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\title[A NEW GENERALIZATION]{\large A NEW GENERALIZATION OF THE EULER TOTIENT FUNCTION}


\author[B. MITTOU]{{\bf BRAHIM MITTOU}\protect\orcidicon{0000-0002-5712-9011}}



\address{Brahim Mittou
\newline\hphantom{iii} Department of Mathematics, University Kasdi Merbah Ouargla, Algeria
\newline\hphantom{iii} EDPNL \& HM Laboratory of ENS Kouba, Algeria}%
\email{\textcolor{blue}{mathmittou@gmail.com, mittou.brahim@univ-ouargla.dz}}%

\thanks{\sc Mittou, B.,
A new generalization of the Euler totient}
\thanks{\copyright \ 2024}
%\thanks{\rm The work is supported by RFFI (grant 03-01-11111)}
\thanks{\it Received January, 1, 2023, Published December, 31, 2023}%


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%\centerline{{\it Communicated by} {\sc P.P. Petrov}
%({\it filled by Editorial board})}


\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} In this paper, for an arithmetic function $f$, we define the related arithmetic function $\varphi_f$  which can be considered as a generalization of the Euler totient function $\varphi$, and we study some of its properties, including multiplicativity, the Dirichlet inverse, and Dirichlet series.\medskip

\noindent{\bf Keywords:} arithmetic function, Euler totient function, Dirichlet product.
 \end{quote}
}

\bigskip

\section{Introduction}
The Euler totient function, denoted by $\varphi(n)$, is a mathematical function that counts the positive integers up to a given integer $n$ that are coprime with $n$; thus,
\begin{equation*}
\varphi(n)=\card\{x\ |\ 1\leq x<n,\ \gcd(x,n)=1\}.
\end{equation*}
It is known that (see, e.g. \cite{1,2}):
$$
\varphi(n)=n\prod_{p|n}\left(1-\dfrac{1}{p} \right),
$$
where the product is taken over all distinct prime factors $p$ of $n$.

Euler totient function is a versatile and essential tool that plays a crucial role in various branches of mathematics and its applications extend beyond number theory into diverse areas of mathematical research and computer science. For this reasons, it has garnered significant attention from many mathematicians, and one form of this interest is to find more than ten generalizations and analogs for it (see \cite[Chapter 3.7]{3}). For example, Schemmel totient function $S_k\ (k\in\mathbb{N})$ is defined as:
\begin{equation*}
S_k(n)=\card\{(x,\ldots,x+k-1)\ |\ 1\leq x+i<n,\ \gcd(x+i,n)=1\ (0\leq i\leq k-1)\},
\end{equation*}
and the Jordan totient function $J_k\ (k\in\mathbb{N})$ is defined as:
\begin{equation*}
J_k(n)=n^k\prod_{p|n}\left(1-\dfrac{1}{p^k} \right).
\end{equation*}
Clearly, we have $S_1(n)=J_1(n)=\varphi(n)$.

In the present paper, we will introduce a new generalization, associated with an arithmetic function $f$, of the Euler totient function. Furthermore, we will discuss some of its properties, including multiplicativity, the Dirichlet inverse, and Dirichlet series. The following arithmetic functions are also well-known and may be used throughout this paper:

$\bullet$ $\mu, \lambda$ are the Möbius and the Liouville functions, respectively.

$\bullet$ $\textbf{\textit{u}}$ is the unit function: $\textbf{\textit{u}}(n)=1$ for all $n$.

$\bullet$ $\textbf{\textit{N}}_\alpha$ ($\alpha\in\mathbb{C}$) is the $\alpha$-th power function: $\textbf{\textit{N}}_\alpha(n)=n^\alpha$, ($\textbf{\textit{N}}_1=\textbf{\textit{N}}$).

$\bullet$ $\psi$ is the Dedekind function: $\psi(n)=n\prod_{p|n}\left(1+\frac{1}{p} \right)$.

\section{Main Result}
Let $f$ be an arithmetic function. Then we define $\varphi_f$ to be the arithmetic function such that:
$$
\varphi_f(n)=n\prod_{p|n}\left(1-\dfrac{f(p)}{p} \right),\quad \varphi_f(1)=1. 
$$
Note that if $f(p)=1$ for all prime numbers $p$, then $\varphi_f=\varphi$. For example:
$$
\varphi_\textbf{\textit{u}}=\varphi,\ \varphi_\mu=\varphi_\lambda=\psi,\text{ and } \varphi_\varphi=-\varphi_\psi.
$$
Let $n$ and $m$ be positive integers and let $d=\gcd(n,m)$. Then
\begin{align*}
\varphi_f(mn)&=mn\prod_{p|mn}\left(1-\dfrac{f(p)}{p} \right)\\
&=\dfrac{\displaystyle m\prod_{p|m}\left(1-\dfrac{f(p)}{p} \right)n\prod_{p|n}\left(1-\dfrac{f(p)}{p} \right)}{\displaystyle \prod_{p|d}\left(1-\dfrac{f(p)}{p} \right)}\\
&=\varphi_f(m)\varphi_f(n)\dfrac{d}{\varphi_f(d)}.
\end{align*} 
In paricular, if $d=1$ then:
$$
\varphi_f(mn)=\varphi_f(m)\varphi_f(n)
$$
which means tha $\varphi_f$ is multiplicative function, so one can have
$$
\varphi_f(n)=\varphi_f\left(\prod_{i=1}^rp_i^{e_i} \right) =\prod_{i=1}^rp_i^{e_i-1}\big(p_i-f(p_i)\big)\quad(n>1).
$$

