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\begin{center}
\vskip 1cm{\LARGE\bf 
Extremal Orders of Certain Functions  \\
\vskip .1in
Associated with Regular Integers (mod $n$)
}
\vskip 1cm
\large
Br\u adu\c t  Apostol\\
``Spiru Haret" Pedagogical High School \\
1 Timotei Cipariu St.  \\
RO --- 620004 Foc\c sani \\
Romania\\
\href{mailto:apo_brad@yahoo.com}{\tt apo\_brad@yahoo.com}\\
\ \\
Lucian Petrescu\\
``Henri Coand\u a" Technical College\\
2 Tineretului St. \\
RO --- 820235 Tulcea \\
Romania\\
\href{mailto:petrescuandreea@yahoo.com}{\tt petrescuandreea@yahoo.com}\\
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\begin{abstract}
Let $V(n)$ denote the number of positive regular integers (mod $n$)
that are $\leq n$, and let $V_k(n)$ be a multidimensional
generalization of the arithmetic function $V(n)$.  We find the
Dirichlet series of $V_k(n)$ and give the extremal orders of some
totients involving arithmetic functions which generalize the
sum-of-divisors function and the Dedekind function. We also give the
extremal orders of other totients regarding arithmetic functions which
generalize the sum of the unitary divisors of $n$ and the unitary
function corresponding to $\phi(n)$, the Euler function. Finally, we
study extremal orders of some compositions, involving the functions
mentioned previously.
\end{abstract}

\section{Introduction}\label{Intro}

Let $n&gt;1$ be an integer. An integer $a$ is called regular (mod $n$) if there exists an integer $x$ such that 
$a^2x \equiv a$ (mod $n$) 
(sequence \seqnum{A143869} in Sloane's {\it Encyclopedia of Integer Sequences}).
Several authors investigated properties of regular integers (mod $n$).
Alkam and Osba \cite{oA07}, using ring-theoretic considerations,
rediscovered some of the statements proved by 
Morgado \cite{jM74, jM72}, who showed that $a&gt;1$  is regular (mod $n$) if and only if $\gcd(a,n)$ is a unitary divisor of $n$. 
T\'oth \cite{lT08} gave direct proofs of some properties, because the proofs of \cite{jM74, jM72} were lengthy and those of \cite{oA07} were ring-theoretical.

Let $ \text{Reg}_n=\lbrace{a: 1\leq a \leq n \text{ and  $a$ is regular (mod $n$)}}\rbrace$, and 
$V(n)=\#\text{Reg}_n$. The function $V$ is multiplicative and $V(p^{\alpha})=\phi(p^{\alpha})+1=p^{\alpha}-p^{\alpha-1}+1$, where $\phi$ is the Euler function. Consequently, $ V(n)=\sum_{d\parallel n} \phi(d)$, for every $n\geq 1$, where $d\parallel n$ means that $d$ is a  unitary divisor of $n$, that is, $d\mid n$ and $ \gcd(d,\frac{n}{d})=1$. Also $\phi(n)&lt;V(n)\leq n$, for every $n&gt;1$, and $V(n)=n$ if and only if $n$ is a squarefree; see \cite{jM72, lT08, oA07}. Thus,
the function $V(n)$ is an analogue of the Euler function $\phi(n)$. 
The function $\phi(n)$ is the sequence \seqnum{A000010} in Sloane's On-Line Encyclopedia of Integer Sequences. Also, the function
$V(n)$ is the sequence \seqnum{A055653}; see \cite{Sl}.

Apostol and 
T\'oth \cite{bA12} considered the multidimensional generalization of the function $V(n)$, $V_k(n)$, where  $k\geq 1$ is a fixed integer. The function $V_k(n)$ is multiplicative and $V_k(p^{\alpha})=\phi_k(p^{\alpha})+1=p^{\alpha k}-p^{(\alpha -1)k}+1$, where $\phi_k$ is the Jordan function of order $k$. Consequently, 
$ V_k(n)=\sum_{d\parallel n}\phi_k(d)$, for every $n\geq 1$. Also $\phi_k(n)&lt;V_k(n)\leq n^k$, for every $n&gt;1$
and $V_k(n)=n^k$ if and only if $n$ is 
squarefree; see \cite{bA12}.      

T\'oth \cite{lT08} proved results concerning the minimal and maximal orders of the functions $V(n)$ and 
$ V(n)/\phi(n)$. Alkam and Osba \cite{oA07}
investigated the minimal order of $V(n)$. S\'andor and 
T\'oth \cite{jS08} and Apostol \cite{Brad1}
studied the extremal orders of compositions of certain functions.

In Section \ref{Prel} we present some notation and results involving
arithmetical functions. Section \ref{Dir} is devoted to the study of
the Dirichlet series of $V_k(n)$. Extremal orders of the function
$V_k(n)$ in connection with $\sigma_k(n)$ and  $\psi_k(n)$ are given in
Section \ref{ExtrOrd}.

