A better method than t-free for Robin's hypothesis

TL;DR

This paper introduces a new, simpler criterion for Robin's inequality, showing that avoiding division by 17th powers of 2 suffices to satisfy the inequality, improving previous t-free conditions.

Contribution

The paper demonstrates that the condition for Robin's inequality can be weakened to N not being divisible by 2^{17}, simplifying the verification process.

Findings

N not divisible by 2^{17} implies Robin's inequality

Simplifies previous t-free conditions for Robin's hypothesis

Provides a new criterion for verifying Robin's inequality

Abstract

For a positive integer t>1, an integer N is called t-free if the exponent of any prime factor of N is less than t. Some works shown if N is t-free, then N satisfies Robin's inequality, for t=5, 7, 11, 16. This article shows that the condition of t-free can be resuced to "N cannot be divided by t-th power of 2". I proved that if N cannot be divided by 17-th power of 2, then N satisfies Robin's inequality.

Equations

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

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

$$\rho (n)<e^{\gamma }\log \log n.$$

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

$$\underset{t\to \infty }{\lim }{2}^t\zeta (t)=1,trunsoverintegers\ge 2.$$

$$\sum _{p\le n}\log \left(\frac{p}{p-1}\right)=\log \log n+\gamma +R(n),$$

$$-\frac{0.005586}{(\log n{)}^2}<R(n)<\frac{0.005586}{(\log n{)}^2}.$$

$$|R(n)|$$

$$=\frac{0.01}{(\log n{)}^2}\left(\frac{1}{2}+\frac{4}{3\log n}\right)<\frac{0.005586}{(\log n{)}^2}.$$

$$N=\prod _{p_i\le p}{p}_i^{a_i}$$

$$T:=\frac{0.005586}{{\left(\log \log \left(10^{(10^{13})}\right)\right)}^2},E(p):=\lfloor \frac{-\log T}{\log p}\rfloor ,$$

$$N=\prod _{p_i\le p_r}{p}_i^{a_i},$$

$$\log T=-12.04,E(2)=\lfloor \frac{12.04}{\log 2}\rfloor =17,$$

$$\rho (N)=\prod _{p_i\le p_r}\frac{p_i-p_i^{-a_i}}{p_i-1}=\prod _{p_i\le p_r}\frac{1-p_i^{-a_i-1}}{1-p_i^{-1}}.$$

$$P(N):=\prod _{p_i\le p_r}\frac{p_i}{p_i-1}=\prod _{p_i\le p_r}\frac{1}{1-p_i^{-1}}.$$

$$S(N):$$

$$=-\sum _{p_i\le p_r}\log (1-p_i^{-a_i-1})=\sum _{p_i\le p_r}\sum _{k=1}^{\infty }\frac{1}{k{p}_i^{(a_i+1)k}},$$

$$\log \rho (N)=\log P(N)-S(N)<\log \log p_r+\gamma +R(p_r)-S(N).$$

$$R(p_r)=\frac{0.005586}{(\log p_r{)}^2}.$$

$$\log \rho (N)<\log \log \log N+\gamma +R(\log N)-S(N),$$

$$S(N)>\sum _{p_i\le p_r}{p}_s^{-a_s-1}\ge p_s^{-a_s-1}\ge p_s^{-E(p_s)}.$$

$$R(\log N)-S(N)$$

$$\le T-p_s^{\frac{\log T}{\log p_s}}=T-e^{(\log p_s)\left(\frac{\log T}{\log p_s}\right)}=T-T=0.$$

$$L:=\prod _{p\le p_r}{p}^{E(p)}.$$

Full text

Robin made a hypothesis [Robin 1984] that the Robin’s inequality

[TABLE]

holds for all integers $n>5040$. Here $\sigma(n)=\sum_{d|n}d$ is the divisor sum function, $\gamma$ is the Euler-Mascheroni constant, log is the nature logarithm.

Robin also proved that his hypothesis is true if and only if Riemann hypothesis is true.

For calculation convenience, we define

[TABLE]

Then Robin’s inequality can also be written as

[TABLE]

Let $N>5040$ be an integer. Write the factorization of N as

[TABLE]

where r is the number of distinct prime factors of N, $p_{i}$ are listed in increasing orders. So $p_{r}$ is the largest prime factor of N.

