Robin's Inequality at Colossally Abundant Numbers and Nicolas' Inequality at Their Largest Prime Factor

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]. Akbary and Friggstad [15] showed that the least counterexample to (1), should one exist, must be a superabundant number, Nazardonyavi and Yakubovich [16] refined this to the smaller class of extremely abundant numbers, and the author proved that its largest prime factor exceeds $e^{31.018}$ [17]. Robin [5] proved that if RH is false, then (1) fails for infinitely many colossally abundant numbers.

A second criterion concerns the primorials $q\# = \prod_{p \le q} p$. For a real number $x \ge2$ put

where $\theta(x) = \sum_{p \le x} \log p$ is the Chebyshev function, so that $c(q) > 0$ is the inequality $\frac{q\#}{\varphi(q\#)} > e^\gamma\cdot\log\log(q\#)$, with $\varphi$ Euler’s totient function. Nicolas [18] proved that $c(x) > 0$ for every $x \ge2$ if RH is true, and that if RH is false, then $c(x) = \Omega_{\pm}(x^{-b})$ for some $0 < b < \frac{1}{2}$. Under RH, Ramanujan’s asymptotic upper bound for $\sigma(n)/n$ [19] contains the term $-\frac{2(\sqrt{2}-1)}{\sqrt{\log n}}$, and Nicolas [20] gave an effective form of this bound, using the structure of colossally abundant numbers and his estimate [21] of $c(x)$ under RH.

In this paper we compare Robin’s inequality at a colossally abundant number with Nicolas’ inequality at its largest prime factor, without assuming RH. For an integer $n \ge16$ put

so that Robin’s inequality holds for $n > 5040$ if and only if $\Delta(n) < 0$. To each colossally abundant number $N$ Nicolas associates a real number $\xi(N)$ [20] Definition 4.3; every prime factor of $N$ is at most $\xi(N)$ § 3. Our main result is the following.

Thus Robin’s function at $N$ and Nicolas’ function at $q$ differ by $\frac{2\sqrt{2}}{\sqrt{q}\log q}$, up to an explicit error that is smaller by a factor $\log q$. Remark 3.5 describes the relation of Theorem 1.1 to the work of Robin and Nicolas, and Remark 3.6 the size of its error term. Theorem 1.1 has the following consequences.

1. Every colossally abundant counterexample to (1) yields a prime $q$ with $\log q > 31.03$ and $c(q) > \frac{2.70}{\sqrt{q}\log q}$ (Corollary 4.2).

2. Conversely, every prime $q$ with $\log q \ge31$ and $c(q) > \frac{2\sqrt{2}}{\sqrt{q}\log q}\left(1-\frac{0.4787}{\log q}\right)$ yields a colossally abundant counterexample to (1) (Corollary 4.3).

3. Nicolas’ theorem $c(x) = \Omega_{+}(x^{-b})$ under the failure of RH implies, through (2), Robin’s theorem that (1) then fails for infinitely many colossally abundant numbers (Corollary 4.4).

4. Under RH, every colossally abundant number $N$ with $\log q \ge31$ satisfies $\Delta(N) < -\frac{0.654}{\sqrt{q}\log q}$ (Corollary 4.6); the constant $2\sqrt{2}-2=2(\sqrt{2}-1)$ obtained from Theorem 1.1 and the main term of $c(q)$ is the constant of Ramanujan’s bound.

5. RH holds if and only if $c(q) < \frac{2.70}{\sqrt{q}\log q}$ for every prime $q \ge10^9$ (Corollary 4.7). This equivalence also follows from Nicolas’ theorem [18] together with his estimate under RH [21, 20], and holds with 2.70 replaced by any constant greater than $2+\tau=2.0461\ldots$; Corollary 4.7 shows that the primes $q$ of item (1) violate it.

For superabundant counterexamples, which need not be colossally abundant, we prove the following (§ 5).

6. Every superabundant number $n$ with largest prime factor $p_k$ satisfies $\log n \ge\psi(p_k)$ (Lemma 5.1, following [16]), and every superabundant counterexample satisfies $c(p_k) > \frac{0.9991}{\sqrt{p_k}\log p_k}$ (Theorem 5.4).

