Robin’s Criterion on Divisibility (II)

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 ${\displaystyle \frac{1}{2}}$. As usual $\sigma \left(n\right)$ is the sum-of-divisors function of n:

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

where $d\mid n$ means the integer d divides n and $d\nmid n$ means the integer d does not divide n. Define $f\left(n\right)$ to be ${\displaystyle \frac{\sigma \left(n\right)}{n}}$. Say $\mathrm{Robin}\left(n\right)$ holds provided

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

The constant $\gamma \approx 0.57721$ 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]:

$$log\left(1+{\displaystyle \frac{1}{t}}\right)<{\displaystyle \frac{1}{t}}.$$

The Ramanujan’s Theorem stated that if the Riemann hypothesis is true, then $\mathrm{Robin}\left(n\right)$ holds for large enough n [2]. Next, we have the Robin’s Theorem:

Proposition 2. 

$\mathrm{Robin}\left(n\right)$ holds for all natural numbers $n>5040$ if and only if the Riemann hypothesis is true [3].

It is known that $\mathrm{Robin}\left(n\right)$ holds for many classes of numbers n. $\mathrm{Robin}\left(n\right)$ 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. 

$\mathrm{Robin}\left(n\right)$ 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 $q^2\nmid n$.

Proposition 4. 

$\mathrm{Robin}\left(n\right)$ holds for all natural numbers $n>5040$ that are square free [4].

In addition, we show that $\mathrm{Robin}\left(n\right)$ holds for some $n>5040$ when ${\displaystyle \frac{{\pi }^2}{6}}·loglog{n}^{\prime }\le loglogn$ such that $n^{\prime }$ 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 $a_1\ge a_2\ge \dots \ge a_m\ge 0$ is called an Hardy-Ramanujan integer [4]. A natural number n is called superabundant precisely when, for all natural numbers $m<n$

$$f\left(m\right)<f\left(n\right).$$

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 $ϵ>0$,

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

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 $q_m>e^{31.018189471}$, $1<{\displaystyle \frac{(1+\frac{1.2762}{log{q}_m})·log\left(1.006479799241\right)}{loglogn}}+{\displaystyle \frac{log{N}_m}{logn}}$, ${(logn)}^{{\beta }_n}<1.000208229291·log\left(N_m\right)$ and $n<{\left(1.006479799241\right)}^m·N_m$, 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 ${\beta }_n={\prod }_{i=1}^m{\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}$ 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 $\mathrm{Robin}\left(n\right)$ fails (i.e. $\mathrm{Robin}\left(n\right)$ 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 $\mathrm{Robin}\left(n\right)$ 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]:

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

Proposition 10. 

We have [8]:

$$\prod _{i=1}^{\infty }{\displaystyle \frac{1}{1-\frac{1}{q_i^2}}}=\zeta \left(2\right)={\displaystyle \frac{{\pi }^2}{6}}.$$

The following is a key Lemma. It gives an upper bound on $f\left(n\right)$ that holds for all natural numbers n. The bound is too weak to prove $\mathrm{Robin}\left(n\right)$ 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,

$$f\left(n\right)<{\displaystyle \frac{{\pi }^2}{6}}·\prod _{i=1}^m{\displaystyle \frac{q_i+1}{q_i}}.$$

Proof. 

Putting together the Propositions 9 and 10 yields the proof:

$$f\left(n\right)<\prod _{i=1}^m\left({\displaystyle \frac{q_i}{q_i-1}}\right)=\prod _{i=1}^m\left({\displaystyle \frac{q_i+1}{q_i}}·{\displaystyle \frac{1}{1-\frac{1}{q_i^2}}}\right)<{\displaystyle \frac{{\pi }^2}{6}}·\prod _{i=1}^m{\displaystyle \frac{q_i+1}{q_i}}.$$

□

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]:

$$\left(\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\right)<{\displaystyle \frac{1771561}{1771560}}·e^{\gamma }·loglogn.$$

Proposition 12. 

$\mathrm{Robin}\left(n\right)$ holds for all natural numbers ${10}^{{10}^{13.11485}}\ge n>5040$ [10].

Theorem 1. 

Suppose $n>5040$. If there exists a prime $q\le 1771559$ with $q\nmid n$, then $\mathrm{Robin}\left(n\right)$ holds.

