Robin’s Inequality and Superabundant Counterexamples

Robin's criterion states that the Riemann Hypothesis holds if and only if \(σ(n) < eγ · n · loglog n\) for every \(n > 5040\). Let \(n'\) be the squarefree kernel of \(n\), \(pₖ\) its largest prime factor, and \(d = eγ/ζ(2) ≈ 1.08276\). We prove that Robin's inequality holds for every \(n > 5040\) with \(d · loglog n' ≤ loglog n\), 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 \(loglog n - log pₖ < C · exp(-c √{log pₖ})\) for absolute constants \(C, c > 0\). Since the least counterexample satisfies \(log n > pₖ\), we deduce that the Riemann Hypothesis is true if there are constants \(c > 0\) and \(0 < α < 2\) with \(c ≥ 0.0562/31.032-α\), and a threshold \(N ≤ 10^{1013.11485}\), such that \(loglog n - log pₖ ≥ c/logα(pₖ)\) for every superabundant number \(n > N\) with \(log n > pₖ\). We also give explicit versions of these bounds: every counterexample satisfies \(pₖ1.000002 > log n\), and every superabundant counterexample satisfies \(-0.010101/(loglog n)³ < loglog n - log pₖ < 0.056201/(loglog n)²\). An enumeration of the superabundant numbers up to \(10⁴⁰\) shows that only \(3\) of the \(244\) exceeding \(5040\) satisfy \(e0.0562 > 1 + (loglog n - log pₖ) · log²(pₖ)\), namely those with \(pₖ > log n\), 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 \(10⁴⁰\) satisfy \(loglog n - log pₖ > 0\), and that for each \(α ∈ \{1, 2, 3, 4, 5\}\) the minimum of \((loglog n - log pₖ) · logα(pₖ)\) over them is positive, ranging from \(0.0223\) for \(α = 1\) to \(6.19\) for \(α = 5\), and is attained at the same superabundant number, whose largest prime factor is \(59\).