Robin’s Criterion on Divisibility (II)

Preprint

Article

Robin’s Criterion on Divisibility (II)

Altmetrics

Downloads

109

Views

118

Comments

Frank Vega^*

This version is not peer-reviewed

Submitted:

24 September 2024

Posted:

25 September 2024

Read the latest preprint version here

Alerts

Abstract

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 $n$ and $\gamma \approx 0.57721$ is the Euler-Mascheroni constant. We show that the Robin inequality is true for all natural numbers $n > 5040$ that are not divisible by some prime between $2$ and $1771559$. We prove that the Robin inequality holds when $\frac{\pi^{2}}{6} \cdot \log\log n' \leq \log\log n$ for some $n>5040$ where $n'$ is the square free kernel of the natural number $n$. The possible smallest counterexample $n > 5040$ of the Robin inequality implies that $q_{m} > e^{31.018189471}$, $1 < \frac{(1 + \frac{1.2762}{\log q_{m}}) \cdot \log(1.006479799241)}{\log \log n}+ \frac{\log N_{m}}{\log n}$, $(\log n)^{\beta_{n}} < 1.000208229291\cdot\log(N_{m})$ and $n < (1.006479799241)^{m} \cdot N_{m}$, where $N_{m} = \prod_{i = 1}^{m} q_{i}$ is the primorial number of order $m$, $q_{m}$ is the largest prime divisor of $n$ and $\beta_{n} = \prod_{i = 1}^{m} \frac{q_{i}^{a_{i}+1}}{q_{i}^{a_{i}+1}-1}$ when $n$ is an Hardy-Ramanujan integer of the form $\prod_{i=1}^{m} q_{i}^{a_{i}}$. By combining these results, we present a 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

1. Introduction

In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part

\frac{1}{2}

. 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 and

d ∤ n

means the integer d does not divide n. Define

f (n)

to be

\frac{σ (n)}{n}

. Say

Robin (n)

holds provided

f (n) < e^{γ} \cdot log log n .

The constant

γ \approx 0.57721

is the Euler-Mascheroni constant and log is the natural logarithm. The following inequality is based on natural logarithms:

Proposition 1.

For

t > 0

[1]:

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

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

Robin (n)

holds for large enough n [2]. Next, we have the Robin’s Theorem:

Proposition 2.

Robin (n)

holds for all natural numbers

n > 5040

if and only if the Riemann hypothesis is true [3].

It is known that

Robin (n)

holds for many classes of numbers n.

Robin (n)

holds for all natural numbers

n > 5040

that are not divisible by 2 [4]. We extend the indivisibility property on the following result:

Proposition 3.

Robin (n)

holds for all natural numbers

n > 5040

that are not divisible by some prime between 3 and 1771559 [5].

We recall that an integer n is said to be square free if for every prime divisor q of n we have

q^{2} ∤ n

Proposition 4.

Robin (n)

holds for all natural numbers

n > 5040

that are square free [4].

In addition, we show that

Robin (n)

holds for some

n > 5040

when

\frac{π^{2}}{6} \cdot log log n^{'} \leq log log n

such that

n^{'}

is the square free kernel of the natural number n [5]. In 1997, Ramanujan’s old notes were published where he defined the generalized highly composite numbers, which include the superabundant and colossally abundant numbers [2]. These numbers were also studied by Leonidas Alaoglu and Paul Erdos (1944) [6]. Let

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

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

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

with

a_{1} \geq a_{2} \geq \dots \geq a_{m} \geq 0

is called an Hardy-Ramanujan integer [4]. A natural number n is called superabundant precisely when, for all natural numbers

m < n

f (m) < f (n) .

Proposition 5.

If n is superabundant, then n is an Hardy-Ramanujan integer [6].

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) .

There is a close relation between the superabundant and colossally abundant numbers.

Proposition 6.

Every colossally abundant number is superabundant [6].

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 7.

The smallest counterexample of the Robin inequality greater than 5040 must be a superabundant number [7].

Suppose that

n > 5040

is the possible smallest counterexample of the Robin inequality, then we prove that

q_{m} > e^{31.018189471}

1 < \frac{(1 + \frac{1.2762}{log q_{m}}) \cdot log (1.006479799241)}{log log n} + \frac{log N_{m}}{log n}

{(log n)}^{β_{n}} < 1.000208229291 \cdot log (N_{m})

and

n < {(1.006479799241)}^{m} \cdot N_{m}

, where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m,

q_{m}

is the largest prime divisor of n and

β_{n} = \prod_{i = 1}^{m} \frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1}

when n is an Hardy-Ramanujan integer of the form

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

(Refer to preliminary results in Vega’s paper [5]).

