10
$\begingroup$

I am looking for an upper-bound of the Euler's totient function $\varphi$ which would be equivalent to the Riemann hypothesis (RH). There is the following Nicolas' criterion about primorial numbers $N_k$ (product of the $k$ first prime numbers):

Nicolas' criterion (MR724536, Théorème 2)

  • If RH is true, then for all $k \ge 1$ $$ N_k / \varphi(N_k) > e^{\gamma}\ln \ln N_k $$
  • If RH is false then there are infinitely many $k$ such that the above inequality holds, and infinitely many $k$ such that it does not.

In particular, RH is equivalent to the following upper-bound: $$\varphi(n)< \frac{n}{e^{\gamma}\ln \ln n}, \text{ for } n=N_k \text{ and } k>1.$$

Now, the above inequality does not hold for $n$ prime greater than $7$. So an upper-bound for all $n$ requires a modification. Let $\omega(n)$ be the number of distinct prime factors of $n$. Then $N_k$ is the smallest number $n$ satisfying $\omega(n)=k$. Consider the following step function $$s=\sum_{k=1}^{\infty}k\chi_{(N_{k-1},N_k]}.$$

Question: Is RH equivalent to the following upper-bound? $$\varphi(n)< \frac{n2^{s(n)-\omega(n)}}{e^{\gamma}\ln \ln n}, \text{ for } n>2.$$

Because $s(n) = \omega(n)$ iff $n$ is primorial (i.e. of the form $N_k$), RH is equivalent to this upper-bound for $n$ primorial, by Nicolas' criterion. So it remains to deduce (from RH) this upper-bound for $n$ non-primorial. Consider the map $\alpha$ defined by $\alpha(N_k)=1$ and for $n$ non-primorial by $$\alpha(n)=\left( \frac{\varphi(n)e^{\gamma}\ln \ln n}{n} \right)^{1/(s(n)-\omega(n))}$$ Then it remains to deduce that $\alpha(n)<2$.

Proposition: There is a sequence $(n_r)$ such that $\alpha(n_r) \to 2$.

Proof: Let $p_r$ be the r-th prime number. Let $q_r$ be the prime number just before $2p_r$ (see A059788), by Bertrand–Chebyshev theorem, $q_r \neq p_r$.
Now consider the number $n_r = N_rq_r/(2p_r)$. Then $s(n_r)-\omega(n_r)=1$. So $$\alpha(n_r) = \frac{\varphi(n_r)e^{\gamma}\ln \ln n_r}{n_r} = \frac{\varphi(N_r)e^{\gamma}\ln \ln N_r}{N_r} \cdot \frac{2(q_r-1)p_r \ln \ln (n_r)}{(p_r-1)q_r\ln \ln (N_r)}.$$ The first component of this multiplication converges to $1_-$, by Rosser-Schoenfeld and Nicolas's criterion (assuming RH). Then $$\alpha(n_r) \sim \frac{2(q_r-1)p_r \ln \ln (n_r)}{(p_r-1)q_r\ln \ln (N_r)} = 2 \cdot \frac{p_rq_r-p_r}{p_rq_r-q_r} \cdot \frac{\ln ( \ln (N_r) - \ln (2p_r/q_r))}{\ln \ln N_r} \to 2. \ \ \square $$ It follows that if the expected upper-bound is true, then it is optimal.

A number $N>2$ is called a champion if $\alpha(N)>1$ and $\forall n<N$ we have $\alpha(n)<\alpha(N)$.
There are exactly $35$ champions $N< 10^{10}$, listed below together with $\alpha(N)$ and the factor decomposition of the ratio with their next primorial.