In Section \ref{UnAnalog} we prove some results regarding $V_k(n)$ and
unitary analogues of the functions $\sigma_k(n)$ and $\phi_k(n)$.

In Section \ref{Comp} we give the extremal orders of some compositions of functions from above.

Section \ref{OpProb} provides other limits of compositions of
arithmetical functions.  We also present some open problems regarding
extremal orders of these compositions.

\section{Preliminaries}\label{Prel}

In what follows let $n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}&gt;1$ be a positive integer and let $k\geq 1$ be an integer.  Throughout the paper
we will use the following  notation:

\begin{itemize}
\item $ p_1, p_2, \ldots$  --- the sequence of the primes;

\item $ \sigma_k(n)$ --- the generalization of $\sigma(n)$, defined by $ \sigma_k(n)=\prod_{i=1}^r\frac{p_i^{(\alpha_i+1)k}-1}{p_i^k-1}$ ;

\item $ \psi_k(n)$ --- the generalization of $\psi(n)$, defined by $ \psi_k(n)=n^k\prod_{p\mid n}(1+\frac{1}{p^k})$ ;

\item $  \zeta(s)$  --- the Riemann zeta function, $ \zeta(s)=\prod_p
\biggl(1-\frac{1}{p^s} \biggr)^{-1}$, $s=\sigma+it\in \mathbb{C}$ and $\sigma&gt;1$;

\item $ \phi(n)$ --- the Euler function, $ \phi(n)=n\prod_{p\mid n}\biggl(1-\frac{1}{p}\biggr)$;

\item $ \phi_k(n)$ --- the Jordan function of order $k$, $ \phi_k(n)=n^k\prod_{p\mid n}\bigl(1-\frac{1}{p^k}\bigr)$;

\item $ \gamma$ --- the Euler constant,
$\gamma= \lim_{\substack{n \to \infty}}(1+\frac{1}{2}+\ldots +\frac{1}{n}-\log n)$;

\item $ \phi^{*}(n)$ --- the unitary analogue of $\phi(n)$,
$ \phi^{*}(n)=\prod_{i=1}^{k}(p_i^{\alpha_i}-1)$;

\item $ \sigma^{*}(n)$ --- the unitary analogue of $\sigma(n)$,
$ \sigma^{*}(n)=\prod_{i=1}^{k}(p_i^{\alpha_i}+1)$.
\end{itemize}

For other arithmetic functions defined by regular integers modulo $n$ we refer to the papers \cite{hT12, lT09}.

Let $f(n)$ be a nonnegative real-valued multiplicative arithmetic function.
Let 
$$ L=L(f):= \limsup_{n\to \infty}\frac{f(n)}{\log \log n}$$
and
$$ \rho(p)=\rho(p,f):=\sup_{\alpha \geq 0}f(p^{\alpha})$$
for primes $p$,
and consider the product $R=R(f):=\prod_p(1-\frac{1}{p})\rho(p)$.

In order to prove the properties below we apply the following results:

\begin{lemma}\label{L:1.1}{\rm(T\'oth and Wirsing \cite[Corollary 1]{lT03}).}
 If $f$ is a nonnegative real-valued multiplicative arithmetic function such that for each prime $p$,
\begin{itemize}
\item[(i)]
$ \rho(p)=\sup_{ \alpha \geq
0}(f(p^{\alpha}))\leq \biggl(1-\frac{1}{p}\biggr)^{-1} $, and
\item[(ii)] there is an exponent $e_p=p^{o(1)} \in\mathbb{N}$ satisfying $ f(p^{e_p})\geq 1+\frac{1}{p}$,
\end{itemize}
then $ \limsup_{n \to \infty}\frac{f(n)}{\log
\log n}=e^{\gamma}\prod_p\biggl(1-\frac{1}{p}\biggr)\rho(p).$
\end{lemma}

\begin{lemma}\label{L:1.2}{\rm(T\'oth and Wirsing \cite[Theorem 1]{lT03}).}
Suppose that $\rho(p)&lt;\infty$ for all primes $p$ and the product $R$ converges unconditionally (i.e., irrespectively of order),
improper limits being allowed, then
$L\leq e^{\gamma}R$.
\end{lemma}

\begin{lemma}\label{L:1.3}{\rm(T\'oth and Wirsing \cite[Theorem 3]{lT03}).}
Suppose that $\rho(p)&lt;\infty$ for all primes $p$, that for each prime $p$ there is an exponent 
$e_p=p^{o(1)}\in \mathbb{N}$ such that $ \prod_pf(p^{e_p})\rho(p)^{-1}&gt;0$ and that the product $R$ converges, improper limits being allowed. Then
$L\geq e^{\gamma}R$.
\end{lemma}

\section{Dirichlet series of $V_k(n)$}\label{Dir}

Apostol and Petrescu \cite{aB11} studied the Dirichlet series of $V_1(n):=V(n)$. In what follows we give the Dirichlet series of $V_k(n)$ for $k\geq 2$ and some results involving the M\"obius function.

