Robin's Inequality and Superabundant Counterexamples

Introduction

The Riemann Hypothesis (RH) is the conjecture that the Riemann zeta function has its zeros only at the negative even integers and at complex numbers with real part $\frac{1}{2}$ [1]. It has been called the “Holy Grail of Mathematics” [2, 3], and numerous equivalent formulations exist [4]. One of particular interest here is Robin’s criterion [5], which asserts that RH is equivalent to the inequality

Here $\gamma\approx0.57721$ is the Euler–Mascheroni constant, $\sigma(n)=\sum_{d\mid n}d$ is the sum-of-divisors function, and $\log$ denotes the natural logarithm. Grönwall’s theorem [6] states that $\limsup_{n\to\infty}\sigma(n)/(n\log\log n)=e^\gamma$, so (1) asserts that this limiting value is never attained beyond 5040. A closely related elementary reformulation in terms of harmonic numbers is due to Lagarias [7], and the extremal integers relevant to both criteria have been studied computationally by Briggs [8] and structurally by Caveney, Nicolas and Sondow [9].

It is known that (1) holds for many classes of integers: odd integers exceeding 9 and squarefree integers exceeding 30 [10], sums of two squares exceeding 720 [11], and $t$-free integers for $t=5,7,11$ and 20 [10, 12, 13, 14]. complementary direction, Akbary and Friggstad [15] showed that the least counterexample to (1), should one exist, must be a superabundant number, and Nazardonyavi and Yakubovich [16] refined this to the smaller class of extremely abundant numbers. The author has also contributed to this line of results [17].

Let $\operatorname{core}(n)$ denote the squarefree kernel of $n$, that is, the product of the distinct primes dividing $n$ [10], and write $n'=\operatorname{core}(n)$. In an earlier version of this work [18] the author showed that (1) holds for every $n>5040$ with $\zeta(2)\cdot\log\log n'\le\log\log n$, and for every $n>5040$ all of whose prime factors $q$ satisfy $q<\sqrt[\beta]{\log n}$, where $\beta=e^\gamma$.

The purpose of the present paper is to replace both constants $\zeta(2)\approx1.64493$ and $e^\gamma\approx1.78107$ by the single smaller constant

Our main results are the following.

Summary of the Main Results

Let $p_k$ denote the largest prime factor of $n$, and call $n$ a counterexample if $n>5040$ and (1) fails for $n$. Besides Theorems 1.1 and 1.2, the main results of this paper are the following.

1. Theorems 1.1 and 1.2 remain valid with $d$ replaced by 1.016 (Theorems 3.4 and 3.5).

2. Every sufficiently large superabundant number $n$ satisfies $p_k>\sqrt[d]{\log n}$ and $\theta(p_k)>\sqrt[d]{\log n}$ (Theorem 4.4), and the least counterexample satisfies $\sqrt[d]{\log n}\le p_k<\log n$ (Corollary 4.2).

3. If the Riemann Hypothesis is false, there are infinitely many superabundant counterexamples (Proposition 5.3).

4. Every counterexample satisfies $p_k>e^{30.54}$ (Proposition 5.1) and $p_k^{1.000002}>\log n$ (Proposition 5.2).

5. Every superabundant counterexample satisfies

(Theorem 5.6), and, in terms of $\log\log n$ alone,

(Corollary 5.7). Moreover, $\log\log n-\log p_k<O\left(\exp\left(-c_2\sqrt{\log p_k}\right)\right)$ for an absolute constant $c_2>0$ (Corollary 5.10). If there are constants $c>0$ and $0<\alpha<2$ with $c\ge0.0562/31.032^{-\alpha}$, and $N\le10^{1013.11485}$, such that every superabundant number $n>N$ with $\log n>p_k$ satisfies $\log\log n-\log p_k\ge\frac{c}{\log^\alpha(p_k)}$, then the Riemann Hypothesis is true (Proposition 5.15).

6. The least counterexample satisfies $\log p_k<\log\log n$ (Theorem 5.11), $1<(\log n)^{\log\log n-\log p_k}<e^{\frac{0.056201}{\log\log n}}$ (Corollary 5.7), and $\log n\cdot\log\log\frac{n}{p_k}>p_k\cdot\log p_k$ (Theorem 5.13).

7. Among the 244 superabundant numbers $5040<n\le10^{40}$, exactly 3 satisfy $e^{0.0562}>1+(\log\log n-\log p_k)\cdot\log^2(p_k)$, namely the 3 with $p_k>\log n$ (Proposition 6.4).

The proofs combine explicit estimates of Rosser and Schoenfeld [19], of Axler [20] and of Aoudjlit, Berkane and Dusart [21] with the numerical verification of (1) for all $5040<n\le10^{1013.11485}$ due to Morrill and Platt [14]. § 4 treats superabundant numbers, § 5 the largest prime factor of a counterexample, and § 6 the computer verification of every numerical claim, including a rigorous enumeration of the 263 superabundant numbers up to $10^{40}$.

