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\journal{Hexagon}

\begin{document}

\begin{frontmatter}

\title{Robin's Inequality and Superabundant Counterexamples}

\author{Frank Vega}

\address{Information Physics Institute, 840 W 67th St, Hialeah, FL 33012, USA}

\ead{vega.frank@gmail.com}

\begin{abstract}
Robin's criterion states that the Riemann Hypothesis holds if and only if $\sigma(n) < e^{\gamma} \cdot n \cdot \log\log n$ for every $n > 5040$. Let $n'$ be the squarefree kernel of $n$, $p_{k}$ its largest prime factor, and $d = e^{\gamma}/\zeta(2) \approx 1.08276$. We prove that Robin's inequality holds for every $n > 5040$ with $d \cdot \log\log n' \leq \log\log n$, and for every $n > 5040$ whose prime factors are all below $(\log n)^{1/d}$, and that both results remain valid with $d$ replaced by $1.016$. Our main results concern superabundant numbers, among which there are infinitely many counterexamples to Robin's inequality if the Riemann Hypothesis is false. Using the classical error term in the prime number theorem, we show that every superabundant counterexample satisfies $\log\log n - \log p_{k} < C \cdot \exp(-c \sqrt{\log p_{k}})$ for absolute constants $C, c > 0$. Since the least counterexample satisfies $\log n > p_{k}$, we deduce that the Riemann Hypothesis is true if there are constants $c > 0$ and $0 < \alpha < 2$ with $c \geq 0.0562/31.03^{2-\alpha}$, and a threshold $N \leq 10^{10^{13.11485}}$, such that $\log\log n - \log p_{k} \geq c/\log^{\alpha}(p_{k})$ for every superabundant number $n > N$ with $\log n > p_{k}$. We also give explicit versions of these bounds: every counterexample satisfies $p_{k}^{1.000002} > \log n$, and every superabundant counterexample satisfies $-0.010101/(\log\log n)^{3} < \log\log n - \log p_{k} < 0.056201/(\log\log n)^{2}$. An enumeration of the superabundant numbers up to $10^{40}$ shows that only $3$ of the $244$ exceeding $5040$ satisfy $e^{0.0562} > 1 + (\log\log n - \log p_{k}) \cdot \log^{2}(p_{k})$, namely those with $p_{k} > \log n$, the largest of which is $10685862914126400$. Testing the hypothesis of the sufficient condition on the same range, we find that all $168$ superabundant numbers between $10685862914126400$ and $10^{40}$ satisfy $\log\log n - \log p_{k} > 0$, and that for each $\alpha \in \{1, 2, 3, 4, 5\}$ the minimum of $(\log\log n - \log p_{k}) \cdot \log^{\alpha}(p_{k})$ over them is positive, ranging from $0.0223$ for $\alpha = 1$ to $6.19$ for $\alpha = 5$, and is attained at the same superabundant number, whose largest prime factor is $59$.
\end{abstract}

\begin{keyword}
Riemann Hypothesis \sep Robin's inequality \sep Sum-of-divisors function \sep Superabundant numbers \sep Squarefree kernel \sep Prime numbers \sep Chebyshev function
\MSC[2020] 11M26 \sep 11A41 \sep 11A25
\end{keyword}

\end{frontmatter}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction}

The Riemann Hypothesis (RH) is the conjecture that the Riemann zeta function has its zeros only at the negative even integers and at complex numbers with real part $\frac{1}{2}$~\cite{Bor+08}. It has been called the ``Holy Grail of Mathematics''~\cite{CO15, CO16}, and numerous equivalent formulations exist~\cite{CO03}. One of particular interest here is Robin's criterion~\cite{Rob84}, which asserts that RH is equivalent to the inequality
\begin{equation}\label{eq:robin}
\sigma(n) < e^{\gamma} \cdot n \cdot \log\log n \qquad \text{for every integer } n>5040.
\end{equation}
Here $\gamma \approx 0.57721$ is the Euler--Mascheroni constant, $\sigma(n) = \sum_{d \mid n} d$ is the sum-of-divisors function, and $\log$ denotes the natural logarithm. Gr{\"o}nwall's theorem~\cite{Gro13} states that $\limsup_{n \to \infty} \sigma(n)/(n \log \log n) = e^{\gamma}$, so \eqref{eq:robin} asserts that this limiting value is never attained beyond $5040$. A closely related elementary reformulation in terms of harmonic numbers is due to Lagarias~\cite{Lag02}, and the extremal integers relevant to both criteria have been studied computationally by Briggs~\cite{Bri06} and structurally by Caveney, Nicolas and Sondow~\cite{CNS12}.

It is known that \eqref{eq:robin} holds for many classes of integers: odd integers exceeding $9$ and squarefree integers exceeding $30$~\cite{Cho+07}, sums of two squares exceeding $720$~\cite{Ban+09}, and $t$-free integers for $t = 5, 7, 11$ and $20$~\cite{Cho+07,SP12,BT15,PM21}. In a complementary direction, Akbary and Friggstad~\cite{AF09} showed that the least counterexample to \eqref{eq:robin}, should one exist, must be a superabundant number, and Nazardonyavi and Yakubovich~\cite{NY14} refined this to the smaller class of extremely abundant numbers. The author has also contributed to this line of results~\cite{FRA22}.

Let $\operatorname{core}(n)$ denote the squarefree kernel of $n$, that is, the product of the distinct primes dividing $n$~\cite{Cho+07}, and write $n' = \operatorname{core}(n)$. In an earlier version of this work~\cite{VEG21} the author showed that \eqref{eq:robin} holds for every $n > 5040$ with $\zeta(2) \cdot \log\log n' \leq \log\log n$, and for every $n > 5040$ all of whose prime factors $q$ satisfy $q < \sqrt[\beta]{\log n}$, where $\beta = e^{\gamma}$. The purpose of the present paper is to replace both constants $\zeta(2) \approx 1.64493$ and $e^{\gamma} \approx 1.78107$ by the single smaller constant
\[
d = \frac{e^{\gamma}}{\zeta(2)} = \frac{6 \cdot e^{\gamma}}{\pi^{2}} \approx 1.08276.
\]
Our main results are the following.

\begin{Theorem}\label{thm:core}
Let $n > 5040$ and $n' = \operatorname{core}(n)$. If $d \cdot \log\log n' \leq \log\log n$, then Robin's inequality holds for $n$.
\end{Theorem}

\begin{Theorem}\label{thm:primes}
Let $n > 5040$. If every prime $q$ that divides $n$ satisfies $q < \sqrt[d]{\log n}$, then Robin's inequality holds for $n$.
\end{Theorem}

\subsection{Summary of the Main Results}

Let $p_{k}$ denote the largest prime factor of $n$, and call $n$ a \emph{counterexample} if $n > 5040$ and \eqref{eq:robin} fails for $n$. Besides Theorems~\ref{thm:core} and~\ref{thm:primes}, the main results of this paper are the following.
\begin{enumerate}
\item[(1)] Theorems~\ref{thm:core} and~\ref{thm:primes} remain valid with $d$ replaced by $1.016$ (Theorems~\ref{thm:core1016} and~\ref{thm:primes1016}).
\item[(2)] Every sufficiently large superabundant number $n$ satisfies $p_{k} > \sqrt[d]{\log n}$ and $\theta(p_{k}) > \sqrt[d]{\log n}$ (Theorem~\ref{thm:largeSA}), and the least counterexample satisfies $\sqrt[d]{\log n} \leq p_{k} < \log n$ (Corollary~\ref{cor:least}).
\item[(3)] If the Riemann Hypothesis is false, there are infinitely many superabundant counterexamples (Proposition~\ref{prop:infSA}).
\item[(4)] Every counterexample satisfies $p_{k} > e^{30.54}$ (Proposition~\ref{prop:Plarge}) and $p_{k}^{1.000002} > \log n$ (Proposition~\ref{prop:Pboot}).
\item[(5)] Every superabundant counterexample satisfies
\[
-\frac{0.0101}{\log^{3}(p_{k})} < \log\log n - \log p_{k} < \frac{0.0562}{\log^{2}(p_{k})},
\]
\[
e^{0.0562} > 1 + (\log\log n - \log p_{k}) \cdot \log^{2}(p_{k})
\]
(Theorem~\ref{thm:window}), and, in terms of $\log\log n$ alone,
\[
-\frac{0.010101}{(\log\log n)^{3}} < \log\log n - \log p_{k} < \frac{0.056201}{(\log\log n)^{2}},
\]
\[
(\log n)^{\log\log n - \log p_{k}} < e^{\frac{0.056201}{\log\log n}}
\]
(Corollary~\ref{cor:windowL}). Moreover, $\log\log n - \log p_{k} < O\left(\exp\left(-c_{2} \sqrt{\log p_{k}}\right)\right)$ for an absolute constant $c_{2} > 0$ (Corollary~\ref{cor:dVP}). If there are constants $c > 0$ and $0 < \alpha < 2$ with $c \geq 0.0562/31.03^{2-\alpha}$, and $N \leq 10^{10^{13.11485}}$, such that every superabundant number $n > N$ with $\log n > p_{k}$ satisfies $\log\log n - \log p_{k} \geq \frac{c}{\log^{\alpha}(p_{k})}$, then the Riemann Hypothesis is true (Proposition~\ref{prop:RHcond}).
\item[(6)] The least counterexample satisfies $\log p_{k} < \log\log n$ (Theorem~\ref{Main}), $1 < (\log n)^{\log\log n - \log p_{k}} < e^{\frac{0.056201}{\log\log n}}$ (Corollary~\ref{cor:windowL}), and $\log n \cdot \log\log \frac{n}{p_{k}} > p_{k} \cdot \log p_{k}$ (Theorem~\ref{thm:lowerx}).
\item[(7)] Among the $244$ superabundant numbers $5040 < n \leq 10^{40}$, exactly $3$ satisfy $e^{0.0562} > 1 + (\log\log n - \log p_{k}) \cdot \log^{2}(p_{k})$, namely the $3$ with $p_{k} > \log n$ (Proposition~\ref{prop:xcount}).
\end{enumerate}

