Robin’s Criterion on Superabundant Numbers

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Robin’s Criterion on Superabundant Numbers

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Frank Vega^*

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09 August 2024

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13 August 2024

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Abstract

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. There are several statements equivalent to the famous Riemann hypothesis. Robin's criterion states that the Riemann hypothesis is true if and only if the inequality $\sigma(n) < e^{\gamma} \cdot n \cdot \log \log n$ holds for all natural numbers $n > 5040$, where $\sigma(n)$ is the sum-of-divisors function of $n$, $\gamma \approx 0.57721$ is the Euler-Mascheroni constant and $\log$ is the natural logarithm. We require the properties of superabundant numbers, that is to say left to right maxima of $n \mapsto \frac{\sigma(n)}{n}$. In this note, using Robin's criterion on superabundant numbers, we prove that the Riemann hypothesis is true.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

MSC: 11M26; 11A41; 11A25

1. Introduction

The Riemann hypothesis stands as one of the most formidable and celebrated unsolved problems in the realm of pure mathematics. Enunciated by Bernhard Riemann in 1859, it centers on the behavior of the Riemann zeta function

ζ (s)

, a complex-valued function of profound significance in number theory. At its core, the hypothesis posits that all non-trivial zeros of the Riemann zeta function lie on a specific line in the complex plane. These non-trivial zeros, points where the function equals zero, are of paramount importance due to their intricate connection to the distribution of prime numbers, those fundamental building blocks of arithmetic.

Prime numbers, as the indivisible integers greater than one, exhibit an erratic and seemingly random pattern. The Riemann hypothesis offers a tantalizing prospect: a potential formula to predict the distribution of these primes with remarkable accuracy. Such a breakthrough would have far-reaching implications, not only for number theory but also for fields as diverse as cryptography, computer science, and physics. The hypothesis has captivated mathematicians for centuries, inspiring countless attempts at proof and generating a wealth of related research [1]. Its resolution would undoubtedly mark a monumental achievement in human understanding of the structure of numbers and the laws governing the universe.

Beyond its theoretical elegance, the Riemann hypothesis carries a substantial monetary reward. It is one of the Clay Mathematics Institute’s Millennium Prize Problems, each offering a million-dollar prize for a correct solution. This financial incentive, combined with the problem’s intrinsic allure, has fueled intense competition and innovation among mathematicians worldwide. While progress has been made in understanding the properties of the Riemann zeta function and its zeros, a definitive proof of the hypothesis remains elusive. Yet, the pursuit of this elusive goal continues to drive mathematical discovery and push the boundaries of human knowledge. We provide a solution to this problem based on the properties of the superabundant numbers over the Robin’s criterion.

2. Background and Ancillary Results

In mathematics, the constant

γ \approx 0.57721

is the Euler-Mascheroni constant which is defined as

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

where log is the natural logarithm and

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

is called the

n^{t h}

harmonic number [2]. The following property is based on this constant:

Proposition 1.

By the Euler-Maclaurin formula,

H_{n} = log n + γ + \frac{1}{2 \cdot n} - ε_{n}

where

0 \leq ε_{n} \leq \frac{1}{8 \cdot n^{2}}

which approaches 0 as n goes to infinity [3].

The following inequalities are based on natural logarithms:

Proposition 2.

For

t > 0

[4]:

\frac{1}{t + 0.5} < log (1 + \frac{1}{t}) .

Proposition 3.

For

x > - 1

[5, pp. 1]:

log (1 + x) \leq x .

As usual

σ (n)

is the sum-of-divisors function of n

\sum_{d ∣ n} d,

where

d ∣ n

means the integer d divides n. Define

f (n)

\frac{σ (n)}{n}

Proposition 4.

Let

\prod_{i = 1}^{r} q_{i}^{a_{i}}

be the representation of n as a product of prime numbers

q_{1} < \dots < q_{r}

with natural numbers

a_{1}, \dots, a_{r}

as exponents. Then [6, Lemma 1 pp. 2],

f (n) = (\prod_{i = 1}^{r} \frac{q_{i}}{q_{i} - 1}) \cdot \prod_{i = 1}^{r} (1 - \frac{1}{q_{i}^{a_{i} + 1}}) .

Definition 1.

We say that

Robin (n)

holds provided that

f (n) < e^{γ} \cdot log log n .

The Ramanujan’s Theorem stated that if the Riemann hypothesis is true, then

Robin (n)

holds for large enough n [7]. Next, we have the Robin’s Theorem:

Proposition 5.

