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\begin{center}
\vskip 1cm{\LARGE\bf Combinatorial Identities Involving the \\
\vskip .1in
M\"obius Function
}
\vskip 1cm
\large
Mohamed El Bachraoui and Mohamed Salim\\
Department of Mathematical Sciences\\
United Arab Emirates University\\
P.O. Box 17551\\
Al-Ain\\
United Arab Emirates\\
\href{mailto:melbachraoui@uaeu.ac.ae}{\tt melbachraoui@uaeu.ac.ae}\\
\href{mailto:msalim@uaeu.ac.ae}{\tt msalim@uaeu.ac.ae}
\end{center}

\vskip .2 in

\begin{abstract}
We derive some identities and inequalities concerning the
M\"obius function. Our main
tool is phi functions for intervals of positive integers and their
unions.
\end{abstract}


\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}


 \section{Introduction} \label{sec:introduction}
 The \emph{M\"obius function} $\mu$ is an important arithmetic function
 in number theory and combinatorics that appears in various
 identities. We mention
 the following identities which are well known and can be found
 in books on elementary number theory and arithmetic functions. Let $n$ be a positive integer. Then
 \[
 \sum_{d|n} \mu(d) =
 \begin{cases}
 1,\ \text{if $n=1$;} \\
 0,\ \text{if $n&gt;1$.}
 \end{cases}
 \]
 If $\tau(n)$ is the number of divisors of $n$, then
 \[
 \sum_{d|n} \mu(d) \tau(n/d) = 1.
 \]
 If $n= \prod_{i=1}^{r} p_i ^{k_i}$ is the prime decomposition of $n$, then
 \[
 \sum_{d|n} \mu(d) \lambda (d) = 2^r,
 \]
 where $\lambda$ denotes the \emph{Liouville lambda} function defined as follows:\\
 If $m=\prod_{i=1}^{s} p_i ^{l_i}$ is the prime decomposition of $m$, then
 \[ \lambda (m) = (-1)^{\sum_{i=1}^{s} l_i}. \]
 %
 For a survey on combinatorial identities we refer to Hall \cite{Hall} and Riordan \cite{Riordan} and their references.
 %
 In this note we shall prove the following two theorems on identities involving the M\"obius mu function.
 %
 \begin{theorem} \label{identities1}
 Let $m$ and $n$ be positive integers such that $n&gt;1$. Then
 we have
 \[ (a)\quad
  \sum_{d|n} \mu(d) 2^{\lfloor \frac{m}{d} \rfloor - \lfloor \frac{m-1}{d} \rfloor} =
  \sum_{d|n} \mu(d) \left( \left\lfloor \frac{m}{d} \right\rfloor -
  \left\lfloor \frac{m-1}{d} \right\rfloor \right) =
  \begin{cases}
  0,\ \text{if $\gcd(m,n) &gt;1$;} \\
  1,\ \text{if $\gcd(m,n) =1$}.
  \end{cases}
  \]
  \begin{multline*} (b)\quad
  \sum_{d|n} \mu(d) 2^{\lfloor \frac{m+1}{d} \rfloor - \lfloor \frac{m-1}{d} \rfloor} = \\
  \begin{cases}
  1,\ \text{if $\gcd(m,n)&gt;1$ and $\gcd(m+1,n)&gt;1$;}\\
  2,\ \text{if $\gcd(m,n)=1$ and $\gcd(m+1,n)&gt;1$ or $\gcd(m,n)&gt;1$ and
           $\gcd(m+1,n)=1$;} \\
  3,\ \text{if $\gcd(m,n)=\gcd(m+1,n)=1$}.
  \end{cases}
  \end{multline*}
 \end{theorem}
 %
 \begin{theorem} \label{identities2}
 Let $m$ and $n$ be positive integers such that $n&gt;1$. Then
 we have:
 \begin{multline*} (a)\quad
 \sum_{d|n} \mu(d) \left(\left\lfloor \frac{m+1}{d} \right\rfloor - \left\lfloor \frac{m-1}{d} \right\rfloor \right)= \\
 \begin{cases}
  0,\ \text{if $\gcd(m,n)&gt;1$ and $\gcd(m+1,n)&gt;1$;}\\
  1,\ \text{if $\gcd(m,n)=1$ and $\gcd(m+1,n)&gt;1$ or $\gcd(m,n)&gt;1$ and
           $\gcd(m+1,n)=1$;} \\
  2,\ \text{if $\gcd(m,n)=\gcd(m+1,n)=1$}.
  \end{cases}
  \end{multline*}
  \begin{multline*} (b)\quad
  \sum_{d|n} \mu(d) \binom{\lfloor \frac{m+1}{d} \rfloor - \lfloor \frac{m-1}{d} \rfloor +1}{2} = \\
  \begin{cases}
  1,\ \text{if $\gcd(m,n)&gt;1$ and $\gcd(m+1,n)&gt;1$;}\\
  2,\ \text{if $\gcd(m,n)=1$ and $\gcd(m+1,n)&gt;1$ or $\gcd(m,n)&gt;1$ and