The proofs combine explicit estimates of Rosser and Schoenfeld~\cite{RS62}, of Axler~\cite{Axl18} and of Aoudjit, Berkane and Dusart~\cite{DU21} with the numerical verification of \eqref{eq:robin} for all $5040 < n \leq 10^{10^{13.11485}}$ due to Morrill and Platt~\cite{PM21}. Section~\ref{sec:SA} treats superabundant numbers, Section~\ref{sec:largest} the largest prime factor of a counterexample, and Section~\ref{sec:computation} the computer verification of every numerical claim, including a rigorous enumeration of the $263$ superabundant numbers up to $10^{40}$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Background and Ancillary Results}

We define the \emph{abundancy index} $I\colon \mathbb{N} \rightarrow \mathbb{Q}$ by $I(n) = \frac{\sigma(n)}{n}$, and we say that \emph{Robin's inequality holds for $n$} if
\[
I(n) < e^{\gamma} \cdot \log\log n.
\]

\begin{Proposition}\label{prop:RH}
Robin's inequality holds for all natural numbers $n > 5040$ if and only if the Riemann Hypothesis is true~\cite{Rob84} [Theorem~1~pp.~188].
\end{Proposition}

As usual, $\varphi$ denotes Euler's totient function. For every $n > 1$ we have $\frac{n'}{\varphi(n')} = \prod_{q \mid n} \frac{q}{q-1}$, where the product runs over the distinct primes dividing $n$.

\begin{Proposition}\label{prop:Choie}
For $n > 1$~\cite{Cho+07} [(2.7)~pp.~362]:
\[
I(n) < \prod_{q \mid n} \frac{q}{q-1} = \frac{n'}{\varphi(n')}.
\]
\end{Proposition}

The next two propositions are explicit estimates of Rosser and Schoenfeld.

\begin{Proposition}\label{prop:RSmertens}
For $x \geq 285$~\cite{RS62} [Theorem~7], see also~\cite{Axl18} [(6.1)]:
\[
\prod_{p \leq x} \left(1 - \frac{1}{p}\right) > \frac{e^{-\gamma}}{\log x} \cdot \left(1 - \frac{1}{2 \log^{2}(x)}\right).
\]
Equivalently, for $x \geq 285$,
\[
\prod_{p \leq x} \frac{p}{p-1} < e^{\gamma} \cdot \log x \cdot \left(1 - \frac{1}{2 \log^{2}(x)}\right)^{-1}.
\]
\end{Proposition}

\begin{Proposition}\label{prop:RSphi}
For every integer $m \geq 3$~\cite{RS62} [Theorem~15]:
\[
\frac{m}{\varphi(m)} < e^{\gamma} \cdot \log\log m + \frac{2.51}{\log\log m}.
\]
\end{Proposition}

Rosser and Schoenfeld in fact prove Proposition~\ref{prop:RSphi} with $2.50637$ in place of $2.51$ for every $m \geq 3$ except $m = 223092870$; the version with $2.51$ holds without exception and is sufficient for our purposes.

\begin{Proposition}\label{prop:MP}
Robin's inequality holds for every integer $n$ with $5040 < n \leq 10^{10^{13.11485}}$~\cite{PM21} [Theorem~5].
\end{Proposition}

The Chebyshev function $\theta(x)$ is defined by $\theta(x) = \sum_{p \leq x} \log p$, where the sum runs over all primes $p \leq x$. For a prime $p$, we write $p\# = \prod_{q \leq p} q$ for the primorial of $p$, so that $\log(p\#) = \theta(p)$. Throughout, $p_{1} = 2 < p_{2} = 3 < p_{3} = 5 < \cdots$ denotes the sequence of all primes.

Ramanujan's unpublished notes~\cite{Ram97} introduced generalized highly composite numbers, encompassing both superabundant and colossally abundant numbers; these were also studied by Alaoglu and Erd\H{o}s~\cite{AE44}. Given the first $k$ consecutive primes $p_{1}, \ldots, p_{k}$, an integer of the form $\prod_{i=1}^{k} p_{i}^{a_{i}}$ with $a_{1} \geq a_{2} \geq \cdots \geq a_{k} \geq 1$ is called a \emph{Hardy--Ramanujan integer}~\cite{Cho+07} [pp.~367]. A natural number $n$ is called \emph{superabundant} if $I(m) < I(n)$ for every natural number $m < n$.

\begin{Proposition}\label{prop:HR}
Every superabundant number is a Hardy--Ramanujan integer~\cite{AE44} [Theorem~1~pp.~450].
\end{Proposition}

\begin{Proposition}\label{prop:AF09}
If $n > 5040$ is the smallest integer such that Robin's inequality does not hold for $n$, then $n$ is superabundant~\cite{AF09} [Theorem~3, p.~273].
\end{Proposition}

\begin{Proposition}\label{prop:largestprime}
If $n > 5040$ is the smallest integer such that Robin's inequality does not hold for $n$, and $p$ denotes the largest prime factor of $n$, then $p < \log n$~\cite{Cho+07} [Lemma~6.1, p.~369].
\end{Proposition}

The following explicit estimates of Axler and of Aoudjit, Berkane and Dusart, and the author's lower bound for the largest prime factor of the least counterexample are used in Section~\ref{sec:largest}.

\begin{Proposition}\label{prop:Axler}
For $x \geq 46{,}909{,}038$~\cite{Axl18} [Proposition~6.1]:
\[
\prod_{p \leq x} \left(1 - \frac{1}{p}\right) > \frac{e^{-\gamma}}{\log x} \cdot \left(1 - \frac{1}{20 \log^{3}(x)} - \frac{3}{16 \log^{4}(x)}\right).
\]
Equivalently, for $x \geq 46{,}909{,}038$,
\[
\prod_{p \leq x} \frac{p}{p-1} < e^{\gamma} \cdot \log x \cdot \left(1 - \frac{1}{20 \log^{3}(x)} - \frac{3}{16 \log^{4}(x)}\right)^{-1}.
\]
\end{Proposition}

\begin{Proposition}\label{prop:Chebyshev}
For $x \geq 7{,}232{,}121{,}212$~\cite{DU21}:
\[
\theta(x) \geq \left(1 - \frac{0.01}{\log^{3}(x)}\right) \cdot x.
\]
\end{Proposition}

\begin{Proposition}\label{prop:Frank}
If $n > 5040$ is the smallest integer for which Robin's inequality fails, then the largest prime factor $p$ of $n$ satisfies $p > e^{31.018189471}$~\cite{FRA22} [Theorem~4.2~pp.~748].
\end{Proposition}

\subsection{Numerical Constants}

We use $\zeta(2) = \frac{\pi^{2}}{6} = 1.6449340668\ldots$~\cite{Edw01}, $\gamma = 0.5772156649\ldots$ and $e^{\gamma} = 1.7810724179\ldots$ The following numerical facts are elementary; they were checked with high-precision arithmetic, and the computed values are listed in Table~\ref{tab:constants} of Section~\ref{sec:computation}.

\begin{Lemma}\label{lem:constants}
Let $d = \frac{e^{\gamma}}{\zeta(2)}$. Then:
\begin{enumerate}
\item[(i)] $d = 1.0827621\ldots$, so that $1.0827 < d < 1.0828$;
\item[(ii)] $e^{\gamma} \cdot (d - 1) = \frac{e^{2\gamma}}{\zeta(2)} - e^{\gamma} = 0.1474054\ldots > 0.1474$;
\item[(iii)] $e^{\gamma} \cdot (d - 1) \cdot \log^{2}(64) > 0.1474 \cdot 17.296 > 2.549 > 2.51$;
\item[(iv)] $\theta(73) = \log(73\#) = 65.876\ldots > 64$;
\item[(v)] $\left(1 - \frac{1}{2 \log^{2}(285)}\right)^{-1} = 1.01589\ldots < 1.016 < d$;
\item[(vi)] $e^{\gamma} \cdot \log 285 \cdot \left(1 - \frac{1}{2 \log^{2}(285)}\right)^{-1} = 10.2275\ldots < 10.23$;
\item[(vii)] $e^{\gamma} \cdot \log 313 = 10.2344\ldots > 10.23$ and $e^{313} < 10^{136}$.
\end{enumerate}
\end{Lemma}

\subsection{Two Auxiliary Lemmas}

\begin{Lemma}\label{lem:mertens}
Let $x \geq 2$. Then:
\begin{enumerate}
\item[(a)] if $x \leq 285$, then $\prod_{p \leq x} \frac{p}{p-1} < 10.23$;
\item[(b)] if $x \geq 285$, then $\prod_{p \leq x} \frac{p}{p-1} < 1.016 \cdot e^{\gamma} \cdot \log x$.
\end{enumerate}
\end{Lemma}

\begin{proof}
Every factor $\frac{p}{p-1}$ exceeds $1$, so the product is nondecreasing in $x$. For (a), Proposition~\ref{prop:RSmertens} at $x = 285$ and Lemma~\ref{lem:constants}~(vi) give
\[
\prod_{p \leq x} \frac{p}{p-1} \leq \prod_{p \leq 285} \frac{p}{p-1} < e^{\gamma} \cdot \log 285 \cdot \left(1 - \frac{1}{2 \log^{2}(285)}\right)^{-1} < 10.23.
\]
For (b), the function $x \mapsto \left(1 - \frac{1}{2 \log^{2}(x)}\right)^{-1}$ is decreasing on $[285, \infty)$, so by Proposition~\ref{prop:RSmertens} and Lemma~\ref{lem:constants}~(v),
\[
\prod_{p \leq x} \frac{p}{p-1} < e^{\gamma} \cdot \log x \cdot \left(1 - \frac{1}{2 \log^{2}(285)}\right)^{-1} < 1.016 \cdot e^{\gamma} \cdot \log x.
\]
\end{proof}

\begin{Lemma}\label{lem:small}
Let $n > 5040$ satisfy $I(n) < 10.23$. Then Robin's inequality holds for $n$.
\end{Lemma}

\begin{proof}
If $\log n \geq 313$, then by Lemma~\ref{lem:constants}~(vii),
\[
e^{\gamma} \cdot \log\log n \geq e^{\gamma} \cdot \log 313 > 10.23 > I(n).
\]
If $\log n < 313$, then $n < e^{313} < 10^{136} < 10^{10^{13.11485}}$ by Lemma~\ref{lem:constants}~(vii), and Robin's inequality holds for $n$ by Proposition~\ref{prop:MP}.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Main Results}

\subsection{A Condition on the Squarefree Kernel}

\begin{proof}[Proof of Theorem~\ref{thm:core}]
Let $n > 5040$ satisfy $d \cdot \log\log n' \leq \log\log n$, where $n' = \operatorname{core}(n)$. By Proposition~\ref{prop:Choie},
\begin{equation}\label{eq:choie}
I(n) < \frac{n'}{\varphi(n')}.
\end{equation}
There are two cases.

