A New Type of Abundant Numbers

TL;DR

This paper introduces a new class of abundant numbers called LR numbers and establishes a connection between these numbers and Robin's hypothesis, showing the hypothesis holds if all LR numbers greater than 5040 satisfy Robin inequality.

Contribution

It defines LR numbers and proves their significance in relation to Robin's hypothesis, providing a new perspective on the problem.

Findings

Robin hypothesis is true iff all LR numbers > 5040 satisfy Robin inequality

LR numbers are a new class of abundant numbers with specific properties

The paper establishes a criterion linking LR numbers to Robin's hypothesis

Abstract

This article defines a new type of abundant numbers, called largest rho-value (abbreviate LR) numbers, and then shows that Robin hypothesis is true if and only if all LR numbers $>5040$ satisfy Robin inequality.

Tables

m	q	k	$z_m$	${\delta }_{q,k}$	$n_m$	$\rho \left(n_m\right)$	$\log \left(n_m\right)$	$G\left(n_m\right)$
1	2	1	2	0.4055	2	1.5000	0.6931	-4.0296
2	3	1	3	0.2877	6	2.0000	1.7918	3.4294
3	5	1	5	0.1823	30	2.4000	3.4012	1.9606
4	2	2	6	0.1542	60	2.8000	4.0943	1.9864
5	7	1	7	0.1335	420	3.2000	6.0403	1.7793
6	11	1	11	0.0870	4,620	3.4909	8.4381	1.6368
7	3	2	12	0.0800	13,860	3.7818	9.5368	1.6770
8	13	1	13	0.0741	180,180	4.0727	12.1017	1.6334
9	2	3	14	0.0690	360,360	4.3636	12.7949	1.7119
10	17	1	17	0.0572	(17#)(3#)2	4.6203	15.6281	1.6807
11	19	1	19	0.0513	(19#)(3#)2	4.8635	18.5725	1.6646
12	23	1	23	0.0426	(23#)(3#)2	5.0750	21.7080	1.6490
13	29	1	29	0.0339	(29#)(3#)2	5.2500	25.0753	1.6295
14	5	2	30	0.0328	$\left(29\#\right)\left(5\#\right)2$	5.4249	26.6847	1.6519
15	2	4	30	0.0328	$\left(29\#\right)\left(5\#\right)2^2$	5.6058	27.3779	1.6937
16	31	1	31	0.0317	$\left(31\#\right)\left(5\#\right)2^2$	5.7866	30.8119	1.6881
17	37	1	37	0.0267	$\left(37\#\right)\left(5\#\right)2^2$	5.9430	34.4228	1.6794
18	3	3	39	0.0253	(37#)(5#)(3#)2	6.0954	35.5214	1.7073
19	41	1	41	0.0241	(41#)(5#)(3#)2	6.2441	39.2350	1.7016
20	43	1	43	0.0230	(43#)(5#)(3#)2	6.3893	42.9962	1.6988

Equations

$$\rho (n):=\frac{\sigma (n)}{n},$$

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

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

$$n=\prod _{i=1}^r{p}_i^{a_i},$$

$$\omega (n):=\sum _{i=1}^r{a}_i.$$

$$S_m:=\{n\in \mathbb{N}|n\ge 2,\omega (n)=m\}.$$

$$Z:=\{z=z_{q,k}:=q+\cdots +q^k|qprime,k\ge 1integer\}.$$

$${\delta }_i={\delta }_{q_i,k_i}:=\log \left(1+\frac{1}{z_i}\right)=\log \left(1+\frac{1}{q_i+\cdots +q_i^{k_i}}\right).$$