Proposition 8.

If the Riemann hypothesis is false, then there are infinitely many colossally abundant numbers

n > 5040

such that

Robin (n)

fails (i.e.

Robin (n)

does not hold) [3].

We can further deduce that

Lemma 1.

If the Riemann hypothesis is false, then there are infinitely many superabundant numbers n such that

Robin (n)

fails.

Proof.

This is a direct consequence of Propositions 2, 6 and 8. □

Putting all together yields a proof of the Riemann hypothesis.

2. A Central Lemma

These are known results:

Proposition 9.

For

n > 1

[4]:

f (n) < \prod_{q ∣ n} \frac{q}{q - 1} .

Proposition 10.

We have [8]:

\prod_{i = 1}^{\infty} \frac{1}{1 - \frac{1}{q_{i}^{2}}} = ζ (2) = \frac{π^{2}}{6} .

The following is a key Lemma. It gives an upper bound on

f (n)

that holds for all natural numbers n. The bound is too weak to prove

Robin (n)

directly, but is critical because it holds for all natural numbers n. Further the bound only uses the primes that divide n and not how many times they divide n.

Lemma 2.

Let

n > 1

and let all its prime divisors be

q_{1} < \dots < q_{m}

. Then,

f (n) < \frac{π^{2}}{6} \cdot \prod_{i = 1}^{m} \frac{q_{i} + 1}{q_{i}} .

Proof.

Putting together the Propositions 9 and 10 yields the proof:

f (n) < \prod_{i = 1}^{m} (\frac{q_{i}}{q_{i} - 1}) = \prod_{i = 1}^{m} (\frac{q_{i} + 1}{q_{i}} \cdot \frac{1}{1 - \frac{1}{q_{i}^{2}}}) < \frac{π^{2}}{6} \cdot \prod_{i = 1}^{m} \frac{q_{i} + 1}{q_{i}} .

□

3. Robin on Divisibility

We know the following Propositions:

Proposition 11.

Let

n > e^{e^{23.762143}}

and let all its prime divisors be

q_{1} < \dots < q_{m}

, then [9]:

(\prod_{i = 1}^{m} \frac{q_{i}}{q_{i} - 1}) < \frac{1771561}{1771560} \cdot e^{γ} \cdot log log n .

Proposition 12.

Robin (n)

holds for all natural numbers

10^{10^{13.11485}} \geq n > 5040

[10].

Theorem 1.

Suppose

n > 5040

. If there exists a prime

q \leq 1771559

with

q ∤ n

, then

Robin (n)

holds.

Proof.

We have that

f (n) < \frac{1771561}{1771560} \cdot e^{γ} \cdot log log (n)

for any number

n > 10^{10^{13.11485}}

since the inequality

10^{10^{13.11485}} > e^{e^{23.762143}}

is satisfied. Note that

f (n) < \frac{n}{φ (n)} = \prod_{q ∣ n} \frac{q}{q - 1}

from the Proposition 9, where

φ (x)

is the Euler’s totient function. Suppose that n is not divisible by some prime

q \leq 1771559

and

n \geq 10^{10^{13.11485}}

. Then,

\begin{matrix} f (n) & < \frac{n}{φ (n)} \\ = \frac{n \cdot q}{φ (n \cdot q)} \cdot \frac{q - 1}{q} \\ < \frac{1771561}{1771560} \cdot \frac{q - 1}{q} \cdot e^{γ} \cdot log log (n \cdot q) \end{matrix}

and

\begin{matrix} \frac{f (n)}{e^{γ} \cdot log log (n)} & < \frac{1771561}{1771560} \cdot \frac{q - 1}{q} \cdot \frac{log log (n \cdot q)}{log log (n)} \\ = \frac{1771561}{1771560} \cdot \frac{q - 1}{q} \cdot \frac{log log (n) + log (1 + \frac{log (q)}{log (n)})}{log log (n)} \\ = \frac{1771561}{1771560} \cdot \frac{q - 1}{q} \cdot (1 + \frac{log (1 + \frac{log (q)}{log (n)})}{log log (n)}) \end{matrix}

\frac{f (n)}{e^{γ} \cdot log log (n)} < \frac{1771561}{1771560} \cdot \frac{q - 1}{q} \cdot (1 + \frac{log (1 + \frac{log (q)}{log (n)})}{log log (n)})

for

n \geq 10^{10^{13.11485}}

. The right hand side is less than 1 for

q \leq 1771559

and

n \geq 10^{10^{13.11485}}

. Therefore,

Robin (n)

holds. □

4. On the Greatest Prime Divisor

We know that

Proposition 13.

