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\begin{center}
\vskip 1cm{\LARGE\bf 
A Gcd-Sum Function Over Regular Integers \\
\vskip .1in
Modulo $n$
}
\vskip 1cm
\large
L\'aszl\'o T\'oth \\
Institute of Mathematics and Informatics \\
University of P\'ecs \\
Ifj\'us\'ag u. 6 \\
7624 P\'ecs \\
Hungary \\
\href{mailto:ltoth@ttk.pte.hu}{\tt ltoth@ttk.pte.hu}\\
\end{center}

\vskip .2 in

\begin{abstract} 
We introduce a gcd-sum function involving regular integers (mod $n$)
and prove results giving its minimal order, maximal order and
average order.
\end{abstract}

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\def\Z{{\Bbb Z}}
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%************************* section 1 *******************************************

\section{Introduction}

Let $n>1$ be an integer with prime
factorization $n=p_1^{\nu_1}\cdots
p_r^{\nu_r}$. An integer $k$ is called {\it regular} (mod $n$) if there
exists an integer $x$ such that $k^2x\equiv k$ (mod $n$), i.e., the
residue class of $k$ is a regular element (in the sense of J. von
Neumann) of the ring $\Z_n$ of residue classes (mod $n$). In
general, an element $k$ of a ring $R$ is said to be (von Neumann)
regular if there is an $x\in R$ such that $k=kxk$. If every $k\in R$
has this property, then $R$ is called a {\it von Neumann regular ring},
cf.\ for example \cite[p.\ 110]{Kap1972}.

It can be shown that $k\ge 1$ is regular (mod $n$) if and only if
for every $i\in \{1,\ldots,r\}$ either $p_i\nmid k$ or
$p_i^{\nu_i}\mid k$. Also, $k\ge 1$ is regular (mod $n$) if and only
if $\gcd (k,n)$ is a unitary divisor of $n$. We recall that $d$ is
said to be a {\it unitary divisor} of $n$ if $d\mid n$ and $\gcd
(d,n/d)=1$, and use the notation $d \mid \mid n$. These and other
characterizations of regular integers are given in our paper
\cite{Tot2008}.

Let $\Reg_n=\{k: 1\le k\le n$ and $k$ is regular (mod $n$)$\}$, and let
$\varrho(n)=\# \Reg_n$ denote the number of regular integers $k$
(mod $n$) such that $1\le k\le n$. The function $\varrho(n)$ was
investigated in paper \cite{Tot2008}. It is multiplicative and
$\varrho(p^{\nu})=\phi(p^{\nu})+1= p^{\nu}-p^{\nu-1}+1$ for every
prime power $p^{\nu}$ ($\nu \ge 1$), where $\phi$ is the Euler
function. Note that $(\varrho(n): n\in \N)$ is the sequence \seqnum{A055653} in Sloane's On-Line
Encyclopedia of Integer Sequences.

In this paper we introduce the function
\begin{equation}
\widetilde{P}(n):=\sum_{k\in \Reg_n} \gcd (k,n). \label{gcd-sum_regular}
\end{equation}

This is analogous to the gcd-sum function, called also Pillai's
arithmetical function,
\begin{equation} P(n):=\sum_{k=1}^n \gcd (k,n), \label{gcd-sum}
\end{equation}
investigated in the recent papers
\cite{Bor2007a,Bor2007b,Bro2001,Bro2007,TanZha2008} of this journal.
This is sequence \seqnum{A018804} in Sloane's Encyclopedia.

Note that the function $P(n)$ was introduced by S. S. Pillai
\cite{Pil1933}, showing that
\begin{equation*}
P(n)=\sum_{d\mid n} d\phi(n/d), \ \ \sum_{d\mid n}
P(d)=n\tau(n)=\sum_{d\mid n} \sigma(d)\phi(n/d),
\end{equation*}
where $\tau(n)$ and $\sigma(n)$ denote, as usual, the number and the
sum of divisors of $n$, respectively.

Note also, that for any arithmetical function $f$,
\begin{equation}
P_f(n):=\sum_{k=1}^n f(\gcd (k,n))=\sum_{d\mid n} f(d)\phi(n/d),
\label{gcd-sum_f}
\end{equation}
which is a result of E. Ces\`{a}ro \cite{Ces1885}. See also the book
\cite[p.\ 127]{Dic1971}.