sage: champions(3,10000000000)
[7, 1.0081297159194946, 2 * 3 * 5 * 7^-1]
[11, 1.1900001764297485, 2 * 3 * 5 * 11^-1]
[13, 1.244431734085083, 2 * 3 * 5 * 13^-1]
[17, 1.3212575912475586, 2 * 3 * 5 * 17^-1]
[19, 1.349881649017334, 2 * 3 * 5 * 19^-1]
[23, 1.395310401916504, 2 * 3 * 5 * 23^-1]
[29, 1.4449400901794434, 2 * 3 * 5 * 29^-1]
[91, 1.4570322036743164, 2 * 3 * 5 * 13^-1]
[119, 1.499191164970398, 2 * 3 * 5 * 17^-1]
[133, 1.5151351690292358, 2 * 3 * 5 * 19^-1]
[143, 1.5473616123199463, 2 * 3 * 5 * 7 * 11^-1 * 13^-1]
[187, 1.5879234075546265, 2 * 3 * 5 * 7 * 11^-1 * 17^-1]
[209, 1.6032350063323975, 2 * 3 * 5 * 7 * 11^-1 * 19^-1]
[1309, 1.6044903993606567, 2 * 3 * 5 * 17^-1]
[1463, 1.61602783203125, 2 * 3 * 5 * 19^-1]
[1547, 1.6261959075927734, 2 * 3 * 5 * 11 * 13^-1 * 17^-1]
[1729, 1.6376748085021973, 2 * 3 * 5 * 11 * 13^-1 * 19^-1]
[2093, 1.6558983325958252, 2 * 3 * 5 * 11 * 13^-1 * 23^-1]
[2261, 1.6681385040283203, 2 * 3 * 5 * 11 * 17^-1 * 19^-1]
[23023, 1.6813620328903198, 2 * 3 * 5 * 23^-1]
[24871, 1.692425012588501, 2 * 3 * 5 * 13 * 17^-1 * 19^-1]
[29029, 1.6975822448730469, 2 * 3 * 5 * 29^-1]
[29393, 1.7114139795303345, 2 * 3 * 5 * 11 * 17^-1 * 19^-1]
[391391, 1.7167580127716064, 2 * 3 * 5 * 23^-1]
[437437, 1.7252918481826782, 2 * 3 * 5 * 17 * 19^-1 * 23^-1]
[493493, 1.7308218479156494, 2 * 3 * 5 * 29^-1]
[7436429, 1.7369874715805054, 2 * 3 * 5 * 23^-1]
[8580495, 1.74891197681427, 2 * 13 * 23^-1]
[9376367, 1.7497258186340332, 2 * 3 * 5 * 29^-1]
[181996815, 1.7533072233200073, 2 * 19 * 31^-1]
[190285095, 1.7621831893920898, 2 * 17 * 29^-1]
[203408205, 1.7683358192443848, 2 * 17 * 31^-1]
[5203883685, 1.7786405086517334, 2 * 23 * 37^-1]
[5277907635, 1.786533236503601, 2 * 19 * 31^-1]
[5898837945, 1.8011537790298462, 2 * 17 * 31^-1]

Note that all these champions are closely related to their next primorial numbers.

Then, the expected upper-bound is checked for $n<10^{10}$.
If it is true in general then it must have infinitely many champions, forced by the sequence $(n_r)$.
See below $[n_r,\alpha(n_r)]$ for $r=11,12$:

[197325643515, 1.8032277291323942]
[7320457889745, 1.8197057337162745]

Code

# %attach SAGE/EulerRH.spyx

from sage.all import *

cpdef g(float x):
    return x/(exp(float(euler_gamma))*ln(ln(x)))

cpdef omega(long n):
    return len(list(factor(n)))

cpdef champions(long m1, long m2):
    cdef long n,o,s,p,pr
    cdef float a,c
    a=1; s=2; p=3; pr=6; n=m1
    while pr<m1:
        p=next_prime(p); pr*=p; s+=1
    while n<=m2:
        #if mod(n,10**7)==0:
            #print(n)
        if n>pr:
            p=next_prime(p); pr*=p; s+=1
        o=omega(n)
        if n<>pr:
            c=(euler_phi(n)/g(float(n)))**(1/float(s-o))
            if c>a:
                a=c
                print([n,a,factor(Integer(pr)/Integer(n))])
        n+=1
$\endgroup$

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