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\begin{center}
\vskip 1cm{\LARGE\bf Extremely Abundant Numbers
\vskip .1in
and the Riemann Hypothesis
}
\vskip 1cm
\large
Sadegh Nazardonyavi and Semyon Yakubovich\\
Departamento de Matem\'{a}tica\\
Faculdade de Ci\^{e}ncias\\
Universidade do Porto\\
4169-007 Porto\\
Portugal\\
\href{mailto:sdnazdi@yahoo.com}{\tt sdnazdi@yahoo.com}\\
\href{mailto:syakubov@fc.up.pt}{\tt syakubov@fc.up.pt}\\
\end{center}

\vskip .2 in

\begin{abstract}
Robin's theorem states that the Riemann hypothesis is equivalent to the
inequality $\sigma(n)&lt;e^\gamma n\log\log n$ for all $n&gt;5040$, where
$\sigma(n)$ is the sum of divisors of $n$ and $\gamma$ is Euler's
constant. It is natural to seek the first integer, if
it exists, that violates this inequality. We introduce the sequence of
extremely abundant numbers, a subsequence of superabundant numbers,
where one might look for this first violating integer. The Riemann
hypothesis is true if and only if there are infinitely many extremely
abundant numbers. These numbers have some connection to the colossally
abundant numbers. We show the fragility of the Riemann hypothesis with
respect to the terms of some supersets of extremely abundant numbers.
\end{abstract}

\section{Introduction}\label{sec-introduction}

There are several statements equivalent to the famous Riemann hypothesis (RH) \cite{Bor}. Some of them are related to the asymptotic behavior of   arithmetic functions. In particular, the well-known Robin theorem (inequality, criterion, etc.) deals with the upper bound of the sum-of-divisors function $\sigma$, which is defined by $\sigma(n):=\sum_{d|n}\, d$. Robin \cite[Theorem 1]{Rob} established an elegant connection between RH and the sum of divisors of $n$ by proving that the RH is true if and only if
\begin{equation}\label{Rob1}
\frac{\sigma(n)}{n\log\log n}&lt;e^\gamma,\qquad\text{for all}\quad n&gt;5040,
\end{equation}
where $\gamma$ is Euler's constant.

Throughout the paper, as in Robin \cite{Rob}, we define the function $f$ by setting
\begin{equation}\label{f(n)}
f(n)=\frac{\sigma(n)}{n\log\log n}.
\end{equation}
Gronwall, in his study of the asymptotic maximal
size for the sum of divisors of $n$ \cite{GR},
found that the order of $\sigma(n)$ is always ``very nearly $n$''
\cite[Theorem 323]{HW}. More precisely, he proved the following theorem.

\begin{theorem}[Gronwall] Let $f$ be defined as in (\ref{f(n)}). Then
\begin{equation}
    \limsup_{n\rightarrow\infty}f(n)=e^\gamma.
\end{equation}
\end{theorem}

Let us call a positive integer $n$ \cite{Al, Ram.N}
\begin{itemize}
    \item[(i)]  \textit{colossally abundant}, if for some $\varepsilon&gt;0$,
   \begin{align}\label{CA-defn}
   \frac{\sigma(n)}{n^{1+\varepsilon}}\geq\frac{\sigma(m)}{m^{1+\varepsilon}},\qquad(m&lt;n)\qquad\text{and}\qquad
    \frac{\sigma(n)}{n^{1+\varepsilon}}&gt;\frac{\sigma(m)}{m^{1+\varepsilon}},\qquad (m&gt;n);
   \end{align}
  \item[(ii)] \textit{generalized superior highly composite}, if there is a positive number $\varepsilon$ such that
  $$
  \frac{\sigma_{-s}(n)}{n^\varepsilon}\geq\frac{\sigma_{-s}(m)}{m^\varepsilon},\qquad(m&lt;n)\qquad\text{and}\qquad \frac{\sigma_{-s}(n)}{n^\varepsilon}&gt;\frac{\sigma_{-s}(m)}{m^\varepsilon},  \qquad(m&gt;n),
  $$
  where $\sigma_{-s}(n)=\sum_{d\mid n}d^{-s}$. The parameter $s$ is assumed to be positive in \cite{Ram.N}. In the case $s=1$, (ii) becomes (i).
\end{itemize}
Ramanujan initiated the study of these classes of numbers in an unpublished part of his 1915 work on highly composite numbers (\cite{Ram.h,Ram.N} and
\cite[pp.\ 78--129, 338--339]{Ram.c}). More precisely, he defined rather general classes of these numbers. For instance, he defined generalized highly composite numbers, containing as a subset superabundant numbers \cite[Section 59]{Ram.h}, and he introduced the generalized superior highly composite numbers, including as a particular case colossally abundant numbers. For more details we refer the reader to \cite{Al, Erd, Ram.N}.

We denote by CA the set of all colossally abundant numbers. We also use CA as an abbreviation for the term ``colossally abundant''. Ramanujan \cite{Ram.N} proved that if $n$ is a generalized superior highly composite number, i.e., a CA number, then under the RH we have
\begin{align*}
\liminf_{n\rightarrow\infty}\left(\frac{\sigma(n)}{n}-e^{\gamma}\log\log n\right)\sqrt{\log n}\geq&amp; -e^\gamma(2\sqrt{2}+\gamma-\log4\pi)\approx-1.558,\\[7pt]
\limsup_{n\rightarrow\infty}\left(\frac{\sigma(n)}{n}-e^{\gamma}\log\log n\right)\sqrt{\log n}\leq&amp; -e^\gamma(2\sqrt{2}-4-\gamma+\log4\pi)\approx-1.393.
\end{align*}
Robin \cite{Rob} also established (independent of the RH) the following inequality
\begin{equation}\label{Rob2}
   f(n)\leq e^\gamma+\frac{0.648214}{(\log\log n)^2}\ ,\qquad (n\geq3),
\end{equation}
where $0.648214\approx(\frac{7}{3}-e^\gamma\log\log12)\log\log12$ and the left-hand side of (\ref{Rob2}) attains its maximum at $n=12$. In the same spirit, Lagarias \cite{Lag} proved the equivalence of the RH to the problem
$$
\sigma(n)\leq e^{H_n}\log H_n+H_n,\qquad(n\geq1),
$$
where $H_n:=\sum_{j=1}^n 1/j$  is the $n$th harmonic number.

Investigating upper and lower bounds for arithmetic functions, Landau \cite[pp.\ 216--219]{Land} obtained the following limits:
$$
\liminf_{n\rightarrow\infty}\frac{\varphi(n)\log\log n}{n}=e^{-\gamma}\ ,\qquad \limsup_{n\rightarrow\infty}\frac{\varphi(n)}{n}=1,
$$
where $\varphi(n)$ is the Euler totient function, which is defined as the number of positive integers not exceeding $n$ that are relatively prime to $n$. It can also be expressed as a product extended over the distinct prime divisors of $n$ \cite[Theorem 2.4]{Apostol} by
$$
\varphi(n)=n\prod_{p\mid n}\left(1-\frac1p\right).
$$
Nicolas \cite{Nic, Nic2} proved that if the RH is true, then we have for all $k\geq2$
\begin{equation}\label{nicolas-ineq}
    \frac{N_k}{\varphi(N_k)\log\log N_k}&gt;e^\gamma,
\end{equation}
where $N_k=\prod_{j=1}^k p_j$ and $p_j$ is the $j$th prime. On the other hand, if the RH is false, then for infinitely many $k$, inequality (\ref{nicolas-ineq}) is true and for infinitely many $k$, inequality (\ref{nicolas-ineq}) is false.

