Subsets of colossally abundant numbers

TL;DR

This paper investigates the properties of colossally abundant numbers and their relation to Robin's hypothesis, establishing that Robin's inequality holds if and only if all CA2 subset numbers greater than 5040 satisfy it.

Contribution

It classifies colossally abundant numbers into three subsets and proves the equivalence between Robin's hypothesis and the Robin inequality holding for CA2 numbers.

Findings

Robin's hypothesis is true if and only if all CA2 numbers > 5040 satisfy Robin inequality.

The paper provides a new classification of colossally abundant numbers into three disjoint subsets.

The results offer a new perspective on the validity of Robin's hypothesis based on subset analysis.

Abstract

Let $G(n)=\sigma (n)/(n \log \log n )$. Robin made hypothesis that $G(n)<e^\gamma$ for all integer $n>5040$. This article divides all colossally abundant numbers in to three disjoint subsets CA1, CA2 and CA3, and shows that Robin hypothesis is true if and only if all CA2 numbers $n>5040$ satisfy Robin inequality.

Tables

index i	$\log n_i$	$P\left(n_i\right)$	is CA1?	is CA2?	$G\left(n_i\right)$
1	0.6931	2	Y	N	-4.0926
2	1.7918	3	Y	N	3.4294
3	2.4849	3	Y	N	2.5634
4	4.0943	5	Y	N	1.9864
5	4.7875	5	Y	N	1.9157
6	5.8861	5	N	Y	1.8335
7	7.8320	7	N	Y	1.8046
8	8.5252	7	N	Y	1.7910
9	10.9231	11	Y	N	1.7512
10	13.4880	13	N	Y	1.7331
11	14.1812	13	N	Y	1.7277
12	15.2798	13	N	Y	1.7235
13	16.8892	13	N	Y	1.7179
14	19.7224	17	N	N	1.7243
15	22.6669	19	N	Y	1.7342
16	25.8023	23	N	Y	1.7374
17	26.4955	23	N	Y	1.7371
18	29.8628	29	N	Y	1.7337
19	33.2968	31	N	Y	1.7340
20	35.2427	31	N	Y	1.7369
21	36.3413	31	N	Y	1.7364
22	39.9522	37	N	Y	1.7375
23	43.6658	41	N	N	1.7380
24	47.4270	43	N	N	1.7403
25	48.1201	43	N	N	1.7406
26	51.9703	47	N	Y	1.7430

K	$x=2^K/K$	$f\left(x\right)$
3	2.67	0.87
4	4.00	1.52
5	6.40	1.94
6	10.67	2.15
7	18.29	2.21
8	32.00	2.16
9	56.89	2.03
10	102.40	1.86
11	186.18	1.67
12	341.33	1.48
13	630.15	1.30
14	1170.29	1.12
15	2184.53	0.97
16	4096.00	0.83

Equations

$$G(n):=\frac{\rho (n)}{\log \log n}.$$

$$G(n)<e^{\gamma },$$

$$F(x,k):=\frac{\log (1+1/(x+x^2+\cdots +x^k))}{\log x}.$$

$$E=\{F(p,k)|primep,integerk\ge 1\}.$$

$$F(x_k,k)=ϵ,1\le k\le K,$$

$$a_p=\{\begin{array}{ll}k,& \text{if xk+1<p≤xk,}\\ 0,& \text{if p>x1.}\end{array}$$

$$n_i:=\prod _p{p}^{a_p}.$$

$$\frac{G(n_{i+1})}{G(n_i)}>{\left(1-\frac{\log q}{3p^2(\log p{)}^2}\right)}^{-1}.$$

$$\frac{G(n_{i+1})}{G(n_i)}<\exp \left(\frac{0.126\log q}{p(\log p{)}^3}\right).$$

$$(x_k+\cdots +x_k^k)\log x_k\ge x_1\log x_1+\left(1-\frac{1}{2x_1}\right)\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}\right).$$

$$(x_k+\cdots +x_k^k)\log x_k<x_1\log x_1+\frac{\log x_1}{2}-\frac{\log x_k}{2}+\frac{\log x_1}{4x_1}.$$

$$(x_k+\cdots +x_k^k)\log x_k\ge x_1\log x_1.$$

$$(x_k+\cdots +x_k^k)\log x_k<x_1\log x_1+\frac{\log x_1}{2}.$$

$$\frac{\log \left(1+\frac{1}{x_k+\cdots +x_k^k}\right)}{\log x_k}=ϵ=\frac{\log \left(1+\frac{1}{x_1}\right)}{\log x_1}.$$