Background and Ancillary Results

We define the abundancy index $I:\mathbb{N}\to\mathbb{Q}$ by $I(n)=\frac{\sigma(n)}{n}$, and we say that Robin’s inequality holds for n if

As usual, $\varphi$ denotes Euler’s totient function. For every $n>1$ we have

where the product runs over the distinct primes dividing $n$.

The next two propositions are explicit estimates of Rosser and Schoenfeld.

Rosser and Schoenfeld in fact prove Proposition 2.4 with 2.50637 in place of 2.51 for every $m\ge3$ except $m=223092870; the version with 2.51 holds without exception and is sufficient for our purposes.

The Chebyshev function $\theta(x)$ is defined by $\theta(x)=\sum_{p\le x}\log p$, where the sum runs over all primes $p\le x$. For a prime $p$, we write $p\#=\prod_{q\le p}q$ for the primorial of $p$, so that $\log(p\#)=\theta(p)$. Throughout, $p_1=2<p_2=3<p_3=5<\cdots$ denotes the sequence of all primes.

Ramanujan’s unpublished notes [22] introduced generalized highly composite numbers, encompassing both superabundant and colossally abundant numbers; these were also studied by Alaoglu and Erdős [23]. Given the first $k$ consecutive primes $p_1,\ldots,p_k$, an integer of the form $\prod_{i=1}^{k}p_i^{a_i}$ with $a_1\ge a_2\ge\cdots\ge a_k\ge1$ is called a Hardy–Ramanujan integer [10] [pp. 367]. A natural number $n$ is called superabundant if $I(m)<I(n)$ for every natural number $m<n$.

The following explicit estimates of Axler and of Aoudjit, Berkane and Dusart, and the author’s lower bound for the largest prime factor of the least counterexample are used in § 5.

Proposition 2.9. For $x \ge46{,}909{,}038$ [20] [Proposition 6.1]:

Equivalently, for $x \ge46{,}909{,}038$,

Proposition 2.10. For $x \ge7{,}232{,}121{,}212$ [21]:

Proposition 2.11. If $n>5040$ is the smallest integer for which Robin’s inequality fails, then the largest prime factor $p$ of $n$ satisfies $p>e^{31.018189471}$ [17] [Theorem 4.2 pp. 748].

Numerical Constants

We use $\zeta(2)=\frac{\pi^{2}}{6}=1.6449340668\ldots$ [24], $\gamma=0.5772156649\ldots$ and $e^{\gamma}=1.7810724179\ldots$. The following numerical facts are elementary; they were checked with high-precision arithmetic, and the computed values are listed in Table 1 of § 6.

Item	Quantity	Computed value
	$\gamma$	$0.577215664901533\ldots$
	$e^{\gamma}$	$1.7810724179902\ldots$
	$\zeta(2)$	$1.64493406684823\ldots$
(i)	$d=e^{\gamma}/\zeta(2)$	$1.08276219326092\ldots$
(ii)	$e^{\gamma}\cdot(d-1)$	$0.147405459669\ldots$
(iii)	$\log^2(64)$	$17.2963085011\ldots$
(iii)	$e^{\gamma}\cdot(d-1)\cdot\log^2(64)$	$2.54957030518\ldots$
(iv)	$\theta(73)$	$65.8767545906\ldots$
(iv)	$\theta(71)$	$61.5862951494\ldots$
(v)	$\left(1-\frac{1}{2\log^2(285)}\right)^{-1}$	$1.01589793074\ldots$
(vi)	$e^{\gamma}\log285\left(1-\frac{1}{2\log^2(285)}\right)^{-1}$	$10.2275448716\ldots$
(vii)	$e^{\gamma}\cdot\log313$	$10.2344040108\ldots$
(vii)	$313/\log10$	$135.9341728\ldots$

Lemma 2.12. Let $d=\frac{e^{\gamma}}{\zeta(2)}$. Then:

(i) $d=1.0827621\ldots$, so that $1.0827<d<1.0828$;

(ii) $e^{\gamma}\cdot(d-1)=\frac{e^{2\gamma}}{\zeta(2)}-e^{\gamma}=0.1474054\ldots>0.1474$;

(iii) $e^{\gamma}\cdot(d-1)\cdot\log^{2}(64)>0.1474\cdot17.296>2.549>2.51$;

(iv) $\theta(73)=\log(73\#)=65.876\ldots>64$;

(v) $\left(1-\frac{1}{2\log^{2}(285)}\right)^{-1}=1.01589\ldots<1.016<d$;

(vi) $e^{\gamma}\cdot\log285\cdot\left(1-\frac{1}{2\log^{2}(285)}\right)^{-1}=10.2275\ldots<10.23$;

(vii) $e^{\gamma}\cdot\log313=10.2344\ldots>10.23$ and $e^{313}<10^{136}$.