For

x \geq 2973

[11]:

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

Theorem 2.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. If

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

q_{m} > e^{31.018189471}

Proof.

According to the Propositions 5 and 7, the primes

q_{1} < \dots < q_{m}

must be the first m consecutive primes and

a_{1} \geq a_{2} \geq \dots \geq a_{m} \geq 0

since

n > 5040

should be an Hardy-Ramanujan integer. From the Theorem 1, we know that necessarily

q_{m} \geq 1771559

. So,

e^{γ} \cdot log log n \leq f (n) < \prod_{q \leq q_{m}} \frac{q}{q - 1} < e^{γ} \cdot (log q_{m} + \frac{0.2}{log (q_{m})})

because of the Propositions 9 and 13. Hence,

log log n - \frac{0.2}{log (q_{m})} < log q_{m} .

However, from the Proposition 12 and Theorem 1, we would obtain that

\begin{matrix} log log n - \frac{0.2}{log (q_{m})} & \geq 13.11485 \cdot log (10) + log log 10 - \frac{0.2}{log (1771559)} \\ > 31.018189471 . \end{matrix}

Since, we have that

log q_{m} > log log n - \frac{0.2}{log (q_{m})} > 31.018189471

then, we would obtain that

q_{m} > e^{31.018189471}

under the assumption that

n > 5040

is the smallest integer such that

Robin (n)

does not hold. □

5. Some Feasible Cases

We can easily prove that

Robin (n)

is true for certain kind of numbers:

Lemma 3.

Robin (n)

holds for

n > 5040

when

q \leq 7

, where q is the largest prime divisor of n.

Proof.

This is an immediate consequence of Theorem 1. □

The next Theorem implies that

Robin (n)

holds for a wide range of natural numbers

n > 5040

Theorem 3.

Let

\frac{π^{2}}{6} \cdot log log n^{'} \leq log log n

for some

n > 5040

such that

n^{'}

is the square free kernel of the natural number n. Then

Robin (n)

holds.

Proof.

Let

n^{'}

be the square free kernel of the natural number n, that is the product of the distinct primes

q_{1}, \dots, q_{m}

. By assumption we have that

\frac{π^{2}}{6} \cdot log log n^{'} \leq log log n .

For all square free

n^{'} \leq 5040

Robin (n^{'})

holds if and only if

n^{'} \notin {2, 3, 5, 6, 10, 30}

[4]. However,

Robin (n)

holds for all

n > 5040

when

n^{'} \in {2, 3, 5, 6, 10, 15, 30}

due to Lemma 3. When

n^{'} > 5040

, we know that

Robin (n^{'})

holds and so

f (n^{'}) < e^{γ} \cdot log log n^{'}

because of the Proposition 4. By the previous Lemma 2:

f (n) < \frac{π^{2}}{6} \cdot \prod_{i = 1}^{m} \frac{q_{i} + 1}{q_{i}} .

So,

\begin{matrix} f (n) & < \frac{π^{2}}{6} \cdot \prod_{i = 1}^{m} \frac{q_{i} + 1}{q_{i}} \\ = \frac{π^{2}}{6} \cdot f (n^{'}) \\ < \frac{π^{2}}{6} \cdot e^{γ} \cdot log log n^{'} \\ \leq e^{γ} \cdot log log n \end{matrix}

according to the formula

f (x)

for the square free numbers [4]. □

6. On Possible Counterexample

For every prime number

p_{n} > 2

, we define the sequence

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

Lemma 4.

As the prime number

p_{n}

increases, the sequence

Y_{n}

is strictly decreasing.

Proof.

This Lemma is obvious. □

In mathematics, the Chebyshev function

θ (x)

is given by

θ (x) = \sum_{p \leq x} log p

where

p \leq x

means all the prime numbers p that are less than or equal to x. We know that

Proposition 14.

For

x \geq 7232121212

[12]:

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

Proposition 15.