Robin (n)

holds for all natural numbers

n > 5040

if and only if the Riemann hypothesis is true [8, Theorem 1 pp. 188].

In 1997, Ramanujan’s old notes were published where he defined the generalized highly composite numbers, which include the superabundant and colossally abundant numbers [7]. Superabundant numbers were also studied by Leonidas Alaoglu and Paul Erdos (1944) [9]. Let

q_{1} = 2, q_{2} = 3, \dots, q_{k}

denote the first k consecutive primes, then an integer of the form

\prod_{i = 1}^{k} q_{i}^{a_{i}}

with

a_{1} \geq a_{2} \geq \dots \geq a_{k} \geq 1

is called a Hardy-Ramanujan integer [10, pp. 367]. A natural number n is called superabundant precisely when, for all natural numbers

m < n

f (m) < f (n) .

We know the following property for the superabundant numbers:

Proposition 6.

If n is superabundant, then n is a Hardy-Ramanujan integer [9, Theorem 1 pp. 450].

In number theory, 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)

. Equivalently,

ν_{p} (n)

is the exponent to which p appears in the prime factorization of n.

Proposition 7.

If n is superabundant and q is the largest prime factor of n, then

ν_{q} (n) = 1

, except when

n \in {4, 36}

[9, Theorem 3 pp. 450].

Several analogues of the Riemann hypothesis have already been proved. Many authors expect (or at least hope) that it is true. However, there are some implications in case of the Riemann hypothesis could be false.

Proposition 8.

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then n must be a superabundant number [11, Theorem 3 pp. 273].

Proposition 9.

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

q < log n

where q is the largest prime factor of n [10, Lemma 6.1 pp. 369].

Proposition 10.

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

q > e^{31.018189471}

where q is the largest prime factor of n [12, Theorem 4.2 pp. 748].

Putting all together yields the proof of the Riemann hypothesis.

3. Main Result

This is a trivial result.

Lemma 1.

For every prime q, the inequality

\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}} < \frac{1}{q} - \frac{3}{8 \cdot {(q + 1)}^{2}}

holds.

Proof.

The inequality

\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}} < \frac{1}{q} - \frac{3}{8 \cdot {(q + 1)}^{2}}

is the same as

\frac{1}{2 \cdot {(q + 1)}^{2}} < \frac{1}{2 \cdot q \cdot (q + 1)}

since

\frac{1}{q} - \frac{1}{2 \cdot q} - \frac{1}{2 \cdot (q + 1)} = \frac{1}{2 \cdot q} - \frac{1}{2 \cdot (q + 1)} = \frac{1}{2 \cdot q \cdot (q + 1)}

and

\frac{1}{8 \cdot {(q + 1)}^{2}} + \frac{3}{8 \cdot {(q + 1)}^{2}} = \frac{4}{8 \cdot {(q + 1)}^{2}} = \frac{1}{2 \cdot {(q + 1)}^{2}} .

In this way, we obtain that

\frac{1}{(q + 1)} < \frac{1}{q}

which is trivially true for every prime q. □

The following is a key Lemma.

Lemma 2.

For every prime q, the inequalities

exp (- \frac{1}{2 \cdot q}) \cdot \frac{exp (H_{q})}{q} \leq e^{γ} \leq exp (\frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) \cdot \frac{exp (H_{q})}{q + 1}

hold.

Proof.

By Proposition 1, we have

H_{q} = log q + γ + \frac{1}{2 \cdot q} - ε_{q}

for every prime q where

0 \leq ε_{q} \leq \frac{1}{8 \cdot q^{2}}

. Therefore, we have

H_{q} \leq log q + γ + \frac{1}{2 \cdot q}

and so,

H_{q} - log q - \frac{1}{2 \cdot q} \leq γ

which is

exp (- \frac{1}{2 \cdot q}) \cdot \frac{exp (H_{q})}{q} \leq e^{γ}

after of applying the exponentiation. By Proposition 1, we obtain

H_{q + 1} = log (q + 1) + γ + \frac{1}{2 \cdot (q + 1)} - ε_{(q + 1)}

for every prime q where

0 \leq ε_{(q + 1)} \leq \frac{1}{8 \cdot {(q + 1)}^{2}}

. So, we could show

H_{q + 1} \geq log (q + 1) + γ + \frac{1}{2 \cdot (q + 1)} - \frac{1}{8 \cdot {(q + 1)}^{2}}

and thus,

H_{q} + \frac{1}{q + 1} - log (q + 1) - \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}} \geq γ

which is

e^{γ} \leq exp (\frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) \cdot \frac{exp (H_{q})}{q + 1}

after of applying the exponentiation where

\frac{1}{q + 1} - \frac{1}{2 \cdot (q + 1)} = \frac{1}{2 \cdot (q + 1)} .