O. Bordell\`{e}s \cite{Bor2007a} showed that
\begin{equation}
P=\mu*(E\cdot \tau), \label{gcd-sum_convo}
\end{equation}
where $*$ denotes the Dirichlet convolution, where $E(n)=n$ and $\mu$ is
the M\"obius function. Then, using this representation he proved the
following asymptotic formula: for every $\varepsilon
>0$,
\begin{equation}
\sum_{n\le x} P(n)=\frac{x^2}{2\zeta(2)}\left(\log x
+2\gamma-\frac1{2}-\frac{\zeta'(2)}{\zeta(2)}\right) + {\cal
O}(x^{1+\theta+\varepsilon}), \label{gcd-sum_asymptotic}
\end{equation}
where $\gamma$ is Euler's constant and $\theta$ is the number
appearing in Dirichlet's divisor problem, that is
\begin{equation}
\sum_{n\le x} \tau(n)=x\log x+(2\gamma-1)x+{\cal
O}(x^{\theta+\varepsilon}). \label{Dirichlet_divisor}
\end{equation}

It is known that $1/4\le \theta \le 131/416 \approx 0.3149$.

Formulae \eqref{gcd-sum_convo} and \eqref{gcd-sum_asymptotic} were
obtained, even in a more general form, in the paper
\cite{ChiSit1985}, where the authors proved also the following
result concerning the maximal order of $P(n)$,
\begin{equation}
\limsup_{n\to \infty} \frac{\log(P(n)/n) \log \log n}{\log n}= \log
2, \label{gcd-sum_maximal}
\end{equation}
which is well known for the function $\tau(n)$ instead of $P(n)/n$.

Asymptotic formulae for generalized gcd-sum functions of type
\eqref{gcd-sum_f}, concerning sets of polynomials with integral
coefficients and regular convolutions were given in our paper
\cite{Tot1998}.

We investigate in what follows arithmetical and asymptotical
properties of the function $\widetilde{P}(n)$. We show that it is
multiplicative and for every $n\ge 1$,
\begin{equation}
\widetilde{P}(n)=n\prod_{p\mid n} \left(2-\frac1{p} \right).
\label{gcd-sum_regular_expl}
\end{equation}

Further, we show that the result \eqref{gcd-sum_maximal} holds also
for the function $\widetilde{P}(n)$, its minimal order is $3n/2$ and
prove an asymptotic formula for its summatory function both without
assuming the Riemann hypothesis and assuming the Riemann hypothesis
(RH), based on a convolution identity analogous to
\eqref{gcd-sum_convo}.

Our results and proofs involve properties of arithmetical functions
defined by unitary divisors and of the unitary convolution. For
background material in this topic we refer to the book
\cite{McC1986}.

As an open question we formulate the following: What is the minimal
order of the Pillai function $P(n)$?

A common generalization of the functions $P(n)$ and $\widetilde{P}(n)$ is outlined
at the end of Section 2.


%********************  section 2  *****************************************

\section{Arithmetical properties}

Let $f$ be an arbitrary arithmetical function and consider the more
general function
\begin{equation*}
\widetilde{P}_f(n):= \sum_{k\in \Reg_n} f(\gcd (k,n)).
\end{equation*}

\begin{prop} \label{prop_1}
For every $n\ge 1$,
\begin{equation}
\widetilde{P}_f(n)=\sum_{d\mid \mid n} f(d) \, \phi(n/d).
\label{gcd-sum_regular_convo1}
\end{equation}
\end{prop}

\begin{proof} The integer $k\ge 1$ is regular iff $\gcd (k,n)\mid \mid n$, 
cf.\ the Introduction, and
obtain
\[
\widetilde{P}_f(n)= \sum_{k=1}^n \sum_{d\mid \mid n} f(d)=
\sum_{d\mid \mid n} f(d) \sum_{\substack{1\le j\le n/d\\(j,n/d)=1}}
1= \sum_{d\mid \mid n} f(d) \phi(n/d).
\]
\end{proof}

\begin{cor} If $f$ is multiplicative, then
$\widetilde{P}_f$ is also multiplicative and
$\widetilde{P}_f(p^\nu)= f(p^{\nu})+p^{\nu}- p^{\nu -1}$ for every
prime power $p^{\nu}$ ($\nu \ge 1$).