For

x \geq 2278382

[12]:

\prod_{q \leq x} \frac{q}{q - 1} \leq e^{γ} \cdot (log x + \frac{0.2}{{log}^{2} (x)}) .

We will prove another important inequality:

Lemma 5.

Let

q_{1}, q_{2}, \dots, q_{m}

denote the first m consecutive primes such that

q_{1} < q_{2} < \dots < q_{m}

and

q_{m} > 7232121212

. Then

\prod_{i = 1}^{m} \frac{q_{i}}{q_{i} - 1} \leq e^{γ} \cdot log (Y_{m} \cdot θ (q_{m})) .

Proof.

From the Proposition 14, we know that

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

In this way, we can show that

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

We know that

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

Consequently, we obtain that

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

Due to the Proposition 15, we prove that

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

when

q_{m} > 7232121212

. □

We use the following Proposition:

Proposition 16.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. Then [9]:

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

The following Theorems have a great significance, because these mean that the possible smallest counterexample of the Robin inequality greater than 5040 must be very close to its square free kernel.

Theorem 4.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. If

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

{(log n)}^{β_{n}} \leq Y_{m} \cdot log (N_{m})

, where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m and

β_{n} = \prod_{i = 1}^{m} \frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1}

Proof.

According to the Propositions 5 and 7, the primes

q_{1} < \dots < q_{m}

must be the first m consecutive primes and

a_{1} \geq a_{2} \geq \dots \geq a_{m} \geq 0

since

n > 5040

should be an Hardy-Ramanujan integer. From the Theorem 2, we know that necessarily

q_{m} > e^{31.018189471}

. From the Proposition 16, we note that

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

However, we know that

\prod_{i = 1}^{m} \frac{q_{i}}{q_{i} - 1} \leq e^{γ} \cdot log (Y_{m} \cdot log (N_{m}))

because of the Lemma 5 when

q_{m} > 7232121212

. If we multiply by

\prod_{i = 1}^{m} (1 - \frac{1}{q_{i}^{a_{i} + 1}})

the both sides of the previous inequality, then we obtain that

f (n) \leq e^{γ} \cdot log (Y_{m} \cdot log (N_{m})) \cdot \prod_{i = 1}^{m} (1 - \frac{1}{q_{i}^{a_{i} + 1}}) .

If n is the smallest integer exceeding 5040 that does not satisfy the Robin inequality, then

e^{γ} \cdot log log n \leq e^{γ} \cdot log (Y_{m} \cdot log (N_{m})) \cdot \prod_{i = 1}^{m} (1 - \frac{1}{q_{i}^{a_{i} + 1}})

because of

e^{γ} \cdot log log n \leq f (n) .

That is the same as

\prod_{i = 1}^{m} \frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1} \cdot log log n \leq log (Y_{m} \cdot log (N_{m}))

(1)

which is equivalent to

{(log n)}^{β_{n}} \leq Y_{m} \cdot log (N_{m})

where

β_{n} = \prod_{i = 1}^{m} \frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1}

. Therefore, the proof is done. □

Theorem 5.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. If

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

{(log n)}^{β_{n}} < 1.000208229291 \cdot log (N_{m})

, where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m and

β_{n} = \prod_{i = 1}^{m} \frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1}

Proof.

From the Theorem 2, we know that necessarily

q_{m} > e^{31.018189471}

. Using the Theorem 4, we obtain that

{(log n)}^{β_{n}} < 1.000208229291 \cdot log (N_{m})

due to Lemma 4 since

Y_{m} < 1.000208229291

whenever

q_{m} > e^{31.018189471}

. □

Theorem 6.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. If

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

n < {(1.006479799241)}^{m} \cdot N_{m}

, where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m.

Proof.

According to the Propositions 5 and 7, the primes

q_{1} < \dots < q_{m}

must be the first m consecutive primes and

a_{1} \geq a_{2} \geq \dots \geq a_{m} \geq 0

since

n > 5040

should be an Hardy-Ramanujan integer. From the Lemma 5, we know that

\prod_{i = 1}^{m} \frac{q_{i}}{q_{i} - 1} \leq e^{γ} \cdot log (Y_{m} \cdot θ (q_{m})) = e^{γ} \cdot log log (N_{m}^{Y_{m}})

for

q_{m} > 7232121212

. In this way, if

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

n < N_{m}^{Y_{m}}

since by the Proposition 9 we have that

e^{γ} \cdot log log n \leq f (n) < \prod_{i = 1}^{m} \frac{q_{i}}{q_{i} - 1} .