           $\gcd(m+1,n)=1$;} \\
  3,\ \text{if $\gcd(m,n)=\gcd(m+1,n)=1$}.
  \end{cases}
  \end{multline*}
 \end{theorem}
%
 Our proofs are combinatorial with phi functions for integer subsets as a main tool.
 %
 For the sake of completeness we include the following result which is a natural
 extension of El Bachraoui \cite[Theorem 2 (a)]{ElBachraoui1} on M\"obius inversion for arithmetical functions in several variables.
 For simplicity we let
 \[
 (\overline{m}_a,\overline{n}_b)= (m_1,m_2,\ldots,m_a,n_1,n_2,\ldots,n_{b})
 \]
 and
 \[
 \left( \frac{\overline{m}_a}{d}, \left[\frac{\overline{n}_b}{d}\right]\right) =
 \left( \frac{m_1}{d},\frac{m_2}{d}, \ldots, \frac{m_a}{d},
 \left[\frac{n_1}{d}\right],\left[\frac{n_2}{d}\right],\ldots,\left[\frac{n_{b}}{d}\right]\right).
 \]
 %
 \begin{theorem} \label{thm:inversion}
 If $F$ and $G$ are arithmetical of $a+b$ variables,
 then
 \[
 G(\overline{m}_a,\overline{n}_b)=\sum_{d|\gcd(m_1,m_2,\ldots,m_a)}F
 \left(\frac{\overline{m}_a}{d},
\left[\frac{\overline{n}_b}{d}\right]\right)\]
if and only if
\[
 F(\overline{m}_a,\overline{n}_b)=\sum_{d|\gcd(m_1,m_2,\ldots,m_a)}\mu(d)G\left(\frac{\overline{m}_a}{d},
 \left[\frac{\overline{n}_b}{d}\right]\right).
 \]
 \end{theorem}
 %
\section{Phi functions} \label{sec:phifunction}
%
 Throughout let $k$, $l$, $m$, $l_1$, $l_2$, $m_1$, $m_2$ and $n$ be positive integers such that $l\leq m$ and
 $l_1\leq m_1 \leq l_2\leq m_2$, let
 $[l,m]=\{l,l+1,\ldots,m\}$, and let $A$ be a nonempty finite set of positive integers.
 The set $A$ is called \emph{relatively prime to $n$}
  if $\gcd(A\cup \{n\}) = \gcd(A,n) = 1$.
  %
  \begin{definition}
  Let
  \[ \Phi(A,n) = \# \{X\subseteq A:\ X\not=
  \emptyset\ \text{and\ } \gcd(X,n) = 1 \}
  \]
  and
  \[ \Phi_k (A,n) = \# \{X\subseteq A:\ \# X=
  k \ \text{and\ } \gcd(X,n) = 1 \}.
  \]
 \end{definition}
 %
 %
 Nathanson, among other things, introduced $\Phi(n)$ and $\Phi_k(n)$ (in our terminology $\Phi([1,n],n)$
 and $\Phi_k([1,n],n)$ respectively) along with their formulas in Nathanson \cite{Nathanson}.
 %\begin{equation}\label{phi1}
 %\begin{split}
 %\Phi(n)  = \Phi([1,n],n) &amp; = \sum_{d|n}\mu(d) 2^{n/d}, \\
 %\Phi_k(n)  = \Phi_k ([1,n],n) &amp; = \sum_{d|n}\mu(d)\binom{n/d}{k}.
 %\end{split}
 %\end{equation}
 Formulas for $\Phi([m,n],n)$ and $\Phi_k([m,n],n)$ can be found in El Bachraoui \cite{ElBachraoui1} and Nathanson and Orosz \cite{Nathanson_Orosz}
  and formulas for $\Phi([1,m],n)$ and $\Phi_k([1,m],n)$ are obtained in
  El Bachraoui \cite{ElBachraoui2}.
 %
 Ayad and Kihel in \cite{Ayad-Kihel1, Ayad-Kihel2} considered extensions to sets in arithmetic progression and obtained
 formulas for $\Phi([l,m],n)$ and $\Phi_k ([l,m],n)$ as consequences.
 Recently the following formulas for $\Phi([1,m_1]\cup [l_2,m_2])$ and $\Phi_k([1,m_1]\cup [l_2,m_2])$ have been found in
 El Bachraoui \cite{ElBachraoui3}.
 %
 %
 \begin{theorem}\label{thm:main1}
 We have
  \[
 \begin{split} (a)\quad
  \Phi([1,m_1]\cup [l_2,m_2],n) &amp;= \sum_{d|n}\mu(d)
   2^{\lfloor \frac{m_1}{d} \rfloor +\lfloor \frac{m_2}{d} \rfloor - \lfloor\frac{l_2 -1}{d} \rfloor}, \\
  (b)\quad
  \Phi_k ([1,m_1]\cup [l_2,m_2],n) &amp;=
   \sum_{d|n} \mu(d)
   \ \binom{\lfloor \frac{m_1}{d} \rfloor +\lfloor \frac{m_2}{d} \rfloor - \lfloor\frac{l_2 -1}{d} \rfloor}{k}.
   \end{split}
  \]
 \end{theorem}
 %
 %
 \section{Phi functions for $[l_1,m_1]\cup [l_2,m_2]$}
 %
 We need two lemmas.
 \begin{lemma} \label{lemma1}
