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\newtheorem{thrm}{Theorem}
\newtheorem{corl}{Corollary}
\newtheorem{lmma}{Lemma}
\newtheorem{defn}{Definition}
\newtheorem{expl}{Example}
\title{On a Certain Kind of Generalized Number-Theoretical M\"obius Function}
\author{
Tom C. Brown\footnote{Department of Mathematics and Statistics, Simon Fraser University, Burnaby, BC Canada V5A 1S6. \texttt{tbrown@sfu.ca}.},
~~Leetsch C. Hsu\footnotemark[2],
~~Jun Wang\footnote{Department of Mathematics, Dalian University of Technology, Dalian 116024, China.}\\
and Peter Jau-Shyong Shiue\footnote{Department of Mathematical Sciences, University of Nevada, Las Vegas, NV 89154-4020, USA.}}
\maketitle
\begin{center}{\small {\bf Citation data:} T.C. {Brown}, C.~Hsu Leetsch, Jun Wang, and Peter J.-S. Shiue, \emph{On a
certain kind of generalized number-theoretical {Moebius} function}, Math.
Scientist \textbf{25} (2000), 72--77.}\bigskip\end{center}
\begin{abstract}The classical M\"obius function appears in many places in number theory
and in combinatorial theory. Several different generalizations of this function
have been studied. We wish to bring to the attention of a wider audience a
particular generalization which has some attractive applications. We give some
new examples and applications, and mention some known results.\\
\noindent Keywords: Generalized M\"obius functions; arithmetical functions; M\"obius-type functions; multiplicative functions\\
\noindent AMS 2000 Subject Classification:\\
Primary 11A25; 05A10\\
Secondary 11B75; 20K99
\end{abstract}
~\\
\begin{center}\it This paper is dedicated to the memory of Professor Gian-Carlo Rota\end{center}
\section{Introduction}
We define a generalized M\"obius function $\mu_\alpha$ for each complex number
$\alpha$. (When $\alpha = 1$, $\mu_1$ is the classical M\"obius function.) We
show that the set of functions $\mu_\alpha$ forms an Abelian group with respect
to the Dirichlet product, and then give a number of examples and applications,
including a generalized M\"obius inversion formula and a generalized Euler
function. Special cases of the generalized M\"obius functions studied here have
been used in \cite{haukkanen1997,hsu1995,hsu+wang1998}. For other generalizations see
\cite{apostol1970,haukkanen1995}. For interesting survey articles, see
\cite{barnabei+brini+rota1986,bender+goldman1975,rota1964}.
Let us recall that the classical M\"obius function $\mu(n)$ is defined for
positive integers $n$ in the following way: $\mu(1) = 1$. If $n$ is not square
free then $\mu(n) = 0$. If $n$ is square free and $r$ is the number of distinct
primes dividing $n$, then $\mu(n) = (-1)^r$ \cite{niven+zuckerman+montgomery1991}.
For any integer $r$, a M\"obius function of order $r$ may be defined by using
binomial coefficients, namely for each positive integer $n$,
\[ \mu_r(n) = \prod_{p|n}\binom{r}{\partial_p(n)}(-1)^{\partial_p(n)} \]
where $p$ runs through all the prime divisors of $n$, and $\partial_p(n)
= \ord_p{n}$ denotes the highest power $k$ of $p$ such that $p^k$ divides
$n$. Obviously $\mu_1(n) = \mu(n)$. For more details, see \cite{hsu1995}.
We now define a generalized M\"obius function $\mu_\alpha$ for each complex
number $\alpha$, by setting
\[ \mu_\alpha(n) = \prod_{p|n}\binom{\alpha}{\partial_p(n)}(-1)^{\partial_p(n)} \]
At the end of the paper, we mention a particularly interesting application of
the case where $\alpha$ is real.