\begin{proposition}\label{P:2.1}
For every $s=\sigma+it \in \mathbb{C}$ with $\sigma&gt;k+1$,
$$\sum_{n\geq 1}\frac{V_k(n)}{n^s}=
\zeta(s-k)\zeta(s)\prod_p\bigl (1-\frac{p^{2s-k}+p^s-p^{s-k}}{p^{3s-k}}\bigr ).$$
\end{proposition}

\bp
Let $ f(n)=\frac{V_k(n)}{n^s}$. 
We have 
$$ \sum_{n\geq 1}|f(n)|\leq 
\sum_{n\geq 1}\frac{1}{n^{\sigma-k}}&lt;\infty,$$
so the series 
$ \sum_{n\geq 1}\frac{V_k(n)}{n^s}$ converges absolutely for $\sigma &gt;k+1$. Since $V_k$ is multiplicative, 
$$ \sum_{n\geq 1}\frac{V_k(n)}{n^s} 
=\prod_p\frac{1}{1-\frac{1}{p^{s-k}}}\cdot \prod_p\frac{1}{1-\frac{1}{p^s}}\cdot
\prod_p\bigl (1-\frac{p^{2s-k}+p^s-p^{s-k}}{p^{3s-k}}\bigr ),$$
and the claim follows.   
\ep

\begin{corollary}\label{C: 2.2}
Let $s=\sigma+it \in \mathbb{C}$, $\sigma&gt;k+1$. Then
$$ \sum_{n\geq 1}\frac{\mu(n)V_k(n)}{n^s}= \prod_p\biggl(1-\frac{1}{p^{s-k}}\biggr)=\frac{1}{\zeta(s-k)}.$$ 
Also,
$$ \sum_{n\geq 1}\frac{\vert\mu(n)\vert V_k(n)}{n^s}=
\prod_p\biggl(1+\frac{1}{p^{s-k}}\biggr)=\frac{\zeta(s-k)}{\zeta(2s-2k)}.$$ 
\end{corollary}

\bp
For$ f(n)=\frac{\mu(n)
V_k(n)}{n^s}$ the series $ \sum_{n\geq 1}\lvert f(n)\rvert$ converges absolutely, so
$$ \sum_{n\geq 1}\frac{\mu(n)V_k(n)}{n^s}=
\prod_p\biggl(1-\frac{1}{p^{s-k}}\biggr)=\frac{1}{\zeta(s-k)}.$$

For the second assertion take $ f(n)=
\frac{\lvert \mu(n)\rvert V_k(n)}{n^s}$.     
\ep

\section{Extremal orders concerning classical generalized arithmetic functions}\label{ExtrOrd}

For the quotient $ \frac{\sigma_k(n)}{V_k(n)}$, we notice
that $ \frac{\sigma_k(n)}{V_k(n)}\geq1$
for every $n\geq 1$.
Since
$$\lim_{\substack{p \to \infty\\ p\text{ prime}}}{\frac{\sigma_k(p)}{V_k(p)}}=
\lim_{\substack{p \to \infty\\ p\text{ prime}}}\frac{p^k+1}{p^k}=1,$$ 
we get $$\liminf_{n \to \infty}\frac{\sigma_k(n)}{V_k(n)}=1;$$
hence the minimal order of $ \frac{\sigma_k(n)}{V_k(n)}$
is 1. Now consider the quotient 
$$\frac{\psi_k(n)}{V_k(n)}.$$ 
Since  $$\frac{\psi_k(n)}{V_k(n)}\geq 1$$
for every $n\geq 1$ and
$$ \frac{\psi_k(p)}{V_k(p)}=\frac{p^k+1}{p^k}$$ for every prime $p$,
it is immediate that
$$ \liminf_{n \to \infty}\frac{\psi_k(n)}{V_k(n)}=1 .$$

Thus, the minimal order of $\frac{\psi_k(n)}{V_k(n)}$ is 1. It is known that 
$$ \limsup_{n \to \infty}\frac{\sigma(n)}{V(n)(\log \log n)^2}=e^{2\gamma}$$
and
$$ \limsup_{n \to \infty}\frac{\psi(n)}{V(n)(\log \log n)^2}=\frac{6}{\pi^2}e^{2\gamma} ;$$ see \cite{Brad1}.
 Proposition \ref{P:3.1} shows that the maximal order of $ \frac{\sigma_k(n)}{V_k(n)}$ and $ \frac{\psi_k(n)}{V_k(n)}$ 
is $ \frac{6}{\pi^2}e^{2\gamma}(\log \log n)^2$.

\begin{proposition}\label{P:3.1} 
For $k\geq 2$,
$$ \limsup_{n \to \infty}\frac{\sigma_k(n)}{V_k(n)(\log \log n)^2}= 
\limsup_{n \to \infty}\frac{\psi_k(n)}{V_k(n)(\log \log n)^2}=\frac{6}{\pi^2}e^{2\gamma}.$$
\end{proposition}

