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\begin{center}
\vskip 1cm{\LARGE\bf 
Explicit Estimates Involving the Primorial\\
\vskip .1in
Integers and Applications }
\vskip 1cm 
\large 
Safia Aoudjit and Djamel Berkane\\
University of Blida\\ 
Department of Mathematics\\
P. O. Box 270, Route de Soumaa\\
Blida\\
Algeria \\
\href{mailto:safiaoudjit9@gmail.com}{\tt safiaoudjit9@gmail.com} \\
\href{mailto:djaber72@univ-blida.dz}{\tt djaber72@univ-blida.dz}\\
\ \\
Pierre Dusart\\
XLIM UMR 7252\\ 
Facult\'{e} des sciences et techniques\\ 
Universit\'{e} de Limoges\\
P. O. Box 123, avenue Albert Thomas\\
87060 Limoges Cedex\\ 
France\\
\href{mailto:pierre.dusart@unilim.fr}{\tt pierre.dusart@unilim.fr}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we propose several explicit bounds for the function that
counts the number of primorial integers less than or equal to a given
positive real number. As applications, we obtain an effective version
of P\'{o}sa's inequality, and a method to estimate the maximal value
of the sum over prime divisors $\sum_{p\mid q}f(p)$ for a positive
decreasing function $f$, when $q$ ranges over all integers less than
$x$. In particular, we improve the upper bound for the maximal value of
$\sum_{p\mid q}\frac{\log p}{p-1}$.
\end{abstract}






\section{Introduction and statement of results}\label{intro}

As usual, let $(p_k )_{k\geq1}$ denote the increasing  sequence of prime numbers, and let $N_k$ be the primorial (prime-factorial) integer of index $k$,
the product of its $k$ first terms. The integers $N_k$ are the terms of the sequence \seqnum {A002110} in the {\it On-line Encyclopedia of Integer Sequences} (OEIS)~\cite{oeis}, and play an important role in number theory from Euclid's proof of the infinity of primes to the remarkable equivalences of the Riemann hypothesis due to Nicolas~\cite{jln} and Robin~\cite{robinrh}.

The approximations linked to $N_k$ have found some unexpected applications in various areas of number theory; see the papers of Betts~\cite{betts}, Planat et al.~\cite{planat} and  Zhang~\cite{Zhang}. However, the most frequently used results concern their logarithm $ \theta(p_k) $ where $\theta$ denotes the Chebyshev function, here we cite the recent papers of Axler~\cite {axler} and Dusart~\cite{dusart18}.

The principal object of this manuscript is to study the function $K(x)$ which counts the number of primorial integers less than or equal to $x$; for instance $K(1)=0$, $K(2)=1$, $K(3)=1$, $K(5)=1$, $K(6)=2$,\ldots and a table of some large values of $x$ is given in Section~\ref{appen}.

The function $K(x)$ is very close to $\pi(x) $ (the usual prime counting function) since it will be shown in what follows that $K(x)\approx\pi(\log x)$. 
It often appears implicitly as an important key in the demonstration of several results; see, for instance, Balazard~\cite{balazard} and Hassani~\cite{hassani}.

We let $\log_i$ denote the $i$-fold iterated logarithm. The asymptotic expansion of $K(x)$ is easy to evaluate. Let us show this using less precise results than those in Balazard~\cite{balazard}:
\begin{theorem}[Balazard]
For every real number $x\geq2$ and every integer $m\geq0$, we have
\begin{equation*}
K(x)  =\frac{\log x}{\log_2x}\left(\sum_{j=0}^{m}\frac{j!}{\log_2^{j} x}+O\left(\frac{1}{\log_2^{m+1}x}\right) \right).
\end{equation*}
\end{theorem}
\begin{proof}
Since for a real $x\geq2$, the integer $K:=K(x)$ is also given by the inequalities
\[
\theta(p_{K})  \leq\log x&lt;\theta(  p_{K+1}),
\]
using the prime number theorem in the forms
\[
\theta(t)=t+O_m\left(\frac{t}{\log^{m+2} t}\right)
\]
and
\[
\pi(t)=\li(t)+O_m \left( \frac{t}{\log^{m+2} t} \right),
\]
where $\li(x)$ indicates the usual logarithmic integral function, one gets successively
\begin{align*}
 p_K  &amp;=\log x + O_m\left(\frac{\log x}{\log_2^{m+2}x} \right),\\
 K(x)  &amp;=\pi(p_{K})  =\li(\log x)+O_m\left(\frac{\log x}{\log_2^{m+2}x} \right).
\end{align*}
Finally, by substituting $\log x$ in the classic asymptotic expansion of $\li(x)$ below
\[
\li(x)=\int_{2}^{x}\frac{dt}{\log t}=\frac{x}{\log x}\left(  0!+\frac{1!}{\log x}+\cdots+\frac{m!}{\log^{m}x}+O\left(\frac{1}{\log^{m+1}x} \right)\right),
\]
we establish the result.
\end{proof}
Consequently, the asymptotic expansion of the function $xK(x)$ corresponds to a type of formula among a large class of expansions related to the iterated logarithms studied in Belbachir and Berkane~\cite{belber} to which an asymptotic expansion for the sum of inverses is determined. The following proposition is a straightforward application of Belbachir and Berkane~\cite[Theorem 1]{belber}:
\begin{proposition}
For every real number $x\geq2$ and every integer $m\geq2$, we have
\begin{align*}
\sum_{2\leq n\leq x}\frac{1}{nK(n)  } &amp;  =\frac{1}{2}\log_2^{2}x-\log_2x-\log_3x+C\\
&amp;  +\frac{\delta_{2}}{\log_2x}+\cdots+\frac{\delta_{m}}{(m-1)\log_2^{m-1}x}+
O\left(\frac{1}{\log_2^{m}x} \right),
\end{align*}
where $C$\ is an absolute constant, and $\left\{  \delta_{j}\right\}  _{j\geq0}$\ is
the sequence \seqnum {A233824} in the OEIS~\cite{oeis} given by the recurrence relation $\delta_{n}+1!\delta_{n-1}+2!\delta_{n-2}+\cdots+(n-1)  !\delta_{1}=n
\cdot n!$.
\end{proposition}

In this article, by comparing $K(x)$ with $\pi(\log x)$ over large and special ranges, our first result (see Theorem~\ref{posa0}) is an effective version of an inequality of P\'{o}sa~\cite{posa}, where he proved that for all $n&gt;1$ there is a $k_n$ such that $N_k&gt;p_{k+1}^n$ for all $n\geq k_n$. We describe an algorithm calculate $k_n$ for every $n$.