Two Auxiliary Lemmas

Lemma 2.13. Let $x\ge2$. Then:

(a) if $x\le285$, then $\prod_{p\le x}\frac{p}{p-1}<10.23$;

(b) if $x\ge285$, then $\prod_{p\le x}\frac{p}{p-1}<1.016\cdot e^{\gamma}\cdot\log x$.

Proof. Every factor $\frac{p}{p-1}$ exceeds 1, so the product is nondecreasing in $x$. For (a), Proposition 2.3 at $x=285$ and Lemma 2.12 (vi) give

For (b), the function $x\mapsto\left(1-\frac{1}{2\log^{2}(x)}\right)^{-1}$ is decreasing on $[285,\infty)$, so by Proposition 2.3 and Lemma 2.12 (v),

Proof. If $\log n \ge313$, then by Lemma 2.12 (vii),

If $\log n < 313$, then $n < e^{313} < 10^{136} < 10^{10^{13.11485}}$ by Lemma 2.12 (vii), and Robin’s inequality holds for $n$ by Proposition 2.5. $\square$

Main Results

A Condition on the Squarefree Kernel

Proof of Theorem 1.1. Let $n > 5040$ satisfy $d \cdot\log\log n' \le\log\log n$, where $n' = \operatorname{core}(n)$. By Proposition 2.2,

There are two cases.

Case I: $\log n' \ge64$. Put $L' = \log\log n'$, so that $L' \ge\log64 > 0$, and in particular $n' \ge3$. By Lemma 2.12 (iii),

that is, $\frac{2.51}{L'} < e^\gamma\cdot(d-1) \cdot L'$. By Proposition 2.4 applied to $m=n'$, and by the hypothesis $d \cdot L' \le\log\log n$,

Together with (2) this gives $I(n) < e^\gamma\cdot\log\log n$.

Case II: $\log n' < 64$. Let $q_1 < \cdots< q_r$ be the distinct primes dividing $n$, so that $r \ge1$ and $n' = q_1 \cdots q_r$. Since $q_i \ge p_i$ for every $i$, we have $n' \ge p_1 \cdots p_r = p_r\#$, hence $\theta(p_r) \le\log n' < 64 < \theta(73)$ by Lemma 2.12 (iv), and so $p_r \le71$. The function $t \mapsto\frac{t}{t-1}$ is decreasing on $(1,\infty)$, so $\frac{q_i}{q_i-1} \le\frac{p_i}{p_i-1}$ for every $i$. By Lemma 2.13 (a),

By (2), $I(n) < 10.23$, and Robin’s inequality holds for $n$ by Lemma 2.14. $\square$

Remark 3.1. Case II of the proof does not use the hypothesis of Theorem 1.1: Robin’s inequality holds for every $n > 5040$ whose squarefree kernel satisfies $\operatorname{core}(n) < e^{64}$. The hypothesis $d \cdot\log\log n' \le\log\log n$ is needed only in Case I.

A Condition on the Prime Factors

Proof of Theorem 1.2. Let $n > 5040$, and suppose that every prime $q$ dividing $n$ satisfies $q < \sqrt[d]{\log n}$. Let $P$ be the largest prime factor of $n$. Then $P^d < \log n$, that is,

Since every prime dividing $n$ is at most $P$ and every factor $\frac{p}{p-1}$ exceeds 1, Proposition 2.2 gives

There are two cases.

Case I: $P \ge285$. By Lemma 2.13 (b), Lemma 2.12 (v) and (3),

Case II: $P < 285$. By Lemma 2.13 (a), $I(n) < 10.23$, and Robin’s inequality holds for $n$ by Lemma 2.14. $\square$

Comparison with the Earlier Constants

Since $d < \zeta(2)$, the hypothesis $\zeta(2) \cdot\log\log n' \le\log\log n$ implies $d \cdot\log\log n' \le\log\log n$ whenever $\log\log n' \ge0$; and when $\log\log n' < 0$ we have $d \cdot\log\log n' < 0 < \log\log n$ for $n > 5040$, so the hypothesis of Theorem 1.1 holds in any case. Likewise, since $d < e^\gamma$ and $\log n > 1$, the condition $q < \sqrt[e^\gamma]{\log n}$ implies $q < \sqrt[d]{\log n}$. Hence Theorems 1.1 and 1.2 contain the two results of the earlier version [18].

The Exponent 1.016

The proofs of Theorems 1.1 and 1.2 extend to the smaller exponent 1.016.

