Robin’s Criterion on Divisibility (II)

1. Introduction

The Riemann hypothesis (RH), concerning the nontrivial zeros of the Riemann zeta function, has been dubbed the “Holy Grail of Mathematics” by many [1,2]. Numerous equivalent formulations exist [3]. One of interest here is Robin’s criterion [4], which states that if RH is true, then the inequality

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

holds for all $n>5040$. Here, $\gamma \approx 0.57721$ is the Euler-Mascheroni constant, $\sigma$ is the divisor sum function, log is the natural logarithm, and n is a natural number. The superabundant and colossally abundant numbers were discovered by Ramanujan [5] and studied later by Alaoglu and Erdős [6]. The falsity of the Riemann hypothesis would imply the existence of infinitely many counterexamples to Robin’s inequality, including superabundant numbers exceeding 5040. Therefore, demonstrating the non-existence of these counterexamples would be enough to verify the Riemann hypothesis. Indeed, this is precisely what we demonstrate in this manuscript. Without a doubt, we provide a solution to this problem grounded in the properties of superabundant numbers and Robin’s criterion.

2. Background and Ancillary Results

In mathematics, the Euler-Mascheroni constant, denoted by $\gamma \approx 0.57721$, is defined as

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

where log represents the natural logarithm and $H_n={\sum }_{k=1}^n{\displaystyle \frac{1}{k}}$ is the $n^{th}$ harmonic number. As usual $\sigma \left(n\right)$ is the divisor sum function of n:

$$\sum _{d\mid n}d$$

where $d\mid n$ means the integer d divides n. We define the abundancy index $I:\mathbb{N}\to \mathbb{Q}$ with $I\left(n\right)={\displaystyle \frac{\sigma \left(n\right)}{n}}$. A precise formula for $I\left(n\right)$ can be derived from its multiplicity:

Proposition 2.1. 

Let $n={\prod }_{i=1}^r{p}_i^{a_i}$ be the prime factorization of n, where $p_1<\dots <p_r$ are distinct prime numbers and $a_1,\dots ,a_r$ are positive integers. Then [7, Lemma 1 (2) pp. 2]:

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

Proposition 2.2. 

For $n>1$ [8, (2.7) pp. 362]:

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

Definition 2.3. 

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

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

The Chebyshev function $\theta \left(x\right)$ is defined as

$$\theta \left(x\right)=\sum _{p\le x}logp$$

where the sum is taken over all prime numbers p less than or equal to x. The following inequality holds for the Chebyshev function:

Proposition 2.4. 

For $x\ge 7232121212$ [9, Lemma 2.7 (4) pp. 19]:

$$\theta \left(x\right)\ge \left(1-{\displaystyle \frac{0.01}{{log}^3\left(x\right)}}\right)·x.$$

Proposition 2.5. 

For $x\ge 2278382$ [9, 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 the Riemann hypothesis implies $\mathsf{Robin}\left(n\right)$ for sufficiently large n [5]. Moreover, Robin’s Theorem establishes a stronger result:

Proposition 2.6. 

$\mathsf{Robin}\left(n\right)$ 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 from 1997 presented generalized highly composite numbers, which encompass both superabundant and colossally abundant numbers [5]. These numbers were also explored by Alaoglu and Erdős in 1944 [6]. 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 termed an Hardy-Ramanujan integer [8, pp. 367]. A natural number n is classified as superabundant if, for all natural numbers $m<n$, the inequality $I\left(m\right)<I\left(n\right)$ holds.

Proposition 2.7. 

If n is superabundant, then n is an Hardy-Ramanujan integer [6, Theorem 1 pp. 450].

Proposition 2.8. 

Let n be a superabundant number. Denoting its largest prime factor by p, we have [6, Theorem 7 pp. 454]:

$$p\sim logn,(n\to \infty ).$$

A number n is said to be colossally abundant if, for some $ϵ>0$,

$${\displaystyle \frac{\sigma \left(n\right)}{n^{1+ϵ}}}\ge {\displaystyle \frac{\sigma \left(m\right)}{m^{1+ϵ}}}\mathit{for}(m>1).$$

Superabundant and colossally abundant numbers are closely related.

Proposition 2.9. 

Every colossally abundant number is superabundant [6, pp. 455].

Several analogues of the Riemann hypothesis have been proven. Many researchers anticipate or hope that the original hypothesis is true. However, there are potential consequences if the Riemann hypothesis were false.

Proposition 2.10. 