Compared to numbers $N_k$ which are the smallest integers that maximize $n/\varphi(n)$, there are integers which play this role for $\sigma(n)/n$ and they are called superabundant numbers. A positive integer $n$ is said to be \textit{superabundant} \cite{Al, Ram.N} if
\begin{equation}\label{sa}
    \frac{\sigma(n)}{n}&gt;\frac{\sigma(m)}{m}\qquad\mbox{for all }\  m&lt;n.
\end{equation}
We will use the symbol SA to denote the set of superabundant numbers and also as an abbreviation for the term ``superabundant''.
Briggs \cite{Briggs} described a computational study of the successive maxima of the relative sum-of-divisors function $\sigma(n)/n$. He also studied the density of these numbers. W\'{o}jtowicz \cite{Wojtowicz} showed that the values of the function $f$ defined in~ (\ref{f(n)}) are close to 0 on a set of asymptotic density 1. Another study on Robin's inequality~ (\ref{Rob1}) is due to Choie et al.\ \cite{Choie}. They have shown that the RH holds true if and only if every natural number divisible by a fifth power greater than 1 satisfies Robin's inequality (\ref{Rob1}).

Akbary and Friggstad \cite{Ak} established the following interesting theorem which enables us to limit our attention to a narrow sequence of positive integers, in order to find a probable counterexample to inequality (\ref{Rob1}).

\begin{theorem}[{\cite[Theorem 3]{Ak}}]
If there is any counterexample to Robin's inequality (\ref{Rob1}), then the least such counterexample is a superabundant number.
\end{theorem}

Unfortunately, to our knowledge, there is no known algorithm (except the formula (\ref{sa}) in the definition) to produce SA numbers. Alaoglu and Erd\H{o}s \cite{Al} proved that
$$
Q(x)&gt; c\,\frac{\log x\log\log x}{(\log\log\log x)^2},
$$
where $Q(x)$ denotes the number of SA numbers not exceeding $x$. Later, Erd\H{o}s and Nicolas \cite{Erd} demonstrated a stronger result that for every $\delta&lt;5/48$ we have
$$
Q(x)&gt;(\log x)^{1+\delta},\qquad(x&gt;x_0).
$$

As a natural question in this direction, it is interesting to introduce and study a set of positive integers to which the first probable violation of inequality (\ref{Rob1}) belongs. Following this aim, we introduce the sequence of \emph{extremely abundant numbers}. We will establish another criterion equivalent to the RH by proving that the RH is true if and only if there are infinitely many extremely abundant numbers. Also, we give a connection between extremely abundant numbers and CA numbers. Moreover, we present approximate formula for the prime factorization of (sufficiently large) extremely abundant numbers.
Finally, we establish the fragility of the RH with
respect to the terms
of certain subsets of SA numbers which are quite close to
the set of extremely abundant numbers.

Before stating the main definition and results of this paper,
we mention recent work of Caveney et al.\ \cite{Caveney}. They defined a positive integer $n$ as an \textit{extraordinary} number if $n$ is composite and $f(n)\geq f(kn)$ for all
$$
k\in \mathbb{N}\cup\{1/p:\ p\ \text{is a prime factor of}\ n\}.
$$
Under these conditions they showed that the smallest extraordinary number is $n = 4$. Then they proved that the RH is true if and only if 4 is the only extraordinary number. For more properties of these numbers and comparisons with SA and CA numbers, we refer the reader to \cite{Caveney2}.

\section{Extremely abundant numbers}

We define a new sequence of positive integers related to the RH. Our
primary contribution and motivation of this definition are  Theorems
\ref{ext} and \ref{rh}.
Let us now state the main definition of this paper.

\begin{definition}\label{extn}
A positive integer $n$ is \textit{extremely abundant} if either $n=10080$, or $n&gt;10080$ and
\begin{equation}\label{extn1}
\frac{\sigma(n)}{n\log\log n}&gt;\frac{\sigma(m)}{m\log\log m},\qquad \text{for all }\ 10080\leq m&lt;n.
\end{equation}
\end{definition}

Here 10080 has been chosen as the smallest SA number greater than 5040. In Table \ref{tabl} we list the first 20 extremely abundant numbers. To find them, we used a list of SA numbers (see~ Proposition \ref{XA-subset-SA}) provided by Kilminster \cite{Kil} and Noe \cite{Noe}.
\begin{remark}
If we choose (instead of 10080) $n_1$ such that $2520&lt;n_1\leq5040$, and define $n$ to be extremely abundant if either $n=n_1$, or $n&gt;n_1$ and
\begin{equation*}
\frac{\sigma(n)}{n\log\log n}&gt;\frac{\sigma(m)}{m\log\log m},\qquad \text{for all }\ n_1\leq m&lt;n,
\end{equation*}
then we have a finite number of elements $n\leq5040$ that satisfy the above inequality. Using inequality (\ref{Rob2}), we have
$$
f(n)&lt;e^\gamma+\frac{0.648214}{(\log\log n)^2}&lt;f(5040),\qquad\text{for some } s_{10308}&lt;n\leq s_{10309},
$$
where $s_k$ denotes the $k$th SA number listed in \cite{Noe}. Checking by computer for values $n$ between $5040$ and $s_{10309}$, we derive a finite set with maximum $5040$. Similarly, one can easily check for $n_1&lt;2520$, and get only sets with finite number of elements.
\end{remark}
Let XA denote the set of all extremely abundant numbers. (We also use XA as an abbreviation for the term ``extremely abundant''.)   Clearly, $XA\neq CA$ (see Table \ref{tabl}). Indeed, we shall prove that infinitely many elements of CA are not in XA and that, if RH holds, then infinitely many elements of XA are in CA. As an elementary result from the definition of XA numbers we have the following proposition.

\begin{proposition}\label{XA-subset-SA}
The inclusion $XA\subset SA$ holds.
\end{proposition}

\begin{proof}
First, $10080\in SA$. Next, if $n&gt;10080$ and $n\in XA$, then for $10080\leq m&lt;n$ we have
$$
\frac{\sigma(n)}{n}=f(n)\log\log n&gt;f(m)\log\log m=\frac{\sigma(m)}{m}.
$$
In particular, for $m=10080$ we get
$$
\frac{\sigma(n)}{n}&gt;\frac{\sigma(10080)}{10080}.
$$
So for $m&lt;10080$, we have
$$
\frac{\sigma(n)}{n}&gt;\frac{\sigma(10080)}{10080}&gt;\frac{\sigma(m)}{m}
$$
since $10080\in SA$. Therefore, $n$ belongs to SA.
\end{proof}

Next, motivating our construction of XA numbers,
we will establish the first interesting result of the paper.

\begin{theorem}\label{ext}
If there is any counterexample to Robin's inequality (\ref{Rob1}), then the least one is an XA number.
\end{theorem}

\begin{proof}
By doing some computer calculations we observe that there is no counterexample to Robin's inequality (\ref{Rob1}) for $5040&lt;n\leq10080$. Now let $n&gt;10080$ be the least counterexample to inequality (\ref{Rob1}). For $m$ satisfying $10080\leq m&lt;n$ we have
$$
f(m)&lt;e^\gamma\leq f(n).
$$
Therefore $n$ is an XA number.
\end{proof}