\emph{Case I: $\log n' \geq 64$.} Put $L' = \log\log n'$, so that $L' \geq \log 64 > 0$, and in particular $n' \geq 3$. By Lemma~\ref{lem:constants}~(iii),
\[
e^{\gamma} \cdot (d - 1) \cdot L'^{2} \geq e^{\gamma} \cdot (d - 1) \cdot \log^{2}(64) > 2.51,
\]
that is, $\frac{2.51}{L'} < e^{\gamma} \cdot (d - 1) \cdot L'$. By Proposition~\ref{prop:RSphi} applied to $m = n'$, and by the hypothesis $d \cdot L' \leq \log\log n$,
\[
\frac{n'}{\varphi(n')} < e^{\gamma} \cdot L' + \frac{2.51}{L'} < e^{\gamma} \cdot L' + e^{\gamma} \cdot (d - 1) \cdot L' = e^{\gamma} \cdot d \cdot L' \leq e^{\gamma} \cdot \log\log n.
\]
Together with \eqref{eq:choie} this gives $I(n) < e^{\gamma} \cdot \log\log n$.

\emph{Case II: $\log n' < 64$.} Let $q_{1} < \cdots < q_{r}$ be the distinct primes dividing $n$, so that $r \geq 1$ and $n' = q_{1} \cdots q_{r}$. Since $q_{i} \geq p_{i}$ for every $i$, we have $n' \geq p_{1} \cdots p_{r} = p_{r}\#$, hence $\theta(p_{r}) \leq \log n' < 64 < \theta(73)$ by Lemma~\ref{lem:constants}~(iv), and so $p_{r} \leq 71$. The function $t \mapsto \frac{t}{t-1}$ is decreasing on $(1, \infty)$, so $\frac{q_{i}}{q_{i}-1} \leq \frac{p_{i}}{p_{i}-1}$ for every $i$. By Lemma~\ref{lem:mertens}~(a),
\[
\frac{n'}{\varphi(n')} = \prod_{i=1}^{r} \frac{q_{i}}{q_{i}-1} \leq \prod_{i=1}^{r} \frac{p_{i}}{p_{i}-1} \leq \prod_{p \leq 71} \frac{p}{p-1} < 10.23.
\]
By \eqref{eq:choie}, $I(n) < 10.23$, and Robin's inequality holds for $n$ by Lemma~\ref{lem:small}.
\end{proof}

\begin{Remark}
Case II of the proof does not use the hypothesis of Theorem~\ref{thm:core}: Robin's inequality holds for every $n > 5040$ whose squarefree kernel satisfies $\operatorname{core}(n) < e^{64}$. The hypothesis $d \cdot \log\log n' \leq \log\log n$ is needed only in Case I.
\end{Remark}

\subsection{A Condition on the Prime Factors}

\begin{proof}[Proof of Theorem~\ref{thm:primes}]
Let $n > 5040$, and suppose that every prime $q$ dividing $n$ satisfies $q < \sqrt[d]{\log n}$. Let $P$ be the largest prime factor of $n$. Then $P^{d} < \log n$, that is,
\begin{equation}\label{eq:Pd}
d \cdot \log P < \log\log n.
\end{equation}
Since every prime dividing $n$ is at most $P$ and every factor $\frac{p}{p-1}$ exceeds $1$, Proposition~\ref{prop:Choie} gives
\[
I(n) < \prod_{q \mid n} \frac{q}{q-1} \leq \prod_{p \leq P} \frac{p}{p-1}.
\]
There are two cases.

\emph{Case I: $P \geq 285$.} By Lemma~\ref{lem:mertens}~(b), Lemma~\ref{lem:constants}~(v) and \eqref{eq:Pd},
\[
I(n) < 1.016 \cdot e^{\gamma} \cdot \log P < e^{\gamma} \cdot d \cdot \log P < e^{\gamma} \cdot \log\log n.
\]

\emph{Case II: $P < 285$.} By Lemma~\ref{lem:mertens}~(a), $I(n) < 10.23$, and Robin's inequality holds for $n$ by Lemma~\ref{lem:small}.
\end{proof}

\subsection{Comparison with the Earlier Constants}

Since $d < \zeta(2)$, the hypothesis $\zeta(2) \cdot \log\log n' \leq \log\log n$ implies $d \cdot \log\log n' \leq \log\log n$ whenever $\log\log n' \geq 0$; and when $\log\log n' < 0$ we have $d \cdot \log\log n' < 0 < \log\log n$ for $n > 5040$, so the hypothesis of Theorem~\ref{thm:core} holds in any case. Likewise, since $d < e^{\gamma}$ and $\log n > 1$, the condition $q < \sqrt[e^{\gamma}]{\log n}$ implies $q < \sqrt[d]{\log n}$. Hence Theorems~\ref{thm:core} and~\ref{thm:primes} contain the two results of the earlier version~\cite{VEG21}.

\begin{Corollary}\label{cor:old}
Let $n > 5040$ and $n' = \operatorname{core}(n)$. If $\zeta(2) \cdot \log\log n' \leq \log\log n$, or if every prime $q$ dividing $n$ satisfies $q < \sqrt[e^{\gamma}]{\log n}$, then Robin's inequality holds for $n$.
\end{Corollary}

\subsection{The Exponent 1.016}

The proofs of Theorems~\ref{thm:core} and~\ref{thm:primes} extend to the smaller exponent $1.016$.

\begin{Lemma}\label{lem:small2}
Let $n > 5040$ satisfy $I(n) < 17.3$. Then Robin's inequality holds for $n$.
\end{Lemma}

\begin{proof}
If $\log n \geq 16600$, then $e^{\gamma} \cdot \log\log n \geq e^{\gamma} \cdot \log 16600 > 17.3 > I(n)$. If $\log n < 16600$, then $n < e^{16600} < 10^{7210} < 10^{10^{13.11485}}$, and Robin's inequality holds for $n$ by Proposition~\ref{prop:MP}.
\end{proof}

\begin{Theorem}\label{thm:core1016}
Let $n > 5040$ and $n' = \operatorname{core}(n)$. If $1.016 \cdot \log\log n' \leq \log\log n$, then Robin's inequality holds for $n$.
\end{Theorem}

\begin{proof}
By Proposition~\ref{prop:Choie}, $I(n) < \frac{n'}{\varphi(n')}$.

\emph{Case I: $\log n' \geq 13000$.} Put $L' = \log\log n' \geq \log 13000 > 0$, so that $n' \geq 3$. Since $e^{\gamma} \cdot 0.016 \cdot \log^{2}(13000) > 2.51$, we have $\frac{2.51}{L'} < 0.016 \cdot e^{\gamma} \cdot L'$, and Proposition~\ref{prop:RSphi} gives
\[
\frac{n'}{\varphi(n')} < e^{\gamma} \cdot L' + \frac{2.51}{L'} < 1.016 \cdot e^{\gamma} \cdot L' \leq e^{\gamma} \cdot \log\log n.
\]

\emph{Case II: $\log n' < 13000$.} Let $q_{1} < \cdots < q_{r}$ be the distinct primes dividing $n$. As in the proof of Theorem~\ref{thm:core}, $n' \geq p_{r}\#$, so $\theta(p_{r}) \leq \log n' < 13000 < \theta(14000)$, hence $p_{r} < 14000$, and by Lemma~\ref{lem:mertens}~(b),
\[
\frac{n'}{\varphi(n')} \leq \prod_{i=1}^{r} \frac{p_{i}}{p_{i}-1} \leq \prod_{p \leq 14000} \frac{p}{p-1} < 1.016 \cdot e^{\gamma} \cdot \log 14000 < 17.3.
\]
Thus $I(n) < 17.3$, and Robin's inequality holds for $n$ by Lemma~\ref{lem:small2}.
\end{proof}

\begin{Theorem}\label{thm:primes1016}
Let $n > 5040$. If every prime $q$ that divides $n$ satisfies $q < \sqrt[1.016]{\log n}$, then Robin's inequality holds for $n$.
\end{Theorem}

\begin{proof}
Let $P$ be the largest prime factor of $n$, so that $1.016 \cdot \log P < \log\log n$, and $I(n) < \prod_{p \leq P} \frac{p}{p-1}$ by Proposition~\ref{prop:Choie}. If $P \geq 285$, Lemma~\ref{lem:mertens}~(b) gives $I(n) < 1.016 \cdot e^{\gamma} \cdot \log P < e^{\gamma} \cdot \log\log n$. If $P < 285$, Lemma~\ref{lem:mertens}~(a) gives $I(n) < 10.23$, and Robin's inequality holds for $n$ by Lemma~\ref{lem:small}.
\end{proof}

\begin{Corollary}\label{cor:d1016}
Theorems~\ref{thm:core1016} and~\ref{thm:primes1016} imply Theorems~\ref{thm:core} and~\ref{thm:primes}.
\end{Corollary}

\begin{proof}
Since $d > 1.016$ by Lemma~\ref{lem:constants}~(v), the hypothesis $d \cdot \log\log n' \leq \log\log n$ implies $1.016 \cdot \log\log n' \leq \log\log n$ when $\log\log n' \geq 0$, and when $\log\log n' < 0$ we have $1.016 \cdot \log\log n' < 0 < \log\log n$ for $n > 5040$. Likewise, $q < \sqrt[d]{\log n}$ implies $q < \sqrt[1.016]{\log n}$, because $\log n > 1$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Superabundant Numbers}\label{sec:SA}