If $n>5040$ is the smallest integer such that $\mathsf{Robin}\left(n\right)$ does not hold, then $p>e^{31.018189471}$ where p is the largest prime factor of n [10, Theorem 4.2 pp. 748].

Proposition 2.11. 

If the Riemann hypothesis is false, then there exist infinitely many colossally abundant numbers $n>5040$ such that $\mathsf{Robin}\left(n\right)$ does not hold [4, Proposition 1 pp. 204].

By combining these results, we present a proof of the Riemann hypothesis.

3. Main Result

Definition 3.1. 

The $\mathit{p}-\mathit{adic}$ order of an integer n is the exponent of the highest power of the prime number p that divides n. It is denoted ${\nu }_p\left(n\right)$.

These are significant findings.

Lemma 3.2. 

For any superabundant number n with largest prime factor p, we have $p\le q^{{\nu }_q\left(n\right)+1}$ for all primes q satisfying $2\le q\le p$.

Proof. 

Suppose $p>q^{{\nu }_q\left(n\right)+1}$. We evaluate $I\left(s\right)$ for $s=n$ and $s=m={\displaystyle \frac{n·q^{{\nu }_q\left(n\right)+1}}{p}}$. Given the multiplicative property of $I\left(s\right)$, we focus on distinct factors. Moreover, n is superabundant and $m<n$. By Proposition 2.1, we obtain:

$$\begin{array}{cc}\hfill 1& <{\displaystyle \frac{I\left(n\right)}{I\left(m\right)}}\hfill \\ \hfill & =\left({\displaystyle \frac{q^{2·{\nu }_q\left(n\right)+2}-q^{{\nu }_q\left(n\right)+1}}{q^{2·{\nu }_q\left(n\right)+2}-1}}\right)·\left(1+{\displaystyle \frac{1}{p}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\left(1+\frac{1}{q^{{\nu }_q\left(n\right)+1}}\right)}}·\left(1+{\displaystyle \frac{1}{p}}\right).\hfill \end{array}$$

Therefore, ${\displaystyle \frac{1}{q^{{\nu }_q\left(n\right)+1}}}<{\displaystyle \frac{1}{p}}$, implying $p<q^{{\nu }_q\left(n\right)+1}$, contradicting our initial assumption. Clearly, we can observe that

$$\begin{array}{cc}\hfill \left({\displaystyle \frac{q^{2·{\nu }_q\left(n\right)+2}-q^{{\nu }_q\left(n\right)+1}}{q^{2·{\nu }_q\left(n\right)+2}-1}}\right)& =\left(1-{\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}-1}{q^{2·{\nu }_q\left(n\right)+2}-1}}\right)\hfill \\ \hfill & =\left(1-{\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}-1}{(q^{{\nu }_q\left(n\right)+1}-1)·(q^{{\nu }_q\left(n\right)+1}+1)}}\right)\hfill \\ \hfill & =\left(1-{\displaystyle \frac{1}{(q^{{\nu }_q\left(n\right)+1}+1)}}\right)\hfill \\ \hfill & =\left({\displaystyle \frac{q^{{\nu }_q\left(n\right)+1}}{(q^{{\nu }_q\left(n\right)+1}+1)}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\left(1+\frac{1}{q^{{\nu }_q\left(n\right)+1}}\right)}}.\hfill \end{array}$$

□

Lemma 3.3. 

If n is a superabundant number and p is a fixed prime, then ${\nu }_p\left(n\right)$ increases without bound as n grows larger.

Proof. 

By joining Proposition 2.8 and Lemma 3.2, we obtain the present result. □

The following simple observations will be used as ancillary results.

Lemma 3.4. 

If the Riemann hypothesis is false, then there exist infinitely many superabundant numbers n such that $\mathsf{Robin}\left(n\right)$ does not hold.

Proof. 

This follows directly from Propositions 2.6, 2.9, and 2.11. □

Definition 3.5. 

For each prime number $p_k>2$, we define the sequence

$$Y_k={\displaystyle \frac{e^{\frac{0.2}{{log}^2\left(p_k\right)}}}{(1-\frac{0.01}{{log}^3\left(p_k\right)})}}.$$

Lemma 3.6. 

Given that $Y_k$ is strictly decreasing, we can verify that $Y_k<1.00021$ whenever $p_k>e^{31.018189471}$.

Proof. 

This Lemma is straightforward. □

The following is a crucial Lemma.

Lemma 3.7. 