As we mentioned in Section \ref{sec-introduction} we will prove an equivalent criterion to the RH for which the proof is based on Robin's inequality (\ref{Rob1}) and Gronwall's theorem. Let $\#A$ denote the cardinality of the set $A$. The second stimulus result is the following theorem. This result also has its own interest that will be discussed in Section \ref{sec-delicacy-of-RH}.

\begin{theorem}\label{rh}
The RH is true if and only if $\# XA=\infty$.
\end{theorem}

\begin{proof}
\emph{Sufficiency.} Assume that RH is not true. Then by Theorem \ref{ext} we have $f(m)\geq e^\gamma$ for some  $m\geq10080$. From Gronwall's theorem, we
know that $M=\sup_{n\geq10080}f(n)$ is finite and there exists $n_0$ such that $f(n_0)=M\geq e^\gamma$ (if $M=e^\gamma$ then set $n_0=m$). An integer $n&gt;n_0$ satisfies $f(n)\leq M=f(n_0)$ and $n$ can not be in XA, so $\#XA\leq n_0$.

\emph{Necessity.} \ On the other hand, if RH is true, then Robin's inequality (\ref{Rob1}) holds. If $\# XA$ is finite, then let $m$ be its largest element. For every $n &gt; m$ the inequality $f(n)\le f(m)$ holds and therefore
$$
\limsup_{n\rightarrow\infty} f(n)\leq f(m)&lt;e^\gamma,
$$
which is a contradiction to Gronwall's theorem.
\end{proof}

There are some primes which cannot be the largest prime factor of any XA number. For example, referring to Table \ref{tabl}, there is no XA number with the largest prime factor $p(n)=149$. Do there exist infinitely many such primes?

\section{Auxiliary lemmas and inequalities}

Chebyshev's functions $\vartheta(x)$ and $\psi(x)$ are defined by
$$
\vartheta(x)=\sum_{p\leq x}\log p,\qquad\psi(x)=\sum_{p^m\leq x}\log p=\sum_{p\leq x}\left\lfloor\frac{\log x}{\log p}\right\rfloor\log p,
$$
where $\lfloor x\rfloor$ denotes the largest integer not exceeding $x$. It is known that the prime number theorem (PNT) (\cite[Theorem 434]{HW} and \cite[Theorem 3, 12]{Ingham}) is equivalent to
\begin{equation}\label{psi(x)-x}
    \psi(x)\sim x.
\end{equation}
In his m\'{e}moir, Chebyshev proved the following lemma that we call
Chebyshev's result.

\begin{lemma}[{\cite[p.\ 379]{Chebyshev}}]
For all $x&gt;1$
\begin{align*}
\vartheta(x)&amp;&lt;\frac65Ax-Ax^{\frac12}+\frac{5}{4\log6}\log^2x+\frac52\log x+2\\
\vartheta(x)&amp;&gt;Ax-\frac{12}{5}Ax^{\frac12}-\frac{5}{8\log6}\log^2x-\frac{15}{4}\log x-3,
\end{align*}
where
$$
A=\log\frac{2^\frac12 3^\frac13 5^\frac15}{30^\frac{1}{30}}\approx0.92129202.
$$
\end{lemma}

We will use the following corollary in the proof of Theorem \ref{p&lt;log(n)-prop}.

\begin{corollary}\label{chebyshev's-result}
We have
$$
\vartheta(x)&gt;\frac x3,\qquad(x\geq3).
$$
\end{corollary}

The next lemma here provides Littlewood's result for oscillation of Chebyshev's $\vartheta$ function.

\begin{lemma}[{\cite[Lemma 4]{Caveney2}}]\label{lemma-4,caveney2}
There exists a constant $c &gt; 0$ such that for infinitely many primes $p$ we have
\begin{equation}\label{theta(p)-p&lt;cp1/2logloglog(p)}
\vartheta(p)&lt;p-c\sqrt{p}\log\log\log p,
\end{equation}
and for infinitely many other primes $p$ we have
$$
\vartheta(p)&gt;p+c\sqrt{p}\log\log\log p.
$$
\end{lemma}

In what follows, we shall frequently use the following elementary inequalities:
\begin{equation}\label{log(1+t)}
    \frac{t}{1+t}&lt;\log(1+t)&lt; t,\qquad (t&gt;0),
\end{equation}
and
\begin{equation}\label{log(1+t)2}
\frac{2t}{2+t}&lt;\log (1+t),\qquad(t&gt;0).
\end{equation}

\section{Some properties of SA, CA and XA numbers}

We divide this section into three subsections, for which we shall exhibit several properties of SA, CA and XA numbers, respectively. We denote by $p(n)$ the largest prime factor of $n$ or, when there is no ambiguity, simply by $p$.

\subsection{SA Numbers}

\begin{proposition}\label{an+1&lt;=2an}
Let $n&lt;n'$ be two consecutive SA numbers. Then
$$
\frac{n'}{n}\leq2.
$$
\end{proposition}

\begin{proof}
Let $n=2^{k_2}\cdots p$. We compare $n$ with $2n$. In fact,
$$
\frac{\sigma(2n)/(2n)}{\sigma(n)/n}=\frac{2^{k_2+2}-1}{2^{k_2+2}-2}&gt;1.
$$
Hence $n'\leq2n$.
\end{proof}

\begin{corollary}
For any positive real number $x\geq1$ there exists at least one SA number $n$ such that $x\leq n&lt;2x$.
\end{corollary}

Alaoglu and Erd\H{o}s \cite{Al} have shown that if $n=2^{k_2}\cdot3^{k_3}\cdots p^{k_p}$ is an SA number, then $k_2\geq k_3\geq\cdots \geq k_p$ and $k_p=1$, except for $n=4,\ 36$.

\begin{proposition}[{\cite[Theorem 2]{Al}}]
Let $n\in SA$, and let $q&lt;r$ be prime factors of $n$ with corresponding exponents $k_q$ and $k_r$. Set
$$
\beta:= \left\lfloor\frac{k_q \log q}{\log r}\right\rfloor.
$$
Then $k_r$ has one of the three values: $\beta-1$, $\beta+1$, $\beta$.
\end{proposition}

As we observe, the above proposition determines the exponent of each prime factor of an SA number with error of at most 1 in terms of a smaller prime factor of that number. In the next lemma we prove a relation between the lower bound of an exponent of a prime factor of $n$ and its largest prime factor $p$.