The second result (see Theorem~\ref{ihas}) is a good improvement of an approximation of the greatest sum over prime divisors $\mathfrak{L}(q)=\sum_{p\mid q}\frac{\log p}{p-1}$ when $q$ ranges over all integers not exceeding $x$. The method employed to prove this theorem can also be applied to evaluate the maximum of sum of type $\mathfrak{L}_f(q)=\sum_{p\mid q}f(p)$ for any positive decreasing function $f$ on $(1,\infty)$. In Section~\ref{max}, we show that the maximal value of $\mathfrak{L}_f(q)$ is $\mathfrak{L}_f(N_{K(x)})$.

These results are follow-ups of fully explicit bounds for the $O$-term in the formula of $K(x)$, mainly Theorems~\ref{up1}--\ref{up5}.

Finally, throughout this paper, $e$ represents Napier's constant, $p$ a prime number and the calculations are made by using {\tt Maple 17}.

\section{Technical lemmas}

The sequence $\log p_k$ has the same asymptotic behavior as its Ces\`{a}ro average $\frac{\log N_k}{k}$, where the first few terms are
\begin{equation*}
\frac{\log N_k}{k}=\log k+\log_2 k -1+\frac{\log_2 k-2}{\log k}-\frac{\log_2^2 k-6\log_2 k+11}{2\log^2 k}+O \left( \frac{\log_2^3 k}{\log^3 k}\right).
\end{equation*}
Robin~\cite {robin} and Dusart~\cite{dusart} obtained fully explicit bounds
for $ p_{k} $ and $ \theta (p_{k}) $, we gathered the ones we need in the following lemma.

\begin{lemma}\label{theta0}
\begin{equation}\label{p1}
    \theta(p_{k})&gt;  k\log k ,\ \forall k\geq3,
\end{equation}
\begin{equation}\label{p2}
    \theta(p_{k})\geq  k(\log k+\log_2 k -a) ,\ \forall k\geq2, \ \text{and} \ a=1.0769,
\end{equation}
\begin{equation}\label{p3}
    \theta(p_{k})\geq  k(\log k+\log_2 k -1+\frac{\log_2 k-2.1454}{\log k}) ,\ \forall k\geq3,
\end{equation}
\begin{equation}\label{p4}
    \theta(p_{k})\leq  k(\log k+\log_2 k -0.9465) ,\ \forall k\geq 14,
\end{equation}
\begin{equation}\label{p5}
    p_{k}\leq  k(\log k+\log_2 k -\frac{1}{2}),\ \forall k\geq20,
\end{equation}
\begin{equation}\label{p6}
    p_k\leq k\log p_k ,\ \forall k\geq 4.
\end{equation}
\end{lemma}
Similarly, the following lemma directly gives an upper bound of $\theta(p_{k+1})$ in terms of $k$ instead of $k+1$. Note that $c=1-\log2$ is an absolute constant.

\begin{lemma}\label{tech3}
We have, when $k\geq2$:
\begin{equation}\label{majo0}
    \theta(p_{k+1})\leq k \left(\log k+\log_2 k -c+\frac{\log_2 k+c}{\log k} \right).
\end{equation}
\end{lemma}

\begin{proof}
From inequality $(\ref{p5})$, by taking the logarithm and using the fact that $\log(1+x)\leq x$ for all $x&gt;0$, we easily obtain
\begin{equation}\label{m}
   \log p_k\leq \log k +\log_2k+\frac{\log_2k-0.5}{\log k};
\end{equation}
this is valid even for $k\geq18$. In particular, replacing $k$ by $2m-1$ and using the inequality $\log(2m-1)&lt;\log(m)+\log(2)$, one gets that for all $m\geq 2$:

\begin{align*}
 \log p_{2m-1} \leq\log m+\log_2 m+\log2+\frac{\log_2 m +\log2-0.5}{\log m}+\frac{\log2}{\log^2 m} .
\end{align*}
Since the right-hand side is a strictly increasing function $M(m)$, the sum of all $\log p_{2m-1}$ until $k-1$ is bounded above by $\int_{2}^{k}M(t)dt$, and we have
\begin{align*}
\sum\limits_{m=1}^{k-1} \log p_{2m-1}\leq k \left(\log k+\log_2k-c+\frac{\log_2k+c}{\log k}\right)- D(k) ,
\end{align*}
where $$D(k)=\left(2\log2-\frac{3}{2}\right)\li(k)+ \left(\frac{3}{2}-2\log2\right)\li(2)-3\log2-2\log_22+4-2\frac{\log_22}{\log2}-\frac{2}{\log2}.$$
Finally, since the function $D(x)$ is negative when $x\geq11$, by applying the logarithm to the inequality obtained in the lemma below, one gets the result for $k\geq11$.   A computer check handles the cases $2\leq k\leq 11$.
\end{proof}

\begin{lemma}\label{tech5}
We have, when $k\geq11$:
\[
N_{k+1}&lt; \prod_{i=1}^{k-1}p_{2i-1}.
\]
\end{lemma}

\begin{proof}
We proceed by induction on $k$. The inequality holds for $k=11$.  
As $k\geq 3$ implies $2k-1\geq k+2$ and then $ p_{k+2}\leq p_{2k-1}$, the lemma follows from the fact that $$N_{k+2}&lt; p_{k+2}\prod_{i=1}^kp_{2i-1}\leq\prod_{i=1}^{k+1}p_{2i-1}.
$$
\end{proof}

\begin{lemma}\label{tech4}
For all $k\geq1$, we have
\[
p_{k+1}\leq \frac{5}{3}p_{k},
\]
and the inequality is sharp at $k=2$.
\end{lemma}