That is the same as

n < N_{m}^{Y_{m} - 1} \cdot N_{m}

. We can check that

q_{m}^{Y_{m} - 1}

is monotonically decreasing for all primes

q_{m} > e^{31.018189471}

. Certainly, the derivative of the function

g (x) = x^{(\frac{e^{\frac{0.2}{{log}^{2} (x)}}}{(1 - \frac{0.01}{{log}^{3} (x)})} - 1)}

is less than zero for all real numbers

x \geq e^{31.018189471}

. Consequently, we would have that

q_{m}^{Y_{m} - 1} < g (e^{31.018189471}) < 1.006479799241

for all primes

q_{m} > e^{31.018189471}

. Moreover, we would obtain that

q_{m}^{Y_{m} - 1} > q_{j}^{Y_{m} - 1}

for every integer

1 \leq j < m

. Finally, we can state that

n < {(1.006479799241)}^{m} \cdot N_{m}

since

N_{m}^{Y_{m} - 1} < {(1.006479799241)}^{m}

when

n > 5040

is the smallest integer such that

Robin (n)

does not hold. □

We know the following results:

Proposition 17.

For

x > 1

[13]:

π (x) \leq (1 + \frac{1.2762}{log x}) \cdot \frac{x}{log x}

where

π (x)

is the prime counting function.

Proposition 18.

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

p < log n

where p is the largest prime divisor of n [4].

Theorem 7.

Let

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

be the representation of n as a product of primes

q_{1} < \dots < q_{m}

with natural numbers as exponents

a_{1}, \dots, a_{m}

. If

n > 5040

is the smallest integer such that

Robin (n)

does not hold, then

1 < \frac{(1 + \frac{1.2762}{log q_{m}}) \cdot log (1.006479799241)}{log log n} + \frac{log N_{m}}{log n}

, where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m.

Proof.

Note that

n < {(1.006479799241)}^{m} \cdot N_{m}

when n is the smallest integer such that

Robin (n)

does not hold. If we apply the logarithm to the both sides, then

log n < m \cdot log (1.006479799241) + log N_{m} .

According to the Proposition 17, we have that

log n < (1 + \frac{1.2762}{log q_{m}}) \cdot \frac{q_{m}}{log q_{m}} \cdot log (1.006479799241) + log N_{m} .

From the Proposition 18, we would have

log n < (1 + \frac{1.2762}{log q_{m}}) \cdot \frac{log n}{log log n} \cdot log (1.006479799241) + log N_{m} .

which is the same as

1 < \frac{(1 + \frac{1.2762}{log q_{m}}) \cdot log (1.006479799241)}{log log n} + \frac{log N_{m}}{log n}

after of dividing by

log n

. □

7. A Conclusive Approach

We use the following results:

Lemma 6.

Let

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

be the representation of a superabundant number

n > 5040

as the product of the first m consecutive primes

q_{1} < \dots < q_{m}

with the natural numbers

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

as exponents. Suppose that

Robin (n)

fails. Then,

β_{n} \leq \frac{log log {(N_{m})}^{Y_{m}}}{log log n},

where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m and

β_{n} = \prod_{i = 1}^{m} (\frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1})

Proof.

Using the inequality (1) and Lemma 1, then this result will be a generalization of Theorem 4 for every possible counterexample

n > 5040

of the Robin’s inequality. □

Proposition 19.

Let

x \geq 11

. For

y > x

[14]:

\frac{log log y}{log log x} < \sqrt{\frac{y}{x}} .

This is the main insight.

Lemma 7.

Let

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

be the representation of a superabundant number

n > 5040

as the product of the first m consecutive primes

q_{1} < \dots < q_{m}

with the natural numbers

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

as exponents. Suppose that

Robin (n)

fails. Then,

β_{n} < \sqrt{\frac{{(N_{m})}^{Y_{m}}}{n}},

where

N_{m} = \prod_{i = 1}^{m} q_{i}

is the primorial number of order m and

β_{n} = \prod_{i = 1}^{m} (\frac{q_{i}^{a_{i} + 1}}{q_{i}^{a_{i} + 1} - 1})

Proof.