\bp
Take $ f(n)=\sqrt{\frac{\sigma_k(n)}{V_k(n)}}$.
Then
$$ f(p^{\alpha})=\sqrt\frac{p^{(\alpha+1)k}-1}{(p^k-1)(p^{\alpha k}-p^{(\alpha-1)k}+1)}
\leq \sqrt{\frac{p+1}{p-1}}=\rho(p)
&lt; \biggl(1-\frac{1}{p}\biggr)^{-1}$$
and
$$ f(p^2)=\sqrt \frac{p^{3k}-1}{(p^k-1)(p^{2k}-p^k+1)}\geq 1+\frac{1}{p}$$
for every prime $p$, so $(ii)$ in  Lemma \ref{L:1.1} is satisfied. We obtain
$$ \limsup_{n \to \infty}\frac{\sqrt{\sigma_k(n)}}{\sqrt{V_k(n)}\log \log n}=
\prod_p{\sqrt{1-\frac{1}{p^{2}}}}e^{\gamma}=\sqrt\frac{6}{\pi^2}e^{\gamma},$$
so
$$ \limsup_{n \to \infty}\frac{\sigma_k(n)}{V_k(n)(\log \log n)^2}=\frac{6}{\pi^2}e^{2\gamma}.$$
Since $\psi_k(n)\leq \sigma_k(n)$ and for the primes $p$ we have $\psi_k(p)=\sigma_k(p)=p^k+1$, the result for 
$ \frac{\psi_k(n)}{V_k(n)(\log \log n)^2}$ follows from the previous one.
\ep

\section{Extremal orders concerning unitary analogues of $\sigma_k$ and $\phi_k$}\label{UnAnalog}

In what follows we consider the functions $\sigma_k^{*}(n)$ and $\phi_k^{*}(n)$,
representing the generalizations for the sum of the unitary divisors
of $n$ and the unitary analogue Euler function, respectively. Let $k\geq 1$ be a fixed integer. We have 
$ \sigma_k^{*}(n)=\sum_{d\parallel n}d^k$ and $\sigma_k^{*}(p^{\alpha})=p^{\alpha k}+1$. Also,
$$ \phi_k^{*}(n):=\sum_{\substack{\gcd(a_1,\ldots,a_k)\in \lbrace{1,2,\ldots,n\rbrace}^k\\\gcd(\gcd(a_1,a_2,\ldots,a_k),n)_{*}=1}}1
=\sum_{d\parallel n}d^k\mu^{*}(\frac{n}{d}),$$
and hence $\phi_k^{*}(p^{\alpha})=p^{\alpha k}-1$.
Note that 
$$\gcd(a,b)_{*}=\max\lbrace d: d\vert a, d\parallel b \rbrace$$
and $\mu^{*}(n)$ is the unitary analogue of the M\"obius function, given by $\mu^{*}(n)=(-1)^{\omega(n)}$,
where $\omega(n)$ is the number of distinct prime factors of $n$. The functions $\sigma_k^{*}(n)$ and $\phi_k^{*}(n)$ are multiplicative. Let
$n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$ be the prime factorisation of $n&gt;1$. We obtain
$$\phi_k^{*}(n)=(p_1^{\alpha_1 k}-1)\cdots(p_r^{\alpha_r k}-1)\quad  \text{and}\quad \sigma_k^{*}(n)=(p_1^{\alpha_1 k}+1)\cdots(p_r^{\alpha_r k}+1).$$

Observe that $\sigma_k^{*}(n)=\sigma_k(n)$ and $\phi_k^{*}(n)=\phi_k(n)$ for all  squarefree $n$. Furthermore, for every
$n\geq 1$,
$$\phi_k(n)\leq \phi_k^{*}(n)\leq n^k\leq \sigma_k^{*}(n)\leq \sigma_k(n).$$
Recall that an integer $n&gt;1$ is called powerful if it is divisible by the square of each of its prime factors. A powerful integer is also called a squarefull integer.
We give extremal orders for the quotients $\frac{\sigma_k^{*}(n)}{V_k(n)}$ and
$\frac{\phi_k^{*}(n)}{V_k(n)}$, the minimal order of $\frac{\phi_k^{*}(n)}{V_k(n)}$
being studied for powerful numbers.
Since $\frac{\sigma_k^{*}(n)}{V_k(n)}\geq 1$ for every $n\geq 1$ and 
$ \lim_{p \to \infty}\frac{\sigma_k^{*}(p)}{V_k(p)}= \lim_{p \to \infty}\frac{p^k+1}{p^k}=1$ for prime numbers $p$,
it follows that
$ \liminf_{n \to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)}=1$.