\begin{proof}
According to Dusart~\cite{dusart}, the interval $[x, \frac{1}{25\log^2( x)}]$ contains at least one prime for all $x\geq 396738$. As $396833$ is prime, we have that, for $p_k\geq396833=p_{33609}$,
\[
p_{k+1}\leq(1+\frac{1}{25\log^2( p_{k})})p_{k}&lt;\frac{4001}{4000}p_{k}&lt;\frac{5}{3}p_{k}.
\]
Finally, by computer, the last inequality is shown to be also valid for $2\leq p_{k}\leq396832$.
\end{proof}

\begin{lemma}
For every positive integer $m$, we have
\begin{equation*}
   \prod_{k=1}^m N_k\leq N_{\frac{m(m+1)}{2}}.
\end{equation*}
\end{lemma}

\begin{proof}
By setting $s:=\sum_{k=1}^mk$, we want to show that $\prod_{k=1}^m N_k$ is less than the primorial of index $s$. This is equivalent to
\begin{equation*}
\sum_{k=1}^m \theta(p_k)\leq \theta(p_s).
\end{equation*}
According to the inequality(\ref{p2}), we have
\begin{align*}
     \theta(p_s)&amp;\geq s(\log s+\log_2 s-a)\\
     &amp;\geq s\left(\log m+\log \left(\frac{m+1}{2}\right)+\log_2m-a \right),
\end{align*}
which implies for $m\geq 14$ that
\begin{equation}\label{nn}
    \theta(p_s)\geq s\left(\log m+\log_2 m+\log\frac{15}{2}-a \right).
\end{equation}
On the other hand, according to the inequality (\ref{p4}), it follows that for $m\geq 14$:
\begin{equation}\label{n}
    \sum_{k=1}^m \theta(p_k)\leq s(\log m+\log_2m-0.9465)+\sum_{k=1}^{13} \theta(p_k).
\end{equation}
However, the left-hand side of the subtraction of the inequality (\ref{n}) from (\ref{nn}) is an increasing function on $m$, which is already positive for $m\geq14$. So, our inequality holds when $m\geq14$ and, we checked $m\leq13$ by computer to end the proof.
\end{proof}

\section{Effective version of P\'{o}sa's inequality}\label{posaineq}

Let us start by giving explicit formulations of the equivalence $K(x)\approx\pi\left(  \log x\right)$.

\begin{lemma}\label{rebitharak(x)2}
For all $\varepsilon&gt;0$, there is a real number $x_0$ such that
\begin{equation*}
\pi \left( \frac{\log x}{1+\varepsilon} \right)\leq K(x)
\end{equation*}
for all $x \geq x_0$.
In particular, $\forall x\geq1$ we have $\pi\left(  \frac{\log x}{1.00000075} \right)\leq K(x).$
\end{lemma}

\begin{proof}
First, a form of the prime number theorem asserts that there exists a decreasing sequence to zero of positive real numbers $(\delta_n)_{n\geq0}$ and a sequence $(u_n)_{n\geq0}$ such that $|\theta(x)-x|&lt;\delta_nx$ for all $x\geq u_n$. Hence, for all $\varepsilon&gt;0$ there exists $n_0(\varepsilon)$ such that $|\theta(x)-x|&lt;\varepsilon x$ for all $x\geq u_{n_0}$. Then, for a given $\varepsilon$, $\theta(x)&lt; (1+\varepsilon) x$ we have
\begin{align*}
    K(x)= \max\left\{k\in \mathbb{N}^*, \ N_k\leq x\right\}&amp;= \max\left\{k\in \mathbb{N}^*, \ \theta(p_k)\leq \log x\right\}\\
    &amp;\geq \max\left\{k\in \mathbb{N}^*, \ p_{k}\geq u_{n_0}, \ (1+\varepsilon)p_k\leq \log x \right\}\\
    &amp;\geq \max\left\{k\in \mathbb{N}^*, \ (1+\varepsilon)p_k\leq \log x \right\}.
\end{align*}
However, the maximum in the last inequality is just $\pi(\log x/(1+\varepsilon))$.

In particular, according to Platt and Trudgian~\cite{platim}
we have $\theta(x)&lt;1.00000075x$ for all $x &gt; 0$. Then, for all $x\geq8$ and further by computer for $x\geq1$ we deduce that
$$
K(x)\geq \pi\left(\frac{\log x}{1.00000075} \right).$$
\end{proof}

\begin{lemma}\label{rebitharak(x)3}
For all $\varepsilon&gt;0$, there is a real number $x_0$ such that
\begin{equation*}
\forall x\geq x_0, \ K(x)\leq \pi \left(\frac{\log x}{1-\varepsilon} \right).
\end{equation*}
In particular, for $x\geq x_{0}$ we have
\[
K(x)\leq \pi \left(\frac{\log x}{\log \alpha} \right), \ 
\]
for the values in Table~\ref{posatable0}.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
  \hline
  $\alpha$ &amp;1.7 &amp; 2 &amp; 2.1 &amp; 2.2 &amp; 2.5  \\\hline
  $x_0$ &amp; 3 &amp; 32 &amp; 3503 &amp; $N_{11}$ &amp; $N_{50}$\\
  \hline
\end{tabular}
\caption{Some values of $x_0$.}
\label{posatable0}
\end{center}
\end{table}
\end{lemma}

\begin{proof}
Similarly to the first part of the above proof, for all $\varepsilon&gt;0$ there exists $n_0(\varepsilon)$ such that $|\theta(x)-x|&lt;\varepsilon x$ for all $x\geq u_{n_0}$. Thus, for a given $\varepsilon$, $\theta(x)&lt; (1-\varepsilon) x$ implies readily that
\begin{align*}
    K(x)&amp;\leq \max\left\{k\in \mathbb{N}^*, \ p_{k}\geq u_{n_0}, \ (1-\varepsilon)p_k\leq \log x \right\}\\
    &amp;\leq \max\left\{k\in \mathbb{N}^*, \ (1-\varepsilon)p_k\leq \log x \right\}.
\end{align*}
However, the maximum in the last inequality is just $\pi(\log x/(1-\varepsilon))$.