$$E:=\{(q,k,{\delta }_{q,k})\}.$$

$$n_m:=\prod _p{p}^{k_p}.$$

$$y_{m,k}+\cdots +y_{m,k}^k=z_m.$$

$$\log \rho (p^k)-\log \rho (p^{k-1})$$

$$=\log \left(\frac{1+\cdots +p^k}{p(1+\cdots +p^{k-1})}\right)$$

$$=\log \left(1+\frac{1}{p+\cdots +p^k}\right)={\delta }_{p,k}.$$

$$\log \rho (p^k)=\sum _{i=1}^k{\delta }_{p,k},\forall k\ge 1.$$

$$n=\prod _p{p}^{a_p}$$

$$\log \rho (n)=\log \prod _p\rho (p^{a_p})=\sum _p\log \rho (p^{a_p})=\sum _p\sum _{i=1}^{a_p}{\delta }_{p,i}.$$

$$z_m^{1/k}-1<y_k<z_m^{1/k}.$$

$$\theta (z_m^{1/k})-\frac{1}{k}\log z_m\le \theta (y_k)\le \theta (z_m^{1/k}).$$

$$(y_k+1{)}^k>y_k+\cdots y_k^k=z_m⟹y_k>z_m^{1/k}-1.$$

$$y_k^k<y_k+\cdots y_k^k=z_m⟹y_k<z_m^{1/k}.$$

$$\theta (y_k)>\theta (z_m^{1/k})-\log (z_m^{1/k})=\theta (z_m^{1/k})-\frac{1}{k}\log z_m.$$

$$\lfloor \frac{\log z_m}{\log p}\rfloor -1\le k_p\le \lfloor \frac{\log z_m}{\log p}\rfloor .$$

$$k_p=\{\begin{array}{ll}k,& y_{k+1}<p\le y_k,\\ 0,& y_1<p.\end{array}$$

$$D(x,k):=\log \left(1+\frac{1}{x+\cdots +x^k}\right).$$

$$W_1:=\sum _{i=1}^{\infty }\left(\frac{1}{z_i}-\log \left(1+\frac{1}{z_i}\right)\right)=0.20208\cdots ,$$

$$W_2:=M+\sum _{z\in Z,z\ne prime}\frac{1}{z}=0.77929\cdots ,$$

$$M:=\gamma +\sum _p\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)=0.26149\cdots .$$

$$-W_1+W_2=\gamma .$$

$$-W_1$$

$$=-\sum _{i=1}^{\infty }\left(\frac{1}{z_i}-\log \left(1+\frac{1}{z_i}\right)\right)$$

$$+\gamma +\sum _p\left(\log \left(1-\frac{1}{p}\right)+\frac{1}{p}\right)+\sum _{z\in Z,z\ne prime}\frac{1}{z}-\gamma$$

Full text

Introduction

Let $n\geq 2$ be an integer, $\sigma(n)=\sum_{d|n}$ is the sum of divisor function. Define

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Robin [Robin 1984] made hypothesis that all integers $n>5040$ satisfy Robin inequality

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where $\gamma$ is the Euler constant.

Write the factorization of n as

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where $p_{i}$ is the i-th prime and $a_{r}\geq 1$. If $p_{i}$ and n are co-prime, we set its exponent $a_{i}=0$.

Define the function of sum of exponents:

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For an integer $m\geq 1$, define a set

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If there exists an element $n_{m}\in S_{m}$ such that $\rho(n_{m})\geq\rho(n),\,\forall\,n\in S_{m}$ , we call $n_{m}$ a largest rho-value (abbreviate LR) number.

We now construct $n_{m}$. Define a set

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We sort Z in increasing order.

Note that Z contains duplicate integers. One known example is $5+5^{2}=2+2^{2}+2^{3}+2^{4}$. If $q+\cdots+q^{k}=p+\cdots+p^{j}$ and $q>p$, then we assign $z_{q,k}$ smaller order than $z_{p,j}$. We will denote by $z_{i}$, $q_{i}$ and $k_{i}$ the numbers related to the i-th element.

Next, we define

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Since $z_{i}$ are increasing, $\delta_{i}$ are decreasing. Define the set of triplets

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Elements in E are ordered by $\delta_{i}$.

Now for a given integer $m\geq 1$, let $E_{m}$ be the subset of first m elements in E. For any prime p, let $k_{p}$ be the largest k among elements $\delta_{p,k}\in E_{m}$. Define

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Theorem 1 will show that $n_{m}$ is an LR number. That is, $\rho(n_{m})\geq\rho(n),\,\forall\,n\in S_{m}$.