In particular, $\widetilde{P}$ is multiplicative and
$\widetilde{P}(p^\nu)= 2p^{\nu}-p^{\nu-1}$ for every prime power
$p^{\nu}$ ($\nu \ge 1$).
\end{cor}

\begin{proof} According to \eqref{gcd-sum_regular_convo1},
$\widetilde{P}_f$ is the unitary convolution of the functions $f$
and $\phi$. It is known that the unitary convolution preserves the
multiplicativity of functions. In particular, for $f(n)=n$ we obtain
that $\widetilde{P}$ is multiplicative and the explicit formula
\eqref{gcd-sum_regular_expl}.
\end{proof}

Let $\omega(n,k)$ denote the number of distinct prime factors of $n$
which do not divide $k$. For $k=1$, $\omega(n):=\omega(n,1)$ is the
number distinct factors of $n$. Also, let $\tau^*(n,k)$ denote the
number of unitary divisors of $n$ which are relatively prime to $k$.
Here $\tau^*(n):=\tau^*(n,1)$ is the number of unitary divisors of
$n$. We have $\tau^*(n,k)=2^{\omega(n,k)}$ and
$\tau^*(n)=2^{\omega(n)}$.

Another representation of $\widetilde{P}$ is given by

\begin{prop} \label{prop_2}
For every $n\ge 1$,
\begin{equation*}
\widetilde{P}(n)=\sum_{de=n}\mu(d)e\cdot 2^{\omega(e,d)}.
%\label{gcd-sum_regular_convo2}
\end{equation*}
\end{prop}

\begin{proof} By Proposition \ref{prop_1},
\begin{align*}
\widetilde{P}(n)=\sum_{\substack{de=n\\(d,e)=1}} d\phi(e) =
\sum_{\substack{de=n\\(d,e)=1}} d \sum_{ab=e} \mu(a)b =
\sum_{\substack{dab=n\\(d,a)=1\\(d,b)=1}} d \mu(a) b \\ =
\sum_{ac=n} \mu(a)c \sum_{\substack{bd=c\\(b,d)=1\\(d,a)=1}} 1
=\sum_{ac=n} \mu(a)c \, \tau^*(c,a).
\end{align*}
\end{proof}

\begin{prop} \label{prop_ineq}
For every $n\ge 1$ we have $\widetilde{P}(n) \le P(n)$, with
equality iff $n$ is square-free, and $2^{\omega(n)}\phi(n)\le
\widetilde{P}(n)\le 2^{\omega(n)}n$, with equality iff $n=1$.
\end{prop}

\begin{proof} This follows at once by \eqref{gcd-sum_regular},
\eqref{gcd-sum} and \eqref{gcd-sum_regular_expl}.
\end{proof}

\begin{rem}\upshape
Let $\ds \widehat{P}(n):=\sum_{k\in \Reg_n} \lcm [k,n]$. Then it
follows, similar to the ``usual'' lcm-sum function that for every
$n\ge 1$,
\begin{equation*}
\widehat{P}(n)=\frac{n}{2}\left(1+\sum_{d\mid \mid n}
d\phi(d)\right).
\end{equation*}
\end{rem}

\begin{rem}\upshape
For every $n\in \N$ let $A(n)$ be an arbitrary nonempty subset of
the set of positive divisors of $n$. For the system of divisors
$A=(A(n): n\in \N)$ and for an arbitrary arithmetical function $f$
consider the following restricted summation of the gcd's:
\begin{equation*}
P_{A,f}(n)= \sum_{\substack{1\le k\le n\\ \gcd (k,n)\in A(n)}} f(\gcd (k,n)).
\end{equation*}

It follows, similar to the proof of Proposition \ref{prop_1}, that
\begin{equation*}
P_{A,f}(n)= \sum_{d\in A(n)} f(d) \phi(n/d).
\end{equation*}

If $A(n)$ is the set of all (positive) divisors of $n$, then we have
the function \eqref{gcd-sum_f} and if $A(n)$ is the set of the
unitary divisors of $n$, then we reobtain
\eqref{gcd-sum_regular_convo1}.

If $A$ is a regular system of divisors of Narkiewicz-type, including the previous
two special cases, then $P_{A,f}$ is the $A$-convolution of the functions $f$ and
$\phi$. It turns out, that $P_{A,f}$ is multiplicative provided $f$ is multiplicative,
cf.\ \cite[Ch.\ 4]{McC1986}.