Let $n$ be a superabundant number with largest prime factor $p_{k}$. By Proposition~\ref{prop:HR}, $n = \prod_{i=1}^{k} p_{i}^{a_{i}}$ with $a_{1} \geq \cdots \geq a_{k} \geq 1$, so the primes dividing $n$ are exactly the primes $p \leq p_{k}$. Therefore
\[
\operatorname{core}(n) = p_{k}\# \qquad \text{and} \qquad \log \operatorname{core}(n) = \theta(p_{k}).
\]
Theorems~\ref{thm:core} and~\ref{thm:primes} thus take the following form.

\begin{Corollary}\label{cor:SA}
Let $n > 5040$ be a superabundant number, and let $p_{k}$ be its largest prime factor. Then Robin's inequality holds for $n$ in each of the following cases:
\begin{enumerate}
\item[(a)] $d \cdot \log \theta(p_{k}) \leq \log\log n$;
\item[(b)] $p_{k} < \sqrt[d]{\log n}$.
\end{enumerate}
\end{Corollary}

\begin{proof}
Part (a) is Theorem~\ref{thm:core} with $\log\log n' = \log \theta(p_{k})$. Part (b) is Theorem~\ref{thm:primes}, since every prime dividing $n$ is at most $p_{k}$.
\end{proof}

Applying Corollary~\ref{cor:SA} to a hypothetical least counterexample gives the following.

\begin{Corollary}\label{cor:least}
Suppose that the Riemann Hypothesis is false. Let $n > 5040$ be the least integer for which Robin's inequality does not hold, and let $p_{k}$ be its largest prime factor. Then
\[
\sqrt[d]{\log n} \leq p_{k} < \log n \qquad \text{and} \qquad \theta(p_{k}) > \sqrt[d]{\log n}.
\]
\end{Corollary}

\begin{proof}
By Propositions~\ref{prop:RH} and~\ref{prop:AF09}, such an $n$ exists and is superabundant. Since Robin's inequality fails for $n$, both conditions of Corollary~\ref{cor:SA} fail. The failure of (b) gives $p_{k} \geq \sqrt[d]{\log n}$, and the failure of (a) gives $d \cdot \log \theta(p_{k}) > \log\log n$, that is, $\theta(p_{k}) > \sqrt[d]{\log n}$. Finally, $p_{k} < \log n$ by Proposition~\ref{prop:largestprime}.
\end{proof}

\begin{Proposition}\label{prop:asymp}
Let $n$ run through the superabundant numbers, and let $p_{k}$ denote the largest prime factor of $n$. Then $p_{k} \sim \log n$~\cite{AE44}. Moreover, $\theta(x) \sim x$ as $x \to \infty$~\cite{Edw01}.
\end{Proposition}

\begin{Theorem}\label{thm:largeSA}
Every sufficiently large superabundant number $n$, with largest prime factor $p_{k}$, satisfies
\[
p_{k} > \sqrt[d]{\log n} \qquad \text{and} \qquad \theta(p_{k}) > \sqrt[d]{\log n}.
\]
\end{Theorem}

\begin{proof}
There are infinitely many superabundant numbers~\cite{AE44}, so $\log n \to \infty$ and, by Proposition~\ref{prop:asymp}, $p_{k} \to \infty$ and $\theta(p_{k}) \sim p_{k} \sim \log n$. Write $p_{k} = \log n \cdot (1 + \delta_{n})$ and $\theta(p_{k}) = \log n \cdot (1 + \varepsilon_{n})$ with $\delta_{n} \to 0$ and $\varepsilon_{n} \to 0$. Since $d > 1$,
\[
\frac{p_{k}^{d}}{\log n} = (1 + \delta_{n})^{d} \cdot (\log n)^{d - 1} \to \infty \qquad \text{and} \qquad \frac{\theta(p_{k})^{d}}{\log n} = (1 + \varepsilon_{n})^{d} \cdot (\log n)^{d - 1} \to \infty,
\]
so $p_{k}^{d} > \log n$ and $\theta(p_{k})^{d} > \log n$ for every sufficiently large superabundant number $n$.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The Largest Prime Factor of a Counterexample}\label{sec:largest}

In this section we say that $n$ is a \emph{counterexample} if $n > 5040$ and Robin's inequality does not hold for $n$, that is, $I(n) \geq e^{\gamma} \cdot \log\log n$. For an integer $n$ with largest prime factor $p_{k}$ we write
\[
L = \log\log n, \qquad P = \log p_{k}, \qquad x = L - P = \log\log n - \log p_{k}.
\]

\subsection{Every Counterexample Has a Large Prime Factor}

\begin{Proposition}\label{prop:Plarge}
Let $n$ be a counterexample with largest prime factor $p_{k}$. Then
\[
p_{k}^{1.016} > \log n \qquad \text{and} \qquad \log p_{k} > 30.54.
\]
In particular $p_{k} > 7{,}232{,}121{,}212 > 46{,}909{,}038$.
\end{Proposition}

\begin{proof}
By Proposition~\ref{prop:MP}, $n > 10^{10^{13.11485}}$, so $L > \log\log\left(10^{10^{13.11485}}\right) = 13.11485 \cdot \log 10 + \log\log 10 > 31.032$. If $p_{k} < 285$, then $I(n) < 10.23$ by Proposition~\ref{prop:Choie} and Lemma~\ref{lem:mertens}~(a), and Robin's inequality would hold for $n$ by Lemma~\ref{lem:small}, a contradiction. Hence $p_{k} \geq 285$, and by Proposition~\ref{prop:Choie} and Lemma~\ref{lem:mertens}~(b),
\[
e^{\gamma} \cdot L \leq I(n) < \prod_{p \leq p_{k}} \frac{p}{p-1} < 1.016 \cdot e^{\gamma} \cdot P.
\]
Thus $1.016 \cdot P > L$, which is the first claim, and $P > \frac{31.032}{1.016} > 30.54$. Finally $e^{30.54} > 7{,}232{,}121{,}212$.
\end{proof}

Proposition~\ref{prop:Plarge} applies to every counterexample, whereas Proposition~\ref{prop:Frank} gives the slightly stronger bound $\log p_{k} > 31.018$ for the least one.

For $P > 0$ we write
\[
\varepsilon(P) = \frac{1}{20 P^{3}} + \frac{3}{16 P^{4}},
\]
which is decreasing in $P$.

\begin{Proposition}\label{prop:Pboot}
Let $n$ be a counterexample with largest prime factor $p_{k}$. Then
\[
p_{k}^{1.000002} > \log n \qquad \text{and} \qquad \log p_{k} > 31.0319.
\]
\end{Proposition}

\begin{proof}
Put $P = \log p_{k}$ and $L = \log\log n$. By Proposition~\ref{prop:Plarge}, $P > 30.54$, so $p_{k} > 46{,}909{,}038$ and $(1 - \varepsilon(P))^{-1} \leq (1 - \varepsilon(30.54))^{-1} < 1.000002$. By Propositions~\ref{prop:Choie} and~\ref{prop:Axler}, and since Robin's inequality fails for $n$,
\[
e^{\gamma} \cdot L \leq I(n) < \prod_{p \leq p_{k}} \frac{p}{p-1} < e^{\gamma} \cdot P \cdot (1 - \varepsilon(P))^{-1} < 1.000002 \cdot e^{\gamma} \cdot P.
\]
Hence $L < 1.000002 \cdot P$, which is $p_{k}^{1.000002} > \log n$. Finally, $L > 31.032$ by Proposition~\ref{prop:MP}, so $P > \frac{31.032}{1.000002} > 31.0319$.
\end{proof}

\subsection{Infinitely Many Superabundant Counterexamples}

A natural number $n$ is called \emph{colossally abundant} if there exists $\varepsilon > 0$ such that $\frac{\sigma(n)}{n^{1+\varepsilon}} \geq \frac{\sigma(m)}{m^{1+\varepsilon}}$ for every natural number $m > 1$~\cite{AE44}.

\begin{Proposition}\label{prop:infSA}
If the Riemann Hypothesis is false, then Robin's inequality fails for infinitely many colossally abundant numbers~\cite{Rob84} [Section~4, Proposition~1]. Since every colossally abundant number is superabundant~\cite{AE44}, there are infinitely many superabundant counterexamples.
\end{Proposition}

\subsection{The Window for Superabundant Counterexamples}

\begin{Theorem}\label{Insight}
Let $n$ be a superabundant counterexample with largest prime factor $p_{k}$. Then
\[
\left(1 - \frac{1}{20 \log^{3}(p_{k})} - \frac{3}{16 \log^{4}(p_{k})}\right)^{-1} > \frac{\log\log n}{\log p_{k}}.
\]
\end{Theorem}

\begin{proof}
By Proposition~\ref{prop:Plarge}, $p_{k} > 46{,}909{,}038$. By Proposition~\ref{prop:HR}, the prime factors of $n$ are exactly the primes $p \leq p_{k}$. By Propositions~\ref{prop:Choie} and~\ref{prop:Axler}, and since Robin's inequality fails for $n$,
\[
e^{\gamma} \cdot (\log p_{k}) \cdot \left(1 - \frac{1}{20 \log^{3}(p_{k})} - \frac{3}{16 \log^{4}(p_{k})}\right)^{-1} > \prod_{p \leq p_{k}} \frac{p}{p-1} > I(n) \geq e^{\gamma} \cdot \log\log n.
\]
Dividing by $e^{\gamma} \cdot \log p_{k} > 0$ gives the claim.
\end{proof}