For the effective upper bounds, the idea is to look at the best practical values of $\alpha&gt;1$ and $k_0(\alpha)$ such as $N_{k_0}&gt;\alpha^{p_{k_0}}$, since this would result in $K(x)$ is not being greater than $\pi(\log x/\log \alpha)$ for all $x\geq N_{k_0}$. Using  inequality (\ref{p3}) combined with inequality (\ref{m}), the problem is reduced to finding $\alpha&gt;1$ and $k_0$ such that
\begin{equation*}
   \left(\log k+\log_2k+\frac{\log_2k}{\log k} \right)(1-\log \alpha)\geq 1+\frac{2.1454}{\log k}-\frac{0.5\log \alpha}{\log k}, \forall k\geq k_0.
\end{equation*}
Starting from a certain $ k_0(\alpha)$, the function on the right is greater than the one on the left as long as $ 1&lt;\alpha&lt; e$. After checking the small values,  we obtain the results in Table~\ref{posatable1}.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
  \hline
  $\alpha$ &amp;1.7 &amp; 2 &amp; 2.1 &amp; 2.2 &amp; 2.5  \\\hline
  $k_0$ &amp; 2 &amp; 4 &amp; 6 &amp; 11 &amp; 50  \\
  \hline
\end{tabular}
\caption{Some values of $k_0$. }
\label{posatable1}
\end{center}
\end{table}

The values of $x_0$ given in Table~\ref{posatable0} are the possible minimums except for $\alpha=2.2$ and $\alpha=2.5$, where the calculations showed that $K(x)$ is smaller than $\pi(\log x/\log \alpha)$ for $x= N_{10}$ and $x= N_{49}$.
\end{proof}
Through an elementary method, this last proposition permits us to get an effective version of a result of P\'{o}sa~\cite{posa} which proved that for all $n&gt;1$ there is a $k_n$ such that $N_k&gt;p_{k+1}^n$ for all $n\geq k_n$.

\begin{theorem}\label{posa0}
For all integer $n\geq1$, there exists $x_0(n)\geq1$ such that for all $x\geq x_0(n),$
\begin{equation*}
\bigl(\forall k\in \mathbb{N}^*,\  K(x)&lt;k&lt;\pi(\sqrt[n]{x})\bigr)\Rightarrow  N_k&gt;p_{k+1}^n.
\end{equation*}
Furthermore, if $k$ verifies $p_{k+1}\geq(\frac{5}{3})^n$:
\begin{equation*}
\forall m\geq k,\  N_m&gt;p_{m+1}^n.
\end{equation*}
\end{theorem}

\begin{proof}
For $n\geq1$ and $x_0(n)=\frac{n^{2n}}{(\log 1.7)^{2n}}$, we can show easily that $\frac{\log x}{\log 1.7}\leq\sqrt[n]{x}$ as long as $x\geq x_0(n)$. Thus, we obtain that $N_k&gt;x\geq p_{k+1}^n$ for all $k$ satisfying the inequalities $$ K(x)&lt;k&lt;\pi(\sqrt[n]{x}\, ), \ x\geq x_0(n).$$

Furthermore, if $k$ fulfils previous inequalities together with $p_{k+1}\geq(5/3)^n$, it implies by induction on $k$ that $N_m&gt;p_{m+1}^n$ for all $m&gt;k$. Indeed, according to Lemma~\ref{tech4}, we have $$p_{m+2}^n\leq (5/3)^np_{m+1}^n &lt;(5/3)^nN_m\leq p_{m+1}N_{m}=N_{m+1}.$$
In practice, using the Maple algorithm below, the calculations of the values of $k_n$ listed in Table~\ref{posatable}, show that the first choice $x_0(n)$ for $n\leq30$ is sufficient.

\begin{verbatim}
// Algorithm1 Computation of K(x).

restart; with(numtheory);
K := proc (L) local s, k;
s := 2; for k from 2 do
if s &lt;= L then s := s*ithprime(k)
else return k-2 end
if end do end proc
\end{verbatim}
\begin{verbatim}
// Algorithm2 Computation of kn.

posa := proc (n) local x0, x, k, R, m, s, i, t;
x0 := floor((n/ln(1.7))^(4*n)); R := (5/3)^n;
for x from x0 to x0+5 do
for k from K(x)+1 to pi(floor(x^(1/n)))-1 do
if R &lt;= ithprime(k+1) then m := k end if end do
end do; for s to m do
if ithprime(s+1)^n &lt; product(ithprime(i), i = 1 .. s) then t := s;
return t end if
end do end proc
\end{verbatim}
\end{proof}

We thus regain the results of Euclid: $N_k&gt;p_{k+1}$ for $k\geq1$ and those of Bonse~\cite{bonse}: $N_k&gt;p_{k+1}^2$ for $k\geq4$ and $N_k&gt;p_{k+1}^3$ for $k\geq5$. The sequence $(k_n)_{n\geq1}$ is sequence \seqnum {A056127} in the OEIS~\cite{oeis}.

P\'{o}sa's inequality has been studied extensively. The reader can find
similar forms 
in papers of Hassani~\cite{hassaniposa} and S\'{a}ndor~\cite{sandor}.

