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\begin{center}
\vskip 1cm{\LARGE\bf The Average Order of the Dirichlet Series of the 
gcd-sum Function
}
\vskip 1cm
\large
Kevin A. Broughan\\
University of Waikato\\
Hamilton, New Zealand\\
\href{mailto:kab@waikato.ac.nz}{\tt kab@waikato.ac.nz} \\
\end{center}

\vskip .2 in

\begin{abstract}
Using a result of Bordell\`es,
we derive the second term and improved error expressions for the
partial sums of the Dirichlet series of the gcd-sum function, for
all real values of the parameter.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}



% text should be lower case, unless caps are necessary for meaning
\def\a{\alpha}
\def\b{\beta}
\def\z{\zeta}
\def\sx{\sum_{n\le \sqrt{x}}}
\def\t{\theta}
\def\e{\epsilon}
\def\g{\gamma}

\section{Introduction}


 The second term and improved error expressions for the
partial sums of the Dirichlet series of the gcd-sum function, first
given in \cite{broughan}, are derived, for all real values of the
parameter, using a result of Bordell\`es \cite{bordelles}.

Let $(m,n)$ represent the positive greatest common divisor of the
positive integers $m$ and $n$, let $g(n):=\sum_{i=1}^n (i,n)$ be the
gcd-sum, let $\tau(n)$ be the number of divisors of $n$, let
$Id(n):=n$ be the identity function, for $\a\in\mathbb{R}$ let
$m_\a(n)=1/n^\a$ be the monomial power function, and let $\phi(n)$
be Euler's phi function. For $x\ge 1$ let
$$
G_\a(x) := \sum_{n\le x} \frac{g(n)}{n^\a}
$$
be the average order.

Let $\t$ be the smallest positive real number such that
$$
\sum_{n\le x} \tau(n)= x\log x + x(2\g-1) + O_\e(x^{\t+\e})~~~~(1)
$$
for all $\e&gt;0$ and $x\ge 1$. For a discussion about the value of
$\t$ and references see \cite{bordelles}. The current best value is
slightly more than 1/4.

The lemma of Bordell\`es \cite{bordelles} enables us to write $g = \mu
* (\tau\cdot Id)$ and this, combined with a best available error for the average order
of $\tau(n)$, enables a further term and greater degree of precision
to be obtained for the $G_\a(x)$ than the expression $g=\phi*Id$
used in \cite{broughan}. This is true for every value of $\a$ except
for $\a=2$ in which case both expressions for $g(n)$ give the same
terms and error form.

In Section 3 we correct some errors in \cite{broughan}.

%=====================================================================================
\section{Average values of the Dirichlet series}

First we give two summations for the sum of divisor function
$\tau(n)$.

\begin{lemma}
For all $\b$ satisfying $\b \ge -\t$ and $\e&gt;0$:
$$
\sum_{n\le x} n^\b \tau(n)= \frac{x^{\b+1}}{\b+1}\log x+
\frac{x^{\b+1}}{\b+1}\big( 2\g-\frac{1}{\b+1}\big)+
O_\e(x^{\t+\b+\e}).
$$
\end{lemma}

\begin{proof}
This follows from (1) and Abel summation as in Lemma 1 of
\cite{bordelles}.
\end{proof}
%----------------------------------------------------------------------

\begin{lemma}
For all $\b$ satisfying $-1&lt; \b &lt; -\t$:
$$
\sum_{n\le x} n^\b \tau(n)= \frac{x^{\b+1}}{\b+1}\log x+
\frac{x^{\b+1}}{\b+1}\big( 2\g-\frac{1}{\b+1}\big)+ O(1)
$$
where the implied constant is absolute.
\end{lemma}
%----------------------------------------------------------------------
Let $c_1 := 2\g -\frac{1}{2}-\frac{\z'(2)}{\z(2)}$. The theorem of
Bordell\`es \cite{bordelles} can be written:
\begin{lemma} For all $\e&gt;0$, as $x\rightarrow\infty$
$$ G_0(x)= \frac{x^2\log x}{2\z(2)}+ c_1 \frac{x^2}{2\z(2)}+
O_\e(x^{1+\t+\e}).
$$
\end{lemma}
It is not needed in what follows, but for completeness the
corresponding sum for the power of the multiplier $-1$ is
\cite{riesel}:
$$
\sum_{n\le x} \frac{\tau(n)}{n}= \frac{\log^2 x}{2} + 2\g\log x +
\g^2-2\g_1 + O(\frac{1}{\sqrt x}),
$$
where $\g_1$ is the first Stieltjes constant.
%========================================================================================
\begin{figure}[H]
\unitlength1cm \center{\includegraphics*[scale=1.0 ,bb=81 3 402
201]{gcd1.eps}} \caption{Error value for $G_0(x)$.}
\end{figure}\par

Figure 1 is a plot of $G_0(x)- \frac{x^2\log x}{2\z(2)}- c_1
\frac{x^2}{2\z(2)}$ for $1\le x\le 1000$. This should be compared
with the corresponding plots for $G_\a$ given below. It shows that,
maybe, another continuous term can be extracted from the error. The
existing computed error is large, but does, on the face of it, go to
zero when divided by $x^2$, consistent with the given value of the
constant $c_1$.