The next result holds for every superabundant number with a sufficiently large prime factor, whether or not Robin's inequality fails.

\begin{Theorem}\label{Lower}
Let $n$ be a superabundant number, and let $p_{k}$ be the largest prime factor of $n$. Assume $p_{k} \geq 7{,}232{,}121{,}212$. Then
\[
\log\log n > \log p_{k} - \frac{0.0101}{\log^{3}(p_{k})}.
\]
\end{Theorem}

\begin{proof}
By Proposition~\ref{prop:HR}, every prime $p \leq p_{k}$ divides $n$, so $n \geq p_{k}\#$ and $\log n \geq \theta(p_{k})$. By Proposition~\ref{prop:Chebyshev},
\[
\log\log n \geq \log \theta(p_{k}) \geq \log p_{k} + \log\left(1 - \frac{0.01}{\log^{3}(p_{k})}\right).
\]
Put $y = \frac{0.01}{\log^{3}(p_{k})}$. Since $\log p_{k} \geq \log 7{,}232{,}121{,}212 > 22.7$, we have $0 < y < 10^{-6}$. For such $y$,
\[
-\log(1 - y) \leq \frac{y}{1 - y} < 1.01 \cdot y,
\]
where the first inequality is $\log(1 + u) \geq \frac{u}{1 + u}$ applied with $u = -y$. Therefore
\[
\log\log n > \log p_{k} - 1.01 \cdot y = \log p_{k} - \frac{0.0101}{\log^{3}(p_{k})},
\]
as claimed.
\end{proof}

\begin{Theorem}\label{thm:window}
Let $n$ be a superabundant counterexample with largest prime factor $p_{k}$, and put $x = \log\log n - \log p_{k}$. Then
\[
-\frac{0.0101}{\log^{3}(p_{k})} < x < \frac{0.0562}{\log^{2}(p_{k})} < 0.0000603,
\]
and consequently
\[
e^{0.0562} > 1 + x \cdot \log^{2}(p_{k}).
\]
\end{Theorem}

\begin{proof}
By Proposition~\ref{prop:Plarge}, $P = \log p_{k} > 30.54$ and $p_{k} > 7{,}232{,}121{,}212$. Write $\varepsilon = \varepsilon(P)$. Multiplying the inequality of Theorem~\ref{Insight} by $P > 0$ gives $L < \frac{P}{1 - \varepsilon}$, that is,
\[
x < \frac{P \cdot \varepsilon}{1 - \varepsilon} = \frac{1}{P^{2}} \cdot \frac{\frac{1}{20} + \frac{3}{16 P}}{1 - \varepsilon(P)} \leq \frac{1}{P^{2}} \cdot \frac{\frac{1}{20} + \frac{3}{16 \cdot 30.54}}{1 - \varepsilon(30.54)} < \frac{0.0562}{P^{2}} < \frac{0.0562}{30.54^{2}} < 0.0000603.
\]
The lower bound is Theorem~\ref{Lower}. Finally, $x \cdot P^{2} < 0.0562$, and since $e^{y} > 1 + y$ for every $y > 0$, we have $1 + x \cdot P^{2} < 1 + 0.0562 < e^{0.0562}$.
\end{proof}

Since every counterexample satisfies $p_{k}^{1.000002} > \log n$, the window of Theorem~\ref{thm:window} can be expressed in terms of $\log\log n$ alone.

\begin{Corollary}\label{cor:windowL}
Let $n$ be a superabundant counterexample with largest prime factor $p_{k}$, and put $x = \log\log n - \log p_{k}$. Then
\[
-\frac{0.010101}{(\log\log n)^{3}} < x < \frac{0.056201}{(\log\log n)^{2}},
\]
and consequently
\[
e^{0.056201} > 1 + x \cdot (\log\log n)^{2} \qquad \text{and} \qquad (\log n)^{x} < e^{\frac{0.056201}{\log\log n}}.
\]
If $n$ is the least counterexample, then moreover
\[
1 < (\log n)^{\log\log n - \log p_{k}} < e^{\frac{0.056201}{\log\log n}}.
\]
\end{Corollary}

\begin{proof}
Put $L = \log\log n$. By Proposition~\ref{prop:Pboot}, $\log p_{k} > \frac{L}{1.000002} > 0$, hence $\frac{1}{\log^{2}(p_{k})} < \frac{1.000002^{2}}{L^{2}}$ and $\frac{1}{\log^{3}(p_{k})} < \frac{1.000002^{3}}{L^{3}}$. Since $0.0562 \cdot 1.000002^{2} < 0.056201$ and $0.0101 \cdot 1.000002^{3} < 0.010101$, Theorem~\ref{thm:window} gives
\[
-\frac{0.010101}{L^{3}} < -\frac{0.0101}{\log^{3}(p_{k})} < x < \frac{0.0562}{\log^{2}(p_{k})} < \frac{0.056201}{L^{2}}.
\]
Multiplying the upper bound by $L^{2} > 0$ gives $x \cdot L^{2} < 0.056201$, and since $e^{y} > 1 + y$ for every $y > 0$, we obtain $1 + x \cdot L^{2} < 1 + 0.056201 < e^{0.056201}$. Since $\log n > 1$, the function $t \mapsto (\log n)^{t}$ is increasing, so $(\log n)^{x} < (\log n)^{\frac{0.056201}{L^{2}}}$; and since $(\log n)^{\frac{1}{L}} = e$, we have $(\log n)^{\frac{0.056201}{L^{2}}} = e^{\frac{0.056201}{L}}$. If $n$ is the least counterexample, then $x > 0$ by Theorem~\ref{Main}, so $(\log n)^{x} > 1$.
\end{proof}

\subsection{The Window under the Error Term of de la Vallée Poussin}

The explicit results above use Proposition~\ref{prop:Axler}. Without explicit constants, the error term in the prime number theorem due to de la Vallée Poussin gives a window that shrinks faster than any power of $\frac{1}{\log p_{k}}$.

\begin{Proposition}\label{prop:dVP}
There is an absolute constant $c_{0} > 0$ such that~\cite{dVP99}
\[
\theta(x) = x + O\left(x \cdot \exp\left(-c_{0} \sqrt{\log x}\right)\right).
\]
\end{Proposition}

\begin{Lemma}\label{lem:MertensdVP}
There is an absolute constant $c_{1} > 0$ such that
\[
\prod_{p \leq x} \frac{p}{p-1} = e^{\gamma} \cdot \log x \cdot \left(1 + O\left(\exp\left(-c_{1} \sqrt{\log x}\right)\right)\right).
\]
\end{Lemma}

\begin{proof}
Write $\theta(t) = t + E(t)$, so that $|E(t)| \leq C \cdot t \cdot \exp(-c_{0} \sqrt{\log t})$ for $t \geq 2$ and some constant $C$, by Proposition~\ref{prop:dVP}. By partial summation,
\[
\sum_{p \leq x} \frac{1}{p} = \sum_{p \leq x} \frac{\log p}{p} \cdot \frac{1}{\log p} = \frac{\theta(x)}{x \log x} + \int_{2}^{x} \theta(t) \cdot \frac{1 + \log t}{t^{2} \log^{2} t} \, dt.
\]
Since $\frac{d}{dt}\left(\log\log t - \frac{1}{\log t}\right) = \frac{1 + \log t}{t \log^{2} t}$, the contribution of the main term $t$ of $\theta(t)$ equals $\frac{1}{\log x} + \log\log x - \frac{1}{\log x} - \log\log 2 + \frac{1}{\log 2}$. The integral of the error term converges absolutely on $[2, \infty)$, and its tail satisfies
\begin{align*}
\int_{x}^{\infty} |E(t)| \cdot \frac{1 + \log t}{t^{2} \log^{2} t} \, dt &\leq 2C \int_{x}^{\infty} \frac{\exp(-c_{0} \sqrt{\log t})}{t \log t} \, dt = 2C \int_{\log x}^{\infty} \frac{e^{-c_{0} \sqrt{u}}}{u} \, du \\
&\leq \frac{2C}{\log x} \cdot \frac{2}{c_{0}^{2}} \left(1 + c_{0} \sqrt{\log x}\right) e^{-c_{0} \sqrt{\log x}},
\end{align*}
where we used $1 + \log t \leq 2 \log t$ for $t \geq e$ and $\int_{v}^{\infty} e^{-c_{0} \sqrt{u}} \, du = \frac{2}{c_{0}^{2}} (1 + c_{0} \sqrt{v}) e^{-c_{0} \sqrt{v}}$. Together with $\frac{|E(x)|}{x \log x} \leq C \exp(-c_{0} \sqrt{\log x})$, this gives a constant $B$ with
\[
\sum_{p \leq x} \frac{1}{p} = \log\log x + B + O\left(\exp\left(-c_{1} \sqrt{\log x}\right)\right)
\]
for every $0 < c_{1} < c_{0}$. Next, put $h(p) = -\log\left(1 - \frac{1}{p}\right) - \frac{1}{p}$, so that $0 < h(p) < \frac{1}{p (p - 1)}$. Then $\sum_{p} h(p) = H$ converges and $\sum_{p > x} h(p) < \sum_{m > x} \frac{1}{m (m - 1)} \leq \frac{1}{x - 1}$. Hence
\[
\log \prod_{p \leq x} \frac{p}{p-1} = \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x} h(p) = \log\log x + B + H + O\left(\exp\left(-c_{1} \sqrt{\log x}\right)\right).
\]
By Mertens' theorem~\cite{Mer74}, $\prod_{p \leq x} \frac{p}{p-1} \sim e^{\gamma} \cdot \log x$, so $B + H = \gamma$. Exponentiating, and using $e^{y} = 1 + O(y)$ for bounded $y$, gives the claim.
\end{proof}

