bibkey: macarevey2026lagarias authors: Andrew MacArevey year: 2026 title: On the Lagarias Inequality and Superabundant Numbers doi: null url: https://arxiv.org/abs/2602.15905v2 claim: The preprint proves monotonicity of the continuous Lagarias comparison and reduces a least Lagarias counterexample to a superabundant integer; it does not identify a FIB source with a superabundant or colossally abundant integer and does not prove RH. strata_touched: [] license: citation-only triage: anchor

Lagarias monotonicity and the superabundant reduction

The source is Andrew MacArevey, On the Lagarias Inequality and Superabundant Numbers, arXiv:2602.15905v2, submitted February 2026. The versioned TeX was inspected for the derivative argument and the superabundant reduction. This is a source applicability check, not a journal-proof audit, Lean verification, or claim of a new RH proof.

The comparison function

Lagarias’s criterion is

$$\sigma (n)\le H_n+e^{H_n}\log H_n(n\ge 1),$$

which is equivalent to RH. The preprint extends the harmonic numbers by

$$H(x)=\psi (x+1)+\gamma$$

and sets

$$L(x)=\frac{H(x)+e^{H(x)}\log H(x)}{x}.$$

Its Lemmas 1–5 use the trigamma series and integral comparisons to obtain

$$\log (x+1)\le H(x)\le 1+\log x,H^{\prime }(x)\ge \frac{1}{x+1}$$

for $x\ge 1$, then lower-bound the numerator of $L^{\prime }(x)$ by

$$G(x)=\frac{x}{x+1}-(1+\log x)+\frac{x}{1+\log x}-\log (1+\log x).$$

For $x\ge e^4$, the elementary estimate $e^{\log x}\ge 2(\log x{)}^2+3\log x+1$ gives $G(x)>0$. Hence $L^{\prime }(x)>0$ on that range. The paper then checks the finite cases $1\le n\le 54$ and concludes that

$$B_n=\frac{H_n+e^{H_n}\log H_n}{n}$$

is strictly increasing. An independent 80-digit decimal recomputation gives a positive minimum finite gap at $n=54$,

$$B_{55}-B_{54}>0.0054605819041062294681,$$

but this numerical check is not a Lean certificate.

The superabundant reduction

Suppose $m$ is the least counterexample to the Lagarias inequality. If $m$ is not superabundant, choose the greatest superabundant $n<m$ and use

$$\frac{\sigma (n)}{n}\ge \frac{\sigma (m)}{m},B_n<B_m.$$

Then failure at $m$ implies failure at $n$, contradicting minimality. Thus a least counterexample, if one exists, is superabundant. This is a reduction of the search class; it does not establish the inequality on that class.

Boundary for the FIB route

Superabundance is defined by the multiplicative ordering of $\sigma (n)/n$ over all smaller integers. A FIB ATOM address such as

$$[null,2,3,25,5]$$

records additive Zeckendorf inclusion and window adjacency. It gives no prime factorization, valuation vector, $\omega (n)$ bound, or comparison against every smaller integer. Therefore the preprint cannot be applied to a FIB-generated family without a new arithmetic bridge proving that the same integer is superabundant (or reducing further to a valid CA source).

The result is useful as a non-duplicative candidate-class filter for the Lagarias formulation. It supplies no signed Robin tail estimate, no FIB-to-divisor-sum transport, and no RH proof.