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\theoremstyle{plain}
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\begin{center}
\vskip 1cm{\Large\bf On Some Products Taken over Prime Numbers
}
\vskip 1cm
\large
Abdelmalek Bedhouche \\
Laboratoire des Math\'ematiques Appliqu\'ees\\
Facult\'e des Sciences Exactes et Informatique\\
Universit\'e de Jijel\\
18000 Jijel\\
Algeria\\
\href{mailto:abdelmalek.bedhouche@univ-bejaia.dz}{\tt abdelmalek.bedhouche@univ-bejaia.dz} \\
\vskip 1cm
Bakir Farhi \\
Laboratoire de Math\'ematiques Appliqu\'ees \\
Facult\'e des Sciences Exactes \\
Universit\'e de Bejaia \\
06000 Bejaia \\
Algeria \\
\href{mailto:bakir.farhi@univ-bejaia.dz}{\tt bakir.farhi@univ-bejaia.dz}
\end{center}

\vskip .2 in

\begin{abstract}
We study expressions of the type $\prod_{p}
p^{\lfloor\frac{x}{f(p)}\rfloor}$, where $x$ is a nonnegative real number,
$f$ is an arithmetic function satisfying some conditions, and the product
is over the primes $p$. We begin by proving that such expressions can be
expressed by using the $\mathrm{lcm}$ function, without reference to
prime numbers; we illustrate this result with several examples. The rest
of the paper is devoted to studying two particular cases related
to $f(m) = m$ and $f(m) = m - 1$. In both cases, we find arithmetic
properties and analytic estimates for the underlying expressions.
\end{abstract}

\section{Introduction and notation}

Throughout this paper, we let $\N^*$ denote the set of positive integers and $\mathscr{P}$ the set of prime numbers. The letter $p$ is reserved for primes. For a given prime number $p$, we let $\vartheta_p$ denote the usual $p$-adic valuation. For $x \in \R$, we let $\lfloor x\rfloor$ denote the integer part of $x$. For $N , b \in \N$, with $b \geq 2$, the expansion of $N$ in base $b$ is 
denoted by $N = \overline{a_k a_{k - 1} \cdots a_1 a_0}_{(b)}$, meaning that $N = a_0 + b a_1 + b^2 a_2 + \cdots + b^k a_k$ (with $k \in \N$, $a_0 , a_1 , \dots , a_k \in \{0 , 1 , \dots , b - 1\}$ and $a_k \neq 0$). In such a context, we let $S_b(N)$ denote the sum of base-$b$ digits of $N$; that is, $S_b(N) := a_0 + a_1 + \cdots + a_k$. Further, we let $\tau$, $\pi$, and $\theta$, respectively,
denote the divisor-counting function, the prime-counting function, and the Chebyshev theta function, defined as follows:
$$
\tau(n) := \sum_{d \mid n} 1 ~~,~~ \pi(x) := \sum_{p \leq x} 1 ~~,\text{ and }~~
\theta(x) := \sum_{p \leq x} \log{p} ~~~~~~~~~~ (\forall n \in \N^* , \forall x \in \R^+) .
$$
It is known that for $n \geq 3$, we have $\tau(n) = n^{O(1/\log\log{n})}$ (see e.g., \cite[Proposition 7.12, page 101]{kon}). So, a fortiori,
\begin{equation}\label{eqrv4}
\tau(n) = O(n^{1/3}) .
\end{equation}
On the other hand, the prime number theorem states that $\pi(x) \sim_{+ \infty} \frac{x}{\log{x}}$. Other equivalent statements are $\theta(x) \sim_{+ \infty} x$ and $\log\lcm(1 , 2 , \dots , n) \sim_{+ \infty} n$ (see e.g., \cite[Chapter 4]{kon}). The weaker estimates $\pi(x) = O(x/{\log{x}})$, $\theta(x) = O(x)$, and $\log\lcm(1 , 2 , \dots , n) = O(n)$ are called Chebyshev's estimates.

In number theory, it is common that a prime factorization of some special numbers $N$ produces, as exponents of each prime $p$, expressions of the form $\lfloor\frac{u_N}{f(p)}\rfloor$ or a sum of such expressions. The most famous example is perhaps the Legendre formula, stating that for natural numbers $n$, we have
\begin{equation}\label{eq24}
n! = \prod_{p} p^{\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \lfloor\frac{n}{p^3}\rfloor + \cdots} ,
\end{equation}
which may be also reformulate in terms of base expansions as follows:
\begin{equation}\label{eq25}
n! = \prod_{p} p^{\frac{n - S_p(n)}{p - 1}} .
\end{equation}
See e.g., \cite[pages 76-77]{moll}. Another famous example is the formula of the least common multiple of the first consecutive positive integers:
\begin{equation}\label{eq26}
\lcm(1 , 2 , \dots , n) = \prod_{p} p^{\lfloor\frac{\log{n}}{\log{p}}\rfloor} ~~~~~~~~~~ (\forall n \in \N) .
\end{equation}
Among other examples which are less known, we can cite the following
\begin{equation}\label{eq27}
\lcm \{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , i_1 + i_2 + \cdots + i_k \leq n\} = \prod_{p} p^{\lfloor\frac{n}{p}\rfloor} ,
\end{equation}
which is pointed out in the book of Cahen and Chabert \cite[page 246]{cah} and also by Farhi \cite{far} in the context of the integer-valued polynomials. Basing on the remark that in Formulas \eqref{eq24}, \eqref{eq26}, and \eqref{eq27}, the right-hand side (which is a product taken over the primes) is interpreted without reference to prime numbers, we may naturally ask if an expression of a general type $\prod_{p} p^{\lfloor\frac{x}{f_1(p)}\rfloor + \lfloor\frac{x}{f_2(p)}\rfloor + \cdots}$, where $x \in \R^+$ and ${(f_i)}_i$ is a sequence of positive functions satisfying some regularity conditions, possesses the same property; that is, it has an interpretation without reference to the primes. In this paper, we study only the case of the products 
$$
\pi_f(x) := \prod_{p} p^{\lfloor\frac{x}{f(p)}\rfloor} ,
$$
for which we affirmatively answer the previous question under some hypotheses on $f$. After giving several applications of our result, we focus our study on the two particular cases $f(p) = p$ and $f(p) = p - 1$. Because in both cases, there is no loss of generality to take $x$ an integer, we are led 
define 
$$
\rho_n := \prod_{p} p^{\lfloor\frac{n}{p}\rfloor} ~~~~\text{and}~~~~ \sigma_n := \prod_{p} p^{\lfloor\frac{n}{p - 1}\rfloor}
$$
for $n \in \N$.  These are,
respectively, 
the sequences \seqnum{A048803} and \seqnum{A091137} of \cite{slo}.

