trureturing — Discovering truth

slug: oeis-a397578-sigma-neighbour-peaks bibkey: ratushnyak2026a397578 doi: null url: https://oeis.org/A397578 triage: theorem motivation_gids:

• D5/S3/Arith/Robin/SigmaNeighbourPeak.result

Problem

Ratushnyak’s OEIS A397578 defines a(n) as the least k satisfying sigma(k) > n sigma(k-1) and sigma(k) > n sigma(k+1), and states: “Conjecture: a(n) exists for all n.” The exact assertion is preregistered in #14990.

Writing $\sigma (k)={\sum }_{d\mid k}d$, the closed claim is

$$\forall n\in \mathbb{N},\exists k\in \mathbb{N},k\ge 2\wedge n\sigma (k-1)<\sigma (k)\wedge n\sigma (k+1)<\sigma (k).$$

The natural domain includes zero. Existence of a witness entails existence of a least witness by well-ordering.

Motivation

The assertion asks whether divisor-sum peaks can simultaneously exceed both neighbours by an arbitrarily large factor. Primorials contain every small prime; their neighbours contain none, separating the sources of large and small abundancy.

Gap

The supplied literature reading identifies the OEIS conjecture without a proof. Erdős (1936) proved that $\sigma (n+1)\ge \sigma (n)$ holds on a set of density one half. Erdős–Győry–Papp proved that chains of comparisons among $\sigma (n+1),\dots ,\sigma (n+4)$ occur infinitely often. Neither result gives the simultaneous arbitrary-ratio assertion. These are supplied preregistration readings, not a fresh external search or an exhaustive originality claim.

Route

Let $P={\prod }_{p\le x}p$ for $x\ge 4$. The identity $\sigma (P)/P={\sum }_{d\mid P}1/d$ includes every term $1/p$ with $p\le x$. Divergence of the prime reciprocal series therefore gives a threshold with $\sigma (P)>6nP$.

For $m=P-1$ or $m=P+1$, no prime at most $x$ divides $m$. The product of the distinct prime factors of $m$ divides $m$, and each factor is at least four. Mathlib’s primorial bound gives

$$4^{\omega (m)}\le m\le 4^x+1<4^{x+1},$$

so $\omega (m)\le x$. Prime-power geometric sums give

$$\frac{\sigma (m)}{m}\le \prod _{p\mid m}\frac{p}{p-1}\le (1+1/x{)}^{\omega (m)}\le (1+1/x{)}^x\le e<3.$$

Thus $n\sigma (P-1)<\sigma (P)$ and $n\sigma (P+1)<\sigma (P)$, since $P+1\le 2P$. For $n=0$, positivity of $\sigma (P)$ supplies the strict inequalities directly.

Falsifier

An index n for which every k fails at least one of the two strict inequalities would refute the assertion. A mismatch in the sum-of-divisors convention, loss of strictness, or an invalid bound on either neighbour’s prime factors would invalidate the proposed proof. A prior resolution of the exact assertion would affect open-problem eligibility without changing the mathematics.

Evidence

D5.S3.Arith.Robin.SigmaNeighbourPeak.result : claim proves the exact quantified assertion. Private lemmas establish the prime-power bound, multiplicative abundancy estimate, rough-number bound, primorial growth and neighbour roughness. The reciprocal divisor identity is reused from GoldenResourceOptimalInteger.reciprocal_divisor_sum. The matching Blueprint result node declares a Proved resolution of this slug. The implementation report supplies the scoped build result.

Triage

• [proved] For every natural n, some primorial k dominates both adjacent divisor sums by factor n.
• [open] The growth rate of the least witness a(n), and whether a(n) is always a highly abundant number.

ASSUMED-UNVERIFIED

No fresh network search was performed because this implementation seat has no network. The OEIS source and literature-gap readings are supplied by preregistration. Completeness of the literature search and exclusive priority remain unverified. The proof asserts existence, with no growth-rate or highly-abundant-minimizer conclusion.