Let us end this section by giving some special
approximations for limited ranges.
\begin{theorem}\label{rangesp}
\begin{itemize}
  \item[(i)] For $1&lt;x\leq\exp( 10^{19})$, we have $\pi(\log x)\leq K(x).$
   \item[(`ii)] For $2310\leq x\leq \exp( 10^{19})$, we have $\frac{\log x}{\log_2 x}(1+\frac{1}{4\log_2 x})\leq K(x).$
  \item[(iii)] For $210\leq x\leq \exp( 10^{19})$, we have $\frac{\log x}{\log_2 x}\leq K(x).$
\end{itemize}
\end{theorem}

\begin{proof}
Let $0&lt;\log x\leq10^{19}$. According to B\"{u}the~\cite{up19}, the primes $p_k$ smaller than $\log x$ verify $\theta(p_k)&lt;p_k$. So,
\begin{align*}
    \pi(\log x)&amp;= \max\left\{k\in \mathbb{N}^*, \ p_{k}\leq \log x \right\}\\
    &amp;\leq \max\left\{k\in \mathbb{N}^*, \ \theta(p_k)\leq \log x \right\}=K(x).
\end{align*}

After a direct computation of the small values, we derive the last two lower bounds from the first result together with the lower bounds for $\pi(x)$, namely (see Dusart~\cite{dusart}):
$$\frac{x}{\log x}\leq\pi(x), \ \text{for} \ x\geq 17\  \text{and}\ \frac{x}{\log x} \left(1+\frac{1}{\log x} \right)\leq\pi(x), \ \text{for} \ x\geq 599.$$
\end{proof}

\section{Explicit bounds for the primorial counting function}\label{boundskx}

First, we give some upper bounds for $K(x)$ by using a classical
method. We will use appropriate bounds for $\theta(p_k)$ (essentially
formulas (\ref{p1}) and (\ref{p2})) while taking advantage of the
fact that $K(x)$ is constant over the intervals $ [N_k, N_ {k + 1}
)$. Thankfully, with the following lemma, we rediscover the estimates
given by Robin~\cite{robin} of the large values of $\omega(n)$.  Here
$\omega(n)$ denotes the number of prime distinct divisors of $n$.

\begin{lemma}\label{tech8}
For all real numbers $x\geq1$, we have $K(x)=\displaystyle \max_{ 1\leq n\leq x} \omega(n).$
Furthermore, if $K(x)=K$, then for all integers $n\leq x$ with $\omega(n)=K$, we have $n\geq N_K$.
\end{lemma}
In other words, $N_{K(x)}$ is the smallest integer less than $x$ whose decomposition into prime numbers is the longest. Thus, $(K(n))_{n\geq1}$ is the sequence \seqnum{A111972} in the OEIS~\cite{oeis}.

\begin{proof}
As $N_k\leq n\leq x &lt; N_{k+1}$ means that $\omega(n)\leq k$ and $K(n)=k$, one sees that $\omega(n)\leq K(n)$
in every interval $ [N_k, N_{k+1})$, which implies that
\begin{align*}
    \max_{ 1\leq n\leq x} \omega(n)=\max_{ 1\leq n&lt; N_{K+1}} \omega(n)=K.
\end{align*}
Let $q_1q_2\cdots q_K$ be an integer less than $x$ such that $q_1&lt;q_2&lt;\ldots&lt;q_K $ are prime numbers. For $K=1$ it is obvious that $q_1\geq p_1$. Now, assuming $q_i\geq p_i$ for $i&lt; K$, it is necessary that $q_K\geq p_K$ otherwise $q_K&lt;q_{K-1}$.
\end{proof}

For to prove our first upper bound, we need the following lemma.

\begin{lemma}\label{techA}
For a large real number $A&gt;0$, if $\kappa(A)$ is a root of the equation $t\log t=A$, then 
\begin{equation*}
    \kappa(A)=\frac{A}{\log A}(1+o(1)),
\end{equation*}
furthermore, for $A&gt;e$:
\[
\frac{A}{\log A}&lt; \kappa(A)\leq \bigl(1+\frac{1}{e}\bigr)\frac{A}{\log A}.
\]
\end{lemma}

\begin{proof}
See Olver~\cite[Theorem 5.~1, Ex.~5.~7]{olver}.
\end{proof}

\begin{theorem}\label{up1}
We have, when $x\geq N_{13}$:
\[
K(x)\leq \bigl(1+\frac{1}{e}\bigr)\ \frac{\log x}{\log_2 x},
\]
and
\[
K(x)\leq1.3841\frac{\log x}{\log_2 x}, \ \text{for} \  x\geq3.
\]
\end{theorem}

\begin{proof}
From inequality (\ref{p1}), one easily deduces that, for $x\geq N_{13}$:$$K(x)\leq \max \left\{k\in \mathbb{N}^*, \ k\log k\leq \log x \right\},$$ however this last set is only a part of the set of roots of inequality $t\log t\leq \log x$. So, according to Lemma~\ref{techA}, one gets
\[
K(x)\leq \kappa(\log x)\leq (1+\frac{1}{e})\ \frac{\log x}{\log_2 x}, \ \forall x\geq N_{13}.
\]

Now, as the function $F_0(x)=\frac{K(x)\log_2 x}{\log x}$ is decreasing over 
every interval $[N_k,\ N_{k+1})$ once $k\geq3$, then, a computer verification can be done only on $N_k$ where $3\leq k\leq12$. This verification shows that the maximum is reached on $N_{9}$ and it is also true for all real numbers $x&lt;N_{3}$ with $F_0(N_{9})\leq1.3841$, which concludes the proof.
\end{proof}

\begin{theorem}\label{up2}
We have, when $x\geq 3 $, the inequality
\[
K(x)\leq\frac{\log x}{\log_2 x}\bigl(1+\frac{1.4575}{\log_2 x}\bigr).
\]
\end{theorem}

\begin{proof}
For $x\geq2$, we consider the function
\begin{equation*}
F_1(x)= \frac{K(x)(\log_2 x)^2}{\log x}-\log_2 x.
\end{equation*}
The function $F_1$ is decreasing over every interval $[N_k,\ N_{k+1})$ as long as $k\geq5$ since $-\log_2 x$ is decreasing and the function $\frac{K(x)(\log_2 x)^2}{\log x}$ decreases when $x\geq1619$. So, $F_1$ reaches its maximum at an integer $N_{k_0}$ where $k_0\geq 5$. On the other hand, for $k\geq 6$, using inequality (\ref{p2}), invoking the decrease of $\frac{(\log(x))^2}{x}$ (valid for $x\geq8$) together with the fact that $\log(1+x)&lt;x$ and $\frac{1}{1+x}\leq1$, we obtain after a long expansion that
\begin{equation*}
    F_1(N_{k})=\frac{k(\log \theta (p_k))^2}{\theta (k)}-\log \theta (p_k))\leq G(k),
\end{equation*}
with $$G(k)=a+ \frac{\log_2k-a}{\log k}+ \frac{1}{\log k} \left(\frac{\log_2k-a}{\log k}\right)^2+\frac{2a}{\log k}\left(\frac{\log_2k-a}{\log k} \right)+ \frac{a^2}{\log k}.$$ The function $G$ is decreasing and smaller than $1.3832444$ for $k\geq \exp(\exp(a+1))\simeq 2922 $. Then, $F_1(N_{k})\leq 1.3833$ for $k\geq 2922$.