It is well-known (see, e.g. \cite[p. 29]{1}) that the Euler totient function $\varphi$ satisfies the following relation:
\begin{equation}\label{equ1}
\varphi=\mu*\textbf{\textit{N}},
\end{equation} 
where $*$ represents the Dirichlet product. The next theorem establishes that $\varphi_f$ satisfies a relation analogous to \eqref{equ1}.

\begin{theorem}\label{th21}
Let $f$ be a multiplicative function. Then
\begin{equation}\label{eq1}
\varphi_f=\mu f*\textbf{N}.
\end{equation}
\end{theorem}

\begin{proof}
We note that the equality holds when $n=1$. So, we assume that $n>1$ and let $p_i$ $(1\leq i\leq r)$ be the distinct prime factors of $n$. Thus
\begin{align*}
\dfrac{\varphi_f(n)}{n}=&\prod_{i=1}^{r}\left(1-\dfrac{f(p_i)}{p_i} \right)\\
=&1-\sum\dfrac{f(p_i)}{p_i}+\sum\dfrac{f(p_i)f(p_j)}{p_ip_j}+\cdots+(-1)^r\dfrac{f(p_1)f(p_2)\cdots f(p_r)}{p_1p_2\cdots p_r}\\
=&1-\sum\dfrac{f(p_i)}{p_i}+\sum\dfrac{f(p_ip_j)}{p_ip_j}+\cdots+(-1)^r\dfrac{f(p_1p_2\cdots p_r)}{p_1p_2\cdots p_r}\\
=&\mu(1)+\mu(p_i)\sum\dfrac{f(p_i)}{p_i}+\mu(p_ip_j)\sum\dfrac{f(p_ip_j)}{p_ip_j}+\cdots+\\
&+\cdots+\mu(p_1p_2\cdots p_r)\dfrac{f(p_1p_2\cdots p_r)}{p_1p_2\cdots p_r}\\
=&\sum_{d|n}\dfrac{\mu f(d)}{d},
\end{align*}
since $\mu(d)=0$ if $d$ is not a square-free integer. Note that, in a term such as $\sum\frac{f(p_ip_j)}{p_ip_j}$, it is implied that we are considering all possible products $p_ip_j$
involving distinct prime factors of $n$. Now we can rewrite above equalities as:
$$
\varphi_f(n)=\sum_{d|n}\mu f(d)\dfrac{n}{d}=(\mu f*\textbf{\textit{N}})(n),
$$
as claimed. The proof is complete.
\end{proof}

\begin{corollary} 
Let $f$ be a completely multiplicative function. Then
\begin{equation}\label{eq2}
\varphi_f=f^{-1}*\textbf{N},
\end{equation}
where $f^{-1}$ is the Dirichlet inverse of $f$.
\end{corollary} 

\begin{proof}
The result follows at once from \eqref{eq1} and the fact that $f^{-1}=\mu f$ (see e.g., \cite[Theorem 2.17]{1}).
\end{proof}

The Dirichlet inverse of the Euler totient function is given by (see, e.g. \cite[p. 37]{1}):
$$
\varphi^{-1}=\textbf{\textit{u}}*\mu\textbf{N}.
$$
The following theorem provides the Dirichlet inverse of $\varphi_f$.
\begin{theorem}
If $f$ is multiplicative function, then 
$$
\varphi_f^{-1}=(\mu f)^{-1}*\mu\textbf{N}.
$$
\end{theorem}

\begin{proof}
It is well-known that \cite{1,2} the identity $(g*h)^{-1}=g^{-1}*h^{-1}$ holds true for all arithmetic functions $g$ and $h$. Furthermore,  $\textbf{\textit{N}}^{-1}=\mu\textbf{\textit{N}}$. It follows by using these facts and Theorem \ref{th21} that:
\begin{align*}
\varphi_f=\mu f*\textbf{\textit{N}}&\Rightarrow \varphi_f^{-1}=(\mu f)^{-1}*\textbf{\textit{N}}^{-1}\\
&\Rightarrow \varphi_f^{-1}=(\mu f)^{-1}*\mu\textbf{\textit{N}}.
\end{align*}
As required, the proof is finished.
\end{proof}

\begin{corollary} 
Let $f$ be a completely multiplicative function and let $\prod_{i=1}^rp_i^{e_i}$ be the prime factorization of the positive integer $n>1$. Then
$$
\varphi_f^{-1}(n)=(f*\mu\textbf{N})(n)=\prod_{i=1}^r f(p_i)^{e_i-1}\big(f(p_i)-p_i\big).
$$
\end{corollary} 