\section{Group-theoretic properties}
Recall that the classical M\"obius function is multiplicative; i.e., if $m$ and
$n$ are relatively prime, then $\mu(mn) = \mu(m)\mu(n)$. It is easily seen that
the definition of $\mu_\alpha$ implies that this property extends to the
M\"obius function of order $\alpha$, giving us the following lemma.
\begin{lmma} For each complex number $\alpha$, $\mu_\alpha$ is a multiplicative
function.\label{l1}\end{lmma}
Next, we recall the definition of the Dirichlet product (or convolution) of two
arithmetic functions $f$ and $g$ (cf. \cite{apostol1970,beumer1962}).
\begin{defn} Given two arithmetic functions $f$ and $g$, the Dirichlet (convolution)
product $f*g$ is again an arithmetic function which is defined by
\[ (f*g)(n) = \sum_{d|n}f(d)g\left(\frac{n}d\right) = \sum_{d|n}f\left(\frac{n}d\right)g(d), \]
where the summations are taken over all positive divisors $d$ of $n$.\end{defn}
Evidently, the product is commutative: $f*g = g*f$. Using a little algebra one
easily shows that the following associative law also holds: $(f*g)*h = f*(g*h)$.
That is, for all positive integers $n$, $((f*g)*h)(n) = (f*(g*h))(n)$. Moreover,
the convolution $f*g$ is a multiplicative function whenever $f$ and $g$ are
multiplicative functions.
\begin{defn} Let
\[ M = \{ \mu_\alpha : \alpha \in\mathbb{C} \} \]
where $\mathbb{C}$ denotes the set of complex numbers. The set $M$ may be
called the set of generalized M\"obius functions of complex order.\end{defn}
\begin{lmma} For any given numbers $\alpha$ and $\beta$ in $\mathbb{C}$, we
have
\[ \mu_\alpha*\mu_\beta = \mu_{\alpha+\beta} \]\label{l2}\end{lmma}
\begin{proof} It is required to show that for all positive integers $n$,
\[ (\mu_\alpha*\mu_\beta)(n) = \sum_{d|n}\mu_\alpha(d)\mu_\beta\left(\frac{n}d\right) = \mu_{\alpha+\beta}(n). \]
Since $\mu_\alpha$ and $\mu_\beta$ are multiplicative (by Lemma \ref{l1}),
the Dirichlet product $\mu_\alpha*\mu_\beta$ is also multiplicative. Thus,
it suffices to consider the case $n = p^k$, where $p$ is prime and $k$ is
a positive integer. We easily find
\begin{align*}
(\mu_\alpha*\mu_\beta)(p^k)
&= \sum_{d|p^k} \mu_\alpha(d)\mu_\beta\left(\frac{p^k}d\right) = \sum_{i=0}^k \mu_\alpha(p^i)\mu_\beta(p^{k-i}) \\
&= \sum_{i=0}^k \binom{\alpha}{i}(-1)^i\binom{\beta}{k - i}(-1)^{k-i} \\
&= (-1)^k\binom{\alpha + \beta}{k} = \mu_{\alpha+\beta}(p^k),
\end{align*}
since the relation $(1 + x)^\alpha(1 + x)^\beta = (1 + x)^{\alpha+\beta}$
implies
\[ \binom{\alpha+\beta}{k} = \sum_{i=0}^k\binom{\alpha}{i}\binom{\beta}{k - i}. \]
\end{proof}
Notice that $\mu_0$ is the M\"obius function of order zero that gives the values
\[ \mu_0(n) = \prod_{p|n}\binom{0}{\partial_p(n)}(-1)^{\partial_p(n)} = \left\{\begin{array}{cc} 1&n=1,\\0&n > 1.\end{array}\right. \]
Let us denote $\mu_0$ by $\delta$. Since from Lemma \ref{l2} we have $\mu_\alpha
*\delta = \delta*\mu_\alpha = \mu_\alpha$ for all $\alpha$, we call it the identity
element with respect to the Dirichlet product operation $*$.
We are now ready to show that $M$ is an Abelian group.
\begin{thrm} $(M,*)$ is an Abelian group with identity element $\delta = \mu_0$.