\begin{lemma}\label{log(p)/log(q)&lt;kq}
Let $n\in SA$, and let $q$ be a prime factor of $n$. Then
$$
\left\lfloor\frac{\log p}{\log q}\right\rfloor\leq k_q.
$$
\end{lemma}

\begin{proof}
If $q = p\;(=p(n))$, then the result is trivial. Let $q &lt; p$ and $k_q=k$. Suppose that $k\leq~\lfloor\log p/\log q\rfloor-1$. Then
\begin{equation}\label{qk&lt;p}
    q^{k+1}&lt;p.
\end{equation}
Now we compare values of $\sigma(\nu)/\nu$, taking $\nu=n$ and $\nu=m=nq^{k+1}/p$. Since $\sigma(\nu)/\nu$ is multiplicative, we restrict our attention to different factors. But $n$ is an SA number and~ $m&lt;n$. Hence
$$
1&lt;\frac{\sigma(n)/n}{\sigma(m)/m}=\frac{q^{2k+2}-q^{k+1}}{q^{2k+2}-1}\left(1+\frac{1}{p}\right)=\frac{1}{1+1/q^{k+1}}\left(1+\frac{1}{p}\right).
$$
Consequently, $p&lt;q^{k+1}$, which contradicts inequality (\ref{qk&lt;p}).
\end{proof}

\begin{proposition}[{\cite[Theorem 5]{Al}}]
Let $n\in SA$. If $k_q=k$ and $q&lt;(\log p)^{\alpha}$, where $\alpha$ is a constant, then
\begin{equation}\label{log-q(k+1)}
\log\frac{q^{k+1}-1}{q^{k+1}-q}&gt;\frac{\log q}{p\log p}\left(1+O\left(\frac{(\log\log p)^2}{\log p\log q}\right)\right),
\end{equation}
\begin{equation}\label{log-q(k+2)}
\log\frac{q^{k+2}-1}{q^{k+2}-q}&lt;\frac{\log q}{p\log p}\left(1+O\left(\frac{(\log\log p)^2}{\log p\log q}\right)\right).
\end{equation}
\end{proposition}

\begin{lemma}\label{cplogp&lt;2k&lt;c'plogp}
Let $n\in SA$, and let $q$ be a fixed prime factor of $n$. Then there exist two positive constants $c$ and $c\,'$ (depending on $q$) such that
$$
c\,p\frac{\log p}{\log q}&lt;q^{k_q}&lt;c\,'p\frac{\log p}{\log q}.
$$
\end{lemma}

\begin{proof}
By inequality (\ref{log(1+t)})
\begin{align*}
\log\frac{q^{k+1}-1}{q^{k+1}-q}=\log\left(1+\frac{q-1}{q^{k+1}-q}\right)&lt;\frac{q-1}{q^{k+1}-q}\leq\frac{1}{q^k}
\end{align*}
and (\ref{log-q(k+1)}), there exists a $c\,'&gt;0$ such that
\begin{equation*}
q^k&lt;c\,'\,\frac{p\log p}{\log q}.
\end{equation*}
On the other hand, again by inequality (\ref{log(1+t)})
\begin{align*}
\log\frac{q^{k+2}-1}{q^{k+2}-q}=\log\left(1+\frac{q-1}{q^{k+2}-q}\right)&gt;\frac{q-1}{q^{k+2}-1}&gt;\frac{1}{2q^{k+1}}
\end{align*}
and (\ref{log-q(k+2)}), there exists a $c&gt;0$ such that
$$
q^k&gt;c\,\frac{p\log p}{\log q}.
$$
\end{proof}

\begin{corollary}
Let $n=2^{k}\cdots p$ be an SA number. Then for $n$ sufficiently large we have
$$
\left\lfloor\frac{k\log2}{\log p}\right\rfloor=1.
$$
\end{corollary}

We showed \cite{Naz} that for sufficiently large $n\in SA$
\begin{equation}\label{log(n)&lt;p(1+.5/log p)}
\log n&lt;p\left(1+\frac{1}{2\log p}\right).
\end{equation}
Our computation on the list of SA numbers in \cite{Noe} suggests that a weaker inequality
\begin{equation}\label{log(n)&lt;p(1+2/3log(p))}
\log n&lt;p\left(1+\frac{2}{3\log p}\right)
\end{equation}
holds for all $n\geq s_{365}$. The product of exponent of a prime factor and the logarithm of the corresponding prime factor of an SA number can be
controlled, on average, by the logarithm of the largest prime factor of that number. More precisely,

\begin{proposition}[{\cite[Theorem 7]{Al}}]\label{p=log}
If $n\in SA$, then
$$
p(n)\sim\log n.
$$
\end{proposition}

The next proposition gives a lower bound of $n\in SA$ in terms of Chebyshev's $\psi$ function compared to the above-mentioned asymptotic relation.

\begin{proposition}
Let $n\in SA$. Then
\begin{equation*}
\psi(p(n))\leq\log n.
\end{equation*}
Moreover,
\begin{equation}\label{psilog}
    \lim_{\substack{n\rightarrow\infty\\n\in SA}}\frac{\psi(p(n))}{\log n}=1.
\end{equation}
\end{proposition}

\begin{proof} In fact, by Lemma \ref{log(p)/log(q)&lt;kq}
$$
\psi(p(n))=\sum_{q\leq p(n)}\left\lfloor\frac{\log p(n)}{\log q}\right\rfloor\log q\leq\sum_{q\leq p(n)}k_q\log q=\log n.
$$
To prove (\ref{psilog}) we appeal to (the equivalent of) the PNT (\ref{psi(x)-x}) and Proposition \ref{p=log}.
\end{proof}

\subsection{CA Numbers}
By the definition of CA numbers (\ref{CA-defn}) it is easily seen that $CA\subset SA$. Here
we give a concise description of the algorithm (essentially borrowed from \cite{Caveney2, Erd, Rob}) to produce CA numbers. For more details on this introduction we refer the reader to \cite{Al, Caveney2, Erd, Rob}.

Let $F$ be defined by
\begin{equation*}\label{F(x,k)}
F(x,k)=\frac{\log(1+1/(x+\cdots+x^k))}{\log x}.
\end{equation*}
For $\varepsilon&gt;0$ we define $x_1=x_1(\varepsilon)$ to be the only number such that
\begin{equation}\label{x(epsilon)}
F(x_1,1)=\varepsilon,
\end{equation}
and $x_k=x_k(\varepsilon)$ (for $k&gt;1$) to be the only number such that
$$
F(x_k,k)=\varepsilon.
$$
Let
$$
E_p=\{F(p,\alpha):\ \alpha\geq1\},\ \ p\ \text{is a prime}
$$
and
$$
E=\bigcup_{p}E_p=\{\varepsilon_1,\,\varepsilon_2,\,\ldots\}.
$$
If $\varepsilon\notin E$, then the function $\sigma(n)/n^{1+\varepsilon}$ attains its maximum at a single point $N_\varepsilon$ whose prime decomposition is
$$
N_\varepsilon=\prod p^{\alpha_p(\varepsilon)},\qquad \alpha_p(\varepsilon)=\left\lfloor\frac{\log\frac{p^{1+\varepsilon}-1}{p^\varepsilon-1}}{\log p}\right\rfloor-1,
$$
or if one prefers
$$
\alpha_p(\varepsilon)=
                        \begin{cases}
                          k, &amp; \text{if $x_{k+1}&lt;p&lt;x_k$, $k\geq1$;} \\
                          0, &amp; \text{if $p&gt;x=x_1$.}
                        \end{cases}
$$
If $\varepsilon\in E$, then at most two $x_k$'s are prime. Hence, there are either two or four CA numbers of parameter $\varepsilon$, defined by
\begin{equation}\label{CA-N=prod-two-or-four}
N_\varepsilon=\prod_{k=1}^K\prod_{\substack{p&lt;x_k\\\text{or}\\p\leq x_k}}p,
\end{equation}
where $K$ is the largest integer such that $x_K\geq2$. In particular, if $N$ is the largest CA number of parameter $\varepsilon$, then
\begin{equation}\label{p(N)=p}
F(p,1)=\varepsilon\Rightarrow p(N)=p,
\end{equation}
where $p(N)$ is the largest prime factor of $N$. Therefore for any $\varepsilon$, formula (\ref{CA-N=prod-two-or-four}) gives all possible values of a CA number $N$ of parameter $\varepsilon$ \cite{Caveney2}.