We begin with the arithmetic study of $\rho_n$ and $\sigma_n$ by establishing several arithmetic properties concerning them; in particular,
we obtain a nontrivial divisor and a nontrivial multiple for $\sigma_n$. Moreover, we determine the $p$-adic valuations of the integers $\frac{\sigma_n}{n!}$ when the prime $p$ is large enough compared to $\sqrt{n}$; we discover that the prime numbers of the form $\lfloor\frac{n}{k} + 1\rfloor$ ($k \in \N^*$, $k < \sqrt{n + 1} + 1$) play a vital role in the arithmetic nature of the $\sigma_n$'s. In another direction, we find asymptotic estimates for $\log{\rho_n}$ and $\log{\sigma_n}$.

\section{An expression of $\pi_f$ using the least common multiple}
Our result of expressing $\pi_f$ in terms of the $\lcm$'s without reference to prime numbers is the following:
\begin{theorem}\label{t1}
Let $f : \N^* \rightarrow \R_+$ be an arithmetic function such that $f(\N^* \setminus \{1\}) \subset \R^*_+$ (i.e., $f$ does not vanish except at $1$ eventually). Consider the set $\N^* \setminus \{1\}$ equipped with the partial order relation ``\,$|$'' of divisibility ($a \mid b \Leftrightarrow a$ divides $b$) and the set $\R^*_+$ equipped with the usual total order relation ``\,$\leq$'', and 
suppose that the map
$$
\begin{array}{rcl}
\widetilde{f} :~ \N^* \setminus \{1\} & \longrightarrow & \R^*_+ \\[2mm]
n & \longmapsto & \dfrac{f(n)}{\log{n}}
\end{array}
$$
is nondecreasing with respect to these two orders. Then for $x \in \R^+$ we
have
$$
\prod_{p} p^{\lfloor\frac{x}{f(p)}\rfloor} = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , f(i_1) + f(i_2) + \cdots + f(i_k) \leq x\} .
$$
\end{theorem}

In order to present a clean proof of Theorem \ref{t1}, we use the following lemma:
\begin{lemma}\label{l1}
Let $f : \N^* \rightarrow \R_+$ as in Theorem \ref{t1}. Then, for prime numbers $p$ and positive integers $a$, we have
$$
\vartheta_p(a) \leq \frac{f(a)}{f(p)} .
$$
\end{lemma}

\begin{proof}
Let $p$ be a prime number and $a$ be a positive integer. Since the inequality of the lemma is trivial when $\vartheta_p(a) = 0$, we may suppose that $\vartheta_p(a) \geq 1$; that is $p \mid a$. So according to our assumptions on $f$, 
we have that $\frac{f(p)}{\log{p}} \leq \frac{f(a)}{\log{a}}$, which translates into $\frac{\log{a}}{\log{p}} \leq \frac{f(a)}{f(p)}$. Hence, $\vartheta_p(a) \leq \frac{\log{a}}{\log{p}} \leq \frac{f(a)}{f(p)}$, as required.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t1}]
Let $x \in \R_+$ be fixed. For a given prime number $p$, the $p$-adic valuation of the left-hand side of the identity of Theorem \ref{t1} is equal to $\lfloor\frac{x}{f(p)}\rfloor$, while the $p$-adic valuation of the right-hand side of the same identity is equal to $\ell_p := \max\{\vartheta_p(i_1 i_2 \dots i_k) ; k \in \N , i_1 , \dots , i_k \in \N^* , f(i_1) + \cdots + f(i_k) \leq x\}$. So, we have to show that $\ell_p = \lfloor\frac{x}{f(p)}\rfloor$ for primes $p$. To do so, we prove the two inequalities $\ell_p \geq \lfloor\frac{x}{f(p)}\rfloor$ and $\ell_p \leq \lfloor\frac{x}{f(p)}\rfloor$.

First, for a given prime number $p$, let us show that $\ell_p \geq \lfloor\frac{x}{f(p)}\rfloor$. By considering the particular natural number $k = \lfloor\frac{x}{f(p)}\rfloor$ and the particular positive integers $i_1 = i_2 = \cdots = i_k = p$, we get $f(i_1) + f(i_2) + \cdots + f(i_k) = k f(p) = \lfloor\frac{x}{f(p)}\rfloor f(p) \leq x$. Thus, according to the definition of $\ell_p$ we have $
\ell_p \geq \vartheta_p\left(i_1 i_2 \cdots i_k\right) = \vartheta_p(p^k) = k = \lfloor\frac{x}{f(p)}\rfloor$, as required.

Now, for a given prime number $p$, let us show that $\ell_p \leq \lfloor\frac{x}{f(p)}\rfloor$. For $k \in \N$ and $i_1 , i_2 , \dots , i_k \in \N^*$, with $f(i_1) + f(i_2) + \cdots + f(i_k) \leq x$, we have
\begin{align*}
\vartheta_p\left(i_1 i_2 \cdots i_k\right) & = \vartheta_p(i_1) + \vartheta_p(i_2) + \cdots + \vartheta_p(i_k) \\
& \leq \frac{f(i_1)}{f(p)} + \frac{f(i_2)}{f(p)} + \cdots + \frac{f(i_k)}{f(p)} ~~~~~~~~~~ (\text{according to Lemma \ref{l1}}) \\
& = \frac{f(i_1) + f(i_2) + \cdots + f(i_k)}{f(p)} \\
& \leq \frac{x}{f(p)} ;
\end{align*}
but since $\vartheta_p(i_1 i_2 \cdots i_k) \in \N$, it follows that: $\vartheta_p\left(i_1 i_2 \cdots i_k\right) \leq \lfloor\frac{x}{f(p)}\rfloor$. The definition of $\ell_p$ concludes that $\ell_p \leq \lfloor\frac{x}{f(p)}\rfloor$, as required. This completes the proof.
\end{proof}

\begin{remark}\label{rmks1}
Let us put ourselves in the situation of Theorem \ref{t1}.
\begin{enumerate}
\item If the map $\widetilde{f}$ is nondecreasing in the usual sense; i.e., with respect to the usual orders of the two sets $\N^* \setminus \{1\}$ and $\R_+^*$, then it remains nondecreasing in the sense imposed by Theorem \ref{t1}. This immediately follows from the implication: $a \mid b \Rightarrow a \leq b$, $\forall a , b \in \N^*$.
\item More generally than the previous item, if the restriction of the map $\widetilde{f}$ on $\N^* \setminus \{1 , 2\}$ is nondecreasing in the usual sense and $\widetilde{f}(2) \leq \widetilde{f}(4)$, then $\widetilde{f}$ is nondecreasing in the sense imposed by Theorem \ref{t1}.
\end{enumerate}
\end{remark}

