REVIEW
A finite arithmetic form of Robin's inequality and its equivalence to the Riemann hypothesis
Number Theory · Submitted 22 Sep 2026 · 23 references
Pith's verdict: New finite-truncation Robin criterion equivalent to RH, clean proof but load-bearing finite check not yet reproducible.
T0 · AI review
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Referee report
The paper
The paper claims
A sharper finite divisor-sum bound is equivalent to Robin's inequality, and hence to the Riemann hypothesis.
The reading
What carries the argument
The central object is the truncated exponential sum P_K(x) := Σ_{j=0}^K x^j / j! evaluated at x = ln ln ln n with K = ω(n). The argument is carried by the structure of a hypothetical minimal counterexample: ungapped prime support rad(n) = q#_K, non-increasing multiplicities, terminal exponent a_K = 1, and the quantitative bounds ⌊ln q_K / ln q_i⌋ ≤ a_i ≤ ⌊3 ln q_K / ln q_i⌋, which reduce the search at fixed K to finitely many exponent tuples. The equivalence with RH pivots on Lemma 21, a strong-tail estimate controlling the difference between the full exponential and its ω(n)-term truncation, combined with Nicolas' conditional bound of Lemma 23.
Extended reading notes
Core claim
The paper establishes that the ω-inequality, σ(n)/n < e^γ Σ_{j=0}^{ω(n)} (ln ln ln n)^j / j! for n > 5040, is equivalent to Robin's inequality and hence to the Riemann hypothesis. The forward direction is immediate because the truncated sum lies strictly below the full exponential e^(ln ln ln n). The reverse direction, RH ⇒ ω-inequality, is proven by combining a conditional bound of Nicolas with a strong-tail estimate: once a minimal counterexample is known to have ω(n) ≥ 21, the truncation error is shown to lie below 6.4×$10^{-15}$ / √(ln n), well inside the 0.095/√(ln n) margin in Nicolas' bound. As a corollary, RH is equivalent to the ω-inequality holding only on colossally abundant numbers.
Load-bearing premise
The equivalence RH ⇒ ω-inequality rests on the finite computations (L.3) up to n ≤ $10^{9}$ and (L.5) over the candidate set N_20 forcing a hypothetical minimal counterexample to have ω(n) ≥ 21, since the strong-tail estimate of Lemma 21 only holds for K ≥ 21 and Nicolas' conditional bound is used as given; if any of those computations missed a counterexample, or Nicolas' bound fails, the central equivalence collapses.
Editorial extensions
If this is right
- The Riemann hypothesis is true if and only if the ω-inequality holds for all colossally abundant numbers strictly greater than 5040.
- If the ω-inequality fails, its minimal counterexample must be a superabundant number whose factorization is supported on the first ω(n) primes, has falling multiplicities, ends in exponent 1, and satisfies ω(n) ≥ 21.
- Every integer-valued truncation order A(n) ≥ ω(n) yields an inequality intermediate between the ω-inequality and Robin's inequality, so any such truncation is again equivalent to RH; the π-inequality is one instance.
- The ω-inequality holds unconditionally for all n > 5040 with ω(n) ≤ 6, for all primorials, for all square-free integers, and for all odd integers.
- If the inequality is false, then RH fails, and the counterexample occurs inside the extremal superabundant class, so verification can legally be restricted to that class.
Reading between the lines
- The equivalence proof effectively moves the computational burden of RH onto a sparse extremal sequence: if the finite search over the candidate tuples N_K could be pushed beyond K = 20 without finding a counterexample, the forcing to larger ω(n) would improve the tail estimate and thereby the scope of the RH-equivariant criterion.
- The observation that no uniform proportional reduction A(n) = ⌊cω(n)⌋ with fixed 0 < c < 1 can succeed, because powers of two become counterexamples, suggests ω(n) is near the minimal natural arithmetic truncation order, though exact pointwise minimality of ω(n) among all arithmetic truncation rules remains open.
- If RH is true, the strong-tail estimate says the gap between Robin's classical bound and the ω-truncated sum is on the order of 10^-15 / √(ln n) for a minimal counterexample candidate — indicating the finite form is an extremely tight refinement of the classical criterion, not a loose one.