Proof. If $\log n \ge16600$, then $e^\gamma\cdot\log\log n \ge e^\gamma\cdot\log16600 > 17.3 > I(n)$. If $\log n < 16600$, then $n < e^{16600} < 10^{7210} < 10^{1013.11485}$, and Robin’s inequality holds for $n$ by Proposition 2.5.

Proof. By Proposition 2.2, $I(n) < \dfrac{n'}{\varphi(n')}$.

Case I: $\log n' \ge13000$. Put $L' = \log\log n' \ge\log13000 > 0$, so that $n' \ge3$. Since $e^\gamma\cdot0.016 \cdot\log^2(13000) > 2.51$, we have $\dfrac{2.51}{L'} < 0.016 \cdot e^\gamma\cdot L'$, and Proposition 2.4 gives

Case II: $\log n' < 13000$. Let $q_1 < \cdots< q_r$ be the distinct primes dividing $n$. As in the proof of Theorem 1.1, $n' \ge p_r\#$, so $\vartheta(p_r) \le\log n' < 13000 < \vartheta(14000)$, hence $p_r < 14000$, and by Lemma 2.13 (b),

Thus $I(n) < 17.3$, and Robin’s inequality holds for $n$ by Lemma 3.3.

Proof. Let $P$ be the largest prime factor of $n$, so that $1.016 \cdot\log P < \log\log n$, and $I(n) < \prod_{p\le P}\dfrac{p}{p-1}$ by Proposition 2.2. If $P \ge285$, Lemma 2.13 (b) gives $I(n) < 1.016 \cdot e^\gamma\cdot\log P < e^\gamma\cdot\log\log n$. If $P < 285$, Lemma 2.13 (a) gives $I(n) < 10.23$, and Robin’s inequality holds for $n$ by Lemma 2.14.

Proof. Since $d > 1.016$ by Lemma 2.12 (v), the hypothesis $d \cdot\log\log n' \le\log\log n$ implies $1.016 \cdot\log\log n' \le\log\log n$ when $\log\log n' \ge0$, and when $\log\log n' < 0$ we have $1.016 \cdot\log\log n' < 0 < \log\log n$ for $n > 5040$. Likewise, $q < \sqrt[d]{\log n}$ implies $q < \sqrt[1.016]{\log n}$, because $\log n > 1$.

Superabundant Numbers

Let $n$ be a superabundant number with largest prime factor $p_k$. By Proposition 2.6, $n = \prod_{i=1}^{k} p_i^{a_i}$ with $a_1 \ge\cdots\ge a_k \ge1$, so the primes dividing $n$ are exactly the primes $p \le p_k$. Therefore

Theorems 1.1 and 1.2 thus take the following form.

Applying Corollary 4.1 to a hypothetical least counterexample gives the following.

The Largest Prime Factor of a Counterexample

In this section we say that $n$ is a counterexample if $n > 5040$ and Robin’s inequality does not hold for $n$, that is, $I(n) \ge e^\gamma\cdot\log\log n$. For an integer $n$ with largest prime factor $p_k$ we write

Every Counterexample Has a Large Prime Factor

Proof. By Proposition 2.5, $n > 10^{1013.11485}$, so $L > \log\log(10^{1013.11485}) = 13.11485 \cdot\log10 + \log\log10 > 31.032$. If $p_k < 285$, then $I(n) < 10.23$ by Proposition 2.2 and Lemma 2.13 (a), and Robin’s inequality would hold for $n$ by Lemma 2.14, a contradiction. Hence $p_k \ge285$, and by Proposition 2.2 and Lemma 2.13 (b),

Thus $1.016 \cdot P > L$, which is the first claim, and $P > \frac{31.032}{1.016} > 30.54$. Finally $e^{30.54} > 7,232,121,212$.

Proposition 5.1 applies to every counterexample, whereas Proposition 2.11 gives the slightly stronger bound $\log p_k > 31.018$ for the least one. For $P > 0$ we write

which is decreasing in $P$.

Proof. Put $P = \log p_k$ and $L = \log\log n$. By Proposition 5.1, $P > 30.54$, so $p_k > 46,909,038$ and $(1-\varepsilon(P))^{-1} \le(1-\varepsilon(30.54))^{-1} < 1.000002$. By Propositions 2.2 and 2.9, and since Robin’s inequality fails for $n$,

Hence $L < 1.000002\cdot P$, which is $p_k^{1.000002} > \log n$. Finally, $L > 31.032$ by Proposition 2.5, so $P > \frac{31.032}{1.000002} > 31.0319$.

Infinitely Many Superabundant Counterexamples

A natural number $n$ is called colossally abundant if there exists $\varepsilon> 0$ such that $\frac{\sigma(n)}{n^{1+\varepsilon}} \ge\frac{\sigma(m)}{m^{1+\varepsilon}}$ for every natural number $m > 1$ [23].