Proof. 

We have that $f\left(n\right)<{\displaystyle \frac{1771561}{1771560}}·e^{\gamma }·loglog\left(n\right)$ for any number $n>{10}^{{10}^{13.11485}}$ since the inequality ${10}^{{10}^{13.11485}}>e^{e^{23.762143}}$ is satisfied. Note that $f\left(n\right)<{\displaystyle \frac{n}{\phi \left(n\right)}}={\prod }_{q\mid n}{\displaystyle \frac{q}{q-1}}$ from the Proposition 9, where $\phi \left(x\right)$ is the Euler’s totient function. Suppose that n is not divisible by some prime $q\le 1771559$ and $n\ge {10}^{{10}^{13.11485}}$. Then,

$$\begin{array}{cc}\hfill f\left(n\right)& <{\displaystyle \frac{n}{\phi \left(n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{n·q}{\phi (n·q)}}·{\displaystyle \frac{q-1}{q}}\hfill \\ \hfill & <{\displaystyle \frac{1771561}{1771560}}·{\displaystyle \frac{q-1}{q}}·e^{\gamma }·loglog(n·q)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\displaystyle \frac{f\left(n\right)}{e^{\gamma }·loglog\left(n\right)}}& <{\displaystyle \frac{1771561}{1771560}}·{\displaystyle \frac{q-1}{q}}·{\displaystyle \frac{loglog(n·q)}{loglog\left(n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1771561}{1771560}}·{\displaystyle \frac{q-1}{q}}·{\displaystyle \frac{loglog\left(n\right)+log(1+\frac{log\left(q\right)}{log\left(n\right)})}{loglog\left(n\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1771561}{1771560}}·{\displaystyle \frac{q-1}{q}}·\left(1+{\displaystyle \frac{log(1+\frac{log\left(q\right)}{log\left(n\right)})}{loglog\left(n\right)}}\right)\hfill \end{array}$$

So

$${\displaystyle \frac{f\left(n\right)}{e^{\gamma }·loglog\left(n\right)}}<{\displaystyle \frac{1771561}{1771560}}·{\displaystyle \frac{q-1}{q}}·\left(1+{\displaystyle \frac{log(1+\frac{log\left(q\right)}{log\left(n\right)})}{loglog\left(n\right)}}\right)$$

for $n\ge {10}^{{10}^{13.11485}}$. The right hand side is less than 1 for $q\le 1771559$ and $n\ge {10}^{{10}^{13.11485}}$. Therefore, $\mathrm{Robin}\left(n\right)$ holds. □

4. On the Greatest Prime Divisor

We know that

Proposition 13. 

For $x\ge 2973$ [11]:

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

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 $\mathrm{Robin}\left(n\right)$ 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 $a_1\ge a_2\ge \dots \ge a_m\ge 0$ since $n>5040$ should be an Hardy-Ramanujan integer. From the Theorem 1, we know that necessarily $q_m\ge 1771559$. So,

$$e^{\gamma }·loglogn\le f\left(n\right)<\prod _{q\le q_m}{\displaystyle \frac{q}{q-1}}<e^{\gamma }·\left(log{q}_m+{\displaystyle \frac{0.2}{log\left(q_m\right)}}\right)$$

because of the Propositions 9 and 13. Hence,

$$loglogn-{\displaystyle \frac{0.2}{log\left(q_m\right)}}<log{q}_m.$$

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

$$\begin{array}{cc}\hfill loglogn-{\displaystyle \frac{0.2}{log\left(q_m\right)}}& \ge 13.11485·log\left(10\right)+loglog10-{\displaystyle \frac{0.2}{log\left(1771559\right)}}\hfill \\ \hfill & >31.018189471.\hfill \end{array}$$

Since, we have that

$$log{q}_m>loglogn-{\displaystyle \frac{0.2}{log\left(q_m\right)}}>31.018189471$$

then, we would obtain that $q_m>e^{31.018189471}$ under the assumption that $n>5040$ is the smallest integer such that $\mathrm{Robin}\left(n\right)$ does not hold. □

5. Some Feasible Cases

We can easily prove that $\mathrm{Robin}\left(n\right)$ is true for certain kind of numbers:

Lemma 3. 