Robin \cite[Proposition 1]{Rob} proved that the maximum order of the function $f$ defined in~ (\ref{f(n)}) is attained by CA numbers. More precisely, if $3\leq N&lt; n&lt; N'$, where $N$ and $N'$ are two successive CA numbers, then

\begin{equation*}\label{f(n)&lt;f(N),f(N')}
f(n)\le \max\{f(N ), f(N\,')\}.
\end{equation*}
In the next proposition we improve the above inequality to a strict one.

\begin{proposition}\label{Robin-prop}
Let $3\leq N&lt; n&lt; N'$, where $N$ and $N'$ are two successive CA numbers. Then
\begin{equation}\label{f(n)&lt;max{f(N),f(N')}}
f(n)&lt;\max\{f(N), f(N')\}.
\end{equation}
\end{proposition}

\begin{proof}
In fact, due to the strict convexity of the function $t\mapsto \varepsilon t-\log\log t$,
Robin's proof extends to the strict inequality (\ref{f(n)&lt;max{f(N),f(N')}}).
\end{proof}

Proposition \ref{Robin-prop} shows that if there is a counterexample to inequality (\ref{Rob1}), then there exists at least one CA number that violates it.

\begin{lemma}\label{N&lt;n&lt;N'-ca-xa}
Let $N&lt;N'$ be two consecutive CA numbers. If there exists an XA number $n&gt;10080$ satisfying $N&lt;n&lt;N'$, then $N'$ is also an XA.
\end{lemma}
\begin{proof}

Let us set
$$
B=\{m\in XA:\ N&lt;m&lt;N'\}.
$$
By assumption $n\in XA$, we have $B\neq\emptyset$. Let $n'=\max B$. Since $n'\in XA$ and $n'&gt;N$, it follows that $f(n')&gt;f(N)$. From inequality (\ref{f(n)&lt;max{f(N),f(N')}}) we must have $f(n')&lt;f(N')$. Hence~ $N'\in~ XA$.
\end{proof}

\begin{remark}\label{N&lt;n&lt;N'-ca-xa-remark}
If $n=10080$, then we have $N=5040$ and $N'=55440$, and
$$
f(N)\approx1.7909,\qquad f(n)\approx1.7558,\qquad f(N')\approx1.7512.
$$
Hence, $f(N')&lt;f(n)&lt;f(N)$ and inequality (\ref{f(n)&lt;max{f(N),f(N')}}) is satisfied with
$$
f(n)&lt;f(N)=\max\{f(N),\ f(N')\}.
$$
Therefore $N'\notin XA$. The point here is that $n=10080$ is the initial XA number, so that it misses the property (\ref{extn1}) of the definition of XA numbers which is used in the proof of Lemma~ \ref{N&lt;n&lt;N'-ca-xa}.
\end{remark}

\begin{theorem}\label{RH-&gt;CA-are-XA}
If RH holds, then there exist infinitely many CA numbers that are also XA.
\end{theorem}

\begin{proof}
If RH holds, then $\#XA=\infty$ by Theorem \ref{rh}. Let $n\in XA$. Since $\#CA=\infty$ \cite{Al, Erd}, there exist two successive CA numbers $N,\ N'$ such that $N&lt;n\leq N'$. If~ $N'=n$ then it is already in XA, otherwise $N'$ belongs to XA via Lemma \ref{N&lt;n&lt;N'-ca-xa}.
\end{proof}

It can be seen that there exist infinitely many CA numbers $N$ for which the largest prime factor $p\ (=p(N))$ is greater than $\log N$. For this purpose we use the following lemma.

\begin{lemma}[{\cite[Lemma 3]{Caveney2}}]\label{log(N)&lt;theta+cx1/2}
Let $N$ be a CA number of parameter $\varepsilon$ with
$$
\varepsilon&lt;F(2, 1) = \log(3/2)/ \log 2
$$
and define $x=x(\varepsilon)$ by (\ref{x(epsilon)}). Then
\begin{itemize}
\item[{\upshape (i)}] for some constant $c &gt;0$
$$
\log N\leq\vartheta(x)+c\sqrt{x}.
$$
\item[{\upshape (ii)}] Moreover, if $N$ is the largest CA number of parameter $\varepsilon$, then
$$
\vartheta(x)\leq\log N\leq\vartheta(x)+c\sqrt{x}.
$$
\end{itemize}
\end{lemma}

\begin{theorem}\label{log(N)&lt;p(N)}
There are infinitely many CA numbers $N_\varepsilon$ such that $\log N_\varepsilon&lt;p(N_\varepsilon)$.
\end{theorem}

\begin{proof}
Let $p$ be sufficiently large satisfying the inequality (\ref{theta(p)-p&lt;cp1/2logloglog(p)}), and let $N_\varepsilon$ be the largest CA number of parameter
$$
\varepsilon=F(p,1).
$$
Then from (\ref{p(N)=p}) it follows that $p(N_\varepsilon) = p$. By part (ii) of Lemma \ref{log(N)&lt;theta+cx1/2} we have
\begin{align*}
\log N_\varepsilon-\vartheta(p)&lt;c\sqrt{p},\qquad(\text{for some}\ c&gt;0).
\end{align*}
On the other hand, by Lemma \ref{lemma-4,caveney2} there exists a constant $c' &gt; 0$ such that
$$
\vartheta(p)-p&lt;-c'\sqrt{p}\log\log\log p,\qquad(c'&gt;0).
$$
Hence
$$
\log N_\varepsilon-p&lt;\{c-c'\log\log\log p\}\sqrt{p}&lt;0,
$$
which is the desired conclusion.
\end{proof}

\subsection{XA Numbers}

We return to XA numbers and present some of their properties. We begin by the first interesting property of the XA numbers whose proof is essentially an application of the definition of XA numbers.

\begin{theorem}\label{p&lt;log(n)-prop}
Let $n\in XA$. Then
$$
p(n)&lt;\log n.
$$
\end{theorem}

\begin{proof}
For $n=10080$ we have
$$
p(10080)=7&lt;9.218&lt;\log(10080).
$$
Let $n&gt;10080$ be an XA number and $m=n/p(n)$. Then $m&gt;10080$, since for all primes $p$ we have $\vartheta(p)&gt;p/3$ (Corollary \ref{chebyshev's-result}). Thus for $n\in SA$ we have
$$
\log n\geq\psi(p(n))\geq\vartheta(p(n))&gt;\frac13p(n)
$$
and $m=n/p(n)&gt;n/(3\log n)&gt;10080$ if $n\geq400,000$. For $n&lt;400,000$ we can check by computation. Hence by Definition \ref{extn} we obtain
$$
1+\frac1{p(n)}=\frac{\sigma(n)/n}{\sigma(m)/m}&gt;\frac{\log\log n}{\log\log m}.
$$
Therefore,
$$
\frac{1}{p(n)}&gt;\frac{\log\log n}{\log\log m}-1=\frac{\log(1+\log {p(n)}/\log m)}{\log\log m}.
$$
Using inequality (\ref{log(1+t)}) we have
$$
\frac{1}{p(n)}&gt;\frac{\log {p(n)}}{\log n\log\log m}&gt;\frac{\log {p(n)}}{\log n\log\log n}\Rightarrow {p(n)}&lt;\log n.
$$
\end{proof}