$$x_k^{ϵ}=1+\frac{1}{x_k+\cdots +x_k^k},x_1^{ϵ}=1+\frac{1}{x_1}.$$

$$\frac{x_k+\cdots +x_k^k}{x_1}$$

$$=\frac{ϵ\log x_1+\frac{(ϵ\log x_1{)}^2}{2!}+\cdots }{ϵ\log x_k+\frac{(ϵ\log x_k{)}^2}{2!}+\cdots }$$

$$=\frac{\log x_1}{\log x_k}\left(\frac{1+\frac{ϵ\log x_1}{2!}+\cdots }{1+\frac{ϵ\log x_k}{2!}+\cdots }\right).$$

$$\frac{1+\frac{ϵ\log x_1}{2!}+\cdots }{1+\frac{ϵ\log x_k}{2!}+\cdots }and1+ϵ\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}+c\right),$$

$$H$$

$$-\left(1+\frac{ϵ\log x_k+\cdots }{2!}\right)\left(1+ϵ\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}+c\right)\right)$$

$$=\sum _{j=1}^{\infty }\frac{(ϵ\log x_1{)}^{j-1}}{j!}-\sum _{j=1}^{\infty }\frac{(ϵ\log x_k{)}^{j-1}}{j!}$$

$$-ϵ\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}+c\right)\sum _{j=1}^{\infty }\frac{(ϵ\log x_k{)}^{j-1}}{j!}$$

$$=\sum _{j=2}^{\infty }\frac{(ϵ\log x_1{)}^{j-1}}{j!}-\sum _{j=2}^{\infty }\frac{(ϵ\log x_k{)}^{j-1}}{j!}$$

$$-\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}+c\right)\sum _{j=2}^{\infty }\frac{{ϵ}^{j-1}(\log x_k{)}^{j-2}}{(j-1)!}$$

$$=-ϵc+\sum _{j=3}^{\infty }\frac{(ϵ\log x_1{)}^{j-1}}{j!}$$

$$-\sum _{j=3}^{\infty }\frac{{ϵ}^{j-1}(\log x_k{)}^{j-2}}{j!}\left(\log x_k+\frac{j\log x_1}{2}-\frac{j\log x_k}{2}+jc\right).$$

$$H$$

$$=\sum _{j=3}^{\infty }\frac{{ϵ}^{j-1}}{j!}\left(((\log x_1{)}^{j-1}-(\log x_k{)}^{j-1})-\frac{j(\log x_k{)}^{j-2}}{2}(\log x_1-\log x_k)\right)$$

$$=\sum _{j=3}^{\infty }\frac{{ϵ}^{j-1}(\log x_1-\log x_k)}{j!}\left(\sum _{m=0}^{j-2}(\log x_1{)}^m(\log x_k{)}^{j-2-m}-\frac{j(\log x_k{)}^{j-2}}{2}\right)$$

$$>0.$$

$$\frac{x_k+\cdots +x_k^k}{x_1}>\frac{\log x_1}{\log x_k}\left(1+ϵ\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}\right)\right).$$

$$ϵ=\frac{\log \left(1+\frac{1}{x_1}\right)}{\log x_1}>\frac{1}{x_1\log x_1}-\frac{1}{2x_1^2\log x_1},$$

$$(x_k+\cdots +x_k^k)\log x_k$$

$$>x_1\log x_1+\left(1-\frac{1}{2x_1}\left(\frac{\log x_1}{2}-\frac{\log x_k}{2}\right)\right).$$

$$H<-ϵc+\sum _{j=3}^{\infty }\frac{(ϵ\log x_1{)}^{j-1}}{j!}.$$

$$\sum _{j=3}^{\infty }\frac{(ϵ\log x_1{)}^{j-1}}{j!}$$

Full text

Introduction

Define $\rho(n)=\sigma(n)/n$, where $\sigma(n)=\sum_{d|n}d$ is the sum of divisor function. Define

[TABLE]

Then Robin hypothesis is: all integers $n>5040$ satisfy Robin inequality

[TABLE]

where $\gamma$ is the Euler constant. Let

[TABLE]

Define a set

[TABLE]

Elements $\epsilon_{i}\in E$ are indexed in decreasing order. Elements in E are called critical parameters, For a given critical parameter $\epsilon_{i}$, we can construct a colossally abundant (abbreviate CA) number as follows: Define $x_{k}$ as the solution of

[TABLE]

where K is the largest integer such that $x_{K}\leq 2$. For each prime define

[TABLE]

and define

[TABLE]

It can be proved that $n_{i}$ is a CA number, and $n_{i}$ will be called the CA number constructed from $\epsilon_{i}$. cf. [Broughan 2017] Section 6.3. For any integer $n\geq 2$, we will write $P(n)$ for the largest prime factor of n.