The Window for Superabundant Counterexamples

Proof. By Proposition 5.1, $p_k > 46,909,038$. By Proposition 2.6, the prime factors of $n$ are exactly the primes $p \le p_k$. By Propositions 2.2 and 2.9, and since Robin’s inequality fails for $n$,

Dividing by $e^\gamma\cdot\log p_k>0$ gives the claim.

The next result holds for every superabundant number with a sufficiently large prime factor, whether or not Robin’s inequality fails.

Proof. By Proposition 2.6, every prime $p \le p_k$ divides $n$, so $n \ge p_k\#$ and $\log n \ge\theta(p_k)$. By Proposition 2.10,

Put $y=\frac{0.01}{\log^3(p_k)}$. Since $\log p_k \ge\log7,232,121,212 > 22.7$, we have $0<y<10^{-6}$. For such $y$,

where the first inequality is $\log(1+u)\ge\frac{u}{1+u}$ applied with $u=-y$. Therefore

as claimed.

Proof. By Proposition 5.1, $P=\log p_k>30.54$ and $p_k>7,232,121,212$. Write $\varepsilon=\varepsilon(P)$. Multiplying the inequality of Theorem 5.4 by $P>0$ gives $L<\frac{P\cdot\varepsilon}{1-\varepsilon}$, that is,

The lower bound is Theorem 5.5. Finally, $x\cdot P^2<0.0562$, and since $e^y>1+y$ for every $y>0$, we have

Since every counterexample satisfies $p_k^{1.000002}>\log n$, the window of Theorem 5.6 can be expressed in terms of $\log\log n$ alone.

Proof. Put $L=\log\log n$. By Proposition 5.2, $\log p_k>\frac{L}{1.000002}>0$, hence

and

Since $0.0562\cdot1.000002^2<0.056201$ and $0.0101\cdot1.000002^3<0.010101$, Theorem 5.6 gives

Multiplying the upper bound by $L^2>0$ gives $x\cdot L^2<0.056201$, and since $e^y>1+y$ for every $y>0$, we obtain

Since $\log n>1$, the function $t\mapsto(\log n)^t$ is increasing, so $(\log n)^x<(\log n)^{\frac{0.056201}{L^2}}$, and since $(\log n)^{\frac{1}{L}}=e$, we have $(\log n)^{\frac{0.056201}{L^2}}=e^{\frac{0.056201}{L}}$. If $n$ is the least counterexample, then $x>0$ by Theorem 5.11, so $(\log n)^x>1$.

The Window under the Error Term of de la Vallée Poussin

The explicit results above use Proposition 2.9. Without explicit constants, the error term in the prime number theorem due to de la Vallée Poussin gives a window that shrinks faster than any power of $\frac{1}{\log p_k}$.

Proof. Write $\theta(t)=t+E(t)$, so that $|E(t)|\leq C\cdot t\cdot\exp\left(-c_0\sqrt{\log t}\right)$ for $t\geq2$ and some constant $C$, by Proposition 5.8. By partial summation,

Since $\frac{d}{dt}\left(\log\log t-\frac{1}{\log t}\right)=\frac{1+\log t}{t\log^2t}$, the contribution of the main term $t$ of $\theta(t)$ equals $\frac{1}{\log x}+\log\log x-\frac{1}{\log x}-\log\log2+\frac{1}{\log2}$. The integral of the error term converges absolutely on $[2,\infty)$, and its tail satisfies

where we used $1+\log t\leq2\log t$ for $t\geq e$ and $\int_{\sqrt{\log x}}^\infty e^{-c_0\sqrt{u}}\,du=\frac{2}{c_0^2}(1+c_0\sqrt{v})e^{-c_0\sqrt{v}}$. Together with $\frac{|E(x)|}{x\log x}\leq C\exp\left(-c_0\sqrt{\log x}\right)$, this gives a constant $B$ with

for every $0<c_1<c_0$. Next, put $h(p)=-\log\left(1-\frac{1}{p}\right)-\frac{1}{p}$, so that $0<h(p)<\frac{1}{p(p-1)}$. Then $\sum_p h(p)=H$ converges and $\sum_{p>x}h(p)<\sum_{m>x}\frac{1}{m(m-1)}\leq\frac{1}{x-1}$. Hence

By Mertens’ theorem [26], $\prod_{p\leq x}\frac{p}{p-1}\sim e^\gamma\cdot\log x$, so $B+H=\gamma$. Exponentiating, and using $e^y=1+O(y)$ for bounded $y$, gives the claim. $\square$

