Re: OT Conjecture about harmonic numbers and Riemann hypothesis

--00000000000061deeb063d428370
Content-Type: text/plain; charset="UTF-8"
Content-Transfer-Encoding: quoted-printable

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=E2=80=AFAM Georgi Guninski <[email protected]>=
 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_fun=
ctions
> ):
>
> Let $\sigma(n)$ denote the sum of divisors on $n$ and $H_n$ denote the
> harmonic numbers.
>
> Define $\Delta(n)=3D H_n+\log{(H_n)} e^{H_n}-\sigma(n)$.
>
> $\lfloor \Delta(n) \rfloor$[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=3Dmpmath.polygamma(0,n+1)+mpmath.euler #harmonic number H_n
>     T=3Dh+mpmath.log(h)*mpmath.exp(h)-sigma(n)
>     return T
>
>
> def delta1main(L=3D10**2,pre=3D20):
>     """
>     Author: Georgi Guninski Mon Aug 25 02:38:30 PM UTC 2025
>     L=3D10^3 Wall time: 1min 32s
>     L=3D10^4 Wall time: 2h 30min
>     """
>     import mpmath
>     mpmath.mp.dps=3Dpre
>     mpmath.mp.pretty=3DTrue
>     for a1 in range(2,L):
>         for a2 in range(2,L):
>             a3=3Da1*a2
>             d1,d2,d3=3D[delta2(i) for i in (a1,a2,a3)]
>             if not (d3>d1 or d3>d2):  print(a1,a2)
>
> ```
>
>

--00000000000061deeb063d428370
Content-Type: text/html; charset="UTF-8"
Content-Transfer-Encoding: quoted-printable

<div dir=3D"auto"><div>Somewhat less off topic: how to make the code fast i=
n Pari?<div dir=3D"auto"><br></div><div dir=3D"auto">I would use forfactore=
d. 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 t=
o N along with retaining the factorization, then go evaluate Delta of the p=
roduct using the known factorization of the product.</div><div dir=3D"auto"=
><br></div><div dir=3D"auto">These two techniques should accelerate the cod=
e considerably.</div><div dir=3D"auto"><br></div><div data-smartmail=3D"gma=
il_signature"><div dir=3D"ltr">Astra mortemque praestare gradatim</div></di=
v><br><div class=3D"gmail_quote"><div dir=3D"ltr" class=3D"gmail_attr">On T=
ue, Aug 26, 2025, 6:39=E2=80=AFAM Georgi Guninski &lt;<a href=3D"mailto:ggu=
[email protected]" rel=3D"noreferrer noreferrer" target=3D"_blank">gguninski=
@gmail.com</a>&gt; wrote:<br></div><blockquote class=3D"gmail_quote" style=
=3D"margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding=
-left:1ex">Apologies for offtopic.<br>
<br>
With latex formatting:=C2=A0 <a href=3D"https://mathoverflow.net/q/499603/1=
2481" rel=3D"noreferrer noreferrer noreferrer noreferrer" target=3D"_blank"=
>https://mathoverflow.net/q/499603/12481</a><br>
<br>
We found conjecture about harmonic numbers related to RH.<br>
<br>
Slightly rephrasing [wikipedia on<br>
RH](<a href=3D"https://en.wikipedia.org/wiki/Riemann_hypothesis#Growth_of_a=
rithmetic_functions" rel=3D"noreferrer noreferrer noreferrer noreferrer" ta=
rget=3D"_blank">https://en.wikipedia.org/wiki/Riemann_hypothesis#Growth_of_=
arithmetic_functions</a>):<br>
<br>
Let $\sigma(n)$ denote the sum of divisors on $n$ and $H_n$ denote the<br>
harmonic numbers.<br>
<br>
Define $\Delta(n)=3D H_n+\log{(H_n)} e^{H_n}-\sigma(n)$.<br>
<br>
$\lfloor \Delta(n) \rfloor$[is OEIS A057641](<a href=3D"https://oeis.org/A0=
57641" rel=3D"noreferrer noreferrer noreferrer noreferrer" target=3D"_blank=
">https://oeis.org/A057641</a>)<br>
<br>
<br>
RH is equivalent to $\Delta(n)&gt;0$ for all integers $n&gt;1$ proved by=C2=
=A0 by<br>
Jeffrey Lagarias in 2002.<br>
<br>
**joro&#39;s conjecture about harmonic numbers** Let $a_1,a_2$ be integers<=
br>
satsifying $a_1&gt;1,a_2&gt;1,(a_1,a_2) \ne (3,4), (a_1,a_2) \ne (4,3)$.<br=
>
We have $\Delta(a_1 a_2) &gt; \Delta(a_1)$ or $\Delta(a_1 a_2)&gt;\Delta(a_=
2)$<br>
<br>
&gt;Q1 Is the conjecture true?<br>
<br>
&gt;Q2 In case Q1 is hopeless, can we get lower bound for the smallest coun=
terexamples by fixing $a_1$ and let $a_2$ vary?<br>
<br>
It holds for $ 1 &lt; a_1,a_2 &lt; 10^4$, but computation is slow.<br>
<br>
We compute $H_n$ as `mpmath.polygamma(0,n+1)+mpmath.euler`<br>
<br>
<br>
```<br>
import mpmath<br>
def delta2(n):<br>
=C2=A0 =C2=A0 h=3Dmpmath.polygamma(0,n+1)+mpmath.euler #harmonic number H_n=
<br>
=C2=A0 =C2=A0 T=3Dh+mpmath.log(h)*mpmath.exp(h)-sigma(n)<br>
=C2=A0 =C2=A0 return T<br>
<br>
<br>
def delta1main(L=3D10**2,pre=3D20):<br>
=C2=A0 =C2=A0 &quot;&quot;&quot;<br>
=C2=A0 =C2=A0 Author: Georgi Guninski Mon Aug 25 02:38:30 PM UTC 2025<br>
=C2=A0 =C2=A0 L=3D10^3 Wall time: 1min 32s<br>
=C2=A0 =C2=A0 L=3D10^4 Wall time: 2h 30min<br>
=C2=A0 =C2=A0 &quot;&quot;&quot;<br>
=C2=A0 =C2=A0 import mpmath<br>
=C2=A0 =C2=A0 mpmath.mp.dps=3Dpre<br>
=C2=A0 =C2=A0 mpmath.mp.pretty=3DTrue<br>
=C2=A0 =C2=A0 for a1 in range(2,L):<br>
=C2=A0 =C2=A0 =C2=A0 =C2=A0 for a2 in range(2,L):<br>
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 a3=3Da1*a2<br>
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 d1,d2,d3=3D[delta2(i) for i in (a=
1,a2,a3)]<br>
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 if not (d3&gt;d1 or d3&gt;d2):=C2=
=A0 print(a1,a2)<br>
<br>
```<br>
<br>
</blockquote></div>
</div></div>

--00000000000061deeb063d428370--