If $n$ is powerful, it is easy to see that
$ \frac{\phi_k^{*}(n)}{V_k(n)}\geq 1$,
taking into account that
$ \frac{\phi_k^{*}(p^{\alpha})}{V_k(p^{\alpha})}\geq 1$ for $\alpha\geq 2$.
For prime numbers $p$, we notice that 
$$ \lim_{p \to
\infty}\frac{\phi_k^{*}(p^2)}{V_k(p^2)}=
\lim_{p\to \infty}\frac{p^{2k}-1}{p^{2k}-p^k+1}=1,$$
which implies that 
$$ \liminf_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)}=1,$$ 
so the minimal
order of $ \frac{\phi_k^{*}(n)}{V_k(n)}$ is 1.

For the maximal orders of these quotients we have
\begin{proposition}\label{P:4.1} 
For $k\geq 1$,
$$\limsup_{n\to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)\log \log n}=\limsup_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)\log \log n}=e^{\gamma}.$$
\end{proposition}

\bp
Take $ f(n)=\frac{\sigma_k^{*}(n)}{V_k(n)}$ in Lemma \ref{L:1.2}, which is a nonnegative real-valued multiplicative arithmetic function. We have
$$ f(p^{\alpha})=\frac{p^{\alpha k}+1}{p^{\alpha k}-p^{(\alpha-1)k}+1}\leq
\biggl(1-\frac{1}{p}\biggr)^{-1}=\rho(p)&lt;\infty$$
and $R=1$, so
$$ \limsup_{n\to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)\log \log n}\leq e^{\gamma}.$$
Now let $ g(n)=\frac{\phi_k^{*}(n)}{V_k(n)}$.
Here
$$ g(p^{\alpha})=\frac{p^{\alpha k}-1}{p^{\alpha k}-p^{(\alpha-1)k}+1}\leq
\biggl(1-\frac{1}{p}\biggr)^{-1}=\rho(p),$$ and
$$ R=\prod_p g(p^1)(\rho(p))^{-1}=\prod_p(p^k+1)\cdot \frac{p-1}{p}&gt;0.$$
Hence, by Lemma \ref{L:1.3} we have 
$$ \limsup_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)\log \log n}\geq e^{\gamma}.$$
It is obvious that $ \phi_k^{*}(n)\leq \sigma_k^{*}(n)$ for every $n\geq 1$. We obtain  
$$ e^{\gamma}\leq \limsup_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)\log \log n}\leq 
\limsup_{n\to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)\log \log n}\leq e^{\gamma},$$ 
which shows that
$$\limsup_{n\to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)\log \log n}=\limsup_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)\log \log n}=e^{\gamma},$$
as desired.
\ep

\begin{corollary}\label{C: 4.2}
The maximal order of both $\frac{\sigma_k^*(n)}{V_k(n)}$
and $\frac{\phi_k^*(n)}{V_k(n)}$ is $e^{\gamma}\log\log
n$.
\end{corollary}

\section{Extremal orders regarding compositions of arithmetical functions}\label{Comp}

We now move to the study of extremal orders of some composite arithmetic functions.
We start with $V_k(V_k(n))$ and $\phi_k(V_k(n))$.

We know that $V_k(n)\leq n^k$ for every $n\geq 1$, so
$$ \frac{V_k(V_k(n))}{n^{k^2}}\leq \frac{(V_k(n))^k}{n^{k^2}}\leq\frac{(n^k)^k}{n^{k^2}}=1$$ 
and
$$ \lim_{\substack{p \to \infty\\
p\textrm{ prime}}}\frac{V_k(V_k(p))}{p^{k^2}}= \lim_{\substack{p \to \infty\\
p\textrm{ prime}}}\frac{V_k(p^k)}{p^{k^2}}= \lim_{\substack{p \to \infty\\
p\textrm{ prime}}}\frac{p^{k^2}-p^{(k-1)k}+1}{p^{k^2}}=1,$$ 
so the maximal order of
$V_k(V_k(n))$ is $n^{k^2}$. Since
$\phi_k(n)\leq n^k$ and $V_k(n)\leq n^k$ for any $n\geq 1$, we have 
$ \frac{\phi_k(V_k(n))}{n^{k^2}}\leq \frac{(V_k(n))^k}{n^{k^2}}\leq 1$. 
But  $ \lim_{\substack{p \to \infty\\ p\text{
prime}}}\frac{\phi_k(V_k(p))}{p^{k^2}}= \lim_{p \to \infty}\frac{p^{k^2}-p^{(k-1)k}}{p^{k^2}}=1$,
so the maximal order of $\phi_k(V_k(n))$ is $n^{k^2}$.

The maximal order of $V(\phi(n))$ was investigated in \cite{Brad1}. Using the general idea of that proof, we
show

\begin{proposition}\label{P:5.1}
The maximal order of $V_k(\phi_k(n))$ is $n^{k^2}$.
\end{proposition}

\bp
We will use Linnik's theorem which states that if $\gcd(t,\ell)=1$, then there exists a prime $p$ such that
$p \equiv \ell$ (mod $t$) and $p\ll t^c$, where $c$ is a constant (one can take $c\leq 11$).