 Let
 \[
 \Psi(l_1,m_1,l_2,m_2, n)= \# \{X \subseteq [l_1,m_1]\cup [l_2,m_2]:\ l_1, l_2\in X\
 \text{and\ } \gcd(X,n)=1 \},
 \]
 \[
 \Psi_k(l_1,m_1,l_2,m_2,n)=\# \{X \subseteq [l_1,m_1]\cup [l_2,m_2]:\ l_1,l_2\in X,\ |X| = k,\
 \text{and\ } \gcd(X,n)=1 \}.
 \]
 Then
 \[
 \text{(a)\quad } \Psi(l_1,m_1,l_2,m_2, n) = \sum_{d|\gcd(l_1,l_2,n)}\mu(d)
  2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor- \frac{l_1+l_2}{d}}, \]
\[
 \text{(b)\quad } \Psi_k (m_1,l_2,m_2, n) = \sum_{d|\gcd(l_1,l_2,n)}\mu(d)
 \binom{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor- \frac{l_1+l_2}{d}}{k-2}.
 \]
 \end{lemma}
 \begin{proof}
 (a) Assume first that $m_2\leq n$. Let $\mathcal{P}(l_1,m_1,l_2,m_2)$ denote the set of subsets of
 $[l_1,m_1]\cup[l_2,m_2]$
 containing $l_1$ and $l_2$ and let $\mathcal{P}(l_1,m_1,l_2,m_2,d)$ be the set of subsets $X$ of
 $[l_1,m_1]\cup[l_2,m_2]$ such that
 $l_1,l_2\in X$ and $\gcd(X,n) = d$. It is clear that the set $\mathcal{P}(l_1,m_1,l_2,m_2)$
 of cardinality $2^{m_1+m_2-l_1-l_2}$  can be
 partitioned using the equivalence relation of having the same $\gcd$ (dividing $l_1$, $l_2$ and $n$).
 Moreover, the mapping
 $A \mapsto \frac{1}{d} X$
 is a one-to-one correspondence between
 $\mathcal{P}(l_1,m_1,l_2,m_2,d)$ and the
 set of subsets $Y$ of
 $[l_1/d, \lfloor m_1/d \rfloor ]\cup [l_2/d,\lfloor m_2/d \rfloor]$
 such that $l_1/d, l_2/d \in Y$ and $\gcd(Y,n/d)= 1$. Then
 \[
  \# \mathcal{P}(l_1,m_1,l_2,m_2,d) =
  \Psi(l_1/d, \lfloor m_1/d \rfloor,l_2 /d,\lfloor m_2/d \rfloor,n/d).
 \]
 Thus
 \[
 2^{m_1+m_2-l_1-l_2} = \sum_{d|(l_1,l_2,n)} \# \mathcal{P}(l_1,m_1,l_2,m_2,d)=
 \sum_{d|(l_1,l_2,n)} \Psi (l_1/d,\lfloor m_1/d \rfloor,l_2 /d,\lfloor m_2/d \rfloor,n/d),
 \]
 which by Theorem \ref{thm:inversion} is equivalent to
 \[
 \Psi(l_1,m_1,l_2,m_2,n) = \sum_{d|(l_1,l_2,n)}\mu(d)
 2^{\lfloor m_1/d \rfloor + \lfloor m_2/d \rfloor - (l_1+l_2)/d}.
 \]
  %
 Assume now that $m_2 &gt;n$ and let $a$ be a positive integer such that $m_2 \leq n^a$.
 As $\gcd(X,n)=1$ if and only if $\gcd(X,n^a)=1$ and $\mu(d) =0$ whenever $d$ has a nontrivial
 square factor, we have
 \[
 \begin{split}
 \Psi(l_1,m_1,l_2,m_2,n) &amp;= \Psi(l_1,m_1,l_2,m_2,n^a) \\
 &amp;=
 \sum_{d|(l_1,l_2,n^a)}\mu(d)
 2^{\lfloor m_1/d \rfloor + \lfloor m_2/d \rfloor - (l_1+l_2)/d} \\
 &amp;=
 \sum_{d|(l_1,l_2,n)}\mu(d)
 2^{\lfloor m_1/d \rfloor + \lfloor m_2/d \rfloor - (l_1+l_2)/d}.
 \end{split}
 \]
 (b) For the same reason as before, we may assume that $m_2 \leq n$.
  Noting that the correspondence
 $X\mapsto \frac{1}{d} X$ defined above preserves the cardinality and using an argument similar to
 the one in part (a), we obtain the following identity
 \[
  \binom{m_1+m_2-l_1-l_2}{k-2}=
  \sum_{d|(l_1,l_2,n)}\Psi_k (l_1/d,\lfloor m_1/d \rfloor,l_2 /d,\lfloor m_2/d \rfloor, n/d )
  \]
  which by Theorem \ref{thm:inversion} is equivalent to
  \[
   \Psi_k (l_1,m_1,l_2,m_2,n) =
   \sum_{d|(l_1,l_2,n)}\mu(d)\binom{\lfloor m_1/d \rfloor + \lfloor m_2/d \rfloor - (l_1+l_2)/d }{k-2}.
 \]
\end{proof}
%
 By arguments similar to the ones in the proof of Lemma \ref{lemma1} we have:
 %
 \begin{lemma} \label{lemma2}
 Let
 \[
 \psi(l,m,n) = \# \{X\subseteq [l,m]:\ l \in X\ \text{and\ } \gcd(X,n)=1 \},
 \]
 \[
 \psi_k(l,m,n) = \# \{X\subseteq [l,m]:\ l \in X,\ \# X=k,\ \text{and\ } \gcd(X,n)=1 \}.
 \]
 Then
 \[
 \psi(l,m,n) = \sum_{d|\gcd(l,n)} \mu(d) 2^{\lfloor m/d \rfloor - l/d} ,
 \]
 \[
 \psi_k(l,m,n) = \sum_{d|\gcd(l,n)} \mu(d) \binom{\lfloor m/d \rfloor - l/d}{k}.