Proof. By Lemma 5.9, there are constants $A>0$ and $c_1>0$ with

for every $t\geq2$; for $t$ in a bounded range this holds after enlarging $A$, because the ratio $\prod_{p\leq t}\frac{p}{p-1}/(e^\gamma\log t)$ is bounded there. Put $P=\log p_k$ and $L=\log\log n$. By Proposition 2.6 and Proposition 2.2, and since Robin’s inequality fails for $n$,

so $L-P<A\cdot P\cdot\exp(-c_1\sqrt{P})$. Put $c_2=\frac{c_1}{2}$. The function $P\mapsto P\exp(-c_2\sqrt{P})$ is bounded on $(0,\infty)$, so $A\cdot P\cdot\exp(-c_1\sqrt{P})\leq C_2\cdot\exp(-c_2\sqrt{P})$ for a suitable constant $C_2$. The lower bound is Theorem 5.5, which applies because $p_k>7,232,121,212$ by Proposition 5.1. The last statement follows from Proposition 5.3.

For the least counterexample the window is one-sided.

Remark 5.12. An earlier version of Theorem 5.11, based on the estimate

of [21] [Lemma 2.7 (5) pp. 19], gave $e^{0.2}>1+x\cdot\log^2(p_k)$. Proposition 2.9 replaces 0.2 by 0.0562.

The proof of Lemma 6.1, p. 369 of [10] also yields the following inequality.

Proof. Combine Proposition 5.3, Theorem 5.6 and Theorem 5.11.

Proof. Suppose that the Riemann Hypothesis is false, and let $n$ be the least counterexample, with largest prime factor $p_k$ and $P=\log p_k$. By Proposition 2.5, $n>10^{13.11485}\ge N$; by Theorem 5.11, $\log n>p_k$; by Proposition 2.7, $n$ is superabundant. Hence the hypothesis applies to $n$, and Theorem 5.6 gives

that is, $P^{2-\alpha}<\frac{0.0562}{c}$, which is the first claim. By Proposition 5.2, $P>31.0319>31.03$, so $c<\frac{0.0562}{P^{2-\alpha}}<\frac{0.0562}{31.03^{2-\alpha}}$. Therefore, if $c\ge\frac{0.0562}{31.03^{2-\alpha}}$, the least counterexample cannot exist, and the Riemann Hypothesis is true by Proposition 2.1.

:::

The precondition $\log n>p_k$ in Proposition 5.15 excludes the superabundant numbers with $p_k>\log n$, for which $\log\log n-\log p_k<0$; examples are $n=27720$, $n=55440$ and $n=10685862914126400$ (Proposition 6.4).

Computational Verification

Every numerical claim of this paper was checked by a self-contained Python 3.12 script, verify_robin_d.py, written with the help of Claude Opus 5.5 (Anthropic) and run by the author, which uses the arbitrary-precision library mpmath with 60 significant decimal digits and exact integer arithmetic wherever a comparison of rational numbers is involved. The script has five parts: it checks the constants of Lemma 2.12, it compares the bounds of Lemma 2.13 with the true values of the Mertens product, it checks the constants of the theorems with the exponent 1.016 and of § 5, it enumerates all superabundant numbers up to $10^{40}$ and tests on each of them the conditions of Corollary 4.1 and the inequality of Theorem 5.11, and it tests on the same numbers the hypothesis of Proposition 5.15. All 33 checks performed by the script passed. We now describe each part and its output in detail.

The Constants of Lemma 2.12

The script computes $\gamma$, $e^\gamma$, $\zeta(2)=\frac{\pi^2}{6}$ (and confirms that this agrees with $\zeta(2)$ evaluated directly to 50 digits) and $d=\frac{e^\gamma}{\zeta(2)}$, and then verifies each item of Lemma 2.12. The computed values are collected in Table 1. Every inequality of Lemma 2.12 holds with a visible margin. The two tightest margins are those of item (iii), where $e^\gamma\cdot(d-1)\cdot\log^2(64)=2.5495\ldots$ exceeds the Rosser–Schoenfeld constant 2.51 by about 0.04, and of item (vii), where $e^\gamma\cdot\log313=10.2344\ldots$ exceeds 10.23 by about 0.004. The script also confirms that $\theta(71)=61.586\ldots<64$, so that 71 is indeed the largest prime that can divide $n'$ in Case II of the proof of Theorem 1.1.

The Mertens Product

The script computes the exact partial products $\prod_{p\le x}\frac{p}{p-1}$ for every prime $x\le2000$ and compares them with the bounds of Lemma 2.13. The results are as follows.

1. The true value is $\prod_{p\le285}\frac{p}{p-1}=10.2257527701\ldots$, which lies below the analytic bound $10.2275\ldots$ obtained from Proposition 2.3, and below $10.23, as Lemma 2.13 (a) asserts. The analytic bound is therefore sharp to within $0.002$ at $x=285$.