Now, from Theorem \ref{t1}, we derive the following corollary in which the condition imposed on $f$ is made simpler.

\begin{corollary}\label{coll1}
Let $f : \N^* \rightarrow \R_+$ be an arithmetic function satisfying $f(\N^* \setminus \{1\}) \subset \R_+^*$. Suppose that the map
$$
\begin{array}{rcl}
\N^* \setminus \{1\} & \longrightarrow & \R_+^* \\[2mm]
n & \longmapsto & \dfrac{f(n)}{n}
\end{array}
$$
is nondecreasing in the usual sense (i.e., with respect to the usual order of $\R$). Then for $x \in \R_+$ we have
$$
\prod_{p} p^{\lfloor\frac{x}{f(p)}\rfloor} = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , f(i_1) + f(i_2) + \cdots + f(i_k) \leq x\} .
$$
\end{corollary}

\begin{proof}
We use Theorem \ref{t1} together with Item 2 of Remark \ref{rmks1}. We remark that $\widetilde{f}$ (defined as in Theorem \ref{t1}) is the product of the two functions: $n \mapsto \frac{f(n)}{n}$ (assumed to be nondecreasing in the usual sense on $\N^* \setminus \{1\}$) and $n \mapsto \frac{n}{\log{n}}$, which is nondecreasing on $\N^* \setminus \{1 , 2\} = \{3 , 4 , 5 , \dots\}$. So, $\widetilde{f}$ is nondecreasing on $\N^* \setminus \{1 , 2\}$ in the usual sense. In addition, we have
$$
\widetilde{f}(2) = \frac{f(2)}{\log{2}} = \frac{f(2)}{2} \cdot \frac{2}{\log{2}} = \frac{f(2)}{2} \cdot \frac{4}{\log{4}} \leq \frac{f(4)}{4} \cdot \frac{4}{\log{4}}
$$
(since $n \mapsto \frac{f(n)}{n}$ is assumed to be nondecreasing in the usual sense on $\N^* \setminus \{1\}$). That is, we have $\widetilde{f}(2) \leq \frac{f(4)}{\log{4}} = \widetilde{f}(4)$. The conclusion follows from Item 2 of Remark \ref{rmks1} and Theorem \ref{t1}.
\end{proof}

\subsection{Some applications}
\begin{enumerate}

\item
By applying Theorem \ref{t1} for $f(m) = \log{m}$, we obtain that for 
$x \in \R_+$, we have
\begin{align*}
\prod_{p} p^{\lfloor\frac{x}{\log{p}}\rfloor} & = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , \log{i_1} + \log{i_2} + \cdots + \log{i_k} \leq x\} \\[-.4cm]
& = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , i_1 i_2 \cdots i_k \leq e^x\} \\
& = \lcm\left(1 , 2 , \dots , \left\lfloor e^x\right\rfloor\right) .
\end{align*}
In particular, by taking $x = \log{n}$ ($n \in \N^*$),
we obtain the following well-known formula:
$$
\prod_{p} p^{\lfloor\frac{\log{n}}{\log{p}}\rfloor} = \lcm\left(1 , 2 , \dots , n\right) .
$$

\item
By applying Corollary \ref{coll1} for the function $f(m) = m$, we obtain in particular that for all $n \in \N$, we have
\begin{multline}\label{eq3}
\prod_{p} p^{\lfloor\frac{n}{p}\rfloor} = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , i_1 + i_2 + \cdots + i_k \leq n\} ,
\end{multline}
which is already pointed out by Cahen and Chabert \cite{cah} and by Farhi \cite{far}.

\medskip

\item
(Generalization of \eqref{eq3}). Let $\alpha \geq 1$. By applying Corollary \ref{coll1} for the function $f(m) = m^{\alpha}$, we obtain in particular that for all $n \in \N$, we have
$$
\prod_{p} p^{\lfloor\frac{n}{p^{\alpha}}\rfloor} = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , i_1^{\alpha} + i_2^{\alpha} + \cdots + i_k^{\alpha} \leq n\} .
$$

\item
For all $n , k \in \N$, with $n \geq k$, let us define, as in \cite{far},
$$
q_{n , k} := \lcm\{i_1 i_2 \cdots i_k ~;~ i_1 , i_2 , \dots , i_k \in \N^* , i_1 + i_2 + \cdots + i_k \leq n\} .
$$
These numbers were studied by Farhi \cite{far} in a context related
to integer-valued polynomials. By applying Corollary \ref{coll1} for the function $f(m) = m - 1$, we obtain that for all $n \in \N$, we have 
\begin{align}
\prod_{p} p^{\lfloor\frac{n}{p - 1}\rfloor} & = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , \nonumber \\[-5mm]
& \hspace*{7.5cm} (i_1 - 1) + (i_2 - 1) + \cdots + (i_k - 1) \leq n\}  \nonumber \\
& = \lcm\{i_1 i_2 \cdots i_k ~;~ k \in \N , i_1 , i_2 , \dots , i_k \in \N^* , i_1 + i_2 + \cdots + i_k \leq n + k\} \nonumber \\
& = \lcm\{q_{n + k , k} ~;~ k \in \N \} , \label{eq5}
\end{align}
which remarkably represents the least common multiple of the $n$\up{th} diagonal of the arithmetic triangle of the $q_{i , j}$'s, beginning as follows (see \cite{far})
\begin{table}[H]
$$
\begin{array}{llllllll}
1 & ~ & ~ & ~ & ~ & ~ & ~ & \\
1 & 1 & ~ & ~ & ~ & ~ & ~ & \\
1 & 2 & 1 & ~ & ~ & ~ & ~ & \\
1 & 6 & 2 & 1 & ~ & ~ & ~ & \\
1 & 12 & 12 & 2 & 1 & ~ & ~ & \\
1 & 60 & 12 & 12 & 2 & 1 & ~ & \\
1 & 60 & 360 & 24 & 12 & 2 & 1 & \\
1 & 420 & 360 & 360 & 24 & 12 & 2 & 1
\end{array}
$$
\caption{The triangle of the $q_{n , k}$'s for $0 \leq k \leq n \leq 7$.}
\end{table}
\end{enumerate}