We divide CA in to 3 disjoint subsets. Let $n_{i}$ be the CA number constructed from $\epsilon_{i}$, and p be the prime succeeding $P(n_{i})$.

$n_{i}$ is called a CA1 number if $\log n_{i}<P(n_{i})$. Theorem 1 shows $G(n_{i})<G(n_{i-1}),\forall\,n_{i}\in CA1,i\geq 3$.

$n_{i}$ is called a CA2 number if $P(n_{i})<\log n_{i}<p$.

$n_{i}$ is called a CA3 number if $p<\log n_{i}$. Let $n_{j}$ be the CA number constructed from F(p,1). Theorem 2 shows that $G(n_{i})<G(n_{j})$.

Corollary 4 shows that Robin hypothesis is true if and only if all CA2 numbers $>5040$ satisfy (RI).

Table 1. CA1 and CA2 numbers in the first 26 CA numbers

So, the smallest CA1 number is $n_{1}=2$; the smallest CA2 number is $n_{6}=360$; the smallest CA3 number is $n_{14}=367\,567\,200$.

We next calculate the bounds of increment for $n_{i}\in CA3$. Let $p>10^{8}$ be the prime succeeding to $P(n_{i})$. Assume $\epsilon_{i+1}=F(q,k)$ for some prime q and integer k. Then

Theorem 3 shows a lower bound

[TABLE]

Theorem 4 shows an upper bound

[TABLE]

I checked the first 5 763 320 CA numbers (i.e. with the largest prime factor up to $10^{8}$). They contain 120 529 CA1 numbers, 5 565 CA2 numbers and 5 637 226 CA3 numbers.

Main Content

Lemma 1.

• Let $\epsilon\in E$ be a critical parameter and $k\geq 1$ be an integer. Let $x_{1}$ and $x_{k}$ be defined by (4). Then*

[TABLE]

[TABLE]

(L1.1) and (L1.2) mean simple version:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence

[TABLE]

Compare

[TABLE]

where c is a to-be-determined real parameter.

[TABLE]

To prove (L1.1), set $c=0$. The lower bound of H is

[TABLE]

Combine (L1.5), (L1.7) and (L1.8), we have

[TABLE]

Since

[TABLE]

we get

[TABLE]

That is, (L1.1) holds.

To prove (L1.2), we have from (L1.7)

[TABLE]

The summation in (L1.10) can be simplified as

[TABLE]

By (L1.3), $\epsilon<1/(x_{1}\log x_{1})$, and we have

[TABLE]

Combine (L1.5), (L1.6) and (L1.12), we get

[TABLE]

∎

Theorem 1. Let $i\geq 3$ be an integer and $n_{i}$ be a CA1 number, $p=P(n_{i})$. Then

[TABLE]

[TABLE]

1. $n_{i}/n_{i-1}=q$. Assume $\epsilon_{i}=F(q,k)$ for some prime q and integer $k\geq 1$.

[TABLE]

By Lemma 1 (L1.1’), we have

[TABLE]

Hence

[TABLE]

2. $n_{i}/n_{i-1}=qr$. Assume $\epsilon_{i}=F(q,k)=F(r,j)$ for some prime q, r and integer $k\geq 1,j\geq 1$. Then we have

[TABLE]

By Lemma 1 (L1.1’), we have

[TABLE]

Hence we get

[TABLE]

∎

Corollary 1. *Let $n_{i}>n_{8}=5040$ be a CA1 number. Let $n_{j}$ be the largest non-CA1 number below $n_{i}$. Then $G(n_{i})<G(n_{j})$. *

[TABLE]

∎

Corollary 2. *Robin hypothesis is true if and only if all non-CA1 numbers $>5040$ satisfy (RI). *

Lemma 2. Let $\epsilon\in E$ be a critical epsilon value. $x_{k}$ are solutions of

[TABLE]