Let $A= \prod_{\substack{k&lt;p\leq x\\p \text{ prime}}}p$.
Since $\gcd(A^2,A+1)=1$, by Linnik's theorem there is a prime number $q$ such that $q \equiv A+1$ (mod $A^2$) and $q\ll (A^2)^c=A^{2c}$,
where $c$ satisfies $c\leq 11$. Also, $q^k \equiv kA+1$ (mod $A^2$). 
Let $q$ be the least prime satisfying the above condition.
We have 
$\phi_k(q)=q^k-1=AB$, where $B=k+sA$, for some $s$. Thus $\gcd(A,B)=1$, so B is free of prime
factors $\leq x$ and $&gt;k$.
Since $V_k(n)$ is multiplicative,
we have
\be\label{Eq:1}
\frac{V_k(\phi_k(q))}{q^{k^2}}=\frac{V_k(AB)}{(AB+1)^k}=\frac{V_k(A)}{A^k}\cdot \frac{V_k(B)}{B^k}\cdot \frac{(AB)^k}{(AB+1)^k}.
\ee
Here $ \frac{(AB)^k}{(AB+1)^k}\to 1$ as $x \to \infty$, so it is sufficient to study 
$ \frac{V_k(A)}{A^k}$ and $ \frac{V_k(B)}{B^k}$.
Clearly,
\be\label{Eq:2}
\frac{V_k(A)}{A^k}=1.
\ee
We have
$ A=\prod_{k&lt;p\leq x}p\leq  \prod_{p\leq x}p=e^{O(x)}$. Since $B\ll A^{10}$ we obtain $B\ll e^{O(x)}$,
so
\be\label{Eq:3}
\log B\ll x.
\ee
If $B= \prod_{i=1}^{r}q_i^{b_i}$ is the prime factorization of $B$,
we obtain, taking into account that $k\geq 1$ is a fixed integer, that
$\log B= \sum_{i=1}^{r}b_i\log q_i&gt;(\log x)\sum_{i=1}^{r}b_i$ for sufficiently large $x$.
But
$ \sum_{i=1}^{r}b_i\geq r$, so $\log B&gt;k\log x$, implying that $ r&lt;\frac{\log B}{\log x}\ll
\frac{x}{\log x}$ (by (\ref{Eq:3})).
Since
$$ \frac{V_k(B)}{B^k}&gt;
 \prod_{i=1}^{r}\biggl(1-\frac{1}{q_i^k} \biggr)\geq \prod_{i=1}^{r}\biggl(1-\frac{1}{q_i} \biggr)&gt;\biggl(1-\frac{1}{x}\biggr)^r\geq
\biggl(1-\frac{1}{x}\biggr)^{O(\frac{x}{\log x})},$$
We obtain
\be\label{Eq:4}
\frac{V_k(B)}{B^k}&gt;1+O\biggl (\frac{1}{\log x} \biggr).
\ee
By \eqref{Eq:1}, \eqref{Eq:2}, \eqref{Eq:4} and $ \frac{(AB)^k}{(AB+1)^k}\to 1$ as $x\to \infty$,
we obtain
\be\label{Eq:5}
\frac{V_k(\phi_k(q))}{q^{k^2}}&gt;1+O\biggl (\frac{1}{\log x} \biggr).
\ee
By relation \eqref{Eq:5}, and since $ \frac{V_k(\phi_k(n))}{n^{k^2}}\leq \frac{(\phi_k(n))^k}{n^{k^2}}\leq 1$, the claim follows.
\ep

The maximal order of $V(\phi^{*}(n))$ is $n$ (see \cite{Brad1}). For the maximal order of $V_k(\phi^{*}(n))$ we give

\begin{proposition}\label{P:5.2}
$$\limsup_{n\to \infty}\frac{V_k(\phi^*(n))}{n^k}=1.$$
\end{proposition}

\bp
We apply the following lemma:

If $a$ is an integer, $a&gt;1$, $p$ is a prime number and $f(n)$ is an arithmetical function satisfying
$\phi(n)\leq f(n)\leq \sigma(n)$,
one has
\be\label{Eq:6}
\lim_{p\to \infty}\frac{f(N(a,p))}{N(a,p)}=1,
\ee
where $ N(a,p)=\frac{a^p-1}{a-1}$
(see, e.g., Suryanarayana \cite{dS77}).

