Robin’s Criterion on Divisibility (II)

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Robin’s Criterion on Divisibility (II)

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Frank Vega^*

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23 October 2024

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25 October 2024

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Abstract

The Riemann hypothesis, renowned for its deep connection to the distribution of prime numbers, remains a central problem in mathematics. Understanding the distribution of primes is crucial for developing efficient algorithms and advancing our knowledge of number theory. The Riemann hypothesis is the assertion that all non-trivial zeros are complex numbers with real part $\frac{1}{2}$. It is considered by many to be the most important unsolved problem in pure mathematics. Several equivalent formulations of the Riemann hypothesis exist. Robin's criterion for the Riemann hypothesis is based on an inequality that divisor sum function $\sigma$ must satisfy at natural numbers greater than 5040. We require the properties of superabundant numbers, that is to say left to right maxima of $n \mapsto \frac{\sigma(n)}{n}$. By using Robin's criterion on superabundant numbers, we present a novel approach that culminates in a complete proof of the Riemann hypothesis. This work is an expansion and refinement of the article "Robin's criterion on divisibility", published in The Ramanujan Journal.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

MSC: Primary 11M26; Secondary 11A41, 11A25

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

σ (n) < e^{γ} \cdot n \cdot log log n

holds for all

n > 5040

. Here,

γ \approx 0.57721

is the Euler-Mascheroni constant,

σ

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 Erdos [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

γ \approx 0.57721

, is defined as

γ = lim_{n \to \infty} (H_{n} - log n),

where log represents the natural logarithm and

H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

is the

n^{t h}

harmonic number. As usual

σ (n)

is the divisor sum function of n:

\sum_{d ∣ n} d

where

d ∣ n

means the integer d divides n. We define the abundancy index

I : N \to Q

with

I (n) = \frac{σ (n)}{n}

. A precise formula for

I (n)

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{matrix} I (n) & = (\prod_{i = 1}^{r} \frac{p_{i}}{p_{i} - 1}) \cdot \prod_{i = 1}^{r} (1 - \frac{1}{p_{i}^{a_{i} + 1}}) \\ = \prod_{i = 1}^{r} I (p_{i}^{a_{i}}) . \end{matrix}

Proposition 2.2.

For

n > 1

[8, (2.7) pp. 362]:

I (n) < \prod_{p ∣ n} \frac{p}{p - 1} .

Definition 2.3.

We say that

Robin (n)

holds provided that

I (n) < e^{γ} \cdot log log n .

The Chebyshev function

θ (x)

is defined as

θ (x) = \sum_{p \leq x} log p

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 \geq 7232121212

[9, Lemma 2.7 (4) pp. 19]:

θ (x) \geq (1 - \frac{0.01}{{log}^{3} (x)}) \cdot x .

Proposition 2.5.

For

x \geq 2278382

[9, Lemma 2.7 (5) pp. 19]:

\prod_{p \leq x} \frac{p}{p - 1} \leq e^{γ} \cdot (log x) \cdot (1 + \frac{0.2}{{log}^{3} (x)}) .

Ramanujan’s Theorem asserts that the Riemann hypothesis implies

Robin (n)

for sufficiently large n [5]. Moreover, Robin’s Theorem establishes a stronger result:

Proposition 2.6.

Robin (n)

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 Erdos 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} \geq a_{2} \geq \dots \geq a_{k} \geq 0

is termed an Hardy-Ramanujan integer [8]. A natural number n is classified as superabundant if, for all natural numbers

m < n

, the inequality

I (m) < I (n)

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 log n, (n \to \infty) .

A number n is said to be colossally abundant if, for some

ϵ > 0

\frac{σ (n)}{n^{1 + ϵ}} \geq \frac{σ (m)}{m^{1 + ϵ}} 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.

n > 5040

is the smallest integer such that

Robin (n)

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

Robin (n)

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

p - adic

order of an integer n is the exponent of the highest power of the prime number p that divides n. It is denoted

ν_{p} (n)

These are significant findings.