$\mathrm{Robin}\left(n\right)$ holds for $n>5040$ when $q\le 7$, where q is the largest prime divisor of n.

Proof. 

This is an immediate consequence of Theorem 1. □

The next Theorem implies that $\mathrm{Robin}\left(n\right)$ holds for a wide range of natural numbers $n>5040$.

Theorem 3. 

Let ${\displaystyle \frac{{\pi }^2}{6}}·loglog{n}^{\prime }\le loglogn$ for some $n>5040$ such that $n^{\prime }$ is the square free kernel of the natural number n. Then $\mathrm{Robin}\left(n\right)$ holds.

Proof. 

Let $n^{\prime }$ be the square free kernel of the natural number n, that is the product of the distinct primes $q_1,\cdots ,q_m$. By assumption we have that

$${\displaystyle \frac{{\pi }^2}{6}}·loglog{n}^{\prime }\le loglogn.$$

For all square free $n^{\prime }\le 5040$, $\mathrm{Robin}\left(n^{\prime }\right)$ holds if and only if $n^{\prime }\notin \{2,3,5,6,10,30\}$ [4]. However, $\mathrm{Robin}\left(n\right)$ holds for all $n>5040$ when $n^{\prime }\in \{2,3,5,6,10,15,30\}$ due to Lemma 3. When $n^{\prime }>5040$, we know that $\mathrm{Robin}\left(n^{\prime }\right)$ holds and so

$$f\left(n^{\prime }\right)<e^{\gamma }·loglog{n}^{\prime }$$

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

$$f\left(n\right)<{\displaystyle \frac{{\pi }^2}{6}}·\prod _{i=1}^m{\displaystyle \frac{q_i+1}{q_i}}.$$

So,

$$\begin{array}{cc}\hfill f\left(n\right)& <{\displaystyle \frac{{\pi }^2}{6}}·\prod _{i=1}^m{\displaystyle \frac{q_i+1}{q_i}}\hfill \\ \hfill & ={\displaystyle \frac{{\pi }^2}{6}}·f\left(n^{\prime }\right)\hfill \\ \hfill & <{\displaystyle \frac{{\pi }^2}{6}}·e^{\gamma }·loglog{n}^{\prime }\hfill \\ \hfill & \le e^{\gamma }·loglogn\hfill \end{array}$$

according to the formula $f\left(x\right)$ for the square free numbers [4]. □

6. On Possible Counterexample

For every prime number $p_n>2$, we define the sequence $Y_n={\displaystyle \frac{e^{\frac{0.2}{{log}^2\left(p_n\right)}}}{(1-\frac{0.01}{{log}^3\left(p_n\right)})}}$.

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 $\theta \left(x\right)$ is given by

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

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

Proposition 14. 

For $x\ge 7232121212$ [12]:

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

Proposition 15. 

For $x\ge 2278382$ [12]:

$$\prod _{q\le x}{\displaystyle \frac{q}{q-1}}\le e^{\gamma }·\left(logx+{\displaystyle \frac{0.2}{{log}^2\left(x\right)}}\right).$$

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

$$\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\le e^{\gamma }·log\left(Y_m·\theta \left(q_m\right)\right).$$

Proof. 

From the Proposition 14, we know that

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

In this way, we can show that

$$\begin{array}{cc}\hfill log\left(Y_m·\theta \left(q_m\right)\right)& \ge log\left(Y_m·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)}}\right)·q_m\right)\hfill \\ \hfill & =log{q}_m+log\left(Y_m·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)}}\right)\right).\hfill \end{array}$$

We know that

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

Consequently, we obtain that

$$log{q}_m+log\left(Y_m·\left(1-{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)}}\right)\right)\ge \left(log{q}_m+{\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}\right).$$

Due to the Proposition 15, we prove that

$$\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\le e^{\gamma }·\left(log{q}_m+{\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}\right)\le e^{\gamma }·log\left(Y_m·\theta \left(q_m\right)\right)$$

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]:

$$f\left(n\right)=\left(\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\right)·\prod _{i=1}^m\left(1-{\displaystyle \frac{1}{q_i^{a_i+1}}}\right).$$

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 $\mathrm{Robin}\left(n\right)$ does not hold, then ${(logn)}^{{\beta }_n}\le Y_m·log\left(N_m\right)$, where $N_m={\prod }_{i=1}^m{q}_i$ is the primorial number of order m and ${\beta }_n={\prod }_{i=1}^m{\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}$.