Table 1. First 20 LR numbers

Here “#” means primorial.

Theorems 2-8 study properties of $n_{m}$. Theorem 9 shows that Robin hypothesis is true if and only if all LR numbers $n_{m}>5040$ satisfy Robin inequality.

We will use $\theta(x)$and $\psi(x)$ for Chebyshev functions.

For an integer $m\geq 1$ define $y_{m,1}:=z_{m}$, and for $k\geq 2$, define $y_{m,k}$ as the solution of

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When m is obvious, we will simply write $y_{k}$ instead of $y_{m,k}$.

Main Content

Theprem 1.

• Let $n_{m}$ be constructed as in (8). Then $\rho(n_{m})\geq\rho(n),\,\forall\,n\in S_{m}$. That is, $n_{m}$ is an LR number.*

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Hence

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For any integer

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in $S_{m}$ we have $\sum_{p}a_{p}=m$ and

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By construction of $n_{m}$, $\log n_{m}=\sum_{p}\sum_{i=1}^{k_{p}}\delta_{p,i}$ is the largest sum of m elements in E. So $\log\rho(n_{m})\geq\log\rho(n)$ for all $n\in S_{m}$. Hence $\rho(n_{m})\geq\rho(n)$ or all $n\in S_{m}$. ∎

Lemma 1. For given $z_{m}$ and $k\geq 2$, we have

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Since there is at most one prime in $(z_{m}^{1/k}-1,\,z_{m}^{1/k}]$, we have

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∎

Lemma 2. For given $n_{m}$, we have for each prime $p<z_{m}$

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$p^{k_{p}+2}\geq 2p^{k_{p}+1}>(1+p^{-1}+\cdots+p^{-k_{p}})p^{k_{p}+1}=p+\cdots+p^{k_{p}+1}>z_{m}$ means $k_{p}+2>\log z_{m}/\log p$. Hence (L2.1) holds. ∎

Theorem 2. Let $m\geq 1$ be an integer, $n_{m}$ and $k_{p}$ be defined as in (8), $y_{k}$ be defined as in (9). Then

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*when $\delta_{m}\neq\delta_{m+1}$.

When $\delta_{q,k}=\delta_{m}=\delta_{m+1}=\delta_{q^{\prime},k^{\prime}}$ and $q>q^{\prime}$, then $k_{q}=k$, $k_{q^{\prime}}=k^{\prime}-1$. *

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First assume $\delta_{m}\neq\delta_{m+1}$. Since $D(x,k)$ decreases in x, $p\leq y_{k}$ means $D(p,k)\geq D(y_{k},k)=D(z_{m},1)$. Hence $D(p,k)=\delta_{p,k}\in E_{m}$. That is, $k_{p}\geq k$. On the other hand, $y_{k+1}<p$ means $D(p,k+1)<D(y_{k+1},k+1)=D(z_{m},1)$. Hence $D(p,k+1)=\delta_{p,k+1}\notin E_{m}$. That is, $k_{p}<k+1$. So $k_{p}=k$.

The $\delta_{m}=\delta_{m+1}$ case is obvious from the requirement ”larger prime has smaller index” when sorting Z. ∎

Define two constants

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here M is the Meissel-Mertens constant

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Theorem 3. **

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∎

Theorem 4. **

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Hence

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Therefore

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The first term is $-W_{1}$. The second term has lower bound 0 and upper bound

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∎

Define

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Theorem 5. **

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Substitute $f(x)=1/x$, we get

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We need to determine the constant. We have, by Mertens Theorem,

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Substitute (5.5) into (5.4), we get

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By theorems 3 and 4,

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∎

We now find bounds for $\psi_{Z}(x)$ in term of $\theta(x)$.

Theorem 6.

• Let $m\geq 2$ be an integer. Then for $z_{m}$ we have*

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where $K=\lfloor\log z_{m}/\log 2\rfloor$ (cf. Lemma 2).