Then $g(t)=g_{\epsilon}(t):=t^{\epsilon}/\log\log t$ has a unique minimum, say $t_{0}$, and $t_{0}$ satisfies

[TABLE]

[TABLE]

Define

[TABLE]

It is obvious that f(t) monotonically increases for $t\in(e,\infty)$, negative near e and positive when t sufficiently large. So $f(t)$ has a unique zero $t_{0}$. $g(t)$ attains minimum at $t_{0}$. Note $x_{1}$ is the solution of $F(x_{1},1)=\log(1+1/x_{1})/\log x_{1}=\epsilon$, Write $t=x_{1}+1/2+d$, where $d=-1/(2\log x_{1})$. We have

[TABLE]

So we get the left inequality of (L2.2). For the right inequality, we have

[TABLE]

here the expansion of $(\log(1+1/x_{1}))^{-1}=x+\frac{1}{2}-\frac{1}{12x_{1}}+\frac{1}{24x_{1}^{2}}-\cdots$ is calculated term wise from the formula $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2x^{2}}+\frac{1}{3x^{3}}-\cdots$. So we have

[TABLE]

∎

Lemma 3.

• Let $\epsilon\in E$ be a critical epsilon value. Let u and $u_{1}<u_{2}$ be positive reals. Then $h(u):=e^{\epsilon u}/\log u$ has a unique minimum at $u_{0}=u_{0}(\epsilon)$ implicitly defined by $\epsilon=1/(u_{0}\log u_{0})$. Assume $u_{0}>40$. Write*

[TABLE]

1) For $u_{0}-\frac{1}{2}<u_{1}<u_{0}$,

[TABLE]

2) For $u_{0}<u_{2}<u_{0}\log u_{0}$,

[TABLE]

3) For $u_{0}<u_{1}<u_{2},\,u_{2}-u_{1}<\log u_{0}$,

[TABLE]

[TABLE]

1. When $u_{0}-\frac{1}{2}<u_{1}<u_{0}$, we have $\log(u_{1}/u_{0})<0$. Hence

[TABLE]

[TABLE]

2. When $u_{0}<u_{2}<u_{0}\log u_{0}$, we have $\log(u_{2}/u_{0})>0$.

[TABLE]

[TABLE]

3. Write $u_{2}=u_{1}+a$ for some real $a<\log u_{0}$.

[TABLE]

Since

[TABLE]

and

[TABLE]

we have

[TABLE]

∎

Lemma 4. *Assume $g(t)=t^{\epsilon}/\log\log t$ takes minimum at $t_{0}=t_{0}(\epsilon)$. Assume $\log t_{0}>40$. Let N and $N_{1}$ be positive integers.

1. If $\log t_{0}-\frac{1}{2}<\log N<\log t_{0}$ and $\log t_{0}+2<\log N_{1}$, then*

[TABLE]

2) If $\log t_{0}<\log N<\log N_{1}$ and $\log N_{1}-\log N<\log\log t_{0}$, then

[TABLE]

[TABLE]

By Lemma 3 (L3.1)

[TABLE]

we have

[TABLE]

2. follows from Lemma 3 (L3.3). ∎

Theorem 2. Let $n_{i}$ be CA3. Let p be the prime succeeding $P(n_{i})$, $n_{j}$ be the CA number constructed from $\epsilon_{j}=F(p,1)$. then

[TABLE]

*where $t_{0}$ is defined as in Lemma 4. *

[TABLE]

[TABLE]

where $g(t)=t^{\epsilon_{j}}/\log\log t$ . By Lemma 2, $g(t)$ attains minimum at $t_{0}$, and

[TABLE]

The smallest CA3 number $n_{14}$ can be directly checked. So, we may start from the next CA3 number $n_{23}$. That is, we may assume $n_{i}\geq n_{23}$ with $p\geq 43$ and $\log p\geq 3.76$.

Case 1) $\log n_{i}<\log t_{0}$. In this case, we have $\log t_{0}<p+\frac{1}{2}$ by (2.3). Hence $\log t_{0}-\frac{1}{2}<p<\log n_{i}<\log t_{0}$, and

[TABLE]

Hence the conditions of Lemma 4 (L4.1) are satisfied and (2.1) holds.