\begin{Corollary}\label{cor:dVP}
There are absolute constants $C_{2} > 0$ and $c_{2} > 0$ such that every superabundant counterexample $n$, with largest prime factor $p_{k}$, satisfies
\[
-\frac{0.0101}{\log^{3}(p_{k})} < \log\log n - \log p_{k} < C_{2} \cdot \exp\left(-c_{2} \sqrt{\log p_{k}}\right).
\]
In particular, if the Riemann Hypothesis is false, then infinitely many superabundant numbers $n$ satisfy $\log\log n - \log p_{k} < O\left(\exp\left(-c_{2} \sqrt{\log p_{k}}\right)\right)$.
\end{Corollary}

\begin{proof}
By Lemma~\ref{lem:MertensdVP}, there are constants $A > 0$ and $c_{1} > 0$ with $\prod_{p \leq t} \frac{p}{p-1} \leq e^{\gamma} \cdot \log t \cdot \left(1 + A \exp\left(-c_{1} \sqrt{\log t}\right)\right)$ for every $t \geq 2$; for $t$ in a bounded range this holds after enlarging $A$, because the ratio $\prod_{p \leq t} \frac{p}{p-1} \big/ (e^{\gamma} \log t)$ is bounded there. Put $P = \log p_{k}$ and $L = \log\log n$. By Proposition~\ref{prop:HR} and Proposition~\ref{prop:Choie}, and since Robin's inequality fails for $n$,
\[
e^{\gamma} \cdot L \leq I(n) < \prod_{p \leq p_{k}} \frac{p}{p-1} \leq e^{\gamma} \cdot P \cdot \left(1 + A \exp\left(-c_{1} \sqrt{P}\right)\right),
\]
so $L - P < A \cdot P \cdot \exp(-c_{1} \sqrt{P})$. Put $c_{2} = \frac{c_{1}}{2}$. The function $P \mapsto P \exp(-c_{2} \sqrt{P})$ is bounded on $(0, \infty)$, so $A \cdot P \cdot \exp(-c_{1} \sqrt{P}) \leq C_{2} \cdot \exp(-c_{2} \sqrt{P})$ for a suitable constant $C_{2}$. The lower bound is Theorem~\ref{Lower}, which applies because $p_{k} > 7{,}232{,}121{,}212$ by Proposition~\ref{prop:Plarge}. The last statement follows from Proposition~\ref{prop:infSA}.
\end{proof}

For the least counterexample the window is one-sided.

\begin{Theorem}\label{Main}
Suppose that the Riemann Hypothesis is false. Let $n > 5040$ be the least integer such that Robin's inequality fails, let $p_{k}$ be the largest prime factor of $n$, and put
\[
x = \log\log n - \log p_{k}.
\]
Then $x > 0$ and
\[
e^{0.0562} > 1 + x \cdot \log^{2}(p_{k}).
\]
\end{Theorem}

\begin{proof}
By Proposition~\ref{prop:AF09}, $n$ is a superabundant counterexample. By Proposition~\ref{prop:largestprime}, $p_{k} < \log n$, hence $L > P$, that is, $x > 0$. The second inequality is part of Theorem~\ref{thm:window}.
\end{proof}

\begin{Remark}
An earlier version of Theorem~\ref{Main}, based on the estimate $\prod_{p \leq x} \frac{p}{p-1} \leq e^{\gamma} \cdot \log x \cdot \left(1 + \frac{0.2}{\log^{3}(x)}\right)$ of~\cite{DU21} [Lemma~2.7~(5)~pp.~19], gave $e^{0.2} > 1 + x \cdot \log^{2}(p_{k})$. Proposition~\ref{prop:Axler} replaces $0.2$ by $0.0562$.
\end{Remark}

The proof of~\cite{Cho+07} [Lemma~6.1, p.~369] also yields the following inequality.

\begin{Theorem}\label{thm:lowerx}
Let $n$ be a superabundant counterexample with largest prime factor $p_{k}$, put $r = \frac{n}{p_{k}}$, and suppose that Robin's inequality holds for $r$; by the proof of~\cite{Cho+07} [Lemma~6.1, p.~369], this is the case when $n$ is the least counterexample. Then
\[
\log n \cdot \log\log r > p_{k} \cdot \log p_{k}.
\]
\end{Theorem}

\begin{proof}
Since $\sigma$ is submultiplicative, $I(n) \leq I(p_{k}) \cdot I(r) = \left(1 + \frac{1}{p_{k}}\right) \cdot I(r)$. Since Robin's inequality fails for $n$ and holds for $r$,
\[
e^{\gamma} \cdot \log\log n \leq I(n) \leq \left(1 + \frac{1}{p_{k}}\right) \cdot I(r) < \left(1 + \frac{1}{p_{k}}\right) \cdot e^{\gamma} \cdot \log\log r,
\]
so $\log\log n < \left(1 + \frac{1}{p_{k}}\right) \cdot \log\log r$, that is,
\[
\log\log n - \log\log r < \frac{\log\log r}{p_{k}}.
\]
On the other hand, $\log r = \log n - \log p_{k}$, so
\[
\log\log n - \log\log r = -\log\left(1 - \frac{\log p_{k}}{\log n}\right) > \frac{\log p_{k}}{\log n},
\]
because $0 < \frac{\log p_{k}}{\log n} < 1$ and $-\log(1 - u) > u$ for $0 < u < 1$. Combining the two inequalities gives $\frac{\log\log r}{p_{k}} > \frac{\log p_{k}}{\log n}$, that is, $\log n \cdot \log\log r > p_{k} \cdot \log p_{k}$.
\end{proof}

\begin{Corollary}\label{cor:Sandwich}
Suppose that the Riemann Hypothesis is false. Then there are infinitely many superabundant counterexamples $n$, and each of them satisfies
\[
\log p_{k} - \frac{0.0101}{\log^{3}(p_{k})} < \log\log n < \log p_{k} + \frac{0.0562}{\log^{2}(p_{k})}
\]
and
\[
e^{0.0562} > 1 + (\log\log n - \log p_{k}) \cdot \log^{2}(p_{k}),
\]
where $p_{k}$ is the largest prime factor of $n$. For the least counterexample, moreover, $\log p_{k} < \log\log n$.
\end{Corollary}

\begin{proof}
Combine Proposition~\ref{prop:infSA}, Theorem~\ref{thm:window} and Theorem~\ref{Main}.
\end{proof}

\begin{Proposition}\label{prop:RHcond}
Suppose that there are constants $c > 0$, $0 < \alpha < 2$ and $N \leq 10^{10^{13.11485}}$ such that every superabundant number $n > N$ with $\log n > p_{k}$, where $p_{k}$ is the largest prime factor of $n$, satisfies
\[
\log\log n - \log p_{k} \geq \frac{c}{\log^{\alpha}(p_{k})}.
\]
If the least counterexample exists, then its largest prime factor $p_{k}$ satisfies $\log p_{k} < \left(\frac{0.0562}{c}\right)^{\frac{1}{2 - \alpha}}$. In particular, if $c \geq \frac{0.0562}{31.03^{2 - \alpha}}$, then the Riemann Hypothesis is true.
\end{Proposition}

\begin{proof}
Suppose that the Riemann Hypothesis is false, and let $n$ be the least counterexample, with largest prime factor $p_{k}$ and $P = \log p_{k}$. By Proposition~\ref{prop:MP}, $n > 10^{10^{13.11485}} \geq N$; by Theorem~\ref{Main}, $\log n > p_{k}$; by Proposition~\ref{prop:AF09}, $n$ is superabundant. Hence the hypothesis applies to $n$, and Theorem~\ref{thm:window} gives
\[
\frac{c}{P^{\alpha}} \leq \log\log n - \log p_{k} < \frac{0.0562}{P^{2}},
\]
that is, $P^{2 - \alpha} < \frac{0.0562}{c}$, which is the first claim. By Proposition~\ref{prop:Pboot}, $P > 31.0319 > 31.03$, so $c < \frac{0.0562}{P^{2 - \alpha}} < \frac{0.0562}{31.03^{2 - \alpha}}$. Therefore, if $c \geq \frac{0.0562}{31.03^{2 - \alpha}}$, the least counterexample cannot exist, and the Riemann Hypothesis is true by Proposition~\ref{prop:RH}.
\end{proof}

The precondition $\log n > p_{k}$ in Proposition~\ref{prop:RHcond} excludes the superabundant numbers with $p_{k} > \log n$, for which $\log\log n - \log p_{k} < 0$; examples are $n = 27720$, $n = 55440$ and $n = 10685862914126400$ (Proposition~\ref{prop:xcount}).

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\section{Computational Verification}\label{sec:computation}

Every numerical claim of this paper was checked by a self-contained Python~3.12 script, \texttt{verify\_robin\_d.py}, written with the help of Claude Opus~5.5 (Anthropic) and run by the author, which uses the arbitrary-precision library \texttt{mpmath} with $60$ significant decimal digits and exact integer arithmetic wherever a comparison of rational numbers is involved. The script has five parts: it checks the constants of Lemma~\ref{lem:constants}, it compares the bounds of Lemma~\ref{lem:mertens} with the true values of the Mertens product, it checks the constants of the theorems with the exponent $1.016$ and of Section~\ref{sec:largest}, it enumerates all superabundant numbers up to $10^{40}$ and tests on each of them the conditions of Corollary~\ref{cor:SA} and the inequality of Theorem~\ref{Main}, and it tests on the same numbers the hypothesis of Proposition~\ref{prop:RHcond}. All $33$ checks performed by the script passed. We now describe each part and its output in detail.