We can now improve each part of \cite[Theorem 4.4]{broughan}:
\begin{theorem} For all $\e&gt;0$ as $x\rightarrow\infty$

(1) If $\a\le 1+\t$:
$$
G_\alpha(x)=\frac{x^{2-\alpha} \log x}{(2-\alpha)\zeta(2)} +
\frac{x^{2-\alpha}}{(2-\a)\z(2)}\big( 2\g-\frac{1}{2-\a}-
\frac{\z'(2)}{\z(2)}\big)+O_\e(x^{\t+1-\a+\e}).
$$

(2) If $1+\t &lt;\a&lt;2$:
$$
G_\alpha(x)=\frac{x^{2-\alpha} \log x}{(2-\alpha)\zeta(2)} +
\frac{x^{2-\alpha}}{(2-\a)\z(2)}\big( 2\g-\frac{1}{2-\a}-
\frac{\z'(2)}{\z(2)}\big)+O(1).
$$

(3) $ G_2(x)= \frac{\log^2 x}{2\z(2)}+\frac{\log
x}{\z(2)}\big(2\gamma-\frac{\z'(2)}{\z(2)}\big)+O(1).$

(4) If $\a&gt;2$:
\begin{eqnarray*}
G_\a(x)&amp;=&amp; \frac{\z(\a-1)^2}{\z(\a)} - \frac{\log x}{(\a-2)\z(2)
x^{\a-2}}\\ &amp;~&amp; - \frac{1}{(\a-2)\z(2)x^{\a-2}}\big(2\g +
\frac{1}{\a-2}-\frac{\z'(2)}{\z(2)}\big) +
O_\e(\frac{1}{x^{\a-1-\t-\e}}).
\end{eqnarray*}

\end{theorem}

\begin{proof}

(1) In this case $\a\le 1+\t$. Let $\b := 1-\a$, $c_2:= 2\g
-1/(1+\b)$. Then, using Lemma 2.1 and the complete multiplicativity
of $m_\a$:
\begin{eqnarray*}
G_\a(x) &amp;=&amp; \sum_{n\le x} \frac{g(n)}{n^\a}= \sum_{n\le x} (m_\a
\cdot g)(n)\\
&amp;=&amp; \sum_{n\le x} ((m_\a\cdot\mu)*(m_\a\cdot Id\cdot \tau))(n)\\
&amp;=&amp; \sum_{d\le x}\frac{\mu(d)}{d^\a}\sum_{e\le x/d}
e^{1-\a}\tau(e)\\
&amp;=&amp; \sum_{d\le x}\frac{\mu(d)}{d^\a}\big[
\frac{1}{\b+1}\frac{x^{\b+1}}{d^{\b+1}}(\log x - \log d) + c_2
\frac{x^{\b+1}}{d^{\b+1}}+ O_\e((\frac{x}{d})^{\t+\b+\e})\big]\\
&amp;=&amp; \frac{x^{\b+1}\log x}{\b+1}\sum_{d\le x}
\frac{\mu(d)}{d^{\a+\b+1}}-\frac{x^{\b+1}}{\b+1}\sum_{d\le x}
\frac{\mu(d)\log d}{d^{\a+\b+1}}\\
&amp;~&amp; + c_2 x^{\b+1}\sum_{d\le x} \frac{\mu(d)}{d^{\a+\b+1}}+
O_\e(x^{\t+\b+\e} \sum_{d\le x} \frac{1}{d^{\t+\a+\b+1}})
\end{eqnarray*}

Therefore
\begin{eqnarray*}G_\a(x)  &amp;=&amp; \frac{x^{2-\a}\log x}{2-\a}[\frac{1}{\z(2)}-\sum_{d&gt;x}
\frac{\mu(d)}{d^2}]-\frac{x^{2-\a}}{2-\a}[\frac{\z'(2)}{\z(2)^2}-\sum_{d&gt;x}
\frac{\mu(d)\log d}{d^2}] \\
&amp;~&amp; + c_2 x^{2-\a}[\frac{1}{\z(2)}-\sum_{d&gt;x}
\frac{\mu(d)}{d^2}]+O_\e(x^{\t+\e+1-\a})\\
&amp;=&amp;\frac{x^{2-\a}\log x}{(2-\a)\z(2)}+
\frac{x^{2-\a}}{(2-\a)\z(2)}[2\g -
\frac{1}{2-\a}-\frac{\z'(2)}{\z(2)}]\\
&amp;~&amp; +O_\e(x^{\t+\e+1-\a})
\end{eqnarray*}

(2) The derivation is the same as in (1), except we use Lemma 2.2
instead of Lemma 2.1.

