Re: OT Conjecture about harmonic numbers and Riemann hypothesis

Somewhat less off topic: how to make the code fast in Pari?

I would use forfactored. This way sigma is evaluated with the factorization already evaluated by a cheap sieve. Secondly I would first compute all the values in a range 0 to N along with retaining the factorization, then go evaluate Delta of the product using the known factorization of the product.

These two techniques should accelerate the code considerably.

Astra mortemque praestare gradatim

On Tue, Aug 26, 2025, 6:39 AM Georgi Guninski <gguninski@gmail.com> wrote:

Apologies for offtopic.

With latex formatting: https://mathoverflow.net/q/499603/12481

We found conjecture about harmonic numbers related to RH.

Slightly rephrasing [wikipedia on
RH](https://en.wikipedia.org/wiki/Riemann_hypothesis#Growth_of_arithmetic_functions):

Let $\sigma(n)$ denote the sum of divisors on $n$ and $H_n$ denote the
harmonic numbers.

Define $\Delta(n)= H_n+\log{(H_n)} e^{H_n}-\sigma(n)$.

$\lfloor \Delta(n) \rfloor$[is OEIS A057641](https://oeis.org/A057641)

RH is equivalent to $\Delta(n)>0$ for all integers $n>1$ proved by by
Jeffrey Lagarias in 2002.

**joro's conjecture about harmonic numbers** Let $a_1,a_2$ be integers
satsifying $a_1>1,a_2>1,(a_1,a_2) \ne (3,4), (a_1,a_2) \ne (4,3)$.
We have $\Delta(a_1 a_2) > \Delta(a_1)$ or $\Delta(a_1 a_2)>\Delta(a_2)$

>Q1 Is the conjecture true?

>Q2 In case Q1 is hopeless, can we get lower bound for the smallest counterexamples by fixing $a_1$ and let $a_2$ vary?

It holds for $ 1 < a_1,a_2 < 10^4$, but computation is slow.

We compute $H_n$ as `mpmath.polygamma(0,n+1)+mpmath.euler`

```
import mpmath
def delta2(n):
h=mpmath.polygamma(0,n+1)+mpmath.euler #harmonic number H_n
T=h+mpmath.log(h)*mpmath.exp(h)-sigma(n)
return T

def delta1main(L=10**2,pre=20):
"""
Author: Georgi Guninski Mon Aug 25 02:38:30 PM UTC 2025
L=10^3 Wall time: 1min 32s
L=10^4 Wall time: 2h 30min
"""
import mpmath
mpmath.mp.dps=pre
mpmath.mp.pretty=True
for a1 in range(2,L):
for a2 in range(2,L):
a3=a1*a2
d1,d2,d3=[delta2(i) for i in (a1,a2,a3)]
if not (d3>d1 or d3>d2): print(a1,a2)

```