Lemma 3.2.

For any superabundant number n with largest prime factor p, we have

p \leq q^{ν_{q} (n) + 1}

for all primes q satisfying

2 \leq q \leq p

Proof.

Suppose

p > q^{ν_{q} (n) + 1}

. We evaluate

I (s)

for

s = n

and

s = m = \frac{n \cdot q^{ν_{q} (n) + 1}}{p}

. Given the multiplicative property of

I (s)

, we focus on distinct factors. Moreover, n is superabundant and

m < n

. By Proposition 2.1, we obtain:

\begin{matrix} 1 & < \frac{I (n)}{I (m)} \\ = (\frac{q^{2 \cdot ν_{q} (n) + 2} - q^{ν_{q} (n) + 1}}{q^{2 \cdot ν_{q} (n) + 2} - 1}) \cdot (1 + \frac{1}{p}) \\ = \frac{1}{(1 + \frac{1}{q^{ν_{q} (n) + 1}})} \cdot (1 + \frac{1}{p}) . \end{matrix}

Therefore,

\frac{1}{q^{ν_{q} (n) + 1}} < \frac{1}{p}

, implying

p < q^{ν_{q} (n) + 1}

, contradicting our initial assumption. Clearly, we can observe that

\begin{matrix} (\frac{q^{2 \cdot ν_{q} (n) + 2} - q^{ν_{q} (n) + 1}}{q^{2 \cdot ν_{q} (n) + 2} - 1}) & = (1 - \frac{q^{ν_{q} (n) + 1} - 1}{q^{2 \cdot ν_{q} (n) + 2} - 1}) \\ = (1 - \frac{q^{ν_{q} (n) + 1} - 1}{(q^{ν_{q} (n) + 1} - 1) \cdot (q^{ν_{q} (n) + 1} + 1)}) \\ = (1 - \frac{1}{(q^{ν_{q} (n) + 1} + 1)}) \\ = (\frac{q^{ν_{q} (n) + 1}}{(q^{ν_{q} (n) + 1} + 1)}) \\ = \frac{1}{(1 + \frac{1}{q^{ν_{q} (n) + 1}})} . \end{matrix}

□

Lemma 3.3.

If n is a superabundant number and p is a fixed prime, then

ν_{p} (n)

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

Robin (n)

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} = \frac{e^{\frac{0.2}{{log}^{2} (p_{k})}}}{(1 - \frac{0.01}{{log}^{3} (p_{k})})} .

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} \frac{p_{i}}{p_{i} - 1} \leq e^{γ} \cdot log (Y_{k} \cdot θ (p_{k})) .

Proof.

Proposition 2.4 implies that

θ (p_{k}) \geq (1 - \frac{0.01}{{log}^{3} (p_{k})}) \cdot p_{k} .

Therefore, it is possible to prove that

\begin{matrix} log (Y_{k} \cdot θ (p_{k})) & \geq log (Y_{k} \cdot (1 - \frac{0.01}{{log}^{3} (p_{k})}) \cdot p_{k}) \\ = log p_{k} + log (Y_{k} \cdot (1 - \frac{0.01}{{log}^{3} (p_{k})})) . \end{matrix}

Thus, we can demonstrate that

\begin{matrix} log (Y_{k} \cdot (1 - \frac{0.01}{{log}^{3} (p_{k})})) & = log (\frac{e^{\frac{0.2}{{log}^{2} (p_{k})}}}{(1 - \frac{0.01}{{log}^{3} (p_{k})})} \cdot (1 - \frac{0.01}{{log}^{3} (p_{k})})) \\ = log (e^{\frac{0.2}{{log}^{2} (p_{k})}}) \\ = \frac{0.2}{{log}^{2} (p_{k})} . \end{matrix}

Consequently, we can establish that

log p_{k} + log (Y_{k} \cdot (1 - \frac{0.01}{{log}^{3} (p_{k})})) \geq (log p_{k} + \frac{0.2}{{log}^{2} (p_{k})}) .