 \]
 \end{lemma}
 %
 We are now ready to prove the main theorem of this section.
 \begin{theorem} \label{main2}
 We have
 \[ (a)\quad
 \Phi([l_1,m_1]\cup[l_2,m_2],n) = \sum_{d|n}\mu(d)
  2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor -
    \lfloor \frac{l_1-1}{d} \rfloor - \lfloor \frac{l_2-1}{d} \rfloor},
 \]
 \[ (b)\quad
 \Phi_k([l_1,m_1]\cup[l_2,m_2],n) = \sum_{d|n}\mu(d)
  \binom{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor -
    \lfloor \frac{l_1-1}{d} \rfloor - \lfloor \frac{l_2-1}{d} \rfloor}{k}.
 \]
 \end{theorem}
 %
 \begin{proof}
 (a)\ Clearly
 \begin{multline*}
 \Phi([l_1,m_1]\cup[l_2,m_2],n) = \\ (1)\quad\quad\quad\qquad\qquad
 \Phi([1,m_1] \cup [l_2,m_2],n) -
 \sum_{i=1}^{l_1 -1} \sum_{j=l_2}^{m_2} \Psi(i,m_1,j,m_2,n) -
 \sum_{i=1}^{l_1-1} \psi(i,m_1,n) = \\
 \sum_{d|n}\mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor -
    \lfloor \frac{l_2-1}{d} \rfloor} -
 \sum_{i=1}^{l_1 -1} \sum_{j=l_2}^{m_2} \sum_{d|(i,j,n)} \mu(d)
    2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor - \frac{i+j}{d}} -
 \sum_{i=1}^{l_1-1} \sum_{d|(i,n)} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor - \frac{i}{d}},
 \end{multline*}
 where the second identity follows by Theorem \ref{thm:main1}, Lemma \ref{lemma1}, and
 Lemma \ref{lemma2}.
 Rearranging the triple summation in identity (1), we get
 \[
 \begin{split}
 \sum_{i=1}^{l_1 -1} \sum_{j=l_2}^{m_2} \sum_{d|(i,j,n)} \mu(d)
    2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor - \frac{i+j}{d}}
 &amp;=
  \sum_{d|n}\sum_{\substack{i=1 \\ d|i}}^{l_1 -1} \sum_{\substack{j=l_2\\d|j}}^{m_2}
  \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor-\frac{i+j}{d}} \\
  &amp;=
 \sum_{d|n} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor}
  \sum_{i=1}^{\lfloor\frac{l_1-1}{d}\rfloor} 2^{-i}
   \sum_{j=\lfloor \frac{l_2-1}{d} \rfloor +1}^{\lfloor \frac{m_2}{d} \rfloor} 2^{-j} \\
 (2) \qquad\qquad\qquad \quad\qquad\qquad\qquad
 &amp;=
  \sum_{d|n} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor - \lfloor\frac{l_2-1}{d}\rfloor}
   (1- 2^{-\lfloor \frac{m_2}{d} \rfloor + \lfloor\frac{l_2 -1}{d} \rfloor})
   \sum_{i=1}^{\lfloor\frac{l_1-1}{d}\rfloor} 2^{-i} \\
 &amp; =
  \sum_{d|n} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor - \lfloor\frac{l_2-1}{d}\rfloor}
  (1- 2^{-\lfloor \frac{m_2}{d} \rfloor + \lfloor\frac{l_2 -1}{d} \rfloor})
  (1- 2^{-\lfloor \frac{l_1-1}{d} \rfloor}).
 \end{split}
 \]
 %
 Similarly the double summation in identity (1) gives
 \[ (3)\qquad\qquad\qquad\qquad
 \sum_{i=1}^{l_1-1} \sum_{d|(i,n)} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor - \frac{i}{d}}
 = \sum_{d|n} \mu(d) 2^{\lfloor \frac{m_1}{d} \rfloor}(1- 2^{-\lfloor \frac{l_1-1}{d} \rfloor}).
 \]
 Combining identities (1), (2), and (3) we find
 \[
 \Phi([l_1,m_1]\cup[l_2,m_2],n) = \sum_{d|n}\mu(d)
 2^{\lfloor \frac{m_1}{d} \rfloor + \lfloor \frac{m_2}{d} \rfloor -
    \lfloor \frac{l_1-1}{d} \rfloor - \lfloor \frac{l_2-1}{d} \rfloor}.
 \]
 This completes the proof of part (a). Part (b) follows by similar ideas.
 %
 \end{proof}
 %
 \begin{corollary}\label{corollary1} (Ayad and Kihel \emph{\cite{Ayad-Kihel2}})
 We have
 \[ \Phi([l,m],n)= \sum_{d|n}2^{\lfloor\frac{m}{d}\rfloor - \lfloor\frac{l-1}{d}\rfloor}, \]
 \[ \Phi_k([l,m],n)= \sum_{d|n}\binom{\lfloor\frac{m}{d}\rfloor - \lfloor\frac{l-1}{d}\rfloor}{k}. \]
 \end{corollary}
 \begin{proof}
 Use Theorem \ref{main2} with $l_1=l$, $m_1=m-1$, and $l_2=m_2=m$.
 \end{proof}