\subsection{The Constants of Lemma~\ref{lem:constants}}

The script computes $\gamma$, $e^{\gamma}$, $\zeta(2) = \frac{\pi^{2}}{6}$ (and confirms that this agrees with $\zeta(2)$ evaluated directly to $50$ digits) and $d = \frac{e^{\gamma}}{\zeta(2)}$, and then verifies each item of Lemma~\ref{lem:constants}. The computed values are collected in Table~\ref{tab:constants}. Every inequality of Lemma~\ref{lem:constants} holds with a visible margin. The two tightest margins are those of item~(iii), where $e^{\gamma} \cdot (d - 1) \cdot \log^{2}(64) = 2.5495\ldots$ exceeds the Rosser--Schoenfeld constant $2.51$ by about $0.04$, and of item~(vii), where $e^{\gamma} \cdot \log 313 = 10.2344\ldots$ exceeds $10.23$ by about $0.004$. The script also confirms that $\theta(71) = 61.586\ldots < 64$, so that $71$ is indeed the largest prime that can divide $n'$ in Case~II of the proof of Theorem~\ref{thm:core}.

\begin{table}[htbp]
\caption{Values computed for Lemma~\ref{lem:constants}. All digits shown are correct.\label{tab:constants}}
\centering
\begin{tabular}{lll}
\toprule
\textbf{Item} & \textbf{Quantity} & \textbf{Computed value} \\
\midrule
 & $\gamma$ & $0.577215664901533\ldots$ \\
 & $e^{\gamma}$ & $1.7810724179902\ldots$ \\
 & $\zeta(2)$ & $1.64493406684823\ldots$ \\
(i) & $d = e^{\gamma}/\zeta(2)$ & $1.08276219326092\ldots$ \\
(ii) & $e^{\gamma} \cdot (d-1)$ & $0.147405459669\ldots$ \\
(iii) & $\log^{2}(64)$ & $17.2963085011\ldots$ \\
(iii) & $e^{\gamma} \cdot (d-1) \cdot \log^{2}(64)$ & $2.54957030518\ldots$ \\
(iv) & $\theta(73)$ & $65.8767545906\ldots$ \\
(iv) & $\theta(71)$ & $61.5862951494\ldots$ \\
(v) & $\left(1 - \frac{1}{2\log^{2}(285)}\right)^{-1}$ & $1.01589793074\ldots$ \\
(vi) & $e^{\gamma} \log 285 \left(1 - \frac{1}{2\log^{2}(285)}\right)^{-1}$ & $10.2275448716\ldots$ \\
(vii) & $e^{\gamma} \cdot \log 313$ & $10.2344040108\ldots$ \\
(vii) & $313 / \log 10$ & $135.9341728\ldots$ \\
\bottomrule
\end{tabular}
\end{table}

\subsection{The Mertens Product}

The script computes the exact partial products $\prod_{p \leq x} \frac{p}{p-1}$ for every prime $x \leq 2000$ and compares them with the bounds of Lemma~\ref{lem:mertens}. The results are as follows.
\begin{enumerate}
\item[(1)] The true value is $\prod_{p \leq 285} \frac{p}{p-1} = 10.2257527701\ldots$, which lies below the analytic bound $10.2275\ldots$ obtained from Proposition~\ref{prop:RSmertens}, and below $10.23$, as Lemma~\ref{lem:mertens}~(a) asserts. The analytic bound is therefore sharp to within $0.002$ at $x = 285$.
\item[(2)] The true value is $\prod_{p \leq 71} \frac{p}{p-1} = 7.82486815925\ldots$, the quantity actually used in Case~II of the proof of Theorem~\ref{thm:core}. The bound $10.23$ used there is thus generous.
\item[(3)] For every prime $x$ with $285 \leq x \leq 2000$ the inequality $\prod_{p \leq x} \frac{p}{p-1} < 1.016 \cdot e^{\gamma} \cdot \log x$ of Lemma~\ref{lem:mertens}~(b) holds. The largest value of the ratio $\prod_{p \leq x} \frac{p}{p-1} \big/ \left(e^{\gamma} \log x\right)$ on this range is $1.014231153\ldots$, attained at $x = 293$, compared with the analytic constant $1.01589\ldots$ of Lemma~\ref{lem:constants}~(v).
\item[(4)] Among the primes $x \leq 2000$, the inequality $\prod_{p \leq x} \frac{p}{p-1} \geq d \cdot e^{\gamma} \cdot \log x$ holds only for $x \leq 23$. Thus the direct argument of Case~I in the proof of Theorem~\ref{thm:primes} already works numerically for every prime $P$ with $29 \leq P \leq 2000$; the threshold $285$ was chosen only because it is the point from which Proposition~\ref{prop:RSmertens} applies.
\end{enumerate}

\subsection{Enumeration of Superabundant Numbers}

To enumerate the superabundant numbers rigorously we use the following reduction to Hardy--Ramanujan integers.

\begin{Lemma}\label{lem:HRmax}
For every natural number $m$ there is a Hardy--Ramanujan integer $m^{*} \leq m$ with $I(m^{*}) \geq I(m)$.
\end{Lemma}

\begin{proof}
For $m = 1$ take $m^{*} = 1$. Otherwise let $q_{1} < \cdots < q_{r}$ be the primes dividing $m$ and $m = \prod_{i=1}^{r} q_{i}^{b_{i}}$. For a prime $p$ and an integer $b \geq 0$ put $g(p, b) = \frac{\sigma(p^{b})}{p^{b}} = \sum_{j=0}^{b} p^{-j}$, so that $I$ is the product of the factors $g(q_{i}, b_{i})$. We proceed in two steps.

\emph{Step 1.} Put $m_{1} = \prod_{i=1}^{r} p_{i}^{b_{i}}$. Since $q_{i} \geq p_{i}$ for every $i$, we have $m_{1} \leq m$, and since $g(p, b)$ is nonincreasing in $p$ for fixed $b$, we have $I(m_{1}) \geq I(m)$.

\emph{Step 2.} Suppose that $p < q$ are primes carrying exponents $a < b$ respectively, and exchange the two exponents. The integer does not increase, because $p^{b} q^{a} \leq p^{a} q^{b}$ is equivalent to $p^{b-a} \leq q^{b-a}$. The abundancy index does not decrease, because $g(p, b) \cdot g(q, a) \geq g(p, a) \cdot g(q, b)$ is equivalent to $\frac{g(p, b)}{g(p, a)} \geq \frac{g(q, b)}{g(q, a)}$, and, with $t = \frac{1}{p}$,
\[
\frac{g(p, b)}{g(p, a)} = 1 + \frac{t^{a+1} + \cdots + t^{b}}{1 + t + \cdots + t^{a}} = 1 + \frac{1 + t + \cdots + t^{b-a-1}}{t^{-a-1} + t^{-a} + \cdots + t^{-1}},
\]
which is increasing in $t$, that is, decreasing in $p$. Applying finitely many such exchanges to $m_{1}$ sorts its exponents into nonincreasing order and produces a Hardy--Ramanujan integer $m^{*} \leq m_{1} \leq m$ with $I(m^{*}) \geq I(m_{1}) \geq I(m)$.
\end{proof}

\begin{Proposition}\label{prop:records}
Let $h_{1} = 1 < h_{2} < h_{3} < \cdots$ be the Hardy--Ramanujan integers in increasing order. Then $n$ is superabundant if and only if $n = h_{j}$ for some $j$ with $I(h_{j}) > I(h_{i})$ for all $i < j$.
\end{Proposition}

\begin{proof}
If $n$ is superabundant, then $n$ is a Hardy--Ramanujan integer by Proposition~\ref{prop:HR}, and $I(n) > I(h_{i})$ for every $h_{i} < n$ by definition. Conversely, let $n = h_{j}$ be such a record, and let $m < n$. By Lemma~\ref{lem:HRmax} there is a Hardy--Ramanujan integer $m^{*} \leq m < n$ with $I(m) \leq I(m^{*})$, and $I(m^{*}) < I(n)$ because $n$ is a record. Hence $I(m) < I(n)$, and $n$ is superabundant.
\end{proof}

The script generates every Hardy--Ramanujan integer $h \leq 10^{40}$ together with $\sigma(h)$, by recursion over the primes with nonincreasing exponents. There are exactly $5{,}754{,}126$ of them. It then sorts them and extracts the records of $I$ as in Proposition~\ref{prop:records}. A comparison $I(h) > I(h')$ is decided exactly through the integer inequality $\sigma(h) \cdot h' > \sigma(h') \cdot h$, so no rounding enters this step. This yields exactly $263$ superabundant numbers not exceeding $10^{40}$. For every superabundant $n$ with $5040 < n \leq 10^{40}$ the script evaluates, with $60$-digit precision, Robin's inequality and the two conditions of Corollary~\ref{cor:SA}. The outcome is summarized in the next proposition.

\begin{Proposition}[Computer-assisted]\label{prop:SAcount}
Among the $263$ superabundant numbers $n \leq 10^{40}$:
\begin{enumerate}
\item[(1)] Robin's inequality holds for every superabundant $n$ with $5040 < n \leq 10^{40}$;
\item[(2)] condition (a) of Corollary~\ref{cor:SA}, namely $d \cdot \log \theta(p_{k}) \leq \log\log n$, holds for exactly $86$ superabundant numbers $n$ with $5040 < n \leq 10^{40}$, and the largest of them is \[ n = 92005279690628304000, \] whose largest prime factor is $p_{k} = 41$;
\item[(3)] condition (b) of Corollary~\ref{cor:SA}, namely $p_{k} < \sqrt[d]{\log n}$, holds for exactly $7$ superabundant numbers $n$ with $5040 < n \leq 10^{40}$, and the largest of them is $n = 48886437600$, whose largest prime factor is $p_{k} = 19$.
\end{enumerate}
\end{Proposition}

Part (1) is already contained in Proposition~\ref{prop:MP}; it serves as an independent consistency check of the enumeration. Condition (a) is equivalent to $\theta(p_{k}) \leq \sqrt[d]{\log n}$, whereas condition (b) is $p_{k} < \sqrt[d]{\log n}$.