Case 2) $\log n_{i}\geq\log t_{0}$. In this case, $\log n_{j}-\log n_{i}<\log t_{0}$. So by Lemma 4 (L4.2), we have

[TABLE]

Since $0.3337(\log n_{j}-\log n_{i})^{2}\geq 0.3337\times(\log 43)^{2}>3.2961$, (2.1) holds. ∎

Corollary 3. *Let $n_{i}$ be a CA3 number. Then there exists $n_{j}\in CA2$ such that $n_{i}<n_{j}$. If $n_{j}$ is the smallest CA2 number above $n_{i}$, then $G(n_{i})<G(n_{j})$. *

We claim that $n_{k-1}$ is CA2. $n_{k-1}$ is not CA1 by minimality of $n_{k}$. If $n_{k-1}$ were CA3, there would exist a prime p such that $P(n_{k-1})<p<\log n_{k-1}$. Then we would have

[TABLE]

This contradicts to $n_{k}\in CA1$. So $n_{k-1}\in CA2$ and we proved the existence of $n_{j}$.

Write $p_{r}=P(n_{i})$, $p_{s}=P(n_{j})$. Let $n_{i_{m}}$ be the CA number generated from parameter $F(p_{m},1)$, $r<m\leq s$. Since $n_{k}$ is the smallest CA1 number above $n_{i}$, and $n_{j}<n_{k}$ is the smallest CA2 number above $n_{i}$, all $n_{i_{m}}<n_{j}$ are CA3. By Theorem 2, we have

[TABLE]

∎

Corollary 4. *Robin hypothesis is true if and only if all CA2 numbers $>5040$ satisfy (RI). *

Under assumption of Theorem 2, is $G(n_{i})<G(n_{i+1})$? Let $\epsilon_{i+1}=F(q,k)$. If $q\geq 3$, Theorem 3 proves $G(n_{i})<G(n_{i+1})$. The case $q=2$ is open. Theorem 3 also shows a lower bound for $G(n_{i+1})/G(n_{i})$ .

Theorem 3. *Let $n_{i}$ be CA3. Let p be the prime succeeding $P(n_{i})$.

1. If $\epsilon_{i+1}=F(q,k)$, $q\geq 3$, then $n_{i+1}/n_{i}=q$, and $G(n_{i})<G(n_{i+1})$.

2. If $\epsilon_{i+1}=F(q,k)$, $q\geq 23$, then $n_{i+1}/n_{i}=q$, and*

[TABLE]

3) If $\epsilon_{i+1}=F(q,k)=F(r,j)$ , $q\geq 23,r\geq 23$, then $n_{i+1}/n_{i}=qr$, and

[TABLE]

1. and 2). Since $\epsilon_{i+1}=F(q,k)$, we have $n_{i+1}=n_{i}q$. Compare $G(n_{i})$ and $G(n_{i+1})$, we have

[TABLE]

Since $\log n_{i}>p>x_{1}$, where $x_{1}$ is defined by (4), Lemma 1 (L1.2’) means

[TABLE]

[TABLE]

When $q\geq 3$, the expression in (3.5) is negative, so (3.4) means $G(n_{i})<G(n_{i+1})$. That is, 1) is true. Now for 2) we have

[TABLE]

It is easy to verify that

[TABLE]

Combine (3.6) and (3.7), we get

[TABLE]

3. Assume $\epsilon_{i+1}=F(q,k)=F(r,j)$, $q\geq 23,r\geq 23$. Then $n_{i+1}=n_{i}qr$. Compare $G(n_{i})$ and $G(n_{i+1})$, we have

[TABLE]

By Lemma 1 (L1.2’), we have

[TABLE]

Then we can proceed with q and r separately as in 2) to prove (3.2). ∎

We will prove Lemmas 5-7, then use them to prove an upper bound for $G(n_{i+1})/G(n_{i})$ in Theorem 4.

Lemma 5. Define

[TABLE]

where $K(x)$ is implicitly defined as the largest integer K satisfying

[TABLE]

*Then

1. $f(x)$ is a piece-wise differentiable function with discontinuous points at $x=\frac{2^{K}}{K}$ for each integer $K\geq 3$.

2. $f(x)$ decreases at differentiable points.

3. $f(x)$ has local maximums at discontinuous points $x=\frac{2^{K}}{K}$. $f\left(\frac{2^{K}}{K}\right)>f\left(\frac{2^{K+1}}{K+1}\right)$, for $K\geq 7$.

4. In particular,*

[TABLE]

1. and 2) are simple.

[TABLE]

So f(x) decreases at all differentiable points.