2. The true value is $\prod_{p\le71}\frac{p}{p-1}=7.82486815925\ldots$, the quantity actually used in Case II of the proof of Theorem 1.1. The bound $10.23$ used there is thus generous.

3. For every prime $x$ with $285\le x\le2000$ the inequality $\prod_{p\le x}\frac{p}{p-1}<1.016\cdot e^{\gamma}\cdot\log x$ of Lemma 2.13 (b) holds. The largest value of the ratio $\prod_{p\le x}\frac{p}{p-1}/(e^{\gamma}\log x)$ on this range is $1.014231153\ldots$, attained at $x=293$, compared with the analytic constant $1.01589\ldots$ of Lemma 2.12 (v).

4. Among the primes $x\le2000$, the inequality $\prod_{p\le x}\frac{p}{p-1}\ge d\cdot e^{\gamma}\cdot\log x$ holds only for $x\le23$. Thus the direct argument of Case I in the proof of Theorem 1.2 already works numerically for every prime $P$ with $29\le P\le2000$; the threshold $285$ was chosen only because it is the point from which Proposition 2.3 applies.

Enumeration of Superabundant Numbers

To enumerate the superabundant numbers rigorously we use the following reduction to Hardy–Ramanujan integers.

Proof. For $m=1$ take $m^*=1$. Otherwise let $q_1<\cdots<q_r$ be the primes dividing $m$ and $m=\prod_{i=1}^{r}q_i^{b_i}$. For a prime $p$ and an integer $b\ge0$ put $g(p,b)=\frac{\sigma(p^b)}{p^b}=\sum_{j=0}^{b}p^{-j}$, so that $I$ is the product of the factors $g(q_i,b_i)$. We proceed in two steps.

Step 1. Put $m_1=\prod_{i=1}^{r}p_i^{b_i}$. Since $q_i\ge p_i$ for every $i$, we have $m_1\le m$, and since $g(p,b)$ is nonincreasing in $p$ for fixed $b$, we have $I(m_1)\ge I(m)$.

Step 2. Suppose that $p<q$ are primes carrying exponents $a<b$ respectively, and exchange the two exponents. The integer does not increase, because $p^bq^a\le p^aq^b$ is equivalent to $p^{b-a}\le q^{b-a}$. The abundance index does not decrease, because $g(p,b)\cdot g(q,a)\ge g(p,a)\cdot g(q,b)$ is equivalent to $\frac{g(p,b)}{g(p,a)}\ge\frac{g(q,b)}{g(q,a)}$, and, with $t=\frac{1}{p}$,

which is increasing in $t$, that is, decreasing in $p$. Applying finitely many such exchanges to $m_1$ sorts its exponents into nonincreasing order and produces a Hardy–Ramanujan integer $m^* \le m_1 \le m$ with $I(m^*) \ge I(m_1) \ge I(m)$.

Proof. If $n$ is superabundant, then $n$ is a Hardy–Ramanujan integer by Proposition 2.6, and $I(n) > I(h_i)$ for every $h_i < n$ by definition. Conversely, let $n = h_j$ be such a record, and let $m < n$. By Lemma 6.1 there is a Hardy–Ramanujan integer $m^* \le m < n$ with $I(m) \le I(m^*)$, and $I(m^*) < I(n)$ because $n$ is a record. Hence $I(m) < I(n)$, and $n$ is superabundant.

The script generates every Hardy–Ramanujan integer $h \le10^{40}$ together with $\sigma(h)$, by recursion over the primes with nonincreasing exponents. There are exactly 5,754,126 of them. It then sorts them and extracts the records of $I$ as in Proposition 6.2. A comparison $I(h) > I(h')$ is decided exactly through the integer inequality $\sigma(h) \cdot h' > \sigma(h') \cdot h$, so no rounding enters this step. This yields exactly 263 superabundant numbers not exceeding $10^{40}$. For every superabundant $n$ with $5040 < n \le10^{40}$ the script evaluates, with 60-digit precision, Robin’s inequality and the two conditions of Corollary 4.1. The outcome is summarized in the next proposition.

Part (1) is already contained in Proposition 2.5; it serves as an independent consistency check of the enumeration. Condition (a) is equivalent to $\theta(p_k) \le\sqrt[d]{\log n}$, whereas condition (b) is $p_k < \sqrt[d]{\log n}$.