We mention here a similar result proved by Choie et al.\ \cite{Choie}.

\begin{proposition}[{\cite[Lemma 6.1]{Choie}}]
Let $t\geq2$ be fixed. Suppose that there exists a $t$-free integer exceeding $5040$ that does not satisfy Robin's inequality (\ref{Rob1}). Let $n$ be the smallest such integer. Then $p(n)&lt;\log n$.
\end{proposition}

In Theorem \ref{RH-&gt;CA-are-XA} we showed that if RH holds, then there exist infinitely many CA numbers that are also XA. Next theorem is a conclusion of Theorems \ref{log(N)&lt;p(N)} and \ref{p&lt;log(n)-prop} which is independent of the RH.

\begin{theorem}
There exist infinitely many CA numbers that are not XA.
\end{theorem}

We conclude this subsection with a result describing the structure of (sufficiently large) XA numbers. More precisely, the next theorem will determine the exponents of the prime factors of a (sufficiently large) XA number with an error at most 1.

\begin{theorem}
Let $n=2^{k_2}\cdots q^{k_q}\cdots p\in XA$. Set
$$
\alpha_q(p)=\left\lfloor\log_q\left(1+(q-1)\frac{p\log p}{q\log q}\right)\right\rfloor.
$$
Then for sufficiently large $n\in XA$ we have $|k_q-\alpha_q(p)|\leq1$.
\end{theorem}

\begin{proof}
Assume that $k_q=k$ and $k-\alpha_q(p)\geq2$. Then we have
\begin{equation}\label{q-to-k&gt;}
q^{k}\geq q^{\alpha_q(p)+2}&gt; q\left(1+(q-1)\frac{p\log p}{q\log q}\right).
\end{equation}
Now let us compare $f(n)$ with $f(m)$, where $m=n/q$. Since $n\in XA$ we must have
$$
\frac{\sigma(n)/n}{\sigma(m)/m}=\frac{q^{k+1}-1}{q^{k+1}-q}&gt;\frac{\log\log n}{\log\log m},
$$
or using inequality (\ref{log(1+t)}),
\begin{equation}\label{q-to-k&lt;}
q^k&lt;1+(q-1)\frac{\log n\log\log m}{q\log q}.
\end{equation}
Comparison of (\ref{q-to-k&gt;}) and (\ref{q-to-k&lt;}) gives that
$$
\log n\log\log m-q p\log p&gt;q\log q.
$$
This contradicts inequality (\ref{log(n)&lt;p(1+.5/log p)}). Now we assume that $k-\alpha_q(p)\leq-2$. Then
\begin{equation}\label{q-to-k+2/q-1&lt;plogp/qlogq}
\frac{q^{k+2}-1}{q-1}\leq\frac{p\log p}{q\log q}.
\end{equation}
Choose $m=nq/p$. Comparing $f(n)$ with $f(m)$ we have
$$
\frac{\sigma(n)/n}{\sigma(m)/m}=\left(1-\frac{q-1}{q^{k+2}-1}\right)\left(1+\frac1p\right)&gt;\frac{\log\log n}{\log\log m}&gt;1+\frac{\log p/q}{\log n\log\log m}.
$$
or simply by (\ref{q-to-k+2/q-1&lt;plogp/qlogq})
\begin{equation*}
-\frac{q\log q}{\log p}\left(1+\frac1p\right)&gt;\frac{p\log p/q}{\log n\log\log m}-1.
\end{equation*}
Hence by (\ref{log(n)&lt;p(1+.5/log p)}) we have
$$
-\frac{q\log q}{\log p}\left(1+\frac1p\right)&gt;\frac{\log p/q}{(\log p)\left(1+\frac{1}{2\log p}\right)\left(1+\frac{1}{2\log^2 p}\right)}-1,
$$
which is false for all $2\leq q\leq p$.
\end{proof}

\begin{remark}
Note that if inequality (\ref{log(n)&lt;p(1+2/3log(p))}) holds for XA numbers $n\geq s_{365}$, then by performing computations for two smaller values of XA numbers, i.e., $s_{20}$ and $s_{356}$ (cf.\ Table \ref{tabl}) we see that the above theorem holds true for all $n\in XA$.
\end{remark}

\section{Fragility of the RH and certain supersets of XA numbers}\label{sec-delicacy-of-RH}

In Theorem \ref{rh} we proved that under the RH the cardinality of the set of XA numbers is infinite. Here we present some interesting theorems which demonstrate the fragility of the RH showing the infinitude of some supersets of XA numbers independent of the RH. These sets are defined by inequalities quite close to that in (\ref{extn1}). The basic inequalities used here to define these sets are (\ref{log(1+t)}) and (\ref{log(1+t)2}).

\begin{lemma}\label{(s(n)/n)/(s(m)/m)&gt;1+log(n/m)/(log(n)loglog(m))}
If $m\geq3$, then there exists $n&gt;m$ such that
\begin{equation*}
\frac{\sigma(n)/n}{\sigma(m)/m}&gt;1+\frac{\log n/m}{\log n\log\log m}.
\end{equation*}
\end{lemma}

\begin{proof}
Let $m\geq3$. Then by inequality (\ref{Rob2})
\begin{equation}\label{3.41}
\frac{\sigma(m)}{m}\leq\left(e^\gamma+\frac{0.648214}{(\log\log m)^2}\right)\log\log m.
\end{equation}
Since for $m'&gt;m$
$$
\frac{\log\log m}{\log\log m'}\left(1+\frac{\log m'/m}{\log m'\log\log m}\right)&lt;1
$$
and the left-hand side is decreasing with respect to $m'$ and tends to zero as $m'\rightarrow\infty$, then for some $m'&gt;m$ we have
\begin{equation}\label{3.42}
\frac{\log\log m}{\log\log m'}\left(1+\frac{\log m'/m}{\log m'\log\log m}\right)\left(e^\gamma+\frac{0.648214}{(\log\log m)^2}\right)=e^\gamma-\varepsilon,
\end{equation}
where $\varepsilon&gt;0$ is arbitrarily small and fixed. Hence by Gronwall's theorem there exists $n\geq m'$ such that
\begin{align*}
\frac{\sigma(n)}{n}&amp;&gt;(e^\gamma-\varepsilon)\log\log n\\
                   &amp;=\frac{\log\log m}{\log\log m'}\left(1+\frac{\log m'/m}{\log m'\log\log m}\right)\left(e^\gamma+\frac{0.648214}{(\log\log m)^2}\right)\log\log n\\
                   &amp;\geq\left(1+\frac{\log n/m}{\log n\log\log m}\right)\frac{\sigma(m)}{m},
\end{align*}
where the last inequality holds by (\ref{3.41}) and (\ref{3.42}).
\end{proof}

\begin{definition}
Let $n_1=10080$, and let $n_{k+1}$ be the least integer greater than $n_k$ such that
$$
\frac{\sigma(n_{k+1})/n_{k+1}}{\sigma(n_k)/n_k}&gt;1+\frac{\log n_{k+1}/n_k}{\log n_{k+1}\log\log n_k},\qquad(k=1,\,2,\,\ldots).
$$
We define $X'$ to be the set of all $n_1$, $n_2$, $n_3$, \ldots.
\end{definition}

