On the Least Colossally Abundant Exception to Robin's Inequality
Bruce Zimov
TL;DR
The paper examines the Robin–RH connection by analyzing consecutive Colossally Abundant numbers to constrain potential counterexamples. It leverages the CA structure, showing that the quotient $n/m$ is a prime or semiprime and that the largest prime factor satisfies $p=\log n(1+o(1))$, which together bound the growth of the abundancy ratio $\frac{G(n)}{G(m)}$ by $1+O\left(\frac{1}{\log n}\right)$. Under the hypothesis that a least CA counterexample exists and RH is false, these bounds imply an upper bound on $G(n)$ that conflicts with the strong lower bound $G(n)>e^{\gamma}\left(1+\frac{c}{(\log n)^b}\right)$ for $0<b<\frac{1}{2}$. Consequently, the least CA counterexample must lie in the band $e^{\gamma}<G(n)<e^{\gamma}\left(1+\frac{c}{(\log n)^b}\right)$ with $0<b<\frac{1}{2}$, narrowing the search and reinforcing the conditional structure around Robin's inequality.
Abstract
Robin's Inequality posits $G(n)<e^γ$ for $n>5040$. Robin also showed that if the Riemann Hypothesis (RH) is false, then $G(n)>e^γ\left(1+\displaystyle\frac{c}{(\log n)^{b}}\right)$ for infinitely many values of $n$. By analyzing the prime or semiprime quotient $\displaystyle\frac{n}{m}$ for consecutive Colossally Abundant (CA) numbers $m$ followed by $n$ (where $m$ satisfies Robin's Inequality and $n$ violates it), we demonstrate that if the Riemann Hypothesis is false, then the least CA counterexample, $n$, must be constrained to the band $e^γ<G(n)<e^γ\left(1+\displaystyle\frac{c}{(\log n)^b}\right)$ where $0 < b < 1/2$, i.e. excluded from the infinite set beyond the higher threshold.