Constants of Section 5 and the Quantity $x \cdot\log^2(p_k)$

For the theorems with the exponent 1.016, the script verifies that $\theta(14000) = 13867.29\ldots> 13000$, that $e^\gamma\cdot0.016 \cdot\log^2(13000) = 2.5571\ldots> 2.51$, that $1.016 \cdot e^\gamma\cdot\log14000 = 17.2756\ldots< 17.3$, that the true value $\prod_{p \le14000} \frac{p}{p-1} = 17.0192\ldots$ is below 17.3, that $e^\gamma\cdot\log16600 = 17.3069\ldots> 17.3$, and that $e^{16600} < 10^{7210}$. The script also verifies the numerical facts used in § 5: that $\log\log(10^{13^{11485}}) = 31.03209\ldots> 31.032$, that $\frac{31.032}{1.016} > 30.54$, that $e^{30.54} > 7,232,121,212 > 46,909,038$, that $(1-\varepsilon(30.54))^{-1} = 1.0000019708\ldots< 1.000002$, that $\frac{e^{\frac{20}{\frac{1-\varepsilon(30.54)}{}}}}{} = 0.0561395\ldots< 0.0562$, that $\frac{0.0562}{30.54^2} = 0.0000602557\ldots< 0.0000603$, that $\frac{31.032}{1.000002} > 31.0319$, that $0.0562 \cdot1.000002^2 < 0.056201$, that $0.0101 \cdot1.000002^3 < 0.010101$, and, for Theorem 5.5, that $\log7,232,121,212 = 22.7017\ldots> 22.7$ and $\frac{0.01}{22.73} = 8.549\ldots\cdot10^{-4} < 10^{-4}$. For every superabundant number $n$ with $5040 < n \le10^{40}$ it computes $x = \log\log n - \log p_k$ and $x \cdot\log(p_k)$, counts the superabundant numbers with $x < 0$, that is, with $p_k > \log n$, and counts those for which the inequality $e^{0.0562} > 1 + x \cdot\log^2(p_k)$ of Theorem 5.11 fails or holds. The output of this part of the script is summarized in the next proposition and in Table 2.

$n$	$p_k$	$x \cdot\log^{2}(p_k)$
278282512406132373381723386382308832000	79	2.172675585
355345361995522876779739093380486662400	83	1.311470417
417423768609198560072585079573463248000	79	2.259925324
484561857266622104699644218246118176000	83	1.379577194
834847537218397120145170159146926496000	79	2.408162468
888363404988807191949347733451216656000	83	1.511996393
1776726809977614383898695466902433312000	83	1.662331969
2665090214966421575848043200353649968000	83	1.749739048
3553453619955228767797390933804866624000	83	1.811518921
5330180429932843151696086400707299936000	83	1.898262198

By Proposition 6.4, for the ten largest superabundant numbers below $10^{40}$, $\log\log n$ exceeds $\log p_k$ by between $\frac{1.31}{\log^{2}(p_k)}$ and $\frac{2.41}{\log^{2}(p_k)}$, and the maximum over the whole range is $\frac{2.86}{\log^{2}(p_k)}$. Table 2 also shows that the largest prime factor of consecutive superabundant numbers need not be monotone: it alternates between 79 and 83 before settling at 83. The value of $x \cdot\log^{2}(p_k)$ is smaller whenever $p_k = 83$, and along the last five entries, where $p_k = 83$ is fixed, it increases from 1.51 to 1.90 as the exponents of the smaller primes grow.

The Hypothesis of Proposition 5.15

For every superabundant number $5040 < n \le10^{40}$, the script computes $x = \log\log n - \log p_k$ and the quantities $x \cdot\log^{\alpha}(p_k)$ for $\alpha= 1, 2, 3, 4, 5$. The output is summarized in the next proposition and in Table 3.

$\alpha$	minimum over the 168 numbers	minimum over the last 50
1	0.02239728649	0.2087437012
2	0.0913257743	0.9120947171
3	0.3723842643	3.985350303
4	1.518410781	17.41378032
5	6.191376815	76.08860503

(3) for every $\alpha\in\{1,2,3,4,5\}$, the minimum of $x \cdot\log^{\alpha}(p_k)$ over these 168 numbers is attained at

with $p_k = 59$, and its value is given in Table 3.

For the ten largest superabundant numbers below $10^{40}$, whose largest prime factors are 79 and 83, the values of $x$ lie between 0.0671 . . . and 0.1261 . . ., and those of $x \cdot\log^{3}(p_k)$ lie between 5.795 . . . and 10.522 . . ..

Acknowledgements

The author thanks Richard J. Lipton for helpful comments on an earlier version of this work.

Statements and Declarations

Declaration of interests

The author declares no competing interests.

Funding sources

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

Declaration of generative AI in scientific writing

The author used Claude Opus 5.5 to write the verification script verify_robin_d.py of § 6 and to generate parts of the text of this manuscript. The experiments were run by the author. After using this tool, the author reviewed and edited the code and the content as needed and takes full responsibility for the content of the publication.

Data availability

The Python script verify_robin_d.py used in § 6 is available in the preprint..