For a given $n \in \N$, let $D_n = {(d_{n , k})}_{k \in \N}$ denote the sequence of the $n$\up{th} diagonal of the above triangle, that is
\begin{equation}
d_{n , k} := q_{n + k , k} = \lcm\{i_1 i_2 \cdots i_k ~;~ i_1 , i_2 , \dots , i_k \in \N^* , i_1 + i_2 + \cdots + i_k \leq n + k\} . \label{eq6}
\end{equation}
In order to simplify Formula \eqref{eq5}, we show that the sequences $D_n$ ($n \in \N$) are all nondecreasing in the divisibility sense and eventually constant. More precisely, we have the following proposition:
\begin{proposition}\label{p1}
For all $n , k \in \N$, we have
$$
d_{n , k} ~~\text{divides}~~ d_{n , k + 1} .
$$
If in addition $k \geq n$, then we have
$$
d_{n , k} = d_{n , n} .
$$
\end{proposition}
\begin{proof}
Let $n , k \in \N$ be fixed. If $i_1 i_2 \cdots i_k$ is a member of the list of $\lcm$ defining $d_{n , k}$, then $1 \cdot i_1 i_2 \cdots i_k$ satisfies $1 + (i_1 + i_2 + \cdots + i_k) \leq 1 + (n + k) = n + (k + 1)$, so the same element is a member of the list of the $\lcm$ defining $d_{n , k + 1}$. Hence, $d_{n , k}$ divides $d_{n , k + 1}$, as required.

Now, let us prove the second part of the proposition. So, suppose that $k \geq n$ and let us prove that $d_{n , k} = d_{n , n}$. It follows from an immediate induction leaning on the result of the first part of the proposition (proved above) that $d_{n , n} \mid d_{n , k}$. So, it remains to prove that $d_{n , k} \mid d_{n , n}$. Let $i_1 , i_2 , \dots , i_k \in \N^*$ such that $i_1 + i_2 + \cdots + i_k \leq n + k$. Let $\ell \in \N$ denote the number of indices $i_r$ ($1 \leq r \leq k$) which are equal to $1$; so we have exactly $(k - \ell)$ indices $i_r$ which are $\geq 2$. Thus, we have
$$
i_1 + i_2 + \cdots + i_k \geq \ell + 2 (k - \ell) = 2 k - \ell .
$$
But since $i_1 + i_2 + \cdots + i_k \leq n + k$, we derive that $2 k - \ell \leq n + k$, which gives $\ell \geq k - n$. This proves that we have at least $(k - n)$ indices $i_r$ which are equal to $1$. By assuming, without loss of generality, that those indices are $i_{n + 1} , i_{n + 2} , \dots , i_k$ (i.e., $i_{n + 1} = i_{n + 2} = \cdots = i_k = 1$), we get
$$
i_1 i_2 \cdots i_n = i_1 i_2 \cdots i_k 
$$
and
$$
i_1 + i_2 + \cdots + i_n = \left(i_1 + i_2 + \cdots + i_k\right) - (k - n) \leq (n + k) - (k - n) = 2 n .
$$
This shows that each product $i_1 i_2 \cdots i_k$ occurring in the definition of $d_{n , k}$ reduces (by permuting the $i_r$'s and eliminate those of them which are equal to $1$) to a product $j_1 j_2 \cdots j_n$ which occurs in the definition of $d_{n , n}$. Consequently $d_{n , k} \mid d_{n , n}$, as required. This completes the proof of the proposition.
\end{proof}

Using Proposition \ref{p1}, for $n \in \N$ we have
\begin{align*}
\lcm\{q_{n + k , k} ~;~ k \in \N\} & = \lcm\{d_{n , k} ~;~ k \in \N\} \\
& = d_{n , n} \\
& = \lcm\{i_1 i_2 \cdots i_n ~;~ i_1 , i_2 , \dots , i_n \in \N^* , i_1 + i_2 + \cdots + i_n \leq 2 n\} .
\end{align*}

\noindent This proves the following interesting corollary, simplifying Formula \eqref{eq5}:
\begin{corollary}\label{coll2}
For $n \in \N$, we have
\begin{equation}
\prod_{p} p^{\lfloor\frac{n}{p - 1}\rfloor} = \lcm\{i_1 i_2 \cdots i_n ~;~ i_1 , i_2 , \dots , i_n \in \N^* , i_1 + i_2 + \cdots + i_n \leq 2 n\} . \tag*{$\square$}
\end{equation}
\end{corollary}

\section[Arithmetic results]{Arithmetic results on the numbers $\rho_n$ and $\sigma_n$}\label{sec3}
A certain number of arithmetic properties concerning the numbers $\rho_n$ and $\sigma_n$ are either immediate or quite easy to prove. We have gathered them in the following proposition:

\begin{proposition}\label{p2}
For natural numbers $n$, we have
\begin{enumerate}
\item[{\rm (i)}] $\rho_n \mid \rho_{n + 1}$, $\sigma_n \mid \sigma_{n + 1}$, and $\rho_n \mid \sigma_n$;
\item[{\rm (ii)}] $\rho_n \mid n!$;
\item[{\rm (iii)}] $n! \mid \sigma_n$ and $\sigma_n \mid (2 n)!$;
\item[{\rm (iv)}] $\sigma_{2 n + 1} = 2 \sigma_{2 n}$.
\end{enumerate}
\end{proposition} 

\begin{proof}
Let $n \in \N$ be fixed. The properties of Item (i) are trivial.

Item (ii) follows because for primes $p$, the term $\lfloor{n/p}\rfloor$ is the first term of $\lfloor{n/p}\rfloor + \lfloor{n/p^2}\rfloor + \cdots$, which is the exact exponent of $p$ in $n!$, which is also $\frac{n - S_p(n)}{p - 1} \leq \frac{n}{p - 1}$; so in particular $\leq \lfloor\frac{n}{p - 1}\rfloor$, giving at the same time the first part of (iii). 

