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Robin's Inequality at Colossally Abundant Numbers and Nicolas' Inequality at Their Largest Prime Factor

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Abstract · v1

Robin's criterion states that the Riemann Hypothesis holds if and only if σ(n)<eγ⋅n⋅log⁡log⁡n\sigma(n) < e^{\gamma} \cdot n \cdot \log\log n for every n>5040n > 5040, and Nicolas' criterion states that it holds if and only if c(q)=log⁡(q#/φ(q#))−γ−log⁡log⁡θ(q)>0c(q) = \log(q\#/\varphi(q\#)) - \gamma - \log\log\theta(q) > 0 for every prime qq, where q#q\# is the primorial of qq and θ\theta is the Chebyshev function. Let NN be a colossally abundant number, let ξ(N)\xi(N) be the real number associated with NN by Nicolas, and let qq be the largest prime not exceeding ξ(N)\xi(N). Writing Δ(N)=log⁡(σ(N)/N)−γ−log⁡log⁡log⁡N\Delta(N) = \log(\sigma(N)/N) - \gamma - \log\log\log N, we prove without any hypothesis that, whenever log⁡q≥31\log q \geq 31, 1.353/(qlog⁡2q)≤Δ(N)−c(q)+22/(qlog⁡q)≤3.970/(qlog⁡2q)1.353/(\sqrt{q} \log^{2} q) \leq \Delta(N) - c(q) + 2\sqrt{2}/(\sqrt{q} \log q) \leq 3.970/(\sqrt{q} \log^{2} q). Thus Robin's inequality at NN and Nicolas' inequality at qq differ by 22/(qlog⁡q)2\sqrt{2}/(\sqrt{q} \log q) up to an explicit error of smaller order. As consequences, every colossally abundant counterexample to Robin's inequality yields a prime qq with c(q)>2.70/(qlog⁡q)c(q) > 2.70/(\sqrt{q} \log q); conversely, every prime qq with log⁡q≥31\log q \geq 31 and c(q)>22(1−0.4787/log⁡q)/(qlog⁡q)c(q) > 2\sqrt{2}(1 - 0.4787/\log q)/(\sqrt{q} \log q) yields a colossally abundant counterexample, so that Nicolas' oscillation theorem implies Robin's theorem on colossally abundant counterexamples; and under the Riemann Hypothesis every colossally abundant number with log⁡q≥31\log q \geq 31 satisfies Δ(N)<−0.654/(qlog⁡q)\Delta(N) < -0.654/(\sqrt{q} \log q), in accordance with the constant 2(2−1)2(\sqrt{2}-1) of Ramanujan's bound. For superabundant counterexamples, using the exponent bound of Nazardonyavi and Yakubovich, which gives log⁡n≥ψ(pk)\log n \geq \psi(p_{k}) for the largest prime factor pkp_{k}, we prove c(pk)>0.9991/(pklog⁡pk)c(p_{k}) > 0.9991/(\sqrt{p_{k}} \log p_{k}), and we show that the least colossally abundant number not smaller than the least counterexample nn is itself a counterexample NN with log⁡N−log⁡ξ(N)≤log⁡n\log N - \log\xi(N) \leq \log n.

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