Finally, for $5\leq k\leq 2921$, we conclude by computer verification over intervals $[N_k,\ N_{k+1})$ that the maximum of $F_1(x)$ is reached at $N_{47}$ with $F_1(N_{47})\leq 1.4575$ and our upper bound is valid for $x&lt; N_5$ as well.
\end{proof}

With the techniques used above, the inequality in the following theorem requires a wider estimate of $\theta(p_k)$. Nevertheless, for this last round, we will use a strong relation between $\pi(x)$ and $\theta(x)$ were given by Robin~\cite{robin}, namely
\begin{equation}\label{robinpitheta}
    \forall x\geq2, \ \ \frac{\pi(x)\log \theta(x)}{\theta(x)}\leq 1+\frac{1}{\log \theta(x)}+\frac{2.89726}{\log^2 \theta(x)}.
\end{equation}

\begin{theorem}\label{up3}
We have, when $x\geq 3 $, the inequality
\begin{equation*}
    K(x)\leq\frac{\log x}{\log_2 x}\left(1+ \frac{1}{\log_2 x}+\frac{2.89726}{\log_2^{2}x}\right).
\end{equation*}
\end{theorem}

\begin{proof}
Similarly, by studying $F_2(N_k)$ which corresponds to the function
\begin{equation*}
F_2(x)= \frac{K(x)(\log_2 x)^3}{\log x}-(\log_2 x)^2-\log_2 x, \ \text{when} \ k\geq10.
\end{equation*}
Since $\pi(p_k)=K(N_k)=k$, formula~(\ref{robinpitheta})~guarantees that $F_2(N_k)\leq 2.89726, \ \forall k\geq10$. Hence, we must now verify that our inequality is valid for $x&lt;N_{10}$.

For $x\geq863$, let $M$ be the strictly increasing function that appears on the right side of our upper bound. A verification by computer over the intervals $[N_k,\ N_{k+1})$, with $5\leq k\leq9$, shows that $M(x)$ is always greater than $K(x)$, by a difference of at least $1.5$. The calculations also show that the upper bound is also true for $x&lt;N_5$, which concludes the proof.
\end{proof}

\begin{theorem}\label{up4}
If the Riemann hypothesis holds, we have
\begin{equation}\label{8}
    K(x)\leq \li^{-1}(\log x)+0.12\sqrt{\log x}, \ \forall x\geq 42,
\end{equation}
where $\li^{-1}$ is the inverse of the logarithmic integral function.
\end{theorem}

\begin{proof}
According to Robin~\cite{robin}, for $k\geq5$ $$\theta(p_k)&gt; \li^{-1}(k)-0.12\sqrt{k\log^3 k},$$ and
as $\li(x)$ is an increasing function, one gets successively
\begin{align*}
    k&amp;\leq \li\bigl(\theta(p_k)\bigr)+\int_{\theta(p_k)}^{\theta(p_k)+0.12\sqrt{k\log^3 k}}\frac{dx}{\log x}\\
    &amp;\leq \li(\theta(p_k))+0.12\frac{\sqrt{k\log^3 k}}{\log \theta(p_k)}.
\end{align*}
Now, applying inequality (\ref{p1}), one gets
$$k\leq \li(\theta(p_k))+0.12\frac{\sqrt{\theta(p_k)}\log k}{\log \theta(p_k)}\leq \li(\theta(p_k))+0.12\sqrt{\theta(p_k)}, \ \forall k\geq5,$$
which is equivalent to $$K(N_k)\leq \li(\log N_k)+0.12\sqrt{\log N_k}, \ \forall k\geq5.$$
Simply, the term on the right side is an increasing function of $k$, which yields that
\begin{equation*}
    K(x)\leq \li^{-1}(\log x)+0.12\sqrt{\log x}, \ \forall x\geq N_5.
\end{equation*}
Finally, with computer verifications, we extend the result to $x\geq42$.
\end{proof}

\begin{theorem}\label{up5}
We have, when $x\geq 2310$
\[
K(x)\geq\frac{\log x}{\log_2 x}\left(1+ \frac{1}{4\log_2 x}\right).
\]
Moreover, for $x\geq 210$
$$K(x)\geq \frac{\log x}{\log_2 x}.$$
\end{theorem}

\begin{proof}
Let $\kappa(x)=\frac{\log x}{\log_2 x}(1+\frac{1}{4\log_2 x})$ and $f_a(x)=k\bigl(\log k+\log_2 k -a\bigr)$. As the function $\frac{\log_2 x-c}{\log x}$ is decreasing for $x\geq8$, inequality~(\ref{majo0}) gives that
$$ \theta(p_{k+1})\leq f(k)=k\bigl(\log k+\log_2 k -c+0.006389\bigr), \ \text{when} \ x\geq 10^{500}.$$ Hence, for $x\geq 10^{500}$, the function $K(x)$ is greater than the minimum of the set
of positive integer solutions of the inequality $f_a(k)&gt;\log x,$ where $a=c-0.006389\leq0.30046382$.