If $n$ is the least counterexample, then the least colossally abundant number $N \ge n$ is a counterexample, and $\log N - \log\xi(N) \le\log n \le\log N$ (Corollary 5.5).

The proofs use the description of colossally abundant numbers by Alaoglu and Erdős [22], Robin [5] and Nicolas [20], explicit estimates for the Chebyshev function of Rosser and Schoenfeld [23], Büthe [24] and Dusart [25], the estimate of Axler [26] for the Mertens product, and the numerical verification of (1) for all $5040<n\le10^{13.11485}$ due to Morrill and Platt [14]. Section 2 collects these results, Section 3 proves Theorem 1.1, Section 4 derives its consequences, Section 5 treats superabundant counterexamples, and Section 6 describes the computer verification of the numerical constants.

Background

We write $I(n)=\frac{\sigma(n)}{n}$ for the abundancy index, and we say that Robin’s inequality holds for $n$ if $I(n)<e^\gamma\cdot\log\log n$. An integer $n>5040$ for which Robin’s inequality does not hold is called a counterexample. We write $\psi(x)=\sum_{p^j\le x}\log p=\sum_{j\ge1}\theta(x^{1/j})$ for the second Chebyshev function, so that $\psi(x)=\sum_{q\le x}\left\lfloor\frac{\log x}{\log q}\right\rfloor\log q$ and $\psi(x)\ge\theta(x)+\theta(\sqrt{x})$. For a prime $p$ we write $p\#=\prod_{q\le p}q$, so that $\log(p\#)=\theta(p)$, and $p_1=2<p_2=3<\cdots$ denotes the sequence of primes. Since $\sum_{p\le x}\log\frac{p}{p-1}$ and $\theta(x)$ change only at primes, $c(x)=c(q)$ when $q$ is the largest prime with $q\le x$.

Robin’s Criterion and Counterexamples

A natural number $n$ is superabundant if $I(m)<I(n)$ for every $m<n$, and colossally abundant if there is $\varepsilon>0$ with $\frac{\sigma(m)}{m^{1+\varepsilon}}\le\frac{\sigma(n)}{n^{1+\varepsilon}}$ for every $m\ge1$ [22]; every colossally abundant number is superabundant [22].

Nicolas’ Criterion

In [18, 21] the statement is given for $\log f(x)=-c(x)$, where $f(x)=e^\gamma\cdot\log\theta(x)\cdot\prod_{p\le x}\left(1-\frac{1}{p}\right)$. Put

where $\rho$ runs over the nontrivial zeros of the Riemann zeta function [20] [(1.3), (1.5)].

Explicit Estimates for Primes

Colossally Abundant Numbers

The Structure of Colossally Abundant Numbers

We recall the description given by Nicolas [20] [Section 4], which goes back to Alaoglu and Erdős [22] and Robin [5]. For real $t>1$ and an integer $j\ge1$ put

which is decreasing in $t$. Let $N$ be a colossally abundant number and $\kappa$ its largest parameter. Then $\xi=\xi(N)>1$ is defined by $F(\xi,1)=\kappa$, and $\xi_j=\xi_j(N)$ for $j\ge2$ by $F(\xi_j,j)=\kappa$, with $\xi_1=\xi$; $K$ is the largest $j$ with $\xi_j\ge2$ [20] [Definition 4.3]. Put $N_\kappa=\prod_{j=1}^{K}\prod_{\xi_{j+1}<p\le\xi_j}p^j$.

In (e), Nicolas states [20] [Lemma 4.10] for $N\ge N^{(0)}$, where $N^{(0)}$ is the colossally abundant number with $\xi(N^{(0)})=10^9+7$. If $\xi>10^9+7$, then $\kappa<F(10^9+7,1)$, and since the parameters of colossally abundant numbers decrease as the numbers increase [20] [(4.4)], we have $N>N^{(0)}$, so (e) applies.

Three Estimates

Throughout this subsection, $N$ is a colossally abundant number with $\lambda=\log\xi\ge31$, where $\xi=\xi(N)$, and we put

By Proposition 3.1 (d), and since $1-\frac{\log2}{2\cdot31}>0.98882$ and $\sqrt{2}\cdot0.98882>1.39838$,

so Lemma 2.10 (b) applies to every t\ge\underaccent{\bar}{\xi}_2 with $\delta=\frac{0.022}{\lambda}$. Moreover \log\underaccent{\bar}{\xi}_2\le\frac{1}{2}(\lambda+\log2), and since $-\log\left(1-\frac{\log2}{2\cdot31}\right)<0.011306$, also \log\underaccent{\bar}{\xi}_2\ge\frac{1}{2}(\lambda+\log2)-0.011306\ge\frac{\lambda}{2}\left(1+\frac{0.670535}{\lambda}\right).

Proof. Put $\ell=\log\xi_2$ and $A_j=\frac{1}{\xi_2^{\ell}}$. Since $t\mapsto t\log t$ is increasing, the bounds on $\xi_2$ and $\ell$ give

and, since

for $\lambda\ge31$,

Since $\frac{\lambda}{2}\le\ell\le\frac{\lambda}{2}\left(1+\frac{\log2}{\lambda}\right)$, we get $A_2=\frac{A_1}{\ell}\le\frac{2M_{\xi}}{\lambda}$, $A_3=\frac{A_2}{\ell}\le\frac{4M_{\xi}}{\lambda^2}$ and

By Lemma 2.11 with $y=\xi_2$ and $3\delta=\frac{0.066}{\lambda}$,

and

where we used $2(1+\delta)\cdot4<8.006$ and $\frac{8.006}{\lambda}\le\frac{8.006}{31}<0.258259$. By Lemma 2.11 with $y=\xi$, using $-(1+\delta)A_2(\xi)+2(1+\delta)A_3(\xi)\le0$,

because $\lambda e^{-\lambda/2}$ is decreasing for $\lambda\ge2$. Since $S_2=\sum_{p>\xi_2}\frac{1}{p^2}-\sum_{p>\xi}\frac{1}{p^2}$, the claim follows.

Proof. Put $D=\log N-\theta(\xi)$. By Proposition 3.1 (c), $\xi_3\ge\xi^{1/3}\ge e^{31/3}>853$, so $K\ge3$, and by Proposition 2.9 (d), $\theta(\xi_3)\ge0.945\cdot\xi^{1/3}>\lambda$, because $0.945\cdot e^{s/3}>s$ for $s\ge31$. Proposition 3.1 (b) and (e) and Lemma 2.10 then give

By (2) and Lemma 2.10 (b),

and

Moreover

because $\lambda e^{-\lambda/6}$ is decreasing for $\lambda\ge6$. Hence

By Proposition 2.9 (b), $\xi\cdot(1-\lambda^{-3})<\theta(\xi)<\xi\cdot(1+\lambda^{-3})$. Hence $\log N=\theta(\xi)+D\le\xi\cdot(1+\lambda^{-3}+\sqrt{2}e^{-\lambda/2})\le\xi\cdot\left(1+\frac{0.0010488}{\lambda}\right)$, so that $\log\log N\le\lambda+\frac{0.0010488}{\lambda}\le\lambda\cdot\left(1+\frac{0.0000339}{\lambda}\right)$ and

Also $\log\theta(\xi)\ge\lambda+\log(1-\lambda^{-3})\ge\lambda-1.0001\lambda^{-3}$, so that

Since $L'(t)=\frac{1}{t\log t}$ is decreasing for $t>1$, the mean value theorem gives

With $\frac{\sqrt{2\xi}}{\xi} = M_\xi$, the lower bound is at least $M_\xi\cdot\frac{1-0.368574/\lambda}{1+0.001085/\lambda}\ge M_\xi\cdot\left(1-\frac{0.369659}{\lambda}\right)$, and the upper bound is at most $M_\xi\cdot\frac{1-0.0538/\lambda}{1-0.00108/\lambda}\le M_\xi\cdot\left(1-\frac{0.05272}{\lambda}\right)$.

Proof. For $p\le\xi$ let $v_p$ be the exponent of $p$ in $N_\kappa$. By Proposition 3.1 (a) and (b), the primes $p\le\xi$ are the primes $p\le q$, and

Every term of the second sum is negative, and $v_p=1$ for $\xi^2<p\le\xi$; since $\log(1-u)\le-u$, the second sum is at most $-S_2$, and the upper bound follows from $I(N)\le I(N_\kappa)$.

For the lower bound we split the second sum into three parts. For $\xi^2<p\le\xi$, $-\log(1-p^{-2})\le\frac{p^{-2}}{1-\xi^{-2}}$, so this part is at least $-S_2-2\xi_2^{-2}S_2$. For $\xi^3<p\le\xi^2$, $v_p=2$, and by Proposition 2.9 (e), since $\xi^3>1418$, this part is at least $-\frac{1}{1-\xi^{-3}}\cdot\frac{0.086}{\xi^{2/3}}\ge-0.0861\cdot\xi^{-2/3}$. For $p\le\xi^3$, if $\xi^{j+1}<p\le\xi^j$ with $j\ge3$, then $p^{j+1}>\xi^{j+1}\ge\xi$ by Proposition 3.1 (c), so $-\log(1-p^{-(j+1)})\le\frac{1}{\xi^{j+1}}$; by Proposition 2.9 (d) and Proposition 3.1 (c), the number of such primes is at most $\pi((3\xi)^{1/3})<\frac{3\cdot1.26\cdot3^{1/3}}{5.4518}\xi^{1/3}$, so this part is at least $-\frac{5.4518}{31}\cdot\frac{\xi^{1/3}}{\xi-1}\ge-0.17587\cdot\xi^{-2/3}$. Together with $I(N)\ge\frac{\xi}{\xi+1}I(N_\kappa)$ and $\log(1+\frac{1}{\xi})\le\frac{1}{\xi}\le e^{-31/3}\cdot\xi^{-2/3}$,

Finally $\xi^{-2/3}=\frac{\lambda^2e^{-\lambda/6}}{\sqrt{2}}\cdot\frac{M_\xi}{\lambda}$, and $\lambda^2e^{-\lambda/6}$ is decreasing for $\lambda\ge12$, so $0.26201\cdot\xi^{-2/3}\le\frac{0.26201\cdot3^{12}\cdot e^{-31/6}}{\sqrt{2}}\cdot\frac{M_\xi}{\lambda}<\frac{1.0155M_\xi}{\lambda}$, while $2\xi_2^{-2}S_2\le2\xi_2^{-2}M_\xi<\frac{10^{-4}\cdot M_\xi}{\lambda}$ by Lemma 3.2 and (2).

Proof of Theorem 1.1

Proof of Theorem 1.1. Put $\xi=\xi(N)$, $\lambda=\log\xi$ and $u=\log q\ge31$, so that $\lambda\ge u\ge31$. Since $q\le\xi$ and the primes $p\le\xi$ are the primes $p\le q$, we have $\theta(\xi)=\theta(q)$ and

By Lemmas 3.2, 3.3 and 3.4,

that is,

Put $M_q=\frac{\sqrt{2}}{\sqrt{q}\,u}$. The prime following $q$ exceeds $\xi$, so by Proposition 2.9 (c), $\xi<q\cdot(1+u^{-3})$ and $\lambda<u+u^{-3}$. Hence $M_{\xi}\le M_q$, $\frac{1}{\lambda}\ge\frac{1-u^{-4}}{u}$, and

Therefore

and

Since $2M_q=\frac{2\sqrt{2}}{\sqrt{q}\log q}$, $0.9574\cdot\sqrt{2}>1.353$ and $2.8070\cdot\sqrt{2}<3.970$, the theorem follows.

In the notation of the proof, Theorem 1.1 is the pair of inequalities

Remark 3.5. The decomposition of $\Delta(N)-c(q)$ used in the proof of Theorem 1.1 follows Nicolas’ proof of [20] Proposition 5.1, in which $\log(\sigma(N)/N)$ is written as $U=U_1+U_2+U_3-U_4-U_5$ [20] [(5.4)], with $-U_5=\sum_{p\le\xi}\log\frac{p}{p-1}$ the Mertens product; a comparison of $\sigma(N)/N$ with this product for colossally abundant numbers $N$ also underlies Robin’s proof of Proposition 2.6 [5] [Section 4]. In [20] Proposition 5.1 the Mertens product is evaluated under RH by [21] [(2.16)], that is, by Proposition 2.8. In Theorem 1.1 it is kept in the form $c(q)+\gamma+\log\log\theta(q)$, so that no hypothesis on the zeros of the Riemann zeta function is used, and the remaining terms are bounded from above and from below with explicit constants for every colossally abundant number with $\log q\ge31$.

Remark 3.6. The two bounds of Theorem 1.1 differ by $\frac{3.970-1.353}{\sqrt{q}\log^2q}$. Consequently, Corollaries 4.2 and 4.3 below do not decide whether Robin’s inequality holds at a colossally abundant number $N$ when $\frac{c(q)}{\sqrt{q}\log q}$ lies between $\frac{2\sqrt{2}}{\sqrt{q}\log q}\left(1-\frac{1.4035}{\log q}\right)$ and $\frac{2\sqrt{2}}{\sqrt{q}\log q}\left(1-\frac{0.4787}{\log q}\right)$. In the proof, the two sides differ by $(2.805819-0.957494)\cdot\frac{M_{\xi}}{\lambda}=1.848325\cdot\frac{M_{\xi}}{\lambda}$; of this, $1.0156$ comes from Lemma 3.4 (the primes $p\le\xi_2$, which have exponent at least 2), $2.43616-1.920374=0.515786$ from Lemma 3.2, and $0.369659-0.05272=0.316939$ from Lemma 3.3.

Consequences

Proof. Put $L=\log\log n$ and $P=\log p_k$. By Proposition 2.3, $n>10^{10^{13.11485}}$, so $L>13.11485\cdot\log10+\log\log10>31.032$. By Proposition 2.2, $e^{\gamma L}\le I(n)<\prod_{p\le p_k}\frac{p}{p-1}$. If $p_k<46,909,038$, then by Proposition 2.9 (f) at $x=46,909,038$, this product is less than $e^\gamma\cdot17.6638\cdot1.000011<e^\gamma\cdot31.032$, a contradiction. Hence $p_k\ge46,909,038$, and Proposition 2.9 (f) gives $L<P\cdot(1-\varepsilon(P))^{-1}$. Since $\varepsilon$ is decreasing and $P\ge\log(46,909,038)$, $(1-\varepsilon(P))^{-1}<1.000011$, so $P>\frac{31.032}{1.000011}>31.0316$; then $(1-\varepsilon(P))^{-1}\le(1-\varepsilon(31.0316))^{-1}<1.000002$, so $L<1.000002\cdot P$ and $P>\frac{31.032}{1.000002}>31.0319$. □

For the least counterexample, the author proved $\log p_k>31.018$ [17] [Theorem 4.2]; Lemma 4.1 applies to every counterexample.

Proof. The largest prime factor $p_k$ of $N$ is at most $\xi(N)$ by Proposition 3.1 (a), so $p_k\le q$ and $\log q>31.0319$ by Lemma 4.1. Since Robin’s inequality fails for $N$, $\Delta(N)\ge0$, and the upper bound in (3) gives the first inequality. The function $u\mapsto2\sqrt{2}\left(1-\frac{1.4035}{u}\right)$ is increasing, and its value at $u=31.0319$ exceeds $2.70$. □

Proof. The lower bound in (3) gives $\Delta(N)>0$, that is, $I(N)>e^\gamma\log\log N$, and $N>5040$ because $q$ divides $N_\kappa$ and $\log N\ge\theta(q)-\log q$ by Proposition 3.1 (b). For $\kappa=F(q,1)$, the number $N_\kappa$ is colossally abundant with largest parameter $\kappa$ by Proposition 3.1 (a), and $\xi(N_\kappa)=q$ because $F(\cdot,1)$ is decreasing. □

By Corollaries 4.2 and 4.3, for colossally abundant numbers $N$ with $\log q\ge31$, Robin’s inequality fails at $N$ when $c(q)\sqrt{q}\log q>2\sqrt{2}\left(1-\frac{0.4787}{\log q}\right)$ and holds at $N$ when $c(q)\sqrt{q}\log q<2\sqrt{2}\left(1-\frac{1.4035}{\log q}\right)$.

Proof. If RH is false, Proposition 2.7 gives $b<\frac12$ and $C>0$ such that $c(x)>Cx^{-b}$ for arbitrarily large $x$. Let $q$ be the largest prime with $q\le x$; then $c(q)=c(x)$ and $q>\frac{x}{2}$, so $c(q)>Cx^{-b}>\frac{4}{\sqrt{x\log(x/2)}}>\frac{2\sqrt{2}}{\sqrt{q}\log q}$ for all sufficiently large such $x$. Corollary 4.3 gives a colossally abundant counterexample $N_{F(q,1)}$ for each such $q$ with $\log q\ge31$, and these numbers are distinct because $\xi(N_{F(q,1)})=q$. □

Remark 4.5. Corollary 4.4 recovers Proposition 2.6 of Robin [5], here obtained from Proposition 2.7. Corollary 4.6 below is an explicit inequality, for colossally abundant numbers with $\log q\ge31$, of the type of Ramanujan’s upper bound for $\sigma(n)/n$ under RH, of which [20] [Theorem 1.1] gives an effective form for every $n\ge2$.

Proof. By (3) and Proposition 2.8, since $q > 10^{9}$,

The number $2\sqrt{2}-2-\tau=2(\sqrt{2}-1)-\tau=0.7822\ldots$ appearing in the proof is the constant obtained from Ramanujan’s bound [19] together with $|S_1(x)|\le\frac{\tau}{\sqrt{x}}$, as in [20] [(1.8)]; here $\sqrt{q}\log q$ plays the role of $\sqrt{\log N\cdot\log\log N}$, since $\log N=q\cdot(1+o(1))$ by the proof of Lemma 3.3 and Proposition 2.9 (c).

Corollary 4.7. The Riemann Hypothesis is true if and only if

for every prime $q\ge10^{9}$.

Proof. The inequality is $c(q)<\frac{2.70}{\sqrt{q}\log q}$. If RH is true, it holds for every prime $q\ge10^{9}$ by Proposition 2.8. If RH is false, Proposition 2.6 gives a colossally abundant counterexample, and Corollary 4.2 gives a prime $q$ with $\log q>31.0319$, so $q>10^{9}$, and $c(q)>\frac{2.70}{\sqrt{q}\log q}$.

By Propositions 2.7 and 2.8, the equivalence of Corollary 4.7 also holds with 2.70 replaced by any constant greater than $2+\tau$. Corollary 4.2 identifies primes that violate it: the primes $q$ attached to colossally abundant counterexamples.

Superabundant Counterexamples

Proof. The first claim is due to Nazardonyavi and Yakubovich [16]; we give the proof for completeness. For a prime $p$ and an integer $b\ge0$ put $g(p,b)=\sum_{j=0}^{b}p^{-j}$. For $i=k$ the claim is $a_k\ge1$. Let $i<k$, write $q=p_i$ and $a=a_i\ge1$, and suppose that $a\le\left\lfloor\frac{\log p_k}{\log q}\right\rfloor-1$. Then $q^{a+1}\le p_k$, and since $q^{a+1}\ne p_k$, we have $q^{a+1}<p_k$. Put $m=\frac{nq^a}{p_k}$, which is an integer with $m<n$, and write $b=a_k\ge1$. Since $I$ is multiplicative,

where we used $g(q,2a+1)=g(q,a)\cdot(1+q^{-(a+1)})$, $g(p_k,b-1)\ge1$ and $q^{a+1}<p_k$. Thus $I(m)>I(n)$ with $m<n$, which contradicts the superabundance of $n$. Finally, by Proposition 2.4 the primes dividing $n$ are exactly the primes $q\le p_k$, so

where $a_q$ denotes the exponent of $q$ in $n$.

Proof. By Proposition 2.2, and since Robin’s inequality fails for $n$, $\frac{n'}{\varphi(n')}>I(n)\ge e^\gamma\cdot\log\log n$. By Proposition 2.3, $\log\log n>31.032$, so $\frac{n'}{\varphi(n')}>55$. Since $\frac{m}{\varphi(m)}\le3$ for every $m\le15$, we get $n'>15$, hence $\log\log n'>0$. Taking logarithms gives $\log\frac{n'}{\varphi(n')}\ge\gamma+\log\log\log n$, and subtracting $\log\log\log n'$ gives the first inequality. The second follows from $n\ge n'$.