Let us prove the second part of (iii). To do so, we use Corollary \ref{coll2}. For $i_1 , i_2 , \dots , i_n \in \N^*$ satisfying $i_1 + i_2 + \cdots + i_n \leq 2 n$, we have that $i_1 i_2 \cdots i_n \mid i_1! i_2! \cdots i_n! \mid (i_1 + i_2 + \cdots + i_n)! \mid (2 n)!$. Thus, $\lcm\{i_1 i_2 \cdots i_n ~;~ i_1 , i_2 , \dots , i_n$ $\in \N^* , i_1 + i_2 + \cdots + i_n \leq 2 n\} \mid (2 n)!$; that is, according to Corollary \ref{coll2}, $\sigma_n \mid (2 n)!$. 

Finally, Item (iv) follows from the fact that $\vartheta_p(\sigma_{2 n + 1}) = \vartheta_p(2 \sigma_{2 n})$ for primes $p$ (distinguish the two cases ``$p$ odd'' and ``$p = 2$''). This completes the proof of the proposition.
\end{proof}

In the following proposition, we shall improve Item (iii) of Proposition \ref{p2}.
\begin{proposition}\label{p3}
For natural numbers $n$, we have
$$
(n + 1)! ~\mid~ \sigma_n ~~\text{and}~~ \sigma_n ~\mid~ n! \, \lcm(1 , 2 , \dots , n , n + 1) .
$$
\end{proposition}

\begin{proof}
Let $n \in \N$ be fixed. We have to show that for primes $p$, one has
\begin{equation}\label{eq7}
\vartheta_p\left((n + 1)!\right) \leq \vartheta_p\left(\sigma_n\right) \leq \vartheta_p\left(n! \, \lcm(1 , 2 , \dots , n , n + 1)\right) .
\end{equation}
Let $p$ be a fixed prime number and let us prove \eqref{eq7}. By setting $e$ the greatest nonnegative integer satisfying $p^e \leq n + 1$, we obtain
$$\vartheta_p(n!) = \sum_{i = 1}^{e} \lfloor\frac{n}{p^i}\rfloor$$
and
$$\vartheta_p((n + 1)!) = \sum_{i = 1}^{e} \lfloor\frac{n + 1}{p^i}\rfloor$$
(according to the Legendre formula), 
$$\vartheta_p(\sigma_n) = \lfloor\frac{n}{p - 1}\rfloor$$
(by definition of $\sigma_n$), and 
$$\vartheta_p(\lcm(1 , 2 , \dots , n + 1)) = e.$$ So \eqref{eq7} reduces to
\begin{equation}\label{eq8}
\sum_{i = 1}^{e} \left\lfloor\frac{n + 1}{p^i}\right\rfloor \leq \left\lfloor\frac{n}{p - 1}\right\rfloor \leq \sum_{i = 1}^{e} \left\lfloor\frac{n}{p^i}\right\rfloor + e .
\end{equation}
On the one hand, we have
$$
\sum_{i = 1}^{e} \left\lfloor\frac{n + 1}{p^i}\right\rfloor \leq \sum_{i = 1}^{e} \frac{n + 1}{p^i} = \frac{n + 1}{p - 1} \left(1 - \frac{1}{p^e}\right) \leq \frac{n}{p - 1}
$$
(since $p^e \leq n + 1$). But since $\sum_{i = 1}^{e} \lfloor\frac{n + 1}{p^i}\rfloor$ is an integer, we derive the inequality
$$
\sum_{i = 1}^{e} \left\lfloor\frac{n + 1}{p^i}\right\rfloor \leq \left\lfloor\frac{n}{p - 1}\right\rfloor,
$$
confirming the left inequality in \eqref{eq8}. On the other hand, by using the refined inequality $\lfloor\frac{a}{b}\rfloor \geq \frac{a + 1}{b} - 1$, which holds for all positive integers $a , b$, we have
\begin{align*}
\left\lfloor\frac{n}{p - 1}\right\rfloor - \sum_{i = 1}^{e} \left\lfloor\frac{n}{p^i}\right\rfloor & \leq \frac{n}{p - 1} - \sum_{i = 1}^{e} \left(\frac{n + 1}{p^i} - 1\right) \\[1mm]
& = \frac{n}{p - 1} - \frac{n + 1}{p - 1} \left(1 - \frac{1}{p^e}\right) + e \\[1mm]
& = \frac{1}{p - 1} \left(\frac{n + 1}{p^e} - 1\right) + e .
\end{align*}
But from the definition of $e$, we have $p^{e + 1} > n + 1$; that is, $\frac{n + 1}{p^e} < p$. By inserting this into the last estimate, we get
$$
\left\lfloor\frac{n}{p - 1}\right\rfloor - \sum_{i = 1}^{e} \left\lfloor\frac{n}{p^i}\right\rfloor < e + 1 .
$$
Next, since $\lfloor\frac{n}{p - 1}\rfloor - \sum_{i = 1}^{e} \lfloor\frac{n}{p^i}\rfloor \in\Z$, we conclude 
$$
\left\lfloor\frac{n}{p - 1}\right\rfloor - \sum_{i = 1}^{e} \left\lfloor\frac{n}{p^i}\right\rfloor \leq e ,
$$
confirming the right inequality of \eqref{eq8}. This completes this proof.
\end{proof}

From Proposition \ref{p3}, we derive an asymptotic estimate for the number $\log{\sigma_n}$ when $n$ tends to infinity.

\begin{corollary}\label{coll3}
We have
$$
\log\sigma_n ~\sim_{+ \infty}~ n \log{n} .
$$
\end{corollary}

\begin{proof}
According to Proposition \ref{p3}, for $n \in \N^*$ we have
$$
\log{(n + 1)!} \leq \log{\sigma_n} \leq \log{(n!)} + \log\lcm(1 , 2 , \dots , n , n + 1) .
$$
Then the asymptotic estimate of the corollary follows from the facts
$$\log{(n + 1)!} \sim_{+ \infty} \log(n!) \sim_{+ \infty} n \log{n}$$
according to Stirling's formula, and 
$$\log\lcm(1 , 2 , \dots , n , n + 1) \sim_{+ \infty} n$$
according to the prime number theorem.
\end{proof}

Note that the asymptotic estimate of the above corollary will be specified in Section \ref{sec4}.