When

n > 5040

is a superabundant number and

Robin (n)

fails, then we have

β_{n} \leq \frac{log log {(N_{m})}^{Y_{m}}}{log log n}

by Lemma 6. We assume that

{(N_{m})}^{Y_{m}} > n > 5040 > 11

since

β_{n} > 1

. Consequently,

\frac{log log {(N_{m})}^{Y_{m}}}{log log n} < \sqrt{\frac{{(N_{m})}^{Y_{m}}}{n}}

by Proposition 19. As result, we obtain that

β_{n} < \sqrt{\frac{{(N_{m})}^{Y_{m}}}{n}} .

□

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. This is the main Theorem.

Theorem 8.

The Riemann hypothesis is true.

Proof.

Under the assumption that the Riemann hypothesis is false, then there would exist infinitely many superabundant numbers n such that

Robin (n)

fails according to Lemma 1. Let

n_{m} > 5040

be a large enough superabundant number (as larger as we want) such that

q_{m}

is the largest prime factor of

n_{m}

. Suppose that

Robin (n_{m})

fails. This implies that

q_{m} > e^{31.018189471}

by Theorem 2. Let

n_{m^{'}} > 5040

be another large enough superabundant number (as larger as we want) such that

n_{m^{'}} > n_{m}

. Suppose that

Robin (n_{m^{'}})

fails too. By Lemma 7, we have

β_{n_{m}} < \sqrt{\frac{{(N_{m})}^{Y_{m}}}{n_{m}}}, β_{n_{m^{'}}} < \sqrt{\frac{{(N_{m^{'}})}^{Y_{m^{'}}}}{n_{m^{'}}}} .

So, we would have

β_{n_{m^{'}}} \cdot β_{n_{m}} < \sqrt{\frac{{(N_{m^{'}})}^{Y_{m^{'}}}}{n_{m^{'}}}} \cdot β_{n_{m}} .

Consequently, we get:

β_{n_{m^{'}}} \cdot β_{n_{m}} < \sqrt{\frac{{(N_{m^{'}})}^{Y_{m^{'}}}}{n_{m^{'}}}} \cdot \sqrt{\frac{{(N_{m})}^{Y_{m}}}{n_{m}}} .

We arrive at:

{(β_{n_{m^{'}}} \cdot β_{n_{m}})}^{2} < \frac{{(N_{m^{'}})}^{Y_{m^{'}}}}{n_{m^{'}}} \cdot \frac{{(N_{m})}^{Y_{m}}}{n_{m}} .

We can see that

{(β_{n_{m^{'}}} \cdot β_{n_{m}})}^{2} > 1 .

However, we claim that

\frac{{(N_{m^{'}})}^{Y_{m^{'}}}}{n_{m^{'}}} \cdot \frac{{(N_{m})}^{Y_{m}}}{n_{m}} \leq 1

which is

log Y_{m} \leq log (\frac{log (n_{m^{'}} \cdot n_{m})}{log ({(N_{m^{'}})}^{\frac{Y_{m^{'}}}{Y_{m}}} \cdot N_{m})}) .

By Proposition 1, we obtain that

\begin{matrix} log Y_{m} & = \frac{0.2}{{log}^{2} (q_{m})} + log (\frac{{log}^{3} (q_{m})}{{log}^{3} (q_{m}) - 0.01}) \\ = \frac{0.2}{{log}^{2} (q_{m})} + log (1 + \frac{0.01}{{log}^{3} (q_{m}) - 0.01}) \\ < \frac{0.2}{{log}^{2} (q_{m})} + \frac{0.01}{{log}^{3} (q_{m}) - 0.01} \end{matrix}

for all

q_{m} > e^{31.018189471}

. Hence, it is enough to show that

\frac{0.2}{{log}^{2} (q_{m})} + \frac{0.01}{{log}^{3} (q_{m}) - 0.01} \leq log (\frac{log (n_{m^{'}} \cdot n_{m})}{log ({(N_{m^{'}})}^{\frac{Y_{m^{'}}}{Y_{m}}} \cdot N_{m})})

which is trivially true under the assumption that

{log}^{2} (q_{m}) \cdot log (\frac{log (n_{m^{'}} \cdot n_{m})}{log ({(N_{m^{'}})}^{\frac{Y_{m^{'}}}{Y_{m}}} \cdot N_{m})}) \geq 0.200322393

as a consequence of

0.200322393 > 0.2 + \frac{0.01 \cdot {log}^{2} (q_{m})}{{log}^{3} (q_{m}) - 0.01} > {log}^{2} (q_{m}) \cdot log Y_{m}

for all

q_{m} > e^{31.018189471}

. In this way, we reach the contradiction

1 < 1

under the assumption that

Robin (n_{m})

fails. This is supported by the fact that

Y_{m}

is strictly decreasing (i.e.