Now, let us show that the function $\kappa(x)=\frac{\log x}{\log_2 x}(1+\frac{1}{4\log_2 x})$ is not in the above set. Indeed, for $x\geq10^{500}$
\begin{align*}
    \log\kappa(x)&amp;\leq\log_2  x-\log_3x+\frac{0.25}{\log_2 x},\  \log_2\kappa(x)\leq  \log_3 x,\\
 \log\kappa(x)+ \log_2\kappa(x)&amp;\leq  \log_2  x+\frac{0.25}{\log_2 x}\leq \log_2  x+\frac{0.25}{\log_2 x},
\end{align*}
so, we obtain the following
\begin{equation*}
    f_a(\kappa(x))\leq\log x(1+\frac{1}{4\log_2 x})(1+\frac{1}{4\log_2^2 x}-\frac{a}{\log_2 x})&lt; \log x.
\end{equation*}

Now suppose that $f(k)&gt; \log x$.  Then $x\to\infty$ implies $k\to\infty$, since $f$ is bounded on every bounded subset of the set of all positive integers. Therefore, for $x\geq10^{500}$, we have $f(k)\le f_a(k)$ and, since $f_a(k)$ is increasing in $k$, we also have $f_a(k)\le f_a(\kappa(x))$ if $k&lt;\kappa(x)$, so that
$$f(k)\le f_a(k)\le f_a(\kappa(x))&lt;\log x,$$
which means that, if $x\geq10^{500}$ and $f(k)\ge \log x$, then $k\ge\kappa(x)$ and so $K(x)\geq \kappa(x)$.

For the values of $x&lt; 10^{500}$, Theorem~\ref{rangesp} is used as $10^{500}&lt; \exp{10^{19}}$. The last lower bound is partly due to the fact that $\frac{\log x}{4\log_2^2 x}\geq1$ when $x\geq 1.7\cdot10^{30}$ and again to Theorem~\ref{rangesp}-(iii).
\end{proof}

\section{Maximal value of sums over prime divisors }\label{max}

Sums over primes and their evaluations are one of the main subjects of multiplicative number theory. Through a special case, we are concerned in this section to give an idea allowing to estimate the maximal value of sum of type $\mathfrak{L}_f(q)=\sum_{p\mid q}f(p)$ for a positive decreasing function $f$ on $(1,\infty)$, when $q$ ranges over all integers not exceeding $x$.

This type of sum appears without estimates in several papers, as in the work of Gordon and Rogers~\cite{rogers} where they derive refined forms of the sums of the divisor function; and in the work of Lehmer~\cite{lehmer} where he studied a generalization of the Euler constant.

Our idea can be summarized as follows. As $f$ is a positive decreasing function, then $\max\limits_{1&lt;q\leq x}\mathfrak{L}_f(q)$ is reached on the smallest integer $q(x)$ less than $x$ among those having the longest decomposition in prime numbers. However,  according to Lemma~\ref{tech8}, we can clearly specify that $q(x)$ is nothing other than $N_{K(x)}$.

Finally, for a given real $x\geq1$, if $K(x)=K$, we deduce
the following
for all positive decreasing functions $f$ on $(1,\infty)$:
\begin{equation*}
   \max\limits_{1&lt;q\leq x}\mathfrak{L}_f(q)=\mathfrak{L}_f(N_{K})=\sum\limits_{\substack{p\leq p_{ K}}}f(p).
\end{equation*}

In particular, there is an approximation of the maximal value of $\mathfrak{L}(q)=\sum_{p\mid q}\frac{\log p}{p-1}$. For a constant $C$, Hassani~\cite{hassani} proved the following result
\begin{equation*}
    \max\limits_{1&lt;q\leq x}\mathfrak{L}(q)\leq\log_2 x+C \ \text{and} \ \ \mathfrak{L}(q)\geq\frac{\log q}{q-1} \ \text{for} \ q\geq2.
\end{equation*}
We propose the following improvements:
\begin{theorem}\label{ihas}
We have, when $x\geq N_7$, the inequality
\[
\max\limits_{1&lt;q\leq x}\mathfrak{L}(q)\leq \log K(x)+\log_2 K(x).
\]
\end{theorem}

\begin{proof}
As in Hassani~\cite{hassani}, we start the estimation of the maximum 
as follows:
\begin{align*}
    \max\limits_{1&lt;q\leq x}\mathfrak{L}(q)&amp;=\sum\limits_{\substack{p\leq p_{ K}}}\frac{\log p}{p}+\sum\limits_{\substack{p\leq p_{ K}}}\frac{\log p}{p(p-1)}\\
    &amp;=\sum\limits_{\substack{p\leq p_{ K}}}\frac{\log p}{p}+\sum\limits_{\substack{p&gt;1}}\frac{\log p}{p(p-1)}-\sum\limits_{\substack{p&gt; p_{ K}}}\frac{\log p}{p(p-1)}.
\end{align*}
However, Rosser and Shoenfeld~\cite{barsho} proved the following:
\begin{equation*}
\forall t\geq32, \ \sum\limits_{\substack{p\leq t}}\frac{\log p}{p}\leq \log t+E+\frac{1}{\log t} \ \text{and} \ \sum\limits_{\substack{p&gt;1}}\frac{\log p}{p(p-1)}+E\approx -0.58.
\end{equation*}
So, recalling that $$p_k\leq k(\log k+\log_2k),$$ once $k\geq 6$, we have for $ K \geq 331$ ($p_K\geq32$ and $\log_2K/\log K&lt; 0.3$) then after checking by hand for $ K \geq 7 $ that
\begin{align*}
        \sum\limits_{\substack{p\leq p_K}}\frac{\log p}{p}+\sum\limits_{\substack{p&gt;1}}\frac{\log p}{p(p-1)}&amp;\leq \log K+\log_2 K+\frac{\log_2 K}{\log K}+\frac{1}{\log p_{12}}-0.58\\
        &amp;\leq \log K+\log_2 K.
\end{align*}
Hence, for $x\geq N_7$ we obtain 
\begin{equation}\label{lq}
    \max\limits_{1&lt;q\leq x}\mathfrak{L}(q)\leq \log K(x)+\log_2 K(x).
\end{equation}
Finally, an computer verifications show that inequality~\ref{lq} is not true for integers smaller than $N_7$.
\end{proof}

\begin{corollary}\label{fin}
We have, when $x\geq 3$, the inequality
\[
\max\limits_{1&lt;q\leq x}\mathfrak{L}(q)\leq \log_2 x+\frac{1.4575}{\log_2x}.
\]
\end{corollary}

