Explicit zero density for the Riemann zeta function

Primes between consecutive powers

“…One of the best explicit estimates is from Kadiri, Lumley, and Ng [22]. Their result builds on Ramaré's [31] explicit version of Ingham's zero-density estimate [19], and is valid for any σ > 1 2…”

New results for witnesses of Robin’s criterion

“…These results could be numerically improved if the lower bound of possible counterexamples is pushed further. From an analytical point of view, it is also most probable that (2.3) could also be tightened if density results for nontrivial zeros of the Riemann zeta function [16] are used. However, in regard to the size of considered integers, as well as of the delicacy of Robin's criterion, it seems quite unlikely that the Riemann hypothesis could be disproven by finding an explicit counterexample.…”

“…Suppose N is a witness for Robin's criterion. Then N is divisible by p if 2 ≤ p ≤ 74419, by p 2 if 2 ≤ p ≤ 271, by p 3 if 2 ≤ p ≤ 41, by p 4 if 2 ≤ p ≤ 13 and by 7 5 , 5 6 , 3 10 and 2 16 .…”

On the error term in the explicit formula of Riemann--von Mangoldt

“…Firstly, we use the smaller value R = 5.5666305 which is the same R appearing in the zero-free region in Section 3 (see equation (3.2)). In the following lemma, we also make small improvements to some of the zero-density estimates in [KLN18]. Lemma 6.1.…”

“…Using Platt and Trudgian's verification of the Riemann hypothesis up to H 0 = 3 • 10 12 [PT21a], and the divisor function estimate in Theorem 2 of [CHT21] to replace (3.13) of [KLN18], we can recalculate the constants in Lemma 4.14 of [KLN18]. Using the notation from [KLN18], we want to optimise over k, µ, α, δ, d, H, and η. We chose H = H 0 − 1, η = 0.2535, k = 1, µ = 1.237, δ = 0.313 and optimised over the other parameters for each σ.…”