Download PDFOpen PDF in browserCurrent versionRobin's Criterion on Superabundant NumbersEasyChair Preprint 9250, version 37 pages•Date: August 15, 2023AbstractRobin's criterion states that the Riemann hypothesis is true if and only if the inequality $\sigma(n) < e^{\gamma} \cdot n \cdot \log \log n$ holds for all natural numbers $n > 5040$, where $\sigma(n)$ is the sumofdivisors function of $n$, $\gamma \approx 0.57721$ is the EulerMascheroni constant and $\log$ is the natural logarithm. We require the properties of superabundant numbers, that is to say left to right maxima of $n \mapsto \frac{\sigma(n)}{n}$. Let $P_{n}$ be equal to $\prod_{q \mid \frac{N_{r}}{6}} \frac{q^{\nu_{q}(n) + 2}  1}{q^{\nu_{q}(n) + 2}  q}$ for a superabundant number $n > 5040$, where $\nu_{p}(n)$ is the $\textit{padic}$ order of $n$, $q_{k}$ is the largest prime factor of $n$ and $N_{r} = \prod_{i = 1}^{r} q_{i}$ is the largest primorial number of order $r$ such that $\frac{N_{r}}{6} < q_{k}^{2}$. In this note, we prove that the Riemann hypothesis is true when $P_{n} \geq Q$ holds for all large enough superabundant numbers $n$, where $Q = \frac{1.2 \cdot (2  \frac{1}{8}) \cdot (3  \frac{1}{3})}{(2  \frac{1}{2^{19}}) \cdot (3  \frac{1}{3^{12}})} \approx 1.0000015809$. We know that $\prod_{q \mid \frac{N_{r}}{6}} (q^{\nu_{q}(n) + 2}  1) \geq Q \cdot \prod_{q \mid \frac{N_{r}}{6}} (q^{\nu_{q}(n) + 2}  q)$ trivially holds for large enough superabundant numbers $n$ and thus, the Riemann hypothesis is true. Keyphrases: Riemann hypothesis, Robin's inequality, Superabundant numbers, prime numbers, sumofdivisors function