We now turn to establish a result evaluating the $p$-adic valuations of the positive integers $\frac{\sigma_n}{n!}$ ($n \in \N^*$) for sufficiently large prime numbers. We discover as a remarkable phenomenon that primes of a special type play a vital role. We have the following theorem:
\begin{theorem}\label{t2}
Let $n$ be a positive integer and $p$ be a prime number such that
$$
\sqrt{n + 1} < p \leq n + 1 .
$$
Then we have
$$
\vartheta_p\left(\frac{\sigma_n}{n!}\right) = \left\lfloor\frac{S_p(n)}{p - 1}\right\rfloor \in \{0 , 1\} .
$$
Furthermore, the equality $\vartheta_p(\frac{\sigma_n}{n!}) = 1$ holds if and only if $S_p(n) \geq p - 1$, which holds if and only if $p$ has the form
$$
p = \left\lfloor\frac{n}{k} + 1\right\rfloor ,
$$
with $k \in \N^*$ and $k < \sqrt{n + 1} + 1$.
\end{theorem}

\begin{proof}
By the definition of $\sigma_n$ and the Legendre formula \eqref{eq25}, we have that
\begin{align}
\vartheta_p\left(\frac{\sigma_n}{n!}\right) & = \vartheta_p\left(\sigma_n\right) - \vartheta_p\left(n!\right) \notag \\[1mm]
& = \left\lfloor\frac{n}{p - 1}\right\rfloor - \frac{n - S_p(n)}{p - 1} \notag \\[1mm]
& = \left\lfloor\frac{n}{p - 1} - \frac{n - S_p(n)}{p - 1}\right\rfloor ~~~~~~~~~~ \left(\text{since } \frac{n - S_p(n)}{p - 1} = \vartheta_p(n!) \in \Z\right) \notag \\[1mm]
& = \left\lfloor\frac{S_p(n)}{p - 1}\right\rfloor . \label{eq9}
\end{align}

Next, let us prove that $\lfloor\frac{S_p(n)}{p - 1}\rfloor \in \{0 , 1\}$. The hypothesis on $p$ insures that $n < p^2 - 1$, which implies that the representation of the positive integer $n$ in base $p$ has the form $n = \overline{a_1 a_0}_{(p)}$, with $a_0 , a_1 \in \{0 , 1 , \dots , p - 1\}$ and $(a_0 , a_1) \neq (p - 1 , p - 1)$. Consequently, we have $S_p(n) = a_0 + a_1 < 2 (p - 1)$, implying that $\frac{S_p(n)}{p - 1} < 2$; hence $\left\lfloor\frac{S_p(n)}{p - 1}\right\rfloor \in \{0 , 1\}$, as required. This achieves the proof of the first part of the theorem, which immediately gives the equivalence between $\vartheta_p(\frac{\sigma_n}{n!}) = 1$ and $S_p(n) \geq p - 1$. Now, let us prove the last part of the theorem. 

Suppose that $S_p(n) \geq p - 1$. As seen above, the representation of $n$ in base $p$ has the form $n = \overline{a_1 a_0}_{(p)} = a_0 + p a_1$, where $a_0 , a_1 \in \{0 , 1 , \dots , p - 1\}$ and $(a_0 , a_1) \neq (p - 1 , p - 1)$. We will show that $k = a_1 + 1$ is suitable for the required form of $p$. By supposition, we have $a_0 + a_1 \geq p - 1$, implying that
$$
p - 1 \leq \frac{a_0 + a_1 p}{a_1 + 1} < p ,
$$
which is equivalent to
$$
\left\lfloor\frac{n}{a_1 + 1}\right\rfloor = p - 1 .
$$
Thus,
$$
p = \left\lfloor\frac{n}{a_1 + 1} + 1\right\rfloor .
$$
Furthermore, we have $a_1 = \lfloor\frac{n}{p}\rfloor \leq \frac{n}{p} < \sqrt{n + 1}$ (since $p > \sqrt{n + 1} > \frac{n}{\sqrt{n + 1}}$). Thus, $k = a_1 + 1$ satisfies the required properties; i.e., $p = \lfloor\frac{n}{k} + 1\rfloor$ and $k < \sqrt{n + 1} + 1$. 

Conversely, suppose that there exists $k \in \N^*$, with $k < \sqrt{n + 1} + 1$, such that $p = \lfloor\frac{n}{k} + 1\rfloor$, and let us show that $S_p(n) \geq p - 1$. Setting $a_0 := n - (k - 1) p$ and $a_1 := k - 1$, we first show that the representation of $n$ in base $p$ is $n = \overline{a_1 a_0}_{(p)}$. Since it is immediate that $n = a_0 + p a_1$, it just remains to prove that $a_0 , a_1 \in \{0 , 1 , \dots , p - 1\}$. Since $k < \sqrt{n + 1} + 1 < p + 1$ then $k - 1 < p$; that is $a_1 \in \{0 , 1 , \dots , p - 1\}$. Next, since $p = \lfloor\frac{n}{k} + 1\rfloor$ then
$$
p \leq \frac{n}{k} + 1 < p + 1 ,
$$
implying that
$$
p - k \leq n - (k - 1) p < p .
$$
Hence
$$
p - k \leq a_0 < p .
$$
But $p - k = (p - 1) - a_1 \geq 0$; thus $a_0 \in \{0 , 1 , \dots , p - 1\}$. We have confirmed that the representation of $n$ in base $p$ is $n = \overline{a_1 a_0}_{(p)}$. Consequently, we have
$$
S_p(n) = a_0 + a_1 = n - (k - 1)(p - 1) .
$$
Then, since $n \geq k (p - 1)$ (because $\frac{n}{k} + 1 \geq \lfloor\frac{n}{k} + 1\rfloor = p$), it follows that $S_p(n) \geq p - 1$, as required. This completes the proof of the theorem.
\end{proof}

\section[Analytic estimates]{Analytic estimates of the numbers $\log{\rho_n}$ and $\log{\sigma_n}$}\label{sec4}
Throughout this section, we let $\ct$ denote the absolute positive constant given by
$$
\ct := \sum_{p} \frac{\log{p}}{p (p - 1)} = 0.755\dots .
$$
Our goal is to find asymptotic estimates for $\log{\rho_n}$ and $\log{\sigma_n}$ as $n$ tends to infinity. The obtained main results are the following:
\begin{theorem}\label{t3}
We have
$$
\log{\rho_n} = n \log{n} - (\ct + 1) n + O\left(\sqrt{n}\right) .
$$
\end{theorem}
\begin{theorem}\label{t4}
We have
$$
\log{\sigma_n} = n \log{n} - n + 2 \sqrt{n} + o(\sqrt{n}) .
$$
\end{theorem}

