infty} \sum_{l>\frac{x}{d\delta}}\frac{\mu(\delta)}{l^{k}\delta^{2k}} \sum_{m\leq l}\frac{1}{m^{k}}\\ &=\sum_{q=1}^{\infty}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta) \sum_{l>\frac{x}{q}}\frac{1}{l^{k}}\sum_{m\leq l}\frac{1}{m^{k}}\\ &=\sum_{q=1}^{\infty}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta) \sum_{l>\frac{x}{q}}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr) \\ &= \sum_{q < x} \frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta) \sum_{l>\frac{x}{q}}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr) + \\ &\qquad + \sum_{q \ge x} \frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta) \sum_{l>\frac{x}{q}}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr) \\ &=: S_1 + S_2, \end{align*} say. We have \begin{align*} S_1 &= \sum_{q\frac{x}{q}}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\ &=\sum_{q\frac{x}{q}}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\ &\ll \sum_{q\geq x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta) \sum_{l=1}^{\infty}\frac{1}{l^{k}} \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\ &\ll\sum_{q\geq x}q^{-k}\sum_{d|q}\frac{|\mu(d)|}{d^{k}}\\ &\ll x^{-k+1}. \end{align*} Therefore we obtain \[ \sum_{n>x}L_{-k}(n)=\frac{\zeta(k)x^{-k+1}\log x}{(k-1)\zeta(k+1)} +O(x^{-k+1}). \] This completes the proof. \end{proof} \begin{thebibliography}{99} \bibitem{alladi} K. Alladi, On generalized Euler functions and related totients, in \textit{New Concepts in Arithmetic Functions}, Matscience Report 83, Madras, 1975. \bibitem{bordelles} O. Bordell\`{e}s, Mean values of generalized gcd-sum and lcm-sum functions, \textit{J. 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Tanigawa and W. Zhai, On the gcd-sum function, \textit{J. Integer Sequences} \textbf{11} (2008), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL11/Tanigawa/tanigawa12.html} {Article 08.2.3}. \bibitem{titchmarsh} E. C. Titchmarsh, \textit{The Theory of the Riemann Zeta-function}, Second Edition, Oxford University Press, 1986. \bibitem{toth} L. T\'{o}th, A survey of gcd-sum functions, \textit{J. Integer Sequences} \textbf{13} (2010), \href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Toth/toth10.html} {Article 10.8.1}. \bibitem{walfisz} A. Walfisz, \textit{Weylsche Exponentialsummen in der neueren Zahlentheorie}, Leipzig BG Teubner, 1963. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25; Secondary 11N37. \noindent \emph{Keywords: } arithmetic function, lcm-sum function, least common multiple. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A018804} and \seqnum{A051193}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 19 2013; revised versions received November 14 2013; November 23 2013; December 19 2013. Published in {\it Journal of Integer Sequences}, December 27 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .