The First Hypothetical Counterexample of Robin's Criterion

1. Introduction

The Riemann Hypothesis (RH), concerning the nontrivial zeros of the Riemann zeta function, has been called the “Holy Grail of Mathematics” [1,2]. Numerous equivalent formulations exist [3]; one of particular interest here is Robin’s criterion [4], which asserts that RH is equivalent to the inequality

$$\sigma \left(n\right)<e^{\gamma }·n·loglogn\mathrm{for}\mathrm{every}\mathrm{integer}n>5040.$$

(1)

Here, $\gamma \approx 0.57721$ is the Euler–Mascheroni constant, $\sigma$ is the divisor sum function, log denotes the natural logarithm, and n is a natural number. Grönwall’s theorem [5] states that

$$\underset{n\to \infty }{lim\; sup}\sigma \left(n\right)/(nloglogn)=e^{\gamma },$$

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 [6], and the extremal integers relevant to both criteria have been studied computationally by Briggs [7] and structurally by Caveney, Nicolas and Sondow [8]; the distribution of superabundant numbers itself goes back to Erdos and Nicolas [9], building on the foundational work of Ramanujan [10] and of Alaoglu and Erdos [11]. A great deal of effort has gone into establishing (1) unconditionally for structured families of integers: odd integers exceeding 9 and squarefree integers exceeding 30 [12], sums of two squares exceeding 720 [13], and t-free integers for $t=5,7,11$ and 20 [12,14,15,16]. In a complementary direction, Akbary and Friggstad [17] showed that the least counterexample to (1), should one exist, must be a superabundant number, and Nazardonyavi and Yakubovich [18] refined this to the smaller class of extremely abundant numbers; further properties of a hypothetical least counterexample, together with improved explicit bounds, are given by Vojak [19]. The author has also contribute to this long list of results concerning to the Robin’s criterion [20].

Among the integers, the superabundant and colossally abundant numbers, introduced by Ramanujan [10] and studied by Alaoglu and Erdos [11], play a central role in the analysis of Robin’s inequality. By a theorem of Akbary and Friggstad [17], if RH is false then the least natural number $n>5040$ for which the Robin’s inequality fails must be superabundant. The argument proceeds by combining Robin’s criterion for superabundant numbers according to the least counterexample, ultimately reducing the problem to a simple proof by contradiction.

2. Background and Ancillary Results

The Euler–Mascheroni constant, denoted $\gamma \approx 0.57721$, is defined as

$$\gamma =\underset{n\to \infty }{lim}\left(H_n-logn\right),$$

where log denotes the natural logarithm and $H_n={\sum }_{k=1}^n{\displaystyle \frac{1}{k}}$ is the $n^{\mathrm{th}}$ harmonic number. As usual, $\sigma \left(n\right)$ denotes the sum of all positive divisors of n:

$$\sigma \left(n\right)=\sum _{d\mid n}d,$$

where $d\mid n$ means that the integer d divides n. We define the abundancy index$I:\mathbb{N}\to \mathbb{Q}$ by $I\left(n\right)={\displaystyle \frac{\sigma \left(n\right)}{n}}$. Since $\sigma$ is multiplicative, $I\left(n\right)$ admits a product representation:

Proposition 1.  

Let $n={\prod }_{i=1}^r{p}_i^{a_i}$ be the prime factorization of n, where $p_1<\dots <p_r$ are distinct primes and $a_1,\dots ,a_r$ are positive integers. Then [21] [Lemma 1 (2) pp. 2]:

$$I\left(n\right)=\left({\displaystyle \prod _{i=1}^r}{\displaystyle \frac{p_i}{p_i-1}}\right)·{\displaystyle \prod _{i=1}^r}\left(1-{\displaystyle \frac{1}{p_i^{a_i+1}}}\right).$$

Proposition 2.  

For $n>1$ [12] [(2.7) pp. 362]:

$$I\left(n\right)<\prod _{p\mid n}{\displaystyle \frac{p}{p-1}}.$$

We introduce the following shorthand for Robin’s inequality:

Definition 1.  

We say that $\mathsf{Robin}\left(n\right)$ holds if

$$I\left(n\right)<e^{\gamma }·loglogn.$$

The following proposition provides the quantitative bounds on the Mertens product ${\prod }_{p\le x}{\displaystyle \frac{p}{p-1}}$ that we will use:

Proposition 3.  

For $x\ge 2,278,382$ [22] [Lemma 2.7 (5) pp. 19]:

$$\prod _{p\le x}{\displaystyle \frac{p}{p-1}}\le e^{\gamma }·(logx)·\left(1+{\displaystyle \frac{0.2}{{log}^3\left(x\right)}}\right).$$

Ramanujan’s theorem asserts that RH implies $\mathsf{Robin}\left(n\right)$ for all sufficiently large n [10]. Robin’s theorem sharpens this to a precise equivalence:

Proposition 4.  

$\mathsf{Robin}\left(n\right)$ holds for all natural numbers $n>5040$ if and only if the Riemann Hypothesis is true [4] [Theorem 1 pp. 188].

Ramanujan’s unpublished notes [10] introduced generalized highly composite numbers, encompassing both superabundant and colossally abundant numbers; these were also studied by Alaoglu and Erdos [11]. Given the first k consecutive primes $p_1=2,p_2=3,\dots ,p_k$, an integer of the form ${\prod }_{i=1}^k{p}_i^{a_i}$ with $a_1\ge a_2\ge \dots \ge a_k\ge 1$ is called a Hardy–Ramanujan integer [12] [pp. 367]. A natural number n is called superabundant if $I\left(m\right)<I\left(n\right)$ for every natural number $m<n$.

Proposition 5.  

Every superabundant number is a Hardy–Ramanujan integer [11] [Theorem 1 pp. 450].

The next three propositions describe the structure of hypothetical counterexamples:

Proposition 6.  

If $n>5040$ is the smallest integer such that $\mathsf{Robin}\left(n\right)$ does not hold, then n is superabundant [17] [Theorem 3, p. 273].

Proposition 7.  

If $n>5040$ is the smallest integer such that $\mathsf{Robin}\left(n\right)$ does not hold, and p denotes the largest prime factor of n, then $p<logn$ [12] [Lemma 6.1, p. 369].

Proposition 8.  

If $n>5040$ is the smallest integer for which $\mathsf{Robin}\left(n\right)$ fails, then the largest prime factor p of n satisfies $p>e^{31.018189471}$ [20] [Theorem 4.2 pp. 748].

Putting all together yields a proof of the Riemann Hypothesis.

3. Main Results

This is the main insight.

Theorem 1.  

Let $n>5040$ be a superabundant number for which $\mathsf{Robin}\left(n\right)$ fails, and let $p_k$ be the largest prime factor of n. Assume $p_k>2,278,382$. Then

$$1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}>{\displaystyle \frac{loglogn}{log{p}_k}}.$$

Proof. 

By Proposition 2 and 3, we obtain

$$e^{\gamma }·(log{p}_k)·\left(1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}\right)\ge \prod _{p\le p_k}{\displaystyle \frac{p}{p-1}}>I\left(n\right)\ge e^{\gamma }loglogn.$$

Hence

$$e^{\gamma }·(log{p}_k)·\left(1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}\right)>e^{\gamma }loglogn$$

implies

$$1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}>{\displaystyle \frac{loglogn}{log{p}_k}}$$

and therefore, the proof is done. □

This is the main theorem.

Theorem 2.  

The Riemann Hypothesis holds.

Proof. 

