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On the Lagarias Inequality and Superabundant Numbers

Andrew MacArevey

TL;DR

The Lagarias inequality provides a criterion equivalent to the Riemann Hypothesis; the paper recasts it via the continuous harmonic extension $H(x)=\psi(x+1)+\gamma$ and the associated $L(x)$ and $B_n$ sequences, deriving $L'(x)=N(x)/x^2$. It then shows $N(x)\ge 0$ for $x\ge 1$ using bounds on $H(x)$ and $H'(x)$, which yields $L'(x)\ge 0$ and hence $B_{n+1}-B_n>0$ for $n\ge 55$ (with direct checks for $1\le n\le 54$). A key consequence is that any counterexample to Lagarias must occur at a superabundant number, significantly narrowing the verification focus and linking the problem to extremal divisor-sum properties, with potential implications for RH evidence.

Abstract

We study the Lagarias inequality, an elementary criterion equivalent to the Riemann Hypothesis. Using a continuous extension of the harmonic numbers, we show that the sequence $B_n=\frac{H_n+e^{H_n}\log(H_n)}{n}$ is strictly increasing for $n\ge 1$. As a consequence, if the Lagarias inequality has counterexamples, then the least counterexample must be a superabundant number; equivalently, it suffices to verify the inequality on the superabundant numbers.

On the Lagarias Inequality and Superabundant Numbers

TL;DR

The Lagarias inequality provides a criterion equivalent to the Riemann Hypothesis; the paper recasts it via the continuous harmonic extension and the associated and sequences, deriving . It then shows for using bounds on and , which yields and hence for (with direct checks for ). A key consequence is that any counterexample to Lagarias must occur at a superabundant number, significantly narrowing the verification focus and linking the problem to extremal divisor-sum properties, with potential implications for RH evidence.

Abstract

We study the Lagarias inequality, an elementary criterion equivalent to the Riemann Hypothesis. Using a continuous extension of the harmonic numbers, we show that the sequence is strictly increasing for . As a consequence, if the Lagarias inequality has counterexamples, then the least counterexample must be a superabundant number; equivalently, it suffices to verify the inequality on the superabundant numbers.
Paper Structure (3 sections, 10 theorems, 46 equations)

This paper contains 3 sections, 10 theorems, 46 equations.

Key Result

Lemma 2.1

Define Then where the numerator $N(x)$ is

Theorems & Definitions (20)

  • Lemma 2.1
  • proof
  • Lemma 2.2
  • proof
  • Lemma 2.3
  • proof
  • Lemma 2.4
  • proof
  • Lemma 2.5
  • proof
  • ...and 10 more