Other special cases for $A$ can also be considered, for ex. $A(n)$ the set of prime
divisors of $n$ or $A(n)$ the set of exponential divisors of $n$.
\end{rem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Asymptotic properties}

\begin{thm} The minimal order of $\widetilde{P}(n)$ is $3n/2$ and the maximal
order of $\log(\widetilde{P}(n)/n)$ is $\log 2 \log n/\log \log n$.
\end{thm}

\begin{proof} From \eqref{gcd-sum_regular_expl} we have
$\widetilde{P}(n)\ge n(3/2)^{\omega(n)}\ge 3n/2$ for every $n\ge 1$,
with equality for $n=2^{\nu}$ ($\nu \ge 1$), giving the minimal
order of $\widetilde{P}(n)$.

For the maximal order take into account \eqref{gcd-sum_maximal},
where the limsup is attained for a sequence of square-free integers
(more exactly for $n_k= \prod_{k/\log^2 k<p\le k} p$, $k\to
\infty$), see \cite[Theorem 4.1]{ChiSit1985}, and use
$\widetilde{P}(n) \le P(n)$ for every $n\ge 1$, with equality iff
$n$ is square-free, by Proposition \ref{prop_ineq}.
\end{proof}

In what follows we prove the following asymptotic formula for the
function $\widetilde{P}$. Let $\psi(n)=n\prod_{p\mid n} (1+1/p)$
denote the Dedekind function,
\begin{equation*}
\alpha(n):=\sum_{p\mid n} \frac{\log p}{p-1}, \quad \beta(n)=
\sum_{p\mid n} \frac{\log p}{p^2-1},
\end{equation*}
\begin{equation*}
\delta(x):=\exp \left(-A(\log x)^{3/5} (\log \log x)^{-1/5}\right),
\end{equation*}
\begin{equation*}
\eta(x):=\exp \left(B \log x (\log \log x)^{-1}\right),
\end{equation*}
where $A$ and $B$ are positive constants and let $\theta$ be the
exponent in \eqref{Dirichlet_divisor}.

\begin{thm} We have
\begin{equation}
\sum_{n\le x} \widetilde{P}(n) = \frac{x^2}{2\zeta(2)} (K_1\log x+
K_2) + {\cal O}(x^{3/2}\delta(x)), \label{P_reg}
\end{equation}
where the constants $K_1$ and $K_2$ are given by
\begin{equation*}
K_1:=\sum_{n=1}^{\infty} \frac{\mu(n)}{n\psi(n)}= \prod_p \left(1-\frac1{p(p+1)}\right),
\end{equation*}
\begin{equation*}
K_2:=K_1\left( 2\gamma-\frac1{2}-\frac{2\zeta'(2)}{\zeta(2)}\right)
- \sum_{n=1}^{\infty} \frac{\mu(n)(\log
n-\alpha(n)+2\beta(n))}{n\psi(n)}.
\end{equation*}

If RH is true, then the error term of \eqref{P_reg} is ${\cal
O}(x^{(7-5\theta)/(5-4\theta)} \eta(x))$.
\end{thm}

\begin{rem}\upshape
Here $K_1\approx 0.7042$. Note that $\ds K_1=\lim_{x\to \infty}
\frac{2}{x^2} \sum_{n\le x} \gamma(n)$, where
$\gamma(n)=\prod_{p\mid n} p$ is the greatest square-free divisor of
$n$. Also, $K_1/\zeta(2)\approx 0.4282$ is the so called ``carefree
constant'', cf.\ \cite[Section 2.5.1]{Fin2003}. For $\theta \approx 0.3149$
one has $(7-5\theta)/(5-4\theta)\approx 1.4505$.
\end{rem}

\begin{proof} We need the following auxiliary results. Let
$\sigma'_s(n)$ be the sum of $s$-th powers of the square-free
divisors of $n$.

\begin{lemma} \label{lemma_2^omega} {\rm (\cite[Theorems 4.3, 5.2]{SurSiv1973})}
If $k\ge 1$ is an integer, then for every $\varepsilon>0$,
\begin{align}
\sum_{n\le x} 2^{\omega(n,k)} = \frac{kx}{\zeta(2)\psi(k)}\left(\log
x+ \alpha(k)-2\beta(k) + 2\gamma -1
-\frac{2\zeta'(2)}{\zeta(2)}\right) \nonumber \\ + {\cal
O}\left(\sigma'_{-1+\varepsilon}(k)\sigma'_{-\theta}(k)
x^{1/2}\delta(x)\right), \label{2^omega}
\end{align} the ${\cal O}$ estimate being uniform in $x$ and $k$.