To establish Theorem \ref{t3}, we need the auxiliary results below. Theorem \ref{t4} is derived from Theorem \ref{t2} and a result of Bordell\`es et al.\ \cite{bor}.  

\begin{lemma}\label{l2}
For $x \geq 1$, we have
$$
\sum_{p > x} \frac{\log{p}}{p (p - 1)} = O\left(\frac{1}{x}\right) .
$$
\end{lemma}
\begin{proof}
Since $\frac{\log{p}}{p (p - 1)} \leq 2 \frac{\log{p}}{p^2}$ (for primes $p$), then it suffices to show that $\sum_{p > x} \frac{\log{p}}{p^2} = O(\frac{1}{x})$. According to the Abel summation formula (see e.g., \cite[Proposition 1.4]{kon}), for positive real numbers $x , y$, with $x < y$ we have
\begin{align*}
\sum_{x < p \leq y} \frac{\log{p}}{p^2} & = \left(\sum_{x < p \leq y} \log{p}\right) \frac{1}{y^2} - \bigintsss_{x}^{y} \left(\sum_{x < p \leq t} \log{p}\right) \left(\frac{1}{t^2}\right)' \, d t \\[2mm]
& = \frac{\theta(y) - \theta(x)}{y^2} + 2 \int_{x}^{y} \frac{\theta(t) - \theta(x)}{t^3} \, d t .
\end{align*}
Then, by setting $y$ to infinity, it follows (since $\theta(y) = O(y)$) that
$$
\sum_{p > x} \frac{\log{p}}{p^2} = 2 \int_{x}^{+ \infty} \frac{\theta(t) - \theta(x)}{t^3} \, d t = 2 \int_{x}^{+ \infty} \frac{\theta(t)}{t^3} \, d t - \frac{\theta(x)}{x^2} .
$$
Using finally $\theta(t) = O(t)$, we get
$$
\sum_{p > x} \frac{\log{p}}{p^2} = O\left(\int_{x}^{+ \infty} \frac{d t}{t^2}\right) + O\left(\frac{1}{x}\right) = O\left(\frac{1}{x}\right) ,
$$
as required. The proof is complete.
\end{proof}

Lemma \ref{l2} above is used in the proof of the following proposition:
\begin{proposition}\label{p10}
For positive integers $n$ we have
$$
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} = \ct \cdot n + O\left(\sqrt{n}\right) .
$$
\end{proposition}
\begin{proof}
Let $n$ be a fixed positive integer. For primes $p$, let $e_p$ denote the greatest nonnegative integer satisfying $p^{e_p} \leq n$; explicitly $e_p = \lfloor\frac{\log{n}}{\log{p}}\rfloor$. So we have $p^{e_p + 1} > n$. On the one hand, we have
$$
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} \leq \sum_{p} \left(\frac{n}{p^2} + \frac{n}{p^3} + \cdots\right) \log{p} = \sum_{p} \frac{n}{p (p - 1)} \log{p} ; 
$$
that is
\begin{equation}\label{eq10}
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} \leq \ct \cdot n .
\end{equation}
On the other hand, we have (according to the definition of the $e_p$'s)
\begin{align*}
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} & = \sum_{p \leq \sqrt{n}} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots + \left\lfloor\frac{n}{p^{e_p}}\right\rfloor\right) \log{p} \\
& \hspace*{-5cm} \geq \sum_{p \leq \sqrt{n}} \left(\left(\frac{n}{p^2} - 1\right) + \left(\frac{n}{p^3} - 1\right) + \cdots + \left(\frac{n}{p^{e_p}} - 1\right)\right) \log{p} \\
& \hspace*{-5cm} = n \sum_{p \leq \sqrt{n}} \left(\frac{1}{p^2} + \frac{1}{p^3} + \cdots + \frac{1}{p^{e_p}}\right) \log{p} - \sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} \\
& \hspace*{-5cm} = n \sum_{p \leq \sqrt{n}} \left(\frac{1}{p (p - 1)} - \frac{1}{p^{e_p} (p - 1)}\right) \log{p} - \sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} \\
& \hspace*{-5cm} = n \sum_{p \leq \sqrt{n}} \frac{\log{p}}{p (p - 1)} - n \sum_{p \leq \sqrt{n}} \frac{\log{p}}{p^{e_p} (p - 1)} - \sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} \\
& \hspace*{-5cm} = n \left(\ct - \sum_{p > \sqrt{n}} \frac{\log{p}}{p (p - 1)}\right) - n \sum_{p \leq \sqrt{n}} \frac{\log{p}}{p^{e_p} (p - 1)} - \sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} ;  
\end{align*}
that is,
\begin{multline}\label{eq28}
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} \geq \ct \, n - n \sum_{p > \sqrt{n}
} \frac{\log{p}}{p (p - 1)} - n \sum_{p \leq \sqrt{n}} \frac{\log{p}}{p^{e_p} (p - 1)} \\
- \sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} .
\end{multline}
But, by using Lemma \ref{l2}, we have
\begin{equation}\label{eq29}
\sum_{p > \sqrt{n}} \frac{\log{p}}{p (p - 1)} = O\left(\frac{1}{\sqrt{n}}\right) .
\end{equation}
Next, by using the fact $p^{e_p} > \frac{n}{p}$ (for primes $p$), we have
\begin{equation}\label{eq30}
\sum_{p \leq \sqrt{n}} \frac{\log{p}}{p^{e_p} (p - 1)} < \frac{1}{n} \sum_{p \leq \sqrt{n}} \frac{p}{p - 1} \log{p} \leq \frac{2}{n} \sum_{p \leq \sqrt{n}} \log{p} = \frac{2}{n} \theta\left(\sqrt{n}\right) = O\left(\frac{1}{\sqrt{n}}\right) ,
\end{equation}
and by using the fact $e_p - 1 < e_p := \lfloor\frac{\log{n}}{\log{p}}\rfloor \leq \frac{\log{n}}{\log{p}}$, we have
\begin{equation}\label{eq31}
\sum_{p \leq \sqrt{n}} \left(e_p - 1\right) \log{p} < \sum_{p \leq \sqrt{n}} \log{n} = (\log{n}) \pi(\sqrt{n}) = O\left(\sqrt{n}\right) .
\end{equation}
Then, by substituting \eqref{eq29}, \eqref{eq30}, and \eqref{eq31} into \eqref{eq28}, we get
\begin{equation}\label{eq13}
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} \geq \ct \, n + O\left(\sqrt{n}\right) .
\end{equation}
Finally, \eqref{eq10} and \eqref{eq13} conclude to
$$
\sum_{p} \left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} = \ct \cdot n + O\left(\sqrt{n}\right) ,
$$
as required.
\end{proof}