Proof. 

According to the Propositions 5 and 7, the primes $q_1<\dots <q_m$ must be the first m consecutive primes and $a_1\ge a_2\ge \dots \ge a_m\ge 0$ 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

$$f\left(n\right)=\left(\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\right)·\prod _{i=1}^m\left(1-{\displaystyle \frac{1}{q_i^{a_i+1}}}\right).$$

However, we know that

$$\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\le e^{\gamma }·log\left(Y_m·log\left(N_m\right)\right)$$

because of the Lemma 5 when $q_m>7232121212$. If we multiply by ${\prod }_{i=1}^m\left(1-{\displaystyle \frac{1}{q_i^{a_i+1}}}\right)$ the both sides of the previous inequality, then we obtain that

$$f\left(n\right)\le e^{\gamma }·log\left(Y_m·log\left(N_m\right)\right)·\prod _{i=1}^m\left(1-{\displaystyle \frac{1}{q_i^{a_i+1}}}\right).$$

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

$$e^{\gamma }·loglogn\le e^{\gamma }·log\left(Y_m·log\left(N_m\right)\right)·\prod _{i=1}^m\left(1-{\displaystyle \frac{1}{q_i^{a_i+1}}}\right)$$

because of

$$e^{\gamma }·loglogn\le f\left(n\right).$$

That is the same as

$$\prod _{i=1}^m{\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}·loglogn\le log\left(Y_m·log\left(N_m\right)\right)$$

(1)

which is equivalent to

$${(logn)}^{{\beta }_n}\le Y_m·log\left(N_m\right)$$

where ${\beta }_n={\prod }_{i=1}^m{\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}$. 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 $\mathrm{Robin}\left(n\right)$ does not hold, then ${(logn)}^{{\beta }_n}<1.000208229291·log\left(N_m\right)$, where $N_m={\prod }_{i=1}^m{q}_i$ is the primorial number of order m and ${\beta }_n={\prod }_{i=1}^m{\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}$.

Proof. 

From the Theorem 2, we know that necessarily $q_m>e^{31.018189471}$. Using the Theorem 4, we obtain that

$${(logn)}^{{\beta }_n}<1.000208229291·log\left(N_m\right)$$

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 $\mathrm{Robin}\left(n\right)$ does not hold, then $n<{\left(1.006479799241\right)}^m·N_m$, 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 $a_1\ge a_2\ge \dots \ge a_m\ge 0$ since $n>5040$ should be an Hardy-Ramanujan integer. From the Lemma 5, we know that

$$\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}\le e^{\gamma }·log\left(Y_m·\theta \left(q_m\right)\right)=e^{\gamma }·loglog\left(N_m^{Y_m}\right)$$

for $q_m>7232121212$. In this way, if $n>5040$ is the smallest integer such that $\mathrm{Robin}\left(n\right)$ does not hold, then $n<N_m^{Y_m}$ since by the Proposition 9 we have that

$$e^{\gamma }·loglogn\le f\left(n\right)<\prod _{i=1}^m{\displaystyle \frac{q_i}{q_i-1}}.$$

That is the same as $n<N_m^{Y_m-1}·N_m$. 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

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

is less than zero for all real numbers $x\ge e^{31.018189471}$. Consequently, we would have that

$$q_m^{Y_m-1}<g\left(e^{31.018189471}\right)<1.006479799241$$

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 $1\le j<m$. Finally, we can state that $n<{\left(1.006479799241\right)}^m·N_m$ since $N_m^{Y_m-1}<{\left(1.006479799241\right)}^m$ when $n>5040$ is the smallest integer such that $\mathrm{Robin}\left(n\right)$ does not hold. □

We know the following results:

Proposition 17. 

For $x>1$ [13]:

$$\pi \left(x\right)\le \left(1+{\displaystyle \frac{1.2762}{logx}}\right)·{\displaystyle \frac{x}{logx}}$$

where $\pi \left(x\right)$ is the prime counting function.

Proposition 18. 