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For (6.2), we have, by Mertens theorem,

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Using Mertens Theorem for large $y_{k}$ and numerical calculation for small $y_{k}$, we have

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Hence

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The cases of $2\leq m\leq 6$ can be numerically verified.

To prove (6.3), set $k=2,\eta_{k}=0.2,x_{k}=3\,594\,641$ in [Dusart 2018] Theorem 4.2, we have for $x\geq 3\,594\,641$,

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Hence

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∎

Theorem 7. Let $m\geq 1$ be an integer. Then

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For the right inequality

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where $K=\lfloor\log z_{m}/\log 2\rfloor$. For the left inequality,

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∎

Theorem 8. **

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Hence

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∎

Lemma 3. * If the Riemann hypothesis is false, then there exists a real b with $0<b<1/2$ and a sequence $(x_{i})$ of reals such that $x_{i}$ is not a prime,*

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and

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By Step (5) of the proof of Theorem 5.29 of [Broughan 2017], there exists a real b with $0<b<1/2$ such that

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where $\Omega_{\pm}$ is the oscillation symbol. That means there exists a sequence $(x_{i})$, $\lim_{i\rightarrow\infty}x_{i}=\infty,$ of reals such that

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Fix an odd i. Then (L3.2) holds for $x_{i}$. We can choose $x_{i}$ so that $K(x_{i})+x_{i}^{-b}$ is a local minimum of $K(x)+x^{-b}$. Define

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Then r(x) changes sign at $x_{i}$. Assume $x_{i}$ lies in an interval $[p,p^{\prime})$ of two consecutive primes.

We claim that $x_{i}\neq p$. For if $x_{i}=p$, then

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If $r(x)$ changed sign at p, $r(x)$ would be positive on the left of $x_{i}$ and negative on the right. That is, $K(p)+p^{-b}$ would be a local maximum, which contradicts $K(x_{i})+x_{i}^{-b}$ being a local minimum.

Therefore, we must have $x_{i}\in(p,p^{\prime})$ and

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Hence (L3.1) holds for all $x_{i}$ with odd i. The sequence $(x_{i})$ in the lemma statement can be set to $(x_{2i+1})$ above. ∎

Lemma 4. * If the Riemann hypothesis is false, then there exists a real b with $0<b<1/2$ and a sequence $(n_{m})$ of LR numbers such that*

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and

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By Lemma 3, there exists a sequence $(x_{i})$ of reals such that $x_{i}$ is not a prime,

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and

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Assume $x_{i}$ is in interval $[z_{m},z_{m+1})\subset[p,p^{\prime})$ with p and p’ consecutive primes. If $x_{i}=z_{m}$, then we are done. So we may assume $x_{i}\in(z_{m},z_{m+1})$. Define

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Then for $x^{-b}\log x<1$ we have

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By [BH 2001] Theorem 1, for sufficiently large x, there is at least one prime in $[x,x+x^{0.525}]$. Hence

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Since $r(x_{i})=0$ and $r(x)<0$ for $x\in(z_{m},x_{i})$, we have

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(L4.5) and (L4.9) mean (L4.1) is true. For (L4.2) we have

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Hence

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So (L4.2) holds. ∎

Lemma 5. **

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So (L5.1) holds. ∎

Theorem 9. * Robin hypothesis is true if and only if all LR numbers $n_{m}>5040$ satisfy*

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If Robin hypothesis is false, then Riemann hypothesis is false. By Theorem 6, we have

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Hence by Theorems 5 and 8 and Lemma 5, we have, for large LR number $n_{m}$,

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By Lemma 4, there exists a real b with $0<b<1/2$ and a sequence $(z_{m})$ such that

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and

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Substitute (9.4) and (9.5) into (9.3), we get

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That is, Robin inequality fails for some large $n_{m}$. ∎

References

[BH 2001] R. C. Baker and G. Harman, The difference between consecutive primes, II, Proc. London Math. Soc. (3) 83 (2001) 532-562.

[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).

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

[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).