If RH is true, then $x^{1/2}\delta(x)$ in the error term of
\eqref{2^omega} can be replaced by $x^{(2-\theta)/(5-4\theta)}
\eta(x)$.
\end{lemma}

Note that
\begin{equation}
\alpha(n)={\cal O}(\log n), \ \ \beta(n)={\cal O}(1), \label{error}
\end{equation}
since $\ds \alpha(n)\le \sum_{p\mid n} \log p= \log \gamma(n)\le
\log n$ and $\ds \beta(n)\ll \sum_{p\mid n} \frac{\log p}{p^2}\le
\sum_p \frac{\log p}{p^2}<\infty$.

\begin{lemma} \label{lemma_n2^omega} For every
$\varepsilon>0$,
\begin{align}
\sum_{n\le x} 2^{\omega(n,k)}n =
\frac{kx^2}{2\zeta(2)\psi(k)}\left(\log x+ \alpha(k)-2\beta(k)+
2\gamma -\frac1{2} -\frac{2\zeta'(2)}{\zeta(2)}\right)   \nonumber
\\ + {\cal O}\left(\sigma'_{-1+\varepsilon}(k)\sigma'_{-\theta}(k)
x^{3/2}\delta(x)\right), \label{n2^omega}
\end{align}

If RH is true, then $x^{3/2}\delta(x)$ in the error term of
\eqref{n2^omega} can be replaced by $x^{(7-5\theta)/(5-4\theta)}
\eta(x)$.
\end{lemma}

\begin{proof}
By partial summation from Lemma \ref{lemma_2^omega}.
\end{proof}

We can now complete the proof of Theorem 2. By Proposition
\ref{prop_2} and Lemma \ref{lemma_n2^omega},
\begin{equation}
\sum_{n\le x} \widetilde{P}(n)=  \sum_{d\le x} \mu(d) \sum_{e\le
x/d} 2^{\omega(e,d)} e \nonumber
\end{equation}
\begin{align*}
= \frac{x^2}{2\zeta(2)} \left( \sum_{d\le x} \frac{\mu(d)}{d\psi(d)}
\left(\log x+2\gamma-\frac1{2}-\frac{2\zeta'(2)}{\zeta(2)} \right)-
\sum_{d\le x} \frac{\mu(d)(\log
d-\alpha(d)+2\beta(d))}{d\psi(d)}\right) \nonumber \\  +  {\cal O}
\left(\sum_{d\le x} \sigma'_{-1+\varepsilon}(d)\sigma'_{-\theta}(d)
(x/d)^{3/2} \delta(x/d)\right).
\end{align*}

For every $\varepsilon>0$ and $x$ sufficiently large,
$x^{\varepsilon}\delta(x)$ is increasing, therefore
\begin{equation*}
(x/d)^{3/2} \delta(x/d)= (x/d)^{3/2-\varepsilon} (x/d)^{\varepsilon}
\delta(x/d)\le (x/d)^{3/2-\varepsilon} x^{\varepsilon} \delta(x)=
x^{3/2} \delta(x)/d^{3/2-\varepsilon}.
\end{equation*}

Furthermore, it is enough to use the inequalities
$\sigma'_{-1+\varepsilon}(d)\le \tau(d)$ (for $\varepsilon<1$) and
$\sigma'_{-\theta}(d)\le \tau(d)$ and then obtain the given formula
using \eqref{error} and the well known estimates
\begin{equation*}
\sum_{d>x} \frac1{d^2}\ll \frac1{x}, \ \ \sum_{d>x} \frac{\log d}{d^2}\ll \frac{\log x}{x}.
\end{equation*}

If we assume RH and in the error term use the property that
$\eta(x)$ is increasing, so $\eta(x/d)\le \eta(x)$ for $d\ge 1$.
\end{proof}

\section{Acknowledgement} The author thanks the referee for helpful
suggestions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords:} regular integers (mod $n$), gcd-sum
function, unitary divisor, Dirichlet divisor problem, Riemann
hypothesis.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A055653},
\seqnum{A018804}.)


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\noindent
Received October 6 2008;
revised version received January 21 2009.  
Published in {\it Journal of Integer Sequences}, February 13 2009.

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