- The paper's method suggests a general recipe: any unconditional verification of Robin's inequality on a class of integers automatically upgrades to the stronger ω-inequality on the same class, provided the judging threshold ω(n) ≥ 21 can be supplied for the class.
Original abstract
Robin's inequality reformulates the Riemann hypothesis as a bound on the sum-of-divisors function $\sigma$. We introduce a finite-truncation version of Robin's inequality determined intrinsically by the arithmetic structure of $n$, \[ \frac{\sigma(n)}{n} < e^\gamma \sum_{j=0}^{\omega(n)} \frac{(\ln\ln\ln n)^j}{j!}, \qquad n>5040, \] where $\omega(n)$ denotes the number of distinct prime factors of $n$, and $\gamma$ is the Euler-Mascheroni constant. The resulting bound is pointwise stronger than Robin's inequality and we prove it unconditionally for integers with $\omega(n)\le 6$, primorials, odd integers and square-free integers. We also prove that this inequality is equivalent to Robin's inequality, and hence to the Riemann hypothesis. Consequently, the Riemann hypothesis is equivalent to the truncated inequality holding for all colossally abundant numbers. We further show that if the inequality fails, its minimal counterexample must be a superabundant number, placing any obstruction within a rigid extremal class of integers.
The compiled record
The paper's own sentences, arranged, kept beside the review and separate from it. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.
The abstract, sentence by sentence
Robin's inequality reformulates the Riemann hypothesis as a bound on the sum-of-divisors function $\sigma$. We introduce a finite-truncation version of Robin's inequality determined intrinsically by the arithmetic structure of $n$, \[ \frac{\sigma(n)}{n} < e^\gamma \sum_{j=0}^{\omega(n)} \frac{(\ln\ln\ln n)^j}{j!}, \qquad n>5040, \] where $\omega(n)$ denotes the number of distinct prime factors of $n$, and $\gamma$ is the Euler-Mascheroni constant. The resulting bound is pointwise stronger than Robin's inequality and we prove it unconditionally for integers with $\omega(n)\le 6$, primorials, odd integers and square-free integers. We also prove that this inequality is equivalent to Robin's inequality, and hence to the Riemann hypothesis. Consequently, the Riemann hypothesis is equivalent to the truncated inequality holding for all colossally abundant numbers. We further show that if the inequality fails, its minimal counterexample must be a superabundant number, placing any obstruction within a rigid extremal class of integers.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Robin's inequality reformulates the Riemann hypothesis as a bound on the sum-of-divisors function $\sigma$. | background 1.00 | 0.06 | 0.05 | asserted 1.00 |
| s1 | We introduce a finite-truncation version of Robin's inequality determined intrinsically by the arithmetic structure of $n$, \[ \frac{\sigma(n)}{n} < e^\gamma \sum_{j=0}^{\omega(n)} \frac{(\ln\ln\ln n)^j}{j!}, \qquad n>5040, \] where $\omega(n)$ denotes the number of distinct prime factors of $n$, and $\gamma$ is the Euler-Mascheroni constant. | contribution 0.98 | 0.94 | 0.12 | asserted 1.00 |
| s2 | The resulting bound is pointwise stronger than Robin's inequality and we prove it unconditionally for integers with $\omega(n)\le 6$, primorials, odd integers and square-free integers. | result 0.98 | 0.95 | 0.14 | asserted 1.00 |
| s3 | We also prove that this inequality is equivalent to Robin's inequality, and hence to the Riemann hypothesis. | result 0.92 | 0.95 | 0.56 | asserted 1.00 |
| s4 | Consequently, the Riemann hypothesis is equivalent to the truncated inequality holding for all colossally abundant numbers. | interpretation 0.61 | 0.87 | 0.37 | asserted 1.00 |
| s5 | We further show that if the inequality fails, its minimal counterexample must be a superabundant number, placing any obstruction within a rigid extremal class of integers. | result 0.95 | 0.93 | 0.09 | asserted 1.00 |
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Figures
Figures reproduced from arXiv:2609.26787 by the authors.
References
Discussion
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