According to $[Morrill;Platt\,2018]$, (RI) holds for all integers $n,5040<n\leq 10^{(10^{13})}$. So, we assume $N>10^{(10^{13})}$.

For a positive integer $t>1$, an integer N is called t-free if N cannot be divided by prime power $p^{t}$ for any prime p. Some works have shown if N is t-free, then N satisfies Robin’s inequality, for t=5 [CLMS 2007], 7 [Solé; Planat 2012], 11 [Broughan;Trudgian 2015], 16 $[Morrill;Platt\,2018]$. $[Morrill;Platt\,2018]$ also shows 25-free integers satisfy (RI). But that is by a different method of numerical calculation as that article said ”We calculated that $R_{25}(N_{177\,244\,758\,016})<1-10^{-16}$”.

According to abundant number theory, any possible counter-example of (RI) has its exponents of primes in decreasing order, so, the condition ”t-free” is basically equivalent to exponent of prime 2 is less than t.

The previous works on t-free used $\zeta(t)$. However, we have $2^{-t}<\zeta(t)<2^{-t+1}$ for integer $t\geq 2$, and

[TABLE]

Hence, using $\zeta(t)$ has almost the same effect as using $2^{-t}$.

This article shows that the condition of t-free can be reduced to ”N cannot be divided by $2^{t}$”. In particular, I proved: if N cannot be divided by $2^{17}$, then N satisfies Robin’s inequality.

Using a method similar to abundant number theory, I can improve the result to $2^{42}$. I will post that improvement later.

**Lemma 1. (Mertens’ third theorem) **

• For any integer $n>7\,713\,133\,853$, we have*

[TABLE]

where $\gamma$ is the Euler-Mascheroni constant, R(n) is the remainder such that

[TABLE]

[TABLE]

∎

Definition. Let N be an integer, p be its largest prime factor. We define a generalized factorization of N as the product

[TABLE]

where $p_{i}$ runs through all primes $p_{i}\leq p$. If a prime $p_{i}<p$ does not divide N, we set $a_{i}=0$, i.e. $p_{i}^{a_{i}}=p_{i}^{0}=1$ if $p_{i}$ is not a factor of N.

*Theorem 1. ** * Let $N>10^{(10^{13})}$ be an integer. Define

[TABLE]

here $\lfloor x\rfloor$ denotes the largest integer $\leq x$. Write the generalized factorization of N as

[TABLE]

*where $p_{i}$ are listed in increasing order and $p_{r}$ is the largest prime factor of N. If $p_{s}^{E(p_{s})}$ does not divide N for some $p_{s}$, then N satisfies (RI).

In particular, since*

[TABLE]

if $2^{17}$ does not divide N, then N satisfies (RI).

[TABLE]

Define

[TABLE]

Define

[TABLE]

Then apply Lemma 1 to $\log P(N)$,

[TABLE]

where

[TABLE]

By method of Lemma 11 of [CLMS 2007] or Theorem 26 of [NY 2014], we may assume $p_{r}<\log N$.

Since $\log\log x+\gamma+R(x)$ is an increasing function in x, we may substitute $p_{r}$ with $\log N$ in (1.4),

[TABLE]

Now,

[TABLE]

Combine (1.5) and (1.7), we have

[TABLE]

Then from (1.6) we get $\log\rho(N)<\log\log\log N+\gamma$. Take exponential and we are done. ∎

*Corollary 1. ** * Let $N>10^{(10^{13})}$ be an integer. Define

[TABLE]

If N is not divisible by L, then N satisfies (RI).

Remark: The size of L can be estimated as $L\approx 10^{(10^{4.868})}$.

References

[Broughan;Trudgian 2015] K.A. Broughan and T. Trudgian. Robin’s inequality for 11-free integers. Integers, 15: Paper No. A12, 5, 2015.

[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.

[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.

[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.

[Solé; Planat 2012] P. Solé and M. Planat. The Robin inequality for 7-free integers. Integers, 12(2):301–309, 2012.