□

This is the main insight.

Lemma 3.

Let

n > 5040

be the possible smallest integer such that

Robin (n)

does not hold. Let q be the largest prime factor of n such that

n = r \cdot q

. Then,

exp (\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) < \frac{log log n}{log log r} .

Proof.

We know this number

n > 5040

must be a superabundant number by Proposition 8. By Proposition 6, let

\prod_{i = 1}^{k} q_{i}^{a_{i}}

be the representation of this superabundant number n as the product of the first k consecutive primes

q_{1} < \dots < q_{k}

with the natural numbers

a_{1} \geq a_{2} \geq \dots \geq a_{k} \geq 1

as exponents. This follows as

\begin{matrix} log log r & = log log (\frac{n}{q_{k}}) \\ = log (log (n) - log q_{k}) \\ = log ((log (n)) \cdot (1 - \frac{log q_{k}}{log n})) \\ = log log (n) + log (1 - \frac{log q_{k}}{log n}) \\ \leq log log (n) - \frac{log q_{k}}{log n} \end{matrix}

since

log (1 - \frac{log q_{k}}{log n}) \leq - \frac{log q_{k}}{log n}

by Proposition 3 because of

- \frac{log q_{k}}{log n} > - 1

by Propositions 9 and 10. For that reason, we have

\frac{log log n}{log log r} \geq \frac{log log n}{log log (n) - \frac{log q_{k}}{log n}} = \frac{\frac{(log n) \cdot log log (n)}{log q_{k}}}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 1} .

We know that

\begin{matrix} log (\frac{\frac{(log n) \cdot log log (n)}{log q_{k}}}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 1}) & = log (1 + \frac{1}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 1}) \\ > \frac{1}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 0.5} \\ = \frac{log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \end{matrix}

because of

log (1 + \frac{1}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 1}) > \frac{1}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 0.5}

by Proposition 2 since

\frac{(log n) \cdot log log (n)}{log q_{k}} - 1 > log (n) - 1 > 0

by Propositions 9 and 10. We arrive at:

\frac{\frac{(log n) \cdot log log (n)}{log q_{k}}}{\frac{(log n) \cdot log log (n)}{log q_{k}} - 1} > exp (\frac{log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}}) .

Hence, it is enough to show that

\frac{log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \geq \frac{1}{q_{k}} - \frac{3}{8 \cdot {(q_{k} + 1)}^{2}}

by Lemma 1. That is equivalent to

1 - \frac{(log n) \cdot log log (n) - 1.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \geq \frac{1}{q_{k}} - \frac{3}{8 \cdot {(q_{k} + 1)}^{2}}

since

\begin{matrix} \frac{log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \\ = \frac{1.5 \cdot log q_{k} - 0.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \\ = \frac{- ((log n) \cdot log log (n) - 1.5 \cdot log q_{k}) + (log n) \cdot log log (n) - 0.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \\ = 1 - \frac{(log n) \cdot log log (n) - 1.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} . \end{matrix}

We only need to show that

1 - \frac{q_{k} \cdot log (q_{k}) - 1.5 \cdot log q_{k}}{q_{k} \cdot log (q_{k}) - 0.5 \cdot log q_{k}} \geq \frac{1}{q_{k}} - \frac{3}{8 \cdot {(q_{k} + 1)}^{2}}

on the basis that

1 - \frac{(log n) \cdot log log (n) - 1.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} \geq \frac{1}{q_{k}} - \frac{3}{8 \cdot {(q_{k} + 1)}^{2}}

and

\frac{(log n) \cdot log log (n) - 1.5 \cdot log q_{k}}{(log n) \cdot log log (n) - 0.5 \cdot log q_{k}} < \frac{q_{k} \cdot log (q_{k}) - 1.5 \cdot log q_{k}}{q_{k} \cdot log (q_{k}) - 0.5 \cdot log q_{k}}

by Proposition 9. This implies that

1 - \frac{2 \cdot q_{k} - 3}{2 \cdot q_{k} - 1} \geq \frac{1}{q_{k}} - \frac{3}{8 \cdot {(q_{k} + 1)}^{2}}