For a superabundant counterexample with largest prime factor $p_k$, Proposition 2.4 gives $n'=p_k\#$, so Lemma 5.2 gives

Experimental Verification

Every numerical inequality between explicit constants used in this paper was checked by a self-contained Python 3.12 script, verify_robin_nicolas.py, written with the help of Claude Opus 5.5 (Anthropic) and run by the author, as in the earlier version of this work. The script uses the arbitrary-precision library mpmath with 60 significant decimal digits, and exact rational arithmetic (Python’s fractions module) for the inequalities that involve only decimal constants, some of which hold with equality. For each inequality it prints the computed value of its left-hand side. The script has three parts, and all 69 checks performed by the script passed: 14 in Part 1, 50 in Part 2 and 5 in Part 3. We now describe the output.

Part 1: Counterexamples

Part 1 checks the constants of Lemmas 4.1, 5.2 and 5.3 and of Theorem 5.4. The computed values are listed in Table 1.

Quantity	Computed value	Bound used
$\log\log(10^{10^{13.11485}})$	31.0320905521…	$> 31.032$
$\log(46{,}909{,}038)$	17.6637209227…	$< 17.6638$
$(1-\varepsilon(\log46{,}909{,}038))^{-1}$	1.0000109986…	$< 1.000011$
$31.032/1.000011$	31.0316586518…	$> 31.0316$
$(1-\varepsilon(31.0316))^{-1}$	1.0000018754…	$< 1.000002$
$31.032/1.000002$	31.0319379361…	$> 31.0319$
$\left(\frac{1}{20}+\frac{16\cdot31.0319}{20}\right)/(1-\varepsilon(31.0319))$	0.0560422746…	$< 0.0562$
$e^{0.0562/31.032}\left(1+\frac{0.0562}{31.03^2}\right)$	1.0000602505…	$< 1.0000603$
$e^{15.515}$	5471153.3376…	$> 5.47\cdot10^6$
$1-1.94/\sqrt{5.47\cdot10^6}$	0.9991705159…	$> 0.99917$
$0.99917/1.0000603$	0.9991097537…	$> 0.9991$
$e^\gamma\cdot31.032$	55.2702392751…	$> 55$
$\max_{m\le15}m/\varphi(m)$	3	$=3$

The starting point is $\log\log(10^{10^{13.11485}}) = 31.0320905\ldots$, the bound for $\log\log n$ coming from Proposition 2.3. At $x = 46{,}909{,}038$, where Proposition 2.9 (f) begins to apply, $\log x = 17.6637\ldots$ and $(1-\varepsilon(\log x))^{-1} = 1.0000109986\ldots$, which lies below 1.000011 by $1.4 \cdot10^{-9}$; this is the smallest margin in Part 1. The two steps of Lemma 4.1 give $\frac{31.032}{1.000011} = 31.03165\ldots$ and $\frac{31.032}{1.000002} = 31.031937\ldots$. In Lemma 5.3, the constant 0.0562 exceeds the computed value 0.0560422…, and the factor 1.0000603 exceeds the computed value 1.0000602505… by $4.9 \cdot10^{-8}$.