Consider the first k consecutive primes $p_1,p_2,\dots ,p_k$ such that $p_1<p_2<\dots <p_k$ and $p_k>7232121212$. It follows that

$$\prod _{i=1}^k{\displaystyle \frac{p_i}{p_i-1}}\le e^{\gamma }·log\left(Y_k·\theta \left(p_k\right)\right).$$

Proof. 

Proposition 2.4 implies that

$$\theta \left(p_k\right)\ge \left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)·p_k.$$

Therefore, it is possible to prove that

$$\begin{array}{cc}\hfill log\left(Y_k·\theta \left(p_k\right)\right)& \ge log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)·p_k\right)\hfill \\ \hfill & =log{p}_k+log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right).\hfill \end{array}$$

Thus, we can demonstrate that

$$\begin{array}{cc}\hfill log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right)& =log\left({\displaystyle \frac{e^{\frac{0.2}{{log}^2\left(p_k\right)}}}{\left(1-\frac{0.01}{{log}^3\left(p_k\right)}\right)}}·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right)\hfill \\ \hfill & =log\left(e^{{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}.\hfill \end{array}$$

Consequently, we can establish that

$$log{p}_k+log\left(Y_k·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(p_k\right)}}\right)\right)\ge \left(log{p}_k+{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}\right).$$

According to Proposition 2.5, we can show that

$$\prod _{i=1}^k{\displaystyle \frac{p_i}{p_i-1}}\le e^{\gamma }·\left(log{p}_k+{\displaystyle \frac{0.2}{{log}^2\left(p_k\right)}}\right)\le e^{\gamma }·log\left(Y_k·\theta \left(p_k\right)\right)$$

when $p_k>7232121212$. □

This is a main insight.

Lemma 3.8. 

For every possible superabundant number $n>5040$ violating $\mathsf{Robin}\left(n\right)$, we have

$${\left(N_k\right)}^{Y_k}>n$$

where $p_k$ is the largest prime factor of n and $N_k={\prod }_{s=1}^k{p}_s$ is the primorial number of order k.

Proof. 

Suppose that $\mathsf{Robin}\left(n\right)$ does not hold. Let $n={\prod }_{i=1}^k{p}_i^{a_i}$ be an Hardy-Ramanujan integer, where the primes $p_1<\dots <p_k$ are the first k consecutive primes, $p_k>e^{31.018189471}$ and $a_1\ge a_2\ge \dots \ge a_k\ge 1$: This follows directly from Propositions 2.7 and 2.10. Given our assumption, we can conclude that

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

(1)

Additionally, we can infer that

$$\prod _{p\le p_k}{\displaystyle \frac{p}{p-1}}\le e^{\gamma }·loglog\left({\left(N_k\right)}^{Y_k}\right)$$

(2)

for all $p_k>e^{31.018189471}$, as per Lemma 3.7 with $log\left({\left(N_k\right)}^{Y_k}\right)=Y_k·\theta \left(p_k\right)$. By combining Proposition 2.2 with the inequalities (1) and (2), we obtain the following result:

$$\begin{array}{cc}\hfill e^{\gamma }·loglog\left({\left(N_k\right)}^{Y_k}\right)& \ge \prod _{p\le p_k}{\displaystyle \frac{p}{p-1}}\hfill \\ \hfill & >I\left(n\right)\hfill \\ \hfill & \ge e^{\gamma }·loglogn.\hfill \end{array}$$

This leads to ${\left(N_k\right)}^{Y_k}>n$ by transitivity. □

This is the main Theorem.

Theorem 3.9. 

The Riemann hypothesis is true.

Proof. 

We use a proof by contradiction, assuming that the Riemann hypothesis is not true. Given Lemma 3.4, there are infinitely many superabundant numbers n that do not satisfy $\mathsf{Robin}\left(n\right)$. Let $n_{k_i}$ be the infinite sequence of superabundant numbers where $\mathsf{Robin}\left(n_{k_i}\right)$ is false and $p_{k_i}$ is the largest prime factor of $n_{k_i}$. Proposition 2.10 additionally implies that $p_{k_i}>e^{31.018189471}$ for all i. Equivalently, Lemma 3.6 implies that $Y_{k_i}<1.00021$. By Lemma 3.3, we conclude that there exists a finite, strictly increasing subsequence $n_{k_j^{\prime }}$ of m superabundant numbers, taken from the general sequence $n_{k_i}$, such that

$${\displaystyle \frac{{\sum }_{j=1}^{m}log{n}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\ge 1.00021,$$

where $N_{k_j^{\prime }}={\prod }_{s=1}^{k_j^{\prime }}{p}_s$ is the primorial number of order $k_j^{\prime }$. Lemma 3.3 guarantees the existence of this subsequence of superabundant numbers $n_{k_j^{\prime }}$ that can be made arbitrarily long, because for every fixed prime p, ${\nu }_p\left(n_{k_j^{\prime }}\right)$ approaches infinity as $n_{k_j^{\prime }}$ grows larger. To illustrate, consider a finite, strictly increasing subsequence $n_{k_j^{\prime }}$ of m superabundant numbers, satisfying the condition that

$${\displaystyle \frac{n_{k_j^{\prime }}}{N_{k_j^{\prime }}}}>N_{k_{j-1}^{\prime }}$$