Since $\phi^{*}(n)\leq n$, it follows that $V_k(\phi^{*}(n))\leq (\phi^{*}(n))^k\leq n^k$,
so
\be\label{Eq:7}
\frac{\sqrt[k]{V_k(\phi^{*}(n))}}{n}\leq 1.
\ee
Obviously, $ \sqrt[k]{V_k(n)}$ meets the conditions of the lemma. We have
\be\label{Eq:8}
\lim_{\substack{p \to \infty\\p\text{ prime}}}\frac{\sqrt[k]{V_k(\phi^{*}(2^p)})}{2^p}=
\lim_{\substack{p \to \infty\\p\text{ prime}}}\frac{\sqrt[k]{V_k(2^p-1)}}{2^p-1}=
\lim_{\substack{p \to \infty\\p\text{ prime}}}\frac{\sqrt[k]{V_k(N(2,p))}}{N(2,p)}=1.
\ee
Now \eqref{Eq:7} and \eqref{Eq:8} imply $ \limsup_{n\to \infty}\frac{\sqrt[k]{V_k(\phi^{*}(n))}}{n}=1$,
and we are done.
\ep

Apostol \cite{Brad1} proved that 
\[ \limsup_{n\to \infty}\frac{\sigma(\phi^{*}(n))}{V(\phi^{*}(n))(\log \log n)^2}=
 \limsup_{n\to \infty}\frac{\sigma(\phi^{*}(n))}{V(\phi^{*}(n))(\log \log \phi^{*}(n))^2}=e^{2\gamma}\]
and
\[ \limsup_{n\to \infty}\frac{\psi(\phi^{*}(n))}{V(\phi^{*}(n))(\log \log n)^2}
 \limsup_{n\to \infty}\frac{\psi(\phi^{*}(n))}{V(\phi^{*}(n))(\log \log \phi^{*}(n))^2}=\frac{6}{\pi^2}e^{2\gamma}.\]
The maximal orders of $ \frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))}$ and
$ \frac{\psi_k(\phi^{*}(n))}{V_k(\phi^{*}(n))}$ are given by

\begin{proposition}\label{P:5.3} 
For $k\geq 2$ we have
\begin{itemize}
\item[(i)]
$ \limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log n)^2}$=
$ \limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log \phi^{*}(n))^2}=\frac{6}{\pi^2}e^{2\gamma}$,

\item[(ii)]
$ \limsup_{n\to \infty}\frac{\psi_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log n)^2}$=
$ \limsup_{n\to \infty}\frac{\psi_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log \phi^{*}(n))^2}=\frac{6}{\pi^2}e^{2\gamma}$. 
\end{itemize}
\end{proposition}

\bp
\noindent (i)
Let  
$$l_1:=\limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log n)^2}
\textrm { and } l_2:= \limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log \phi^{*}(n))^2}.$$
Since
$\phi^{*}(n)\le n$ for every $n \ge 1$, we have
\begin{eqnarray*}
l_1 &amp;= &amp;\limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log n)^2} \\
&amp;\le &amp; 
l_2= \limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log \phi^{*}(n))^2} \\
&amp; \le &amp; 
\limsup_{m\to \infty}\frac{\sigma_k(m)}{V_k(m)(\log \log m)^2}=\frac{6}{\pi^2}e^{2\gamma},
\end{eqnarray*}
by Proposition \ref{P:3.1}.
Since $\gcd(n,1)=1$, by Linnik's theorem, there exists a prime number
$p$ such that $p \equiv 1$ (mod $n$) and $p\ll n^c$.  Let $p_n$ be the
least prime such that $p_n \equiv 1$ (mod $n$), for every $n$.  Then
$n\mid p_n-1$ and $p_n\ll n^c$, so $\log \log p_n\sim \log \log n$.

Observe that $a\mid b$ implies $ \frac{\sigma_k(a)}{V_k(a)}\leq \frac{\sigma_k(b)}{V_k(b)}$.
If $p^{\beta}\mid p^{\alpha}$ $(\beta\leq \alpha)$, it is easy to see that 
$ \frac{\sigma_k(p^{\beta})}{V_k(p^{\beta})}\leq \frac{\sigma_k(p^{\alpha})}{V_k(p^{\alpha})}$.
The general case follows, taking into account that $ \frac{\sigma_k(n)}{V_k(n)}$ is multiplicative. So,
\[ \frac{\sigma_k(\phi^{*}(p_n))}{V_k(\phi^{*}(p_n))(\log \log p_n)^2}=
\frac{\sigma_k(p_n-1)}{V_k(p_n-1)(\log \log p_n)^2}\sim
\frac{\sigma_k(p_n-1)}{V_k(p_n-1)(\log \log n)^2}.\]
On the other hand, 
$$\frac{\sigma_k(p_n-1)}{V_k(p_n-1)(\log \log n)^2}\geq
\frac{\sigma_k(n)}{V_k(n)(\log \log n)^2}.$$
Therefore,
$$ \limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))(\log \log n)^2}\geq
\limsup_{n\to \infty}\frac{\sigma_k(\phi^{*}(p_n))}{V_k(\phi^{*}(p_n))(\log \log p_n)^2}$$
$$\geq\limsup_{n\to \infty}\frac{\sigma_k(n)}{V_k(n)(\log \log n)^2}=\frac{6}{\pi^2}e^{2\gamma}.$$
We obtain
$ \frac{6}{\pi^2}e^{2\gamma}\leq l_1\leq l_2\leq \frac{6}{\pi^2}e^{2\gamma}$, 
and hence 
$ l_1=l_2=\frac{6}{\pi^2}e^{2\gamma}$.\\[12 pt]