\begin{proof}
Using the upper bound of Theorem~\ref{up2}, we obtain successively
\begin{align*}
    \log K(x)&amp;\leq \log_2 x-\log_3 x+\frac{1.4575}{\log_2x}\\
    \log_2 K(x)&amp;\leq \log_3 x+\log (1-\frac{\log_3 x}{\log_2 x}+\frac{1.4575}{\log_2^2x})\\
    \log K(x)+\log_2 K(x)&amp;\leq \log_2 x+\frac{1.4575}{\log_2x}, \ x\geq 2219.
\end{align*}
Combining this with inequality~(\ref{lq}), this gives the result for $x\geq N_7$. We conclude the proof using computer verifications for the small values.
\end{proof}

\begin{corollary}\label{fin2}
We have, when $x\geq 43$, the inequality
\[
\max\limits_{1&lt;q\leq x}\mathfrak{L}(q)\leq \log_2 x+\log (1.3841).
\]
\end{corollary}

\begin{proof}
Similarly to the previous proof, but using the second upper bound of Theorem~\ref{up1}, we obtain that, for  $x\geq 3$
$$\log K(x)\leq \log_2 x-\log_3 x+\log (1.3841).$$ And as $K(x)\leq \log x$ by definition, we get that $\log_2 K(x)\leq \log_3 x$. Therefore, using inequality~(\ref{lq}) gives the result for $x\geq N_7$; then computer verification gives the result for $x\geq43$.
\end{proof}

\begin{figure}[H]
\centering
\includegraphics[scale=0.60]{lq.png}
\caption{Graph of points $(q,\mathfrak{L}(q))$  for $1\leq q \leq2\cdot10^6$.
The upper bound of Theorem~\ref{ihas} (resp., Corollary~\ref{fin2}) is shown in red (resp., in blue).}
\label{fig1}
\end{figure}

\section{Minimal value of sums over prime divisors}\label{min}

Concerning $\min\limits_{1&lt;q\leq x}\mathfrak{L}(q)$, we obtain the following optimal inequality
$$ \forall x\geq 2, \ \min\limits_{1&lt;q\leq x}\mathfrak{L}(q)\geq l(x).$$ Indeed, as the minimal value of $\mathfrak{L}(q)$ is reached on the largest prime number less than $x$, we can deduce that if $\pi(x)=r$, then
$$\min\limits_{1&lt;q\leq x}\mathfrak{L}(q)=l(p_r)\geq l(x),$$
since $l$ is a decreasing function. According to the proof of Lemma~\ref{tech4}, this lower bound is the best since we already have that, for $x$ not large enough ($\geq 396833$)
$$
    x&lt;p_{r+1}\Leftrightarrow  \frac{p_r}{x}&gt; \frac{p_r}{p_{r+1}}\geq \frac{4000}{4001},
$$
which implies $$ l(x)\leq \min\limits_{1&lt;q\leq x}\mathfrak{L}(q)&lt; l(\frac{4000}{4001}x).$$


\section{Appendix}\label{appen}

In Table~\ref{posatable} we give the values of $K(10^n)$ and the terms of
the sequence 
$$k_n=\min \{k, p_1\cdots p_k&gt; p_{k+1}^n\}$$ for $n\leq 30$.
\begin{table}[H]
\begin{center}
\begin{tabular}{|c|c|c||c|c|c||c|c|c|}
  \hline
  $n$ &amp; $K(10^n)$&amp; $k_n$&amp;$n$ &amp; $K(10^n)$&amp; $k_n$&amp;$n$ &amp; $K(10^n)$&amp; $k_n$ \\\hline
  1 &amp; 2 &amp;  2&amp;11 &amp; 10 &amp; 16&amp;21 &amp; 16 &amp;  29\\\hline
  2 &amp; 3 &amp;  4&amp;12 &amp; 11 &amp;  18&amp;22 &amp; 17 &amp;  30\\\hline
  3 &amp; 4 &amp;  5&amp;13 &amp; 12 &amp;  19&amp;23 &amp; 17 &amp;  32 \\\hline
  4 &amp; 5 &amp;  7 &amp;14 &amp; 12 &amp;  20&amp;24 &amp; 18 &amp;  33\\\hline
  5 &amp; 6 &amp;  8&amp;15 &amp; 13 &amp;  21&amp;25 &amp; 19 &amp;  34 \\\hline
  6 &amp; 7 &amp;  10&amp;16 &amp; 13 &amp;  23&amp;26 &amp; 19 &amp;  35\\\hline
  7 &amp; 8 &amp;  11&amp;17 &amp; 14 &amp;  24&amp;27 &amp; 20 &amp;  36\\\hline
  8 &amp; 8 &amp;  13&amp;18 &amp; 15 &amp;  25&amp;28 &amp; 20 &amp;  38\\\hline
  9 &amp; 9 &amp;  14&amp;19 &amp; 15 &amp;  26&amp;29 &amp; 21 &amp;  39\\\hline
  10 &amp; 10 &amp; 15&amp;20 &amp; 16 &amp;  28&amp;30 &amp; 21 &amp;  40\\\hline
\end{tabular}
\end{center}
\caption{Some values of $K(10^n)$ and $k_n=\min \{k, p_1\cdots p_k&gt; p_{k+1}^n\}$.}
\label{posatable}
\end{table}
We can also give $K(10^{40})=26$, $K(10^{50})=31$, $K(10^{60})=36$, $K(10^{100})=53$, $K(10^{200})=92$, $K(10^{300})=128$, $K(10^{10^3})=350$ and, $K(10^{10^4})=2584$.


\section{Acknowledgments}
\label{sec:8}
The second author would like to thank Professor H. Belbachir for fruitful discussions about an early version of this paper.

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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:  Primary
11A25; Secondary 11A41, 11L20, 11N64.

\noindent {\it Keywords}: primorial integer, sum over prime divisors,
P\'{o}sa's inequality, arithmetic function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000040},
\seqnum{A002110},
\seqnum{A056127},
\seqnum{A111972}, and
\seqnum{A233824}.)

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Received February 6 2021;
revised versions received February 17 2021; February 19 2021; July 11 2021;
July 28 2021.
Published in {\it Journal of Integer Sequences}, July 28 2021.

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