3. Because $f(x)$ adds an extra summand 2 at point $x=\frac{2^{K}}{K}$, it is discontinuous there. To show $f(x)$ decreases from one discontinuous point to next, let $x_{s}=2^{K(x_{s})}/K(x_{s})$, i.e. $(K(x_{s})x_{s})^{1/K(x_{s})}=2$. Then the next discontinuous point is $x_{t}:=2^{K(x_{t})}/K(x_{t})$ where $K(x_{t})=K(x_{s})+1$. So we have

[TABLE]

Now we want to show $f(x_{s})>f(x_{t})$.

[TABLE]

When $K(x_{t})\geq 8$,

[TABLE]

[TABLE]

For $x_{t}<2868$ and $K\geq 7$, $f(2^{K}/K)>f(2^{K+1}/(K+1))$ can be directly calculated:

4. Direct calculation shows $f\left(\frac{2^{31}}{31}\right)=0.10923475$. ∎

Lemma 6.

• Let $\theta(x)$ and $\psi(x)$ be Chebyshev functions. Define*

[TABLE]

where K is the largest integer k such that $(kx)^{1/k}\geq 2$. Then

[TABLE]

[TABLE]

Setting $k=2,\eta_{2}=0.01$ in Theorem 4.2 of [Dusart 2018], we have,

[TABLE]

Combine (L6.3) and (L6.4), we get

[TABLE]

By (L6.5) and Lemma 5, we have

[TABLE]

∎

Lemma 7. Let n be a CA number and $p=P(n)$. Then

[TABLE]

[TABLE]

where K is the largest integer k such that $(kx)^{1/k}\geq 2$ and $\psi_{0}$ is defined as in Lemma 6. By Lemma 6, we have

[TABLE]

∎

Theorem 4. Let $n_{i}$ be CA3. Let p be the prime succeeding $P(n_{i})$ and $p>10^{8}$. Then $p<\log n_{i}$, Assume $n_{i+1}=n_{i}q$ and $\epsilon_{i+1}=F(q,k)$ for some prime q and integer k. Then

[TABLE]

[TABLE]

By Lemma 7,

[TABLE]

$c:=1+\frac{0.06323}{(\log p)^{2}}$, for $p>10^{8}$. Hence

[TABLE]

By Lemma 2, $p<\log t_{0}$. So (4.2) means

[TABLE]

∎

References

[Briggs 2006] K. Briggs. Abundant numbers and the Riemann hypothesis. Experiment. Math., 15(2):251–256, 2006.

[Broughan 2017] K. Broughan, Equivalents of the Riemann Hypothesis Vol 1. Cambridge Univ. Press. (2017)

[CLMS 2007] Y. -J. Choie, N. Lichiardopol, P. Moree, and P. Solé. On Robin’s criterion for the Riemann hypothesis. J. Théor. Nombres Bordeaux, 19(2):357–372, 2007.

[CNS 2012] G. Caveney, J.-L. Nicolas, and J. Sondow, On SA, CA, and GA numbers, Ramanujan J. 29 (2012), 359–384.

[Dusart 1998] P. Dusart Sharper bounds for $\psi$, $\theta$, $\pi$, $p_{k}$, Rapport de recherche $n\,1998-06$, Laboratoire d’Arithmétique de Calcul formel et d’Optimisation

[Dusart 2018] P. Dusart. Explicit estimates of some functions over primes. Ramanujan J., 45(1):227–251, 2018.

[EN 1975] P. Erdös and J.-L Nicolas, Repartition des nombres superabundants. Bulletin de la S. M., tome 103 (1975), p. 65-90

[Morrill;Platt 2018] T. Morrill, D. Platt. Robin’s inequality for 25-free integers and obstacles to analytic improvement

https://arxiv.org/abs/1809.10813

[NY 2014] S. Nazardonyavi and S. Yakubovich. Extremely Abundant Numbers and the Riemann Hypothesis. Journal of Integer Sequences, Vol. 17 (2014), Article 14.2.8

[PT 2016] D.J. Platt and Tim Trudgian. On the first sign change of $\theta(x)-x$. Math. Comp. 85 (2016), 1539-1547

[Robin 1984] G. Robin. Grandes valeurs de la fonction somme des diviseurs et hypothése de Riemann. Journal de mathématiques pures et appliquées. (9), 63(2):187–213, 1984.

[Wu 2019] X. Wu. Properties of counterexample of Robin hypothesis.

https://arxiv.org/abs/1901.09832