We are now able to prove Theorem \ref{t3}.

\begin{proof}[Proof of Theorem \ref{t3}]
For sufficiently large integers $n$, we have, according to Legendre's formula,
\begin{align*}
\log{\rho_n} = \sum_{p} \left\lfloor\frac{n}{p}\right\rfloor \log{p} & = \sum_{p}\left(\left\lfloor\frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor + \cdots\right) \log{p} \\
& - \sum_{p}\left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} \\[2mm]
& = \log(n!) - \sum_{p}\left(\left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots\right) \log{p} .
\end{align*}
The weaker form of Stirling's approximation formula $\log(n!) = n \log{n} - n + O(\log{n})$ and Proposition \ref{p10} imply that
$$
\log{\rho_n} = n \log{n} - (\ct + 1) n + O(\sqrt{n}) ,
$$
as required.
\end{proof}

We now turn to estimate $\log{\sigma_n}$. To do so, we rely on Theorem \ref{t2} and a result of Bordell\`es et al.\ \cite{bor} (already conjectured by Kellner \cite{kell}), part of which is recalled below:
\begin{theorem}[Corollary 1.6 of \cite{bor}]\label{tbor}
We have
$$
\sum_{\begin{subarray}{c}
p > \sqrt{n} \\
S_p(n) \geq p
\end{subarray}} 1 = \frac{2 \sqrt{n}}{\log{n}} + o\left(\frac{\sqrt{n}}{\log{n}}\right)
$$
as $n \rightarrow + \infty$.
\end{theorem}

\begin{proof}[Proof of Theorem \ref{t4}]
For a given positive integer $n$, according to Theorem \ref{t2} we have
$$
\frac{\sigma_n}{n!} = \prod_{\begin{subarray}{c}
\sqrt{n + 1} < p \leq n + 1 \\
S_p(n) \geq p - 1
\end{subarray}} \!\!\! p = \prod_{\begin{subarray}{c}
\sqrt{n + 1} < p \leq n + 1 \\
S_p(n) = p - 1
\end{subarray}} \!\!\! p \cdot \prod_{\begin{subarray}{c}
p > \sqrt{n + 1} \\
S_p(n) \geq p
\end{subarray}} p
$$
(remark that $S_p(n) \geq p$ implies $p \leq n$). Thus,
\begin{equation}\label{eqrv1}
\log{\sigma_n} = \log(n!) + \sum_{\begin{subarray}{c}
\sqrt{n + 1} < p \leq n + 1 \\
S_p(n) = p - 1
\end{subarray}} \!\!\!\log{p} + \sum_{\begin{subarray}{c}
p > \sqrt{n + 1} \\
S_p(n) \geq p
\end{subarray}} \log{p} .
\end{equation}
Now, on the one hand, we remark that $n \equiv \restmod{S_p(n)}{(p - 1)}$ (for 
primes $p$), so for a prime $p$ satisfying $\sqrt{n + 1} < p \leq n + 1$, the condition $S_p(n) = p - 1$ is equivalent to $(p - 1) \mid n$. Consequently,
\begin{equation}\label{eqrv2}
\sum_{\begin{subarray}{c}
\sqrt{n + 1} < p \leq n + 1 \\
S_p(n) = p - 1
\end{subarray}} \!\!\!\log{p} \leq \sum_{d \mid n} \log(d + 1) \leq \tau(n) \log(n + 1) = O\left(n^{1/3} \log{n}\right)
\end{equation}
(by \eqref{eqrv4}). On the other hand, by using Theorem \ref{tbor}, we have
\begin{equation}\label{eqrv3}
\sum_{\begin{subarray}{c}
p > \sqrt{n + 1} \\
S_p(n) \geq p
\end{subarray}} \log{p} = \sum_{\begin{subarray}{c}
p > \sqrt{n} \\
S_p(n) \geq p
\end{subarray}} \log{p} + O\left(\log{n}\right) = \left(\sum_{\begin{subarray}{c}
p > \sqrt{n} \\
S_p(n) \geq p
\end{subarray}} 1\right) \log{n} + O\left(\log{n}\right) = 2 \sqrt{n} + o\left(\sqrt{n}\right) . 
\end{equation}
Then, by inserting \eqref{eqrv2} and \eqref{eqrv3} together with the Stirling approximation formula $\log(n!) = n \log{n} - n + O(\log{n})$ into \eqref{eqrv1}, we conclude that
$$
\log{\sigma_n} = n \log{n} - n + 2 \sqrt{n} + o(\sqrt{n}) ,
$$
as required.
\end{proof}

\section{Acknowledgment}
The authors acknowledge support from the Algerian DGRSDT (Direction G\'en\'erale de la Recherche Scientifique et du D\'eveloppement Technologique). The authors are also grateful to the anonymous reviewer for indicating them the two references \cite{bor} and \cite{kell} and for having shown them the argument of proving Theorem \ref{t4}, which they initially stated in the form of a conjecture.


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B. C. Kellner, On a product of certain primes, {\it J. Number Theory} {\bf 179} (2017), 149--164.
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J.-M. De Koninck and F. Luca, {\it Analytic Number Theory: Exploring the Anatomy of Integers}, Graduate Studies in Mathematics, Vol.\ {\bf 134}, American Mathematical Society, 2012.
\bibitem{moll}
Victor H. Moll, {\it Numbers and Functions. From a Classical-Experimental Mathematician's Point of View}, Student Mathematical Library, Vol.\ {\bf 65}, American Mathematical Society, 2012.
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\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A41; Secondary 11A05, 11N37.

\noindent \emph{Keywords}: prime factorization, least common multiple, prime number of a special form, asymptotic estimate of arithmetic functions.

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\noindent (Concerned with sequences
\seqnum{A048803} and
\seqnum{A091137}.)

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\noindent
Received July 26 2022; 
revised version received  November 1 2022; December 15 2022.
Published in {\it Journal of Integer Sequences}, January 2 2023.

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