(3)

for $1<j\le m$ and

$${\displaystyle \frac{log\left(\frac{N_{k_m^{\prime }}·N_{k_1^{\prime }}}{n_{k_1^{\prime }}}\right)}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\le 2-1.00022,$$

(4)

where $n_{k_1^{\prime }}$ and $n_{k_m^{\prime }}$ are the minimum and maximum elements of the subsequence, respectively. Utilizing inequalities (3) and (4), we deduce that:

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\sum }_{j=1}^{m}log{n}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}& ={\displaystyle \frac{\left({\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}\right)+{\sum }_{j=1}^{m}log\left(\frac{n_{k_j^{\prime }}}{N_{k_j^{\prime }}}\right)}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & >{\displaystyle \frac{log\left(\frac{n_{k_1^{\prime }}}{N_{k_1^{\prime }}}\right)+\left({\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}\right)+{\sum }_{j=1}^{m-1}log{N}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & ={\displaystyle \frac{log\left(\frac{n_{k_1^{\prime }}}{N_{k_1^{\prime }}}\right)+\left(log{N}_{k_m^{\prime }}\right)+2·{\sum }_{j=1}^{m-1}log{N}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & ={\displaystyle \frac{2·{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}-\left(\left(log{N}_{k_m^{\prime }}\right)-log\left(\frac{n_{k_1^{\prime }}}{N_{k_1^{\prime }}}\right)\right)}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & ={\displaystyle \frac{2·{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}-{\displaystyle \frac{log\left(\frac{N_{k_m^{\prime }}·N_{k_1^{\prime }}}{n_{k_1^{\prime }}}\right)}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & =2-{\displaystyle \frac{log\left(\frac{N_{k_m^{\prime }}·N_{k_1^{\prime }}}{n_{k_1^{\prime }}}\right)}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}\hfill \\ \hfill & \ge 2-(2-1.00022)\hfill \\ \hfill & =1.00022.\hfill \end{array}$$

This yields the trivial inequality $1.00022>1.00021$. While this is just one example, many such cases may exist. Therefore, by Lemmas 3.6 and 3.8, it suffices to show that the inequality

$${\displaystyle \frac{{\sum }_{j=1}^{m}log{n}_{k_j^{\prime }}}{{\sum }_{j=1}^{m}log{N}_{k_j^{\prime }}}}>Y_{k_1^{\prime }}$$

leads to a contradiction, because $1.00021$ is always greater than $Y_{k_1^{\prime }}$ when $n_{k_1^{\prime }}$ is the smallest element in the subsequence. Definitely, that’s the same as saying

$$\sum _{j=1}^{m}log{n}_{k_j^{\prime }}>Y_{k_1^{\prime }}·\sum _{j=1}^{m}log{N}_{k_j^{\prime }}$$

and it is clear that

$$\begin{array}{cc}\hfill Y_{k_1^{\prime }}·\sum _{j=1}^{m}log{N}_{k_j^{\prime }}& =\sum _{j=1}^{m}log{\left(N_{k_j^{\prime }}\right)}^{Y_{k_1^{\prime }}}\hfill \\ \hfill & >\sum _{j=1}^{m}log{\left(N_{k_j^{\prime }}\right)}^{Y_{k_j^{\prime }}}\hfill \\ \hfill & >\sum _{j=1}^{m}log{n}_{k_j^{\prime }}\hfill \end{array}$$

due to the fact that $Y_{k_j^{\prime }}$ decreases as the subsequence $n_{k_j^{\prime }}$ grows, and ${\left(N_{k_j^{\prime }}\right)}^{Y_{k_j^{\prime }}}>n_{k_j^{\prime }}$ according to Lemmas 3.6 and 3.8. Hence, our initial assumption has been contradicted. This proof by contradiction, combined with Proposition 2.6, establishes the truth of the Riemann hypothesis. □