\label{t1}\end{thrm}
\begin{proof} By Lemma \ref{l2} we see that $M$ is closed with respect to the
operation $*$. Moreover, we also have
\[ \mu_\alpha*\mu_\beta = \mu_\beta*\mu_\alpha \qquad (\alpha,\beta\in\mathbb{C}), \]
\[ (\mu_\alpha*\mu_\beta)*\mu_\gamma = \mu_\alpha*(\mu_\beta*\mu_\gamma) \qquad (\alpha,\beta,\gamma\in\mathbb{C}), \]
\[ \mu_\alpha*\delta = \delta*\mu_\alpha = \mu_\alpha\qquad \mu_\alpha*\mu_{-\alpha} = \mu_{-\alpha}*\mu_\alpha = \delta \qquad (\alpha\in\mathbb{C}), \]
Thus, the theorem is proved.\end{proof}
Of course, if $G$ is any additive subgroup of $\mathbb{C}$, then $M_G = \{\mu_\alpha : \alpha\in G\}$
is a subgroup of $M$.
\section{Corollaries, examples and applications}
\begin{corl} (Generalized M\"obius inversion formulae).) For all $\alpha\in\mathbb{C}$
and arithmetic functions $f,g$,
\[ \left[\forall n\in\mathbb{N}~~f(n) = \sum_{d|n}\mu_\alpha\left(\frac{n}d\right)g(d)\right]
\Leftrightarrow \left[\forall n\in\mathbb{N}~~g(n) = \sum_{d|n}\mu_{-\alpha}\left(\frac{n}d\right)f(d)\right]. \]
\label{c1}\end{corl}
\begin{proof} In fact, this is equivalent to the statement
\[ f = \mu_\alpha*g \Leftrightarrow g = \mu_{-\alpha}*f, \]
which follows from
\[ f = \mu_{\alpha}*g \Leftrightarrow \mu_{-\alpha}*f = \mu_{-\alpha}*\mu_\alpha*g = \delta*g = g. \]
\end{proof}
Evidently, Corollary \ref{c1} with $\alpha = 1$ implies the classical M\"obius inversion
formulae $(f = \mu*g \Leftrightarrow g = \mu_{-1}*f$), since $\mu_1 = \mu$ and $\mu_{-1}\equiv 1$:
\[ \mu_{-1}(n) = \prod_{p|n} \binom{-1}{\partial_p(n)}(-1)^{\partial_p(n)}
= \prod_{p|n} \frac{(\partial_p(n))!}{(\partial_p(n))!} = 1. \]
Note here that the M\"obius $\mu$-function and $\mu_{-1}\equiv1$ are inverses
of each other under convolution.
\begin{corl} For all $n\in\mathbb{N}$ and $\alpha\in\mathbb{C}$,
\[ \sum_{d|n} \mu_\alpha(d) = \mu_{\alpha-1}(n). \]\label{c2}\end{corl}
This is equivalent to the statement $(\mu_{-1}*\mu_\alpha)(n) = \mu_{\alpha-1}(n)$.