Part 2: The Proof of Theorem 1.1

Part 2 checks the constants of Lemmas 2.10, 3.2, 3.3 and 3.4 and of the proof of Theorem 1.1, all at the threshold $\lambda= 31$, where $e^{31} = 29048849665247.4\ldots$; each quantity checked at $\lambda= 31$ is monotone in $\lambda$ in the direction stated in the proofs. Table 2 lists the main values. In Lemma 2.10 (b), the relative error $\frac{0.022}{\lambda}$ for $\theta(t)$ on $t\ge\xi_2$ comes from Büthe’s bound, with computed value 0.0219063… at $\lambda=31$, whereas Dusart’s bound contributes at most 0.00105. In Lemma 3.3, the contribution $1.397905\cdot31\cdot e^{-31/6}=0.2471636\ldots$ of the primes with exponent at least 3 reduces the coefficient 0.301 coming from $\xi_2$ to 0.0538. In Lemma 3.4, the primes with exponent 2 and those with exponent at least 3 contribute $0.262002530\ldots\cdot\xi^{-2/3}$, which is $1.0154801\ldots\cdot\frac{M_\xi}{\lambda}$ at $\lambda=31$. Twelve of the checks in Part 2 are identities or inequalities between the decimal constants of the proofs, such as $0.066+0.370148+2=2.436148$ and $0.313233-0.066+1.9314-0.258259=1.920374$; they are verified in exact rational arithmetic. The smallest positive margins in Part 2 are those of $\frac{1}{31^2}+\frac{1}{31^3}=0.000553858\ldots<0.000554$, of $2\left(\frac{\log2}{2}-0.011306\right)=0.6705351805\ldots\ge0.670535$, and of $2\left(1-\frac{0.370148+18}{31^2}\right)=1.9314003\ldots>1.9314$, with differences $1.4\cdot10^{-7}$, $1.8\cdot10^{-7}$ and $3.1\cdot10^{-7}$. The final constants of Theorem 1.1 are $0.9574\cdot\sqrt{2}=1.3539680\ldots>1.353$ and $2.8070\cdot\sqrt{2}=3.9696974\ldots<3.970$.

Quantity	Computed value	Bound used
$1-\frac{\log2}{62}$	0.9888202067 . . .	$>0.98882$
$1.94\cdot31\cdot e^{-31/4}/1.18253$	0.0219062995 . . .	$<0.022$
$-\log\left(1-\frac{\log2}{62}\right)$	0.0112427568 . . .	$<0.011306$
$0.670535-\frac{\log2}{2}-\frac{\log2}{62}\cdot\frac{0.670535}{31}$	0.3164649670 . . .	$>0.316464$
$0.316464-\frac{0.316464^2}{31}$	0.3132333721 . . .	$>0.313233$
$2\left(1-\frac{0.370148+\log2}{31}\right)$	1.9314003109 . . .	$>1.9314$
$1.0022\cdot31\cdot e^{-31/2}/\sqrt{2}$	$4.0760229901\ldots\cdot10^{-6}$	$<4.08\cdot10^{-6}$
$1.397905\cdot31\cdot e^{-31/6}$	0.2471636095 . . .	$<0.24717$
$\frac{1}{31^2}+\sqrt{2}\cdot31\cdot e^{-31/2}$	0.0010487169 . . .	$<0.0010488$
$0.0861+0.17587+e^{-31/3}$	0.2620025305 . . .	$<0.26201$
$0.26201\cdot31^2\cdot e^{-31/6}/\sqrt{2}$	1.0154801390 . . .	$<1.0155$
$\frac{1}{2\cdot31^2}+\frac{1}{31^3}$	0.0005538585 . . .	$<0.000554$
$0.957494\cdot(1-31^{-4})$	0.9574929632 . . .	$>0.9574$
$0.9574\cdot\sqrt{2}$	1.3539680646 . . .	$>1.353$
$2.8070\cdot\sqrt{2}$	3.9696974696 . . .	$<3.970$

Part 3: The Corollaries

Part 3 checks the constants of Proposition 2.8 and Corollaries 4.2, 4.6 and 4.7. The script computes $\tau=2+\gamma-\log(4\pi)=0.0461914179\ldots$, so that $2+\tau=2.0461914179\ldots<2.0462$, and $2\sqrt{2}-2-\tau=0.7822357068\ldots$, the constant discussed after Corollary 4.6. The constant of Corollary 4.2 is $2\sqrt{2}\left(1-\frac{1.4035}{31.0319}\right)=2.7005040047\ldots>2.70$, and the constant of Corollary 4.6 is $2+\tau-2\sqrt{2}+\frac{2\sqrt{2}\cdot1.4035}{31}=-0.654180497\ldots<-0.654$. The constant 2.70 of Corollary 4.7 exceeds the bound 2.0462 of Proposition 2.8 by 0.6538. Finally, $e^{31.0319}=2.999\ldots\cdot10^{13}>10^9$, so the primes $q$ of Corollary 4.2 lie in the range of Proposition 2.8.

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_nicolas.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_nicolas.py used in § 6 is available in the preprint.