If $n>5040$ is the smallest integer such that $\mathrm{Robin}\left(n\right)$ does not hold, then $p<logn$ 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 $\mathrm{Robin}\left(n\right)$ does not hold, then $1<{\displaystyle \frac{(1+\frac{1.2762}{log{q}_m})·log\left(1.006479799241\right)}{loglogn}}+{\displaystyle \frac{log{N}_m}{logn}}$, where $N_m={\prod }_{i=1}^m{q}_i$ is the primorial number of order m.

Proof. 

Note that $n<{\left(1.006479799241\right)}^m·N_m$ when n is the smallest integer such that $\mathrm{Robin}\left(n\right)$ does not hold. If we apply the logarithm to the both sides, then

$$logn<m·log\left(1.006479799241\right)+log{N}_m.$$

According to the Proposition 17, we have that

$$logn<\left(1+{\displaystyle \frac{1.2762}{log{q}_m}}\right)·{\displaystyle \frac{q_m}{log{q}_m}}·log\left(1.006479799241\right)+log{N}_m.$$

From the Proposition 18, we would have

$$logn<\left(1+{\displaystyle \frac{1.2762}{log{q}_m}}\right)·{\displaystyle \frac{logn}{loglogn}}·log\left(1.006479799241\right)+log{N}_m.$$

which is the same as

$$1<{\displaystyle \frac{\left(1+\frac{1.2762}{log{q}_m}\right)·log\left(1.006479799241\right)}{loglogn}}+{\displaystyle \frac{log{N}_m}{logn}}$$

after of dividing by $logn$. □

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 $a_1\ge a_2\ge \dots \ge a_m\ge 1$ as exponents. Suppose that $\mathrm{Robin}\left(n\right)$ fails. Then,

$${\beta }_n\le {\displaystyle \frac{loglog{\left(N_m\right)}^{Y_m}}{loglogn}},$$

where $N_m={\prod }_{i=1}^m{q}_i$ is the primorial number of order m and ${\beta }_n={\prod }_{i=1}^m\left({\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}\right)$.

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 $x\ge 11$. For $y>x$ [14]:

$${\displaystyle \frac{loglogy}{loglogx}}<\sqrt{{\displaystyle \frac{y}{x}}}.$$

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 $a_1\ge a_2\ge \dots \ge a_m\ge 1$ as exponents. Suppose that $\mathrm{Robin}\left(n\right)$ fails. Then,

$${\beta }_n<\sqrt{{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n}}},$$

where $N_m={\prod }_{i=1}^m{q}_i$ is the primorial number of order m and ${\beta }_n={\prod }_{i=1}^m\left({\displaystyle \frac{q_i^{a_i+1}}{q_i^{a_i+1}-1}}\right)$.

Proof. 

When $n>5040$ is a superabundant number and $\mathrm{Robin}\left(n\right)$ fails, then we have

$${\beta }_n\le {\displaystyle \frac{loglog{\left(N_m\right)}^{Y_m}}{loglogn}}$$

by Lemma 6. We assume that ${\left(N_m\right)}^{Y_m}>n>5040>11$ since ${\beta }_n>1$. Consequently,

$${\displaystyle \frac{loglog{\left(N_m\right)}^{Y_m}}{loglogn}}<\sqrt{{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n}}}$$

by Proposition 19. As result, we obtain that

$${\beta }_n<\sqrt{{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n}}}.$$

□

In number theory, 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 ${\nu }_p\left(n\right)$. Equivalently, ${\nu }_p\left(n\right)$ 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 $\mathrm{Robin}\left(n\right)$ 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 $\mathrm{Robin}\left(n_m\right)$ fails. This implies that $q_m>e^{31.018189471}$ by Theorem 2. Let $n_{m^{\prime }}>5040$ be another large enough superabundant number (as larger as we want) such that $n_{m^{\prime }}>n_m$. Suppose that $\mathrm{Robin}\left(n_{m^{\prime }}\right)$ fails too. By Lemma 7, we have

$${\beta }_{n_m}<\sqrt{{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n_m}}},{\beta }_{n_{m^{\prime }}}<\sqrt{{\displaystyle \frac{{\left(N_{m^{\prime }}\right)}^{Y_{m^{\prime }}}}{n_{m^{\prime }}}}}.$$

So, we would have

$${\beta }_{n_{m^{\prime }}}·{\beta }_{n_m}<\sqrt{{\displaystyle \frac{{\left(N_{m^{\prime }}\right)}^{Y_{m^{\prime }}}}{n_{m^{\prime }}}}}·{\beta }_{n_m}.$$