Note that the case $\alpha = 1$ gives the classical identity of Gauss
\[ \sum_{d|n} \mu(d) = \mu_0(n) = \delta(n). \]
\begin{corl} Let $f$ be a completely multiplicative function such that
$f(mn) = f(n)f(m)$ for all positive integers $m$ and $n$, and let $r$
be a positive integer. Then the $r$-times convolution of $\mu_rf$ with
$f$ satisfies
\[ (\mu_rf)*f*f*\cdots*f = \mu_0f, \]
where $(\mu_\alpha f)(n) = \mu_\alpha(n)f(n)$.\label{c3}\end{corl}
This follows easily form Corollary \ref{c2} and induction on $r$. Indeed
we have
\[ ((\mu_rf)*f)(n) = \sum_{d|n}\mu_r(d)f(d)f\left(\frac{n}d\right) = f(n)\sum_{d|n}\mu_r(d) = (\mu_{r-1}f)(n). \]
Moreover, it may be of interest to note that
\[ \mu_{-2}(n) = (\mu_{-1}*\mu_{-1})(n) = \sum_{d|n}1 = \tau(n), \]
where $\tau(n)$ denotes the number of positive divisors of $n$. Thus, $\tau
= \mu_{-2}$. Consequently, from $\mu_{-2}*\mu_1 = \mu_{-1}$ and $\mu_{-2}*\mu_2
= \mu_0$, we may obtain the identities
\[ \sum_{d|n}\tau(d)\mu\left(\frac{n}d\right) = 1 \text{ and } \sum_{d|n}\tau(d)\mu_2\left(\frac{n}d\right) = \delta(n). \]
\begin{expl} Let $\sigma_r(n)$ denote the sum of the $r$th powers of
the divisors of $n$. The well-known identity
\[ n^r = \sum\mu(d)\sigma_r\left(\frac{n}d\right) \]
can be proved very simply in the following way. Let $i_r(n) = n^r$. Then,
since $\mu_{-1}\equiv1$, we have $i_r*\mu_{-1}=\sigma_r$, and hence
\[ (\mu*\sigma_r)(n) = (\mu*i_r*\mu_{-1})(n) = (i_r*\mu*\mu_{-1})(n) = (i_r*\delta)(n) = i_r(n) = n^r. \]
\label{e1}
\end{expl}
\begin{expl} Euler's $\varphi$-function may be written as $\varphi
= i_1*\mu_1$. Moreover, using $\tau = \mu_{-2}$ we can easily
prove the identity
\[ \sigma(n) = \sum_{d|n} \varphi(d)\tau\left(\frac{n}d\right). \]
In fact, these statements follow easily from the relations
\[ \varphi(n) = \sum_{d|n} d\mu\left(\frac{n}d\right) = (i_1*\mu_1)(n) \]
and $\varphi*\tau = (i_1*\mu_1)*\mu_{-2} = i_1*\mu_{-1} = \sigma$ (see
Example \ref{e1}).\label{e2}\end{expl}
\begin{expl}Fix a positive integer $r\geq1$, and define $\varphi_r = i_1*\mu_r$.
Then, if $n$ is `$r$-powerful', that is, $\partial_p(n)\geq r$ for every
prime divisor $p$ of $n$, we have
\[ \varphi_r(n) = n\prod_{p|n}\left(1 - \frac1p\right)^r. \]
This may be verified as follows:
\[ \varphi_r(n) = \sum_{d|n} d\mu_r\left(\frac{n}d\right) = n\sum_{d|n}\frac{\mu_r(d)}{d}
= n\prod_{p|n}\sum_{j=0}^r\binom{r}{j}\left(-\frac1p\right)^j = n\prod_{p|n}\left(1-\frac1p\right)^r. \]
\end{expl}
Note that, if $r = 1$, then $\varphi_1 = \varphi$ is the classical
Euler function. Thus, $\varphi_r$ may be called the generalized Euler
function of order $r$. This function has a similar meaning to that of
$\varphi$, in that $\phi_r$ counts the number of integers $a$, $1\leq
a\leq n$, such that $a$ is `$r$th-degree prime to $n$'. (This means
that for each prime divisor $p$ of $n$, there are $a_0,a_1,\ldots,
a_{r-1}$ with $0 0$, using his exponential and logarithmic operators $\Exp$
and $\Log$, as $f^\alpha = \Exp(\alpha\Log f)$. Here $(\Log f)(1) = \log f(1)$
and
\[ (\Log f)(n) = (\log n)^{-1}\sum_{d|n} f(d)f^{(-1)}\left(\frac{n}d\right) \log d \qquad \text{for }n> 1,\]
where $f^{(-1)}$ is the Dirichlet inverse of $f$, and $\Exp = (\Log)^{-1}$.
Recently Haukkanen \cite{haukkanen1997} proved the following result: if $f$
is a completely multiplicative function and $\alpha$ is a real number, then
$f^\alpha = \mu_{-\alpha}f$. This result shows that the representation
problem for $f^\alpha$ can be nicely solved by using $\mu_\alpha$.
\end{expl}
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