\frac{Y_{m^{'}}}{Y_{m}} < 1

m^{'} > m

{lim}_{m^{'} \to \infty} Y_{m^{'}} = 1

, and we can always be able to take both superabundant numbers

n_{m}

and

n_{m^{'}} > n_{m}

as larger as we want. Furthermore, for every fixed prime q,

ν_{q} (n)

goes to infinity as long as n goes to infinity whenever n is superabundant [6,14]. For that reason, we can definitely assure that the inequalities

n_{m^{'}} > n_{m}, {log}^{2} (q_{m}) \cdot log (\frac{log (n_{m^{'}} \cdot n_{m})}{log ({(N_{m^{'}})}^{\frac{Y_{m^{'}}}{Y_{m}}} \cdot N_{m})}) \geq 0.200322393

can simultaneously hold for every superabundant number

n_{m^{'}}

greater than some threshold. Accordingly,

Robin (n)

holds for all large enough superabundant numbers

n > 5040

. By Lemma 1, this contradicts the fact that there exist infinitely many superabundant numbers n, such that

Robin (n)

fails if the Riemann hypothesis were false. By reductio ad absurdum, we prove that the Riemann hypothesis is true. □

Acknowledgments

Many thanks to Michel Planat, Patrick Solé and Richard J. Lipton for their support.

References

Nicolas, J.L. The sum of divisors function and the Riemann hypothesis. The Ramanujan Journal 2022, 58, 1113–1157. [Google Scholar] [CrossRef]
Nicolas, J.L.; Robin, G. Highly Composite Numbers by Srinivasa Ramanujan. The Ramanujan Journal 1997, 1, 119–153. [Google Scholar] [CrossRef]
Robin, G. Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann. J. Math. pures appl 1984, 63, 187–213. [Google Scholar]
Choie, Y.; Lichiardopol, N.; Moree, P.; Solé, P. On Robin’s criterion for the Riemann hypothesis. Journal de Théorie des Nombres de Bordeaux 2007, 19, 357–372. [Google Scholar] [CrossRef]
Vega, F. Robin’s criterion on divisibility. The Ramanujan Journal 2022, 59, 745–755. [Google Scholar] [CrossRef]
Alaoglu, L.; Erdos, P. On Highly Composite and Similar Numbers. Transactions of the American Mathematical Society 1944, 56, 448–469. [Google Scholar] [CrossRef]
Akbary, A.; Friggstad, Z. Superabundant numbers and the Riemann hypothesis. The American Mathematical Monthly 2009, 116, 273–275. [Google Scholar] [CrossRef]
Ayoub, R. Euler and the Zeta Function. The American Mathematical Monthly 1974, 81, 1067–1086. [Google Scholar] [CrossRef]
Hertlein, A. Robin’s Inequality for New Families of Integers. Integers 2018, 18. [Google Scholar]
Platt, D.J.; Morrill, T. Robin’s inequality for 20-free integers. INTEGERS: Electronic Journal of Combinatorial Number Theory 2021. [Google Scholar]
Dusart, P. Estimates of some functions over primes without RH. arXiv preprint 2010, arXiv:1002.0442. [Google Scholar]
Aoudjit, S.; Berkane, D.; Dusart, P. On Robin’s criterion for the Riemann Hypothesis. Notes on Number Theory and Discrete Mathematics 2021, 27, 15–24. [Google Scholar] [CrossRef]
Dusart, P. The k^th prime is greater than k(lnk+lnlnk-1) for k≥2. Mathematics of Computation 1999, 68, 411–415. [Google Scholar] [CrossRef]
Nazardonyavi, S.; Yakubovich, S. Superabundant numbers, their subsequences and the Riemann hypothesis. arXiv preprint 2013, arXiv:1211.2147v3. [Google Scholar]

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

MDPI Initiatives

Important Links

Choose an area of interest and we will send you notifications of new preprints at your preferred frequency.

Disclaimer

Robin’s Criterion on Divisibility (II)

Abstract

1. Introduction

2. A Central Lemma

3. Robin on Divisibility

4. On the Greatest Prime Divisor

5. Some Feasible Cases

6. On Possible Counterexample

7. A Conclusive Approach

Acknowledgments

References

MDPI Initiatives

Important Links

Subscribe