Consequently, we get:

$${\beta }_{n_{m^{\prime }}}·{\beta }_{n_m}<\sqrt{{\displaystyle \frac{{\left(N_{m^{\prime }}\right)}^{Y_{m^{\prime }}}}{n_{m^{\prime }}}}}·\sqrt{{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n_m}}}.$$

We arrive at:

$${({\beta }_{n_{m^{\prime }}}·{\beta }_{n_m})}^2<{\displaystyle \frac{{\left(N_{m^{\prime }}\right)}^{Y_{m^{\prime }}}}{n_{m^{\prime }}}}·{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n_m}}.$$

We can see that

$${({\beta }_{n_{m^{\prime }}}·{\beta }_{n_m})}^2>1.$$

However, we claim that

$${\displaystyle \frac{{\left(N_{m^{\prime }}\right)}^{Y_{m^{\prime }}}}{n_{m^{\prime }}}}·{\displaystyle \frac{{\left(N_m\right)}^{Y_m}}{n_m}}\le 1$$

which is

$$log{Y}_m\le log\left({\displaystyle \frac{log(n_{m^{\prime }}·n_m)}{log({\left(N_{m^{\prime }}\right)}^{\frac{Y_{m^{\prime }}}{Y_m}}·N_m)}}\right).$$

By Proposition 1, we obtain that

$$\begin{array}{cc}\hfill log{Y}_m& ={\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}+log\left({\displaystyle \frac{{log}^3\left(q_m\right)}{{log}^3\left(q_m\right)-0.01}}\right)\hfill \\ \hfill & ={\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}+log\left(1+{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)-0.01}}\right)\hfill \\ \hfill & <{\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}+{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)-0.01}}\hfill \end{array}$$

for all $q_m>e^{31.018189471}$. Hence, it is enough to show that

$${\displaystyle \frac{0.2}{{log}^2\left(q_m\right)}}+{\displaystyle \frac{0.01}{{log}^3\left(q_m\right)-0.01}}\le log\left({\displaystyle \frac{log(n_{m^{\prime }}·n_m)}{log({\left(N_{m^{\prime }}\right)}^{\frac{Y_{m^{\prime }}}{Y_m}}·N_m)}}\right)$$

which is trivially true under the assumption that

$${log}^2\left(q_m\right)·log\left({\displaystyle \frac{log(n_{m^{\prime }}·n_m)}{log({\left(N_{m^{\prime }}\right)}^{\frac{Y_{m^{\prime }}}{Y_m}}·N_m)}}\right)\ge 0.200322393$$

as a consequence of

$$0.200322393>0.2+{\displaystyle \frac{0.01·{log}^2\left(q_m\right)}{{log}^3\left(q_m\right)-0.01}}>{log}^2\left(q_m\right)·log{Y}_m$$

for all $q_m>e^{31.018189471}$. In this way, we reach the contradiction $1<1$ under the assumption that $\mathrm{Robin}\left(n_m\right)$ fails. This is supported by the fact that $Y_m$ is strictly decreasing (i.e. ${\displaystyle \frac{Y_{m^{\prime }}}{Y_m}}<1$ if $m^{\prime }>m$), ${lim}_{m^{\prime }\to \infty }{Y}_{m^{\prime }}=1$, and we can always be able to take both superabundant numbers $n_m$ and $n_{m^{\prime }}>n_m$ as larger as we want. Furthermore, for every fixed prime q, ${\nu }_q\left(n\right)$ 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

$$n_{m^{\prime }}>n_m,{log}^2\left(q_m\right)·log\left({\displaystyle \frac{log(n_{m^{\prime }}·n_m)}{log({\left(N_{m^{\prime }}\right)}^{\frac{Y_{m^{\prime }}}{Y_m}}·N_m)}}\right)\ge 0.200322393$$

can simultaneously hold for every superabundant number $n_{m^{\prime }}$ greater than some threshold. Accordingly, $\mathrm{Robin}\left(n\right)$ 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 $\mathrm{Robin}\left(n\right)$ fails if the Riemann hypothesis were false. By reductio ad absurdum, we prove that the Riemann hypothesis is true. □