\noindent (ii) The proof is similar to the proof of (i), taking into account that
$a\mid b$ implies $ \frac{\psi_k(a)}{V_k(a)}\leq \frac{\psi_k(b)}{V_k(b)}$ and
$ \limsup_{n \to \infty}\frac{\psi_k(n)}{V_k(n)(\log \log n)^2}=\frac{6}{\pi^2}e^{2\gamma}$,
by Proposition \ref{P:3.1}. 
\ep

So, the maximal orders of both $
\frac{\sigma_k(\phi^{*}(n))}{V_k(\phi^{*}(n))}$ and $
\frac{\psi_k(\phi^{*}(n))}{V_k(\phi^{*}(n))}$ are
$ \frac{6}{\pi^2}e^{2\gamma}(\log\log n)^2$. 

In a similar manner, since
$$  \limsup_{n\to \infty}\frac{\sigma_k^{*}(n)}{V_k(n)\log \log n}= \limsup_{n\to \infty}\frac{\phi_k^{*}(n)}{V_k(n)\log \log n}=e^{\gamma}$$
(using Proposition 
\ref{P:4.1}), the fact that $a\mid b$ implies
$ \frac{\sigma_k^{*}(a)}{V_k(a)}\leq \frac{\sigma_k^{*}(b)}{V_k(b)}$ and
$ \frac{\phi_k^{*}(a)}{V_k(a)}\leq \frac{\phi_k^{*}(b)}{V_k(b)}$, respectively,
 it can be shown that
$$ \limsup_{n\to \infty}\frac{\sigma_k^{*}(\phi^{*}(n))}{V_k(\phi^{*}(n))\log \log n}=
 \limsup_{n\to \infty}\frac{\sigma_k^{*}(\phi^{*}(n))}{V_k(\phi^{*}(n))\log \log \phi^{*}(n)}=e^{\gamma}$$
and
$$ \limsup_{n\to \infty}\frac{\phi_k^{*}(\phi^{*}(n))}{V_k(\phi^{*}(n))\log \log n}=
 \limsup_{n\to \infty}\frac{\phi_k^{*}(\phi^{*}(n))}{V_k(\phi^{*}(n))\log \log \phi^{*}(n)}=e^{\gamma}.$$

\section{Open Problems}\label{OpProb}

\begin{openproblem} \text{Note that}
 $$ \liminf_{n\to \infty}\frac{V_k(\phi(n))}{n^k}=
\liminf_{n\to \infty}\frac{V_k(\phi^{*}(n))}{n^k}=\liminf_{n\to \infty}\frac{\phi_k^{*}(V(n))}{n^k}=0.$$
For $n_k=p_1\cdots p_r$ (the product of the first $r$ primes), we have
$$ \frac{V_k(\phi(n_r))}{n_r^k}= \frac{V_k((p_1-1)\cdots (p_r-1))}{p_1^k\cdots p_r^k}\leq
\frac{(p_1-1)^k\cdots (p_r-1)^k}{p_1^k\cdots p_r^k}=\bigl ((1-\frac{1}{p_1})\cdots (1-\frac{1}{p_r})\bigr)^k,$$
so
$$ \lim_{r\to \infty}\frac{V_k(\phi(n_r))}{n_r^k}=
\lim_{r\to \infty}\bigl((1-\frac{1}{p_1})\cdots (1-\frac{1}{p_r})\bigr)^k=0,$$ similarly the other relations. What are the minimal orders
for the $V_k(\phi(n))$, $V_k(\phi^{*}(n))$, and $\phi_k^{*}(V(n))$ ?
\end{openproblem}

\begin{openproblem} Taking $n_r=p_1\cdots p_r$  (the product of the first $r$ primes),
$$ \frac{\sigma_k^{*}(V(n_r))}{n_r^k}=\frac{\sigma_k^{*}(p_1\cdots p_r)}{p_1^k\cdots p_r^k}=
\frac{(p_1^k+1)\cdots (p_r^k+1)}{p_1^k\cdots p_r^k}=\bigl((1+\frac{1}{p_1})\cdots (1+\frac{1}{p_r})\bigr)^k\to \infty$$
as $r\to \infty$,
so
$ \limsup_{n\to \infty}\frac{\sigma_k^{*}(V(n))}{n^k}=\infty$. What is the maximal order for $\sigma_k^{*}(V(n))$ ?
\end{openproblem}

\section{Acknowledgments}

The authors would like to thank the referees for helpful suggestions.


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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11N37.

\noindent {\it Keywords}: arithmetical function, composition, regular integer (mod $n$), extremal order.

\bigskip
\hrule
\bigskip

\noindent(Concerned with sequences 
\seqnum{A000010},
\seqnum{A000203},
\seqnum{A055653}, and
\seqnum{A143869}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 25 2013;
revised versions received  July 9 2013; July 30 2013; August 5 2013.
Published in {\it Journal of Integer Sequences}, August 8 2013.

\bigskip
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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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