 \section{Proofs of the main results}
 %
 \noindent
 \emph{Proof of Theorem \ref{identities1}.\ }
(a) Apply Corollary~\ref{corollary1} to $l=m$ and $k=1$. \\
(b) Apply Corollary~\ref{corollary1} to the interval $[m,m+1]$. \\


 \noindent
 %
 \emph{Proof of Theorem \ref{identities2}.\ }
 (a) Apply Theorem \ref{main2}(b) to $l_1=m$, $m_1=m+1$, $l_2=m_2=n$, and $k=1$ and use the fact that
 $\sum_{d|n} \mu(d) = 0$ whenever $n&gt;1$. \\
 (b) Apply Theorem \ref{main2}(b) to $l_1=m$, $m_1=m+1$, $l_2=m_2=n$, and $k=2$.

\vspace{1cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \section{Acknowledgment} The authors are grateful to the referee
 for valuable comments and interesting suggestions.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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 \bibitem{Riordan}
 J. Riordan,
 {\it Combinatorial Identities}, Wiley, 1968.

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11B05, 11B75.

\noindent \emph{Keywords: } Combinatorial identities,
M\"obius function, relatively prime sets, phi functions,
M\"obius inversion formula.

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\noindent (Concerned with sequence
\seqnum{A0xxxxx}.)

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\noindent
Received March 2 2010;
revised version received  August 7 2010.
Published in {\it Journal of Integer Sequences}, September 3 2010.

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\noindent
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