According to Proposition 2.5, we can show that

\prod_{i = 1}^{k} \frac{p_{i}}{p_{i} - 1} \leq e^{γ} \cdot (log p_{k} + \frac{0.2}{{log}^{2} (p_{k})}) \leq e^{γ} \cdot log (Y_{k} \cdot θ (p_{k}))

when

p_{k} > 7232121212

. □

This is a main insight.

Lemma 3.8.

For every possible superabundant number

n > 5040

violating

Robin (n)

, we have

{(N_{k})}^{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

Robin (n)

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} \geq a_{2} \geq \dots \geq a_{k} \geq 0

: This follows directly from Propositions 2.7 and 2.10. Given our assumption, we can conclude that

I (n) \geq e^{γ} \cdot log log n .

(1)

Additionally, we can infer that

\prod_{p \leq p_{k}} \frac{p}{p - 1} \leq e^{γ} \cdot log log ({(N_{k})}^{Y_{k}})

(2)

for all

p_{k} > e^{31.018189471}

, as per Lemma 3.7 with

log ({(N_{k})}^{Y_{k}}) = Y_{k} \cdot θ (p_{k})

. By combining Proposition 2.2 with the inequalities (1) and (2), we obtain the following result:

\begin{matrix} e^{γ} \cdot log log ({(N_{k})}^{Y_{k}}) & \geq \prod_{p \leq p_{k}} \frac{p}{p - 1} \\ > I (n) \\ \geq e^{γ} \cdot log log n . \end{matrix}

This leads to

{(N_{k})}^{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

Robin (n)

. Let

n_{k_{i}}

be the infinite sequence of superabundant numbers where

Robin (n_{k_{i}})

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, increasing subsequence

n_{k_{j}^{'}}

of m superabundant numbers, taken from the general sequence

n_{k_{i}}

, such that

\frac{\sum_{j = 1}^{m} log n_{k_{j}^{'}}}{\sum_{j = 1}^{m} log N_{k_{j}^{'}}} \geq 1.00021,

where

N_{k_{j}^{'}} = \prod_{s = 1}^{k_{j}^{'}} p_{s}

is the primorial number of order

k_{j}^{'}

. Lemma 3.3 guarantees the existence of this subsequence of superabundant numbers

n_{k_{j}^{'}}

that can be made arbitrarily long, because for every fixed prime p,

ν_{p} (n_{k_{j}^{'}})

approaches infinity as

n_{k_{j}^{'}}

grows larger. Let

n_{k_{1}^{'}}

be the smallest element in this subsequence. Therefore, by Lemmas 3.6 and 3.8, it suffices to show that the inequality

\frac{\sum_{j = 1}^{m} log n_{k_{j}^{'}}}{\sum_{j = 1}^{m} log N_{k_{j}^{'}}} > Y_{k_{1}^{'}}

is a contradiction since

1.00021 > Y_{k_{1}^{'}}

. Definitely, that’s the same as saying

\sum_{j = 1}^{m} log n_{k_{j}^{'}} > Y_{k_{1}^{'}} \cdot \sum_{j = 1}^{m} log N_{k_{j}^{'}}

and it is clear that

\begin{matrix} Y_{k_{1}^{'}} \cdot \sum_{j = 1}^{m} log N_{k_{j}^{'}} & = \sum_{j = 1}^{m} log {(N_{k_{j}^{'}})}^{Y_{k_{1}^{'}}} \\ > \sum_{j = 1}^{m} log {(N_{k_{j}^{'}})}^{Y_{k_{j}^{'}}} \\ > \sum_{j = 1}^{m} log n_{k_{j}^{'}} \end{matrix}

due to the fact that

Y_{k_{j}^{'}}

decreases as the subsequence

n_{k_{j}^{'}}

grows, and

{(N_{k_{j}^{'}})}^{Y_{k_{j}^{'}}} > n_{k_{j}^{'}}

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. □

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