\begin{proof}
Because $f$ is completely multiplicative we have $f^{-1}=\mu f$, so 
\begin{align*}
\varphi_f^{-1}(n)&=\big((f^{-1})^{-1}*\mu\textbf{\textit{N}}\big)(n)\\
&=(f*\mu\textbf{\textit{N}})(n)\\
&=\sum_{d|n}d\mu(d)f\left( \dfrac{n}{d}\right).
\end{align*}
We know that $\varphi_f^{-1}$ is multiplicative, which means that it suffices to compute $\varphi_f^{-1}(p_i^{e_i})$ to determine $\varphi_f^{-1}(n)$. Thus
\begin{align*}
\varphi_f^{-1}(p_i^{e_i})&=\sum_{d|p_i^{e_i}}d\mu(d)f\left( \dfrac{p_i^{e_i}}{d}\right)\\
&=\sum_{k=0}^{e_i}p_i^k\mu(p_i^k)f\left( \dfrac{p_i^{e_i}}{p_i^k}\right)\\
&=f(p_i^{e_i})-p_i f(p_i^{e_i-1})\\
&=f(p_i)^{e_i-1}\big(f(p_i)-p_i\big), 
\end{align*}
since $f$ is completely multiplicative. Hence 
$$
\varphi_f^{-1}(n)=\prod_{i=1}^r f(p_i)^{e_i-1}\big(f(p_i)-p_i\big).
$$
This completes the proof.
\end{proof}

\begin{example}
\begin{enumerate}
\item If $f=\textbf{u}$, then 
$$
\varphi_f^{-1}(n)=\varphi^{-1}(n)=\prod_{i=1}^r(1-p_i).
$$
\item If $f=\lambda$, then 
$$
\varphi_f^{-1}(n)=\psi^{-1}(n)=\prod_{i=1}^r(-1)^{e_i}(1+p_i).
$$
\item If $f=\textbf{\textit{N}}_\alpha$, then 
$$
\varphi_f^{-1}(n)=n^\alpha\prod_{i=1}^r\left( 1-\dfrac{1}{p_i^{\alpha-1}}\right).
$$
\end{enumerate}
\end{example}

The Dirichlet series associated with an arithmetic function $f$ is an infinite series of the form \cite{1,2}:
$$
D(f,s)=\sum_{n=1}^\infty\dfrac{f(n)}{n^s}\quad(s\in\mathbb{C}).
$$
The most famous example is the Riemann $\zeta$ function $\zeta(s)=D(\textbf{\textit{u}},s)=\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^s}$ $(\Re(s)>1)$. Since $\varphi_f=\mu f*\textbf{\textit{N}}$, we have according to \cite[Theorem 11.5]{1}:
\begin{equation}\label{eq3}
D(\varphi_f,s)=D(\mu f,s)D(\textbf{\textit{N}},s),
\end{equation}
which holds in the half-plane where both series on the right-hand side of (\ref{eq3}) converge absolutely. In particular, if $f$ is completely multiplicative, then we have the following theorem:
\begin{theorem}
Let $D(f,s)$ be the Dirichlet series associated with a completely multiplicative function $f$. Suppose that $D(f,s)$ is converge absolutely in the half-plane $\Re(s)>a$, then:
\begin{equation*}
D(\varphi_f,s)=\dfrac{\zeta(s-1)}{D(f,s)}\ \big(\Re(s)>\max\{2,a\}\big). 
\end{equation*}
\end{theorem}

\begin{proof}
This follows at once from (\ref{eq2}), (\ref{eq3}), and \cite[Examples 2,3 p. 229]{1}.
\end{proof}

\begin{example}
\begin{enumerate}
\item If $f=\textbf{u}$, then 
$$
D(\varphi_f,s)=D(\varphi,s)=\sum_{n=1}^\infty\dfrac{\varphi(n)}{n^s}=\dfrac{\zeta(s-1)}{\zeta(s)}\ \big(\Re(s)>2\big).
$$
\item If $f=\lambda$, then 
$$
D(\varphi_f,s)=D(\psi,s)=\sum_{n=1}^\infty\dfrac{\psi(n)}{n^s}=\dfrac{\zeta(s)\zeta(s-1)}{\zeta(2s)}\ \big(\Re(s)>2\big).
$$
\item If $f=\textbf{\textit{N}}_\alpha$, then 
$$
D(\varphi_f,s)=\sum_{n=1}^\infty\dfrac{\varphi_f(n)}{n^s}=\dfrac{\zeta(s-1)}{\zeta(s-\alpha)}\ \Big(\Re(s)>\max\{2,1+\Re(\alpha)\}\Big).
$$
\item If $f=\chi$ is a Dirichlet character, then 
$$
D(\varphi_f,s)=\sum_{n=1}^\infty\dfrac{\varphi_f(n)}{n^s}=\dfrac{\zeta(s-1)}{L(s,\chi)}\ \big(\Re(s)>2\big),
$$
where $L(s,\chi)$ is the well know Dirichlet $L$-function associated with $\chi$.
\end{enumerate}
\end{example}


\bigskip
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\end{thebibliography}
\end{document}