Suppose that the Riemann Hypothesis is false. Let $n>5040$ be the least counterexample of the Robin’s criterion. By Proposition 6, the number n must be superabundant. Let $p_k$ be the largest prime factor of n. By Proposition 8 and Theorem 1, we would have

$$1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}>{\displaystyle \frac{loglogn}{log{p}_k}},$$

since $p_k>e^{31.018189471}>2,278,382$. That is the same as

$${\left(1+{\displaystyle \frac{0.2}{{log}^3\left(p_k\right)}}\right)}^{{log}^3\left(p_k\right)}>{\left({\displaystyle \frac{loglogn}{log{p}_k}}\right)}^{{log}^3\left(p_k\right)}$$

where we can rewrite the expression in terms of the ratio $t={\displaystyle \frac{x}{y}}$ for $x=0.2$ and $y={log}^3\left(p_k\right)$:

$${\left(1+{\displaystyle \frac{x}{y}}\right)}^y={\left[{\left(1+t\right)}^{{\displaystyle \frac{1}{t}}}\right]}^x.$$

Define the auxiliary function $h\left(t\right)={(1+t)}^{1/t}$ for $t>0$. Taking the derivative of $logh\left(t\right)$ shows that $h\left(t\right)$ is strictly decreasing on $(0,\infty )$:

• ${lim}_{t\to 0^{+}}h\left(t\right)=e$
• $h\left(1\right)=2$
• ${lim}_{t\to \infty }h\left(t\right)=1$.

So, we have

$$e\ge {\left(1+t\right)}^{{\displaystyle \frac{1}{t}}}$$

for $p_k>e^{31.018189471}$. Consequently,

$$e^{0.2}>{\left({\displaystyle \frac{loglogn}{log{p}_k}}\right)}^{{log}^3\left(p_k\right)}.$$

Using a variant of Bernoulli’s inequality, we get

$${\left({\displaystyle \frac{loglogn}{log{p}_k}}\right)}^{{log}^3\left(p_k\right)}={\left(1+{\displaystyle \frac{loglogn}{log{p}_k}}-1\right)}^{{log}^3\left(p_k\right)}\ge {\left(1+loglogn-log{p}_k\right)}^{{\displaystyle \frac{{log}^3\left(p_k\right)}{log\left(p_k\right)}}}={\left(1+loglogn-log{p}_k\right)}^{{log}^2\left(p_k\right)}$$

since ${(1+x)}^r\le 1+rx$ for $x>0$ and $0<r<1$. Certainly, we have $loglogn-log{p}_k>0$ by Proposition 7. By the original Bernoulli’s inequality, we arrive at

$${\left(1+loglogn-log{p}_k\right)}^{{log}^2\left(p_k\right)}\ge 1+(loglogn-log{p}_k)·{log}^2\left(p_k\right)$$

since ${(1+x)}^r\ge 1+rx$ for $x>0$ and $r>1$. We have

$$loglogn-log{p}_k=log{\displaystyle \frac{logn}{p_k}}=loglog{n}^{1/p_k}=log(1+log{n}^{1/p_k}-1)\ge {\displaystyle \frac{log{n}^{1/p_k}-1}{log{n}^{1/p_k}}}=1-{\displaystyle \frac{1}{log{n}^{1/p_k}}}$$

since $log(1+x)\ge x/(x+1)$ for $x>-1$. By Proposition 7, we can further deduce that

$$log{n}^{1/p_k}>log{n}^{1/log\left(n\right)}=e$$

and thus

$$1-{\displaystyle \frac{1}{log{n}^{1/p_k}}}>1-{\displaystyle \frac{1}{e}}.$$

Therefore,

$$e^{0.2}>1+\left(1-{\displaystyle \frac{1}{e}}\right)·{log}^2\left(p_k\right).$$

Hence, it is enough to show that

$$e^{0.2}<1.2215$$

and

$$1+\left(1-{\displaystyle \frac{1}{e}}\right)·{log}^2\left(p_k\right).>1+\left(1-{\displaystyle \frac{1}{e}}\right)·31.{018189471}^2>608$$

for $p_k>e^{31.018189471}$ where

$$1.2215>608$$

is trivially false. Since the negation of the Riemann Hypothesis would imply the existence of a trivial contradiction, then the Riemann Hypothesis must be true. □