Timeline for Exact formula for partial sums of Liouville function $L(n)$ (OEIS sequence A002819)
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9, 2021 at 9:35 | comment | added | user257465 | Related facebook.com/groups/229783678691987/permalink/315436670126687/… | |
| May 26, 2021 at 16:20 | history | edited | Vincent Granville | CC BY-SA 4.0 |
some update to my code
|
| May 26, 2021 at 15:05 | answer | added | Mats Granvik | timeline score: 3 | |
| May 23, 2021 at 0:55 | comment | added | KConrad | The identity $L(x) = \sum_{n \leq x} \mu(n)\lfloor{\sqrt{x/n}\rfloor}$ is a consequence of the identity $\lambda(n) = \sum_{d^2 \mid n} \mu(n/d^2)$ for $n \geq 1$ since it implies $L(x) = \sum_{d \leq \sqrt{x}} M(x/d^2)$, and swapping the order of a double sum there implies $L(x) = \sum_{n \leq x} \mu(n)\lfloor{\sqrt{x/n}\rfloor}$. | |
| May 23, 2021 at 0:11 | history