One can easily show that
\begin{equation}\label{3.43}
XA\subset X'\subset SA.
\end{equation}
Our first result towards the fragility of the RH is the following theorem.

\begin{theorem}\label{3.57}
The set $X'$ has an infinite number of elements.
\end{theorem}

\begin{proof}
If the RH is true, then the cardinality of $X'$ is infinite by (\ref{3.43}). If RH is not true, then by Theorem \ref{rh} there exists $m_0\geq10080$ such that
$$
\frac{\sigma(m_0)/m_0}{\sigma(m)/m}\geq\frac{\log\log m_0}{\log\log m},\qquad\text{for all}\ m\geq10080.
$$
By Lemma \ref{(s(n)/n)/(s(m)/m)&gt;1+log(n/m)/(log(n)loglog(m))} there exists $m'&gt;m_0$ such that $m'$ satisfies
$$
\frac{\sigma(m')/m'}{\sigma(m_0)/m_0}&gt;1+\frac{\log m'/m_0}{\log m'\log\log m_0}.
$$
Let $n$ be the least number greater than $m_0$ for which
$$
\frac{\sigma(n)/n}{\sigma(m_0)/m_0}&gt;1+\frac{\log n/m_0}{\log n\log\log m_0}.
$$
Hence $n\in X'$.
\end{proof}

The following lemma can be proved in the same manner as Lemma \ref{(s(n)/n)/(s(m)/m)&gt;1+log(n/m)/(log(n)loglog(m))}.

\begin{lemma}
If $m\geq3$, then there exists $n&gt;m$ such that
\begin{equation*}
\frac{\sigma(n)/n}{\sigma(m)/m}&gt;1+\frac{2\log n/m}{(\log m+\log n)\log\log m}.
\end{equation*}
\end{lemma}

We continue our approach towards the fragility of the RH via (the stronger) inequality (\ref{log(1+t)2}) defining a smaller superset of XA numbers as follows.

\begin{definition}
Let $n_1=10080$, and let $n_{k+1}$ be the least integer greater than $n_k$, such that
$$
\frac{\sigma(n_{k+1})/n_{k+1}}{\sigma(n_k)/n_k}&gt;1+\frac{2\log n_{k+1}/n_k}{(\log n_k+\log n_{k+1})\log\log n_k},\qquad(k=1,\,2,\,\ldots).
$$
We define $X''$ to be the set of all $n_1$, $n_2$, $n_3$, \ldots.
\end{definition}

By elementary inequality (\ref{log(1+t)2}) and
$$
\frac{t}{1+t}&lt;\frac{2t}{2+t},\qquad(t&gt;0)
$$
one can easily show the inclusion $XA\subset X''\subset X'$. The following theorem is a refinement of
Theorem \ref{3.57} with a similar proof.

\begin{theorem}
The set $X''$ has an infinite number of elements.
\end{theorem}

We calculated the number of elements in $XA$, $X'$ and $X''$ up to the $300,000$th element of SA in \cite{Noe}. Note that
$$
\#XA=9240,\qquad\#X'=9535,\qquad\#X''=9279
$$
and
$$
\#(X'-XA)=295,\qquad\#(X''-XA)=39.
$$
It might be interesting to look at the list of elements of $X''-XA$ up to $s_{300,000}$:
\begin{align*}
X''-XA&amp;=\{s_{55}, s_{62}, s_{91}, s_{106}, s_{116}, s_{127}, s_{128}, s_{137}, s_{138}, s_{149}, s_{181}, s_{196}, s_{212}, s_{219},\\
&amp;\hspace{22pt}s_{224}, s_{231}, s_{232}, s_{246}, s_{247}, s_{259}, s_{260}, s_{263}, s_{272}, s_{273}, s_{276}, s_{288}, s_{294},\\
&amp;\hspace{22pt}s_{299}, s_{305}, s_{311}, s_{317}, s_{330}, s_{340}, s_{341}, s_{343}, s_{354}, s_{65343}, s_{271143}, s_{271151}\}
\end{align*}
Note that the second XA number is $s_{356}$ (see Table \ref{tabl}) and
only three out of 39 elements in the set $X''-XA$ up to $s_{300,000}$,
namely $s_{65343}$, $s_{271143}$ and $s_{271151}$, are greater than $s_{356}$.

\section{Numerical experiments}

We present here some numerical results, mainly for the set of XA
numbers (sorted in increasing order) up to 13770th element, which is
less than $C_1:=s_{500,000}$, based on the list provided by Noe~\cite{Noe}.

\begin{property}
Let $n=2^{k_2}\cdots q^{k_q}\cdots r^{k_r}\cdots p\in XA$, where $2\leq q&lt;r\leq p$. Then for $10080&lt;n\leq C_1$
\begin{itemize}
  \item[{\upshape (i)}] $\log n&lt;q^{k_q+1}$,
  \item[{\upshape (ii)}] $r^{k_r}&lt;q^{k_q+1}&lt;r^{k_r+2}$,
  \item[{\upshape (iii)}] $q^{k_q}&lt;k_qp$,
  \item[{\upshape (iv)}] $q^{k_q}\log q&lt;\log n\log\log n&lt;q^{k_q+2}$.
  \end{itemize}
\end{property}

\begin{property}
Let $n=2^{k_2}\cdots x_k^k\cdots p\in XA$, where $2&lt;x_k&lt;p$ is the greatest prime factor of exponent $k$. Then
$$
\sqrt{p}&lt;x_2&lt;\sqrt{2p},\qquad\text{for}\ \ 10080&lt;n\leq C_1.
$$
\end{property}

\begin{property}
Let $n=2^{k_2}\cdots q^{k_q}\cdots p$ and $n'=2^{k'_2}\cdots q^{k'_q}\cdots p'$  be two consecutive XA numbers. Then for $10080&lt;n&lt;n'\leq C_1$
$$
|k_q-k'_q|\leq1,\qquad \mbox{for all}\ \ 2\leq q\leq p'.
$$
\end{property}

\begin{property}\label{d-p}
If $m$, $n$ are XA numbers, then for $10080\leq m&lt;n\leq C_1$
\begin{itemize}
  \item[{\upshape (i)}] $p(m)\leq p(n)$,
  \item[{\upshape (ii)}] $d(m)\leq d(n)$.
\end{itemize}
\end{property}

\begin{remark}
We note that Property \ref{d-p} is not true for SA numbers. For instance,
$$
s_{47}=(19\sharp)(3\sharp)^22,\qquad s_{48}=(17\sharp)(5\sharp)(3\sharp)2^3,
$$
$$
p(s_{47})=19&gt;17=p(s_{48}).
$$
and
$$
s_{173}=(59\sharp)(7\sharp)(5\sharp)(3\sharp)^22^3,\qquad s_{174}=(61\sharp)(7\sharp)(3\sharp)^22^2,
$$
$$
d(s_{173})=5308416&gt;5160960=d(s_{174}).
$$
\end{remark}

\begin{property}
If $n$, $n'\in XA$ are consecutive, then for $10080\leq n&lt;n'&lt;C_1$
$$
\frac{n'}{n}&gt;1+c\,\frac{(\log\log n)^2}{\log n},\qquad(0&lt;c\leq4),
$$
$$
\frac{n'}{n}&gt;1+c\,\frac{(\log\log n)^2}{\sqrt{\log n}},\qquad(0&lt;c\leq0.195).
$$
\end{property}

\begin{property}
If $n$, $n'\in XA$ are consecutive, then for $10080\leq n&lt;n'&lt;C_1$
$$
\frac{f(n')}{f(n)}&lt;1+\frac{1}{p\,'},
$$
where $p\,'$ is the largest prime factor of $n'$.
\end{property}