(3) This is given in the proof of \cite[Theorem 4.4, Case
2]{broughan}.

(4) Let $\a&gt;2$. Then, using \cite[Theorem 4.1]{broughan} and Abel
summation:
\begin{eqnarray*}
G_\a(x)&amp;=&amp; \frac{\z(\a-1)^2}{\z(\a)}-\sum_{n&gt;x} \frac{g(n)}{n^\a}\\
&amp;=&amp; \frac{\z(\a-1)^2}{\z(\a)}-
\lim_{y\rightarrow\infty}\frac{G_0(y)}{y^\a} + \frac{G_0(x)}{x^\a}-
\a\int_x^\infty \frac{G_0(t)}{t^{\a+1}} dt\\
&amp;=&amp;\frac{\z(\a-1)^2}{\z(\a)} + \frac{G_0(x)}{x^\a}-
\a\int_x^\infty \frac{G_0(t)}{t^{\a+1}} dt\\
&amp;=&amp; \frac{\z(\a-1)^2}{\z(\a)}-\frac{\log x}{(\a-2)\z(2) x^{\a-2}}\\
&amp;~&amp; - \frac{1}{2(\a-2)\z(2) x^{\a-2}}[2
c_1+\frac{\a}{\a-2}]+O_\e(\frac{1}{x^{\a-1-\t-\e}}),
\end{eqnarray*}
and the result follows after substituting for $c_1$.
\end{proof}

%==================================================================================================
\begin{figure}[H]
\unitlength1cm \center{\includegraphics* [scale=1.0 ,bb=81 3 402
201]{gcd2.eps}} \caption{Error value for $G_\a(x)$ with $\a=1$.}
\end{figure}\par

Figure 2 is a plot of $G_1(x)- \frac{x\log x}{\z(2)}-
\frac{x}{\z(2)}\big(2\g-1-\frac{\z'(2)}{\z(2)}\big)$ for $1\le x\le
1000$.



\begin{figure}[H]
\unitlength1cm \center{\includegraphics*[scale=1.0 ,bb=81 3 402
201]{gcd3.eps}} \caption{Error value for $G_\a(x)$ with $\a=3/2$.}
\end{figure}\par


Figure 3 is a plot of $$G_\alpha(x)-\frac{x^{2-\alpha} \log
x}{(2-\alpha)\zeta(2)} - \frac{x^{2-\alpha}}{(2-\a)\z(2)}\big(
2\g-\frac{1}{2-\a}- \frac{\z'(2)}{\z(2)}\big)$$ for $\a=3/2$ and
$1\le x\le 1000$.
%% End of article:


\section{Errors in ``The gcd-sum function"}
Corrections to some errors in \cite{broughan}. Page 7 line 8: the
second $g$ should be $Id$, page 9 line 3: 5.1 should be 4.1, page 12
line 2: 5.4 should be 4.3, page 12 line 3: 5.3 should be 4.2, page
13 line 13 and line -1: $e,d$ should be $e.d$, page 14 line -1: 5.4
should be 4.4.

\section{Acknowledgments}

The support of the University of Waikato is
gratefully acknowledged. The plots were produced using the
Mathematica functions ListPlot, and DivisorSigma. Helpful
suggestions made by a referee are also gratefully acknowledged.


\begin{thebibliography}{9}
\bibitem{bordelles}
O. Bordell\'es. \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL10/Bordelles/bordelles90.html}{A note on the average order of the gcd-sum
function}, {\it J. Integer Sequences}, {\bf 10} (2007) Article
07.3.3.

\bibitem{broughan}
K. A. Broughan. \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL4/BROUGHAN/gcdsum.html}{The gcd-sum function}, {\it J. Integer
Sequences}, {\bf 4} (2001), Article 01.2.2.

\bibitem{riesel}
H. Riesel and R. C. Vaughan.  On sums of primes, {\it Arkiv f\"or
Mathematik} {\bf 21} (1983), 45--74.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A05; Secondary 11A25, 11M06, 11N37, 11N56.

\noindent \emph{Keywords: } 
greatest common divisor, Dirichlet series, gcd-sum.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A018804}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 28, 2007;
revised version received  April 13 2007.
Published in {\it Journal of Integer Sequences}, April 13 2007.

\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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