which is

(2 \cdot q_{k} - 1) - (2 \cdot q_{k} - 3) \geq \frac{2 \cdot q_{k} - 1}{q_{k}} - \frac{3 \cdot (2 \cdot q_{k} - 1)}{8 \cdot {(q_{k} + 1)}^{2}}

and

2 \geq 2 - \frac{1}{q_{k}} - \frac{3 \cdot (2 \cdot q_{k} - 1)}{8 \cdot {(q_{k} + 1)}^{2}}

due to

\frac{q_{k} \cdot log (q_{k}) - 1.5 \cdot log q_{k}}{q_{k} \cdot log (q_{k}) - 0.5 \cdot log q_{k}} = \frac{q_{k} - 1.5}{q_{k} - 0.5} = \frac{2 \cdot q_{k} - 3}{2 \cdot q_{k} - 1}

and

(2 \cdot q_{k} - 1) - (2 \cdot q_{k} - 3) = 2, \frac{2 \cdot q_{k} - 1}{q_{k}} = 2 - \frac{1}{q_{k}} .

Since the inequality

2 \geq 2 - \frac{1}{q_{k}} - \frac{3 \cdot (2 \cdot q_{k} - 1)}{8 \cdot {(q_{k} + 1)}^{2}}

trivially holds, then the proof is done. □

This is the main theorem.

Theorem 1.

The Riemann hypothesis is true.

Proof.

Let

n > 5040

be the possible smallest integer such that

Robin (n)

does not hold. We know this number must be a superabundant number by Proposition 8. Let q be the largest prime factor of n such that

n = q \cdot r

. Under our assumption, we have

f (n) \geq e^{γ} \cdot log log n .

By Lemma 2, we have

exp (- \frac{1}{2 \cdot q}) \cdot \frac{exp (H_{q})}{q} \leq e^{γ}

and so,

f (n) \cdot exp (\frac{1}{2 \cdot q}) \geq \frac{exp (H_{q})}{q} \cdot log log n

after making a distribution. By Propositions 4 and 7, we deduce that

f (r) \cdot \frac{q + 1}{q} \cdot exp (\frac{1}{2 \cdot q}) \geq \frac{exp (H_{q})}{q} \cdot log log n .

where

f (q) = \frac{q + 1}{q}

. That would be

f (r) \cdot exp (\frac{1}{2 \cdot q}) \geq \frac{exp (H_{q})}{q + 1} \cdot log log n .

By Lemma 2, we can see that

e^{γ} \cdot exp (- \frac{1}{2 \cdot (q + 1)} - \frac{1}{8 \cdot {(q + 1)}^{2}}) \leq \frac{exp (H_{q})}{q + 1} .

We arrive at:

f (r) \cdot exp (\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) \geq e^{γ} \cdot log log n

after making a distribution. We know that

Robin (r)

holds under the assumption that

n = q \cdot r

and whenever

n > 5040

would be the smallest integer such that

Robin (n)

does not hold. By our supposition, we get

(e^{γ} \cdot log log r) \cdot exp (\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) \geq e^{γ} \cdot log log n

which is

exp (\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) \geq \frac{log log n}{log log r} .

However, this contradicts the fact that

exp (\frac{1}{2 \cdot q} + \frac{1}{2 \cdot (q + 1)} + \frac{1}{8 \cdot {(q + 1)}^{2}}) < \frac{log log n}{log log r} .

holds by Lemma 3. Consequently, we obtain a contradiction under the assumption of the existence of the smallest integer

n > 5040

such that

Robin (n)

does not hold. By Reductio ad absurdum, we prove that the Riemann hypothesis is true according to Proposition 5. □

4. Conclusions

The Riemann hypothesis is far more than a mathematical curiosity. Its implications reverberate across diverse scientific domains. A proof would illuminate not only the mysterious patterns of prime numbers but could also revolutionize fields as disparate as cryptography and particle physics. For instance, understanding the precise distribution of primes is paramount for the security protocols underlying modern digital communication. A proven Riemann hypothesis could potentially unlock more efficient methods of prime number generation, bolstering the defenses of our digital world. Beyond this, the hypothesis may hold clues about the underlying structure of the universe. Some physicists believe that its resolution could shed light on the distribution of energy levels in complex systems, a fundamental question in their field. In essence, the Riemann hypothesis serves as a bridge connecting seemingly disparate areas of knowledge. Its solution could catalyze breakthroughs that reshape our understanding of the natural world.

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Robin’s Criterion on Superabundant Numbers

Abstract

1. Introduction

2. Background and Ancillary Results

3. Main Result

4. Conclusions

References

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