We have checked the following properties up to $C_2=s_{250,000}$ and up to 8150th element of XA numbers which is less than $C_2$.

\begin{property}
If $n$, $n'\in SA$ are consecutive, then
$$
\frac{\sigma(n')/n'}{\sigma(n)/n}&lt;1+\frac{1}{p\,'},\qquad(n'&lt;C_2),
$$
where $p\,'$ is the largest prime factor of $n'$.
\end{property}

The number of distinct prime factors of a number $n$ is denoted by $\omega(n)$ (see \cite{Sandor}).

\begin{property}
If $n$, $n'\in XA$ are consecutive, then for $10080\leq n&lt;n'&lt;C_2$, then
$$
\frac{n}{\omega(n)}\leq\frac{n'}{\omega(n')}.
$$
\end{property}

The comparison of the sets CA and XA is given. We calculated them up to
$C=s_{1,000,000}$ from the list of SA in \cite{Noe}.
\begin{align*}
\#&amp;\{n\in XA: n&lt;C\}=24875,\\ \#&amp;\{n\in CA: n&lt;C\}=21187,\\ \#&amp;\{n\in
CA\cap XA: n&lt;C\}=20468,\\ \#&amp;\{n\in XA\setminus CA: n&lt;C\}=4407,\\
\#&amp;\{n\in CA\setminus XA: n&lt;C\}=719, \end{align*} We conclude this
paper with another remark on choosing the first element of XA as
10080.

\begin{remark}
If we replace $n_1=10080$, the initial number in the definition of XA numbers, by $n_1=665280$, we do not need to pose the condition (i.e., $n&gt;n_1$) in Lemma \ref{N&lt;n&lt;N'-ca-xa}. Indeed, if we choose the initial number $n_1=665280$, then $N=55440&lt;665280&lt;720720=N'$, where in this case $N'$ is also an XA number. Therefore we do not need Remark \ref{N&lt;n&lt;N'-ca-xa-remark}. Moreover, if we choose $n_1=665280$, then there are 37 more XA numbers.
\end{remark}

\begin{table}[!h]
\begin{tabular}{||c|c|c|c|c|c|c||}
\hline
  &amp; $n$          &amp;Type&amp; $f(n)$    &amp; $p(n)$ &amp; $\log n$ &amp; $k_2$\\
  \hline
  1&amp;$\ \quad s_{20}=(7\sharp)(3\sharp)2^3=10080$ &amp; &amp; $1.75581$ &amp; $7$ &amp; $9.21831$ &amp; $5$\\[3pt]
  2&amp;$s_{356}=(113\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp;c &amp; $1.75718$ &amp; $113$ &amp; $126.444$ &amp; $8$\\
  3&amp;$s_{368}=(127\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp;c &amp; $1.75737$ &amp; $127$ &amp; $131.288$ &amp; $8$\\
  4&amp;$s_{380}=(131\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp;c &amp; $1.75764$ &amp; $131$ &amp; $136.163$ &amp; $8$\\
  5&amp;$s_{394}=(137\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp;c &amp; $1.75778$ &amp; $137$ &amp; $141.083$ &amp; $8$\\
  6&amp;$s_{408}=(139\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp;c &amp; $1.75821$ &amp; $139$ &amp; $146.018$ &amp; $8$\\
  7&amp;$s_{409}=(139\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^4$&amp;c &amp; $1.75826$ &amp; $139$ &amp; $146.711$ &amp; $9$\\
  8&amp;$s_{438}=(151\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^3$&amp; &amp; $1.75831$ &amp; $151$ &amp; $156.039$ &amp; $8$\\
  9&amp;$s_{440}=(151\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^4$&amp;c &amp; $1.75849$ &amp; $151$ &amp; $156.732$ &amp; $9$\\
  10&amp;$s_{444}=(151\sharp)(13\sharp)(7\sharp)(3\sharp)^2 2^4$&amp;c &amp; $1.75860$ &amp; $151$ &amp; $158.678$ &amp; $9$\\
  11&amp;$s_{455}=(157\sharp)(13\sharp)(5\sharp)(3\sharp)^2 2^4$&amp; &amp; $1.75864$ &amp; $157$ &amp; $161.788$ &amp; $9$\\
  12&amp;$s_{458}=(157\sharp)(13\sharp)(7\sharp)(3\sharp)^2 2^3$&amp; &amp; $1.75866$ &amp; $157$ &amp; $163.041$ &amp; $8$\\
  13&amp;$s_{459}=(157\sharp)(13\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.75892$ &amp; $157$ &amp; $163.734$ &amp; $9$ \\
  14&amp;$s_{476}=(163\sharp)(13\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.75914$ &amp; $163$ &amp; $168.828$ &amp; $9$ \\
  15&amp;$s_{486}=(163\sharp)(17\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;  &amp; $1.75918$ &amp; $163$ &amp; $171.661$ &amp; $9$ \\
  16&amp;$s_{493}=(167\sharp)(13\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.75943$ &amp; $167$ &amp; $173.946$ &amp; $9$ \\
  17&amp;$s_{502}=(167\sharp)(17\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.75966$ &amp; $167$ &amp; $176.779$ &amp; $9$ \\
  18&amp;$s_{519}=(173\sharp)(17\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.76006$ &amp; $173$ &amp; $181.933$ &amp; $9$ \\
  19&amp;$s_{537}=(179\sharp)(17\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.76038$ &amp; $179$ &amp; $187.120$ &amp; $9$ \\
  20&amp;$s_{555}=(181\sharp)(17\sharp)(7\sharp)(3\sharp)^2 2^4$ &amp;c  &amp; $1.76089$ &amp; $181$ &amp; $192.318$ &amp; $9$ \\
  \hline
\end{tabular}
\centering
\caption{First 20 extremely abundant numbers ($p_k\sharp:=\prod_{j=1}^k p_j$ and c represents a colossally abundant number).}\label{tabl}
\end{table}

\section{Acknowledgments}

We thank J. C. Lagarias, C. Calderon, J. Stopple, and M. Wolf for useful
discussion and sending us some relevant references. Our sincere thanks
to J.-L. Nicolas for careful reading of the manuscript, helpful
comments, and worthwhile suggestions which improved the
presentation of the paper. The work of the first author is supported by
the Calouste Gulbenkian Foundation, under Ph.D.\ grant number
CB/C02/2009/32. Research partially funded by the European Regional
Development Fund through the programme COMPETE and by the Portuguese
Government through the FCT under the project PEst-C/MAT/UI0144/2011.

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\bigskip
\hrule
\bigskip

\noindent 2010 \emph{Mathematics Subject Classification}:
Primary 11A25; Secondary 11N37, 11Y70, 11K31.

\noindent \emph{Keywords: }
extremely abundant number,
superabundant number,
colossally abundant number, Robin's theorem, Chebyshev's function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A004394},
\seqnum{A004490}, and
\seqnum{A217867}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 1 2013;
revised versions received  August 16 2013; January 15 2014.
Published in {\it Journal of Integer Sequences}, February 7 2014.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}




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