\subsection{Constants of Section~\ref{sec:largest} and the Quantity \texorpdfstring{$x \cdot \log^{2}(p_{k})$}{x log2 pk}}

For the theorems with the exponent $1.016$, the script verifies that $\theta(14000) = 13867.29\ldots > 13000$, that $e^{\gamma} \cdot 0.016 \cdot \log^{2}(13000) = 2.5571\ldots > 2.51$, that $1.016 \cdot e^{\gamma} \cdot \log 14000 = 17.2756\ldots < 17.3$, that the true value $\prod_{p \leq 14000} \frac{p}{p-1} = 17.0192\ldots$ is below $17.3$, that $e^{\gamma} \cdot \log 16600 = 17.3069\ldots > 17.3$, and that $e^{16600} < 10^{7210}$. The script also verifies the numerical facts used in Section~\ref{sec:largest}: that $\log\log\left(10^{10^{13.11485}}\right) = 31.03209\ldots > 31.032$, that $\frac{31.032}{1.016} > 30.54$, that $e^{30.54} > 7{,}232{,}121{,}212 > 46{,}909{,}038$, that $(1 - \varepsilon(30.54))^{-1} = 1.0000019708\ldots < 1.000002$, that $\frac{\frac{1}{20} + \frac{3}{16 \cdot 30.54}}{1 - \varepsilon(30.54)} = 0.0561395\ldots < 0.0562$, that $\frac{0.0562}{30.54^{2}} = 0.0000602557\ldots < 0.0000603$, that $\frac{31.032}{1.000002} > 31.0319$, that $0.0562 \cdot 1.000002^{2} < 0.056201$, that $0.0101 \cdot 1.000002^{3} < 0.010101$, and, for Theorem~\ref{Lower}, that $\log 7{,}232{,}121{,}212 = 22.7017\ldots > 22.7$ and $\frac{0.01}{22.7^{3}} = 8.549\ldots \cdot 10^{-7} < 10^{-6}$. For every superabundant number $5040 < n \leq 10^{40}$ it computes $x = \log\log n - \log p_{k}$ and $x \cdot \log^{2}(p_{k})$, counts the superabundant numbers with $x < 0$, that is, with $p_{k} > \log n$, and counts those for which the inequality $e^{0.0562} > 1 + x \cdot \log^{2}(p_{k})$ of Theorem~\ref{Main} fails or holds.
The output of this part of the script is summarized in the next proposition and in Table~\ref{tab:xlog}.

\begin{Proposition}[Computer-assisted]\label{prop:xcount}
Among the $244$ superabundant numbers $n$ with $5040 < n \leq 10^{40}$, and with $p_{k}$ the largest prime factor of $n$ and $x = \log\log n - \log p_{k}$:
\begin{enumerate}
\item[(1)] exactly $3$ satisfy $x < 0$, that is, $p_{k} > \log n$;
\item[(2)] exactly $3$ satisfy $e^{0.0562} > 1 + x \cdot \log^{2}(p_{k})$, namely the $3$ numbers of part~(1), which are $n = 27720$ and $n = 55440$ with $p_{k} = 11$ and $n = 10685862914126400$ with $p_{k} = 37$, with $x \cdot \log^{2}(p_{k}) = -0.4173\ldots$, $-0.0403\ldots$ and $-0.0325\ldots$, respectively;
\item[(3)] the remaining $241$ satisfy $e^{0.0562} \leq 1 + x \cdot \log^{2}(p_{k})$, and the largest of them is \[ n = 5330180429932843151696086400707299936000, \] with $p_{k} = 83$, $x = 0.097216398\ldots$ and $x \cdot \log^{2}(p_{k}) = 1.8982622\ldots$;
\item[(4)] the maximum of $x \cdot \log^{2}(p_{k})$ is $2.855801894\ldots$, attained at \[ n = 10567690344536672406900888090467424000, \] with $p_{k} = 73$, and the minimum is $-0.4173252461\ldots$, attained at $n = 27720$ with $p_{k} = 11$.
\end{enumerate}
\end{Proposition}

\begin{table}[htbp]
\caption{The quantity $x \cdot \log^{2}(p_{k})$, with $x = \log\log n - \log p_{k}$, for the ten largest superabundant numbers $n \leq 10^{40}$. The window of Theorem~\ref{thm:window} requires $x \cdot \log^{2}(p_{k}) < 0.0562$.\label{tab:xlog}}
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{rrr}
\toprule
$n$ & $p_{k}$ & $x \cdot \log^{2}(p_{k})$ \\
\midrule
$278282512406132373381723386382308832000$ & $79$ & $2.172675585$ \\
$355345361995522876779739093380486662400$ & $83$ & $1.311470417$ \\
$417423768609198560072585079573463248000$ & $79$ & $2.259925324$ \\
$484561857266622104699644218246118176000$ & $83$ & $1.379577194$ \\
$834847537218397120145170159146926496000$ & $79$ & $2.408162468$ \\
$888363404988807191949347733451216656000$ & $83$ & $1.511996393$ \\
$1776726809977614383898695466902433312000$ & $83$ & $1.662331969$ \\
$2665090214966421575848043200353649968000$ & $83$ & $1.749739048$ \\
$3553453619955228767797390933804866624000$ & $83$ & $1.811518921$ \\
$5330180429932843151696086400707299936000$ & $83$ & $1.898262198$ \\
\bottomrule
\end{tabular}%
}
\end{table}

By Proposition~\ref{prop:xcount}, for the ten largest superabundant numbers below $10^{40}$, $\log\log n$ exceeds $\log p_{k}$ by between $\frac{1.31}{\log^{2}(p_{k})}$ and $\frac{2.41}{\log^{2}(p_{k})}$, and the maximum over the whole range is $\frac{2.86}{\log^{2}(p_{k})}$. Table~\ref{tab:xlog} also shows that the largest prime factor of consecutive superabundant numbers need not be monotone: it alternates between $79$ and $83$ before settling at $83$. The value of $x \cdot \log^{2}(p_{k})$ is smaller whenever $p_{k} = 83$, and along the last five entries, where $p_{k} = 83$ is fixed, it increases from $1.51$ to $1.90$ as the exponents of the smaller primes grow.

\subsection{The Hypothesis of Proposition~\ref{prop:RHcond}}

For every superabundant number $5040 < n \leq 10^{40}$, the script computes $x = \log\log n - \log p_{k}$ and the quantities $x \cdot \log^{\alpha}(p_{k})$ for $\alpha = 1, 2, 3, 4, 5$. The output is summarized in the next proposition and in Table~\ref{tab:alpha}.

\begin{Proposition}[Computer-assisted]\label{prop:alphacount}
Among the $244$ superabundant numbers $5040 < n \leq 10^{40}$, with $x = \log\log n - \log p_{k}$:
\begin{enumerate}
\item[(1)] exactly $3$ satisfy $x \leq 0$, namely $n = 27720$ with $x = -0.0725795\ldots$, $n = 55440$ with $x = -0.0070194\ldots$, and $n = 10685862914126400$ with $x = -0.0024977\ldots$; these $3$ numbers have $p_{k} > \log n$ and are therefore excluded by the precondition $\log n > p_{k}$ of Proposition~\ref{prop:RHcond};
\item[(2)] there are exactly $168$ superabundant numbers $n$ with $10685862914126400 < n \leq 10^{40}$, and all of them satisfy $x > 0$;
\item[(3)] for every $\alpha \in \{1, 2, 3, 4, 5\}$, the minimum of $x \cdot \log^{\alpha}(p_{k})$ over these $168$ numbers is attained at
\[
n = 58144272988663390205476800,
\]
with $p_{k} = 59$, and its value is given in Table~\ref{tab:alpha}.
\end{enumerate}
\end{Proposition}

\begin{table}[htbp]
\caption{Minimum of $x \cdot \log^{\alpha}(p_{k})$, with $x = \log\log n - \log p_{k}$, over the $168$ superabundant numbers $10685862914126400 < n \leq 10^{40}$, and over the $50$ largest superabundant numbers $n \leq 10^{40}$, where it is attained at $p_{k} = 79$.\label{tab:alpha}}
\centering
\begin{tabular}{rrr}
\toprule
$\alpha$ & \textbf{minimum over the $168$ numbers} & \textbf{minimum over the last $50$} \\
\midrule
$1$ & $0.02239728649$ & $0.2087437012$ \\
$2$ & $0.0913257743$ & $0.9120947171$ \\
$3$ & $0.3723842643$ & $3.985350303$ \\
$4$ & $1.518410781$ & $17.41378032$ \\
$5$ & $6.191376815$ & $76.08860503$ \\
\bottomrule
\end{tabular}
\end{table}

For the ten largest superabundant numbers below $10^{40}$, whose largest prime factors are $79$ and $83$, the values of $x$ lie between $0.0671\ldots$ and $0.1261\ldots$, and those of $x \cdot \log^{3}(p_{k})$ lie between $5.795\ldots$ and $10.522\ldots$.

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\section*{Acknowledgements}

The author thanks Richard J. Lipton for helpful comments on an earlier version of this work.

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\section*{Statements and Declarations}

\subsection*{Declaration of interests}

The author declares no competing interests.

\subsection*{Funding sources}

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

\subsection*{Declaration of generative AI in scientific writing}

The author used Claude Opus 5.5 to write the verification script \texttt{verify\_robin\_d.py} of Section~\ref{sec:computation} and to generate parts of the text of this manuscript. The experiments were run by the author. After using this tool, the author reviewed and edited the code and the content as needed and takes full responsibility for the content of the publication.

\subsection*{Data availability}

The Python script \texttt{verify\_robin\_d.py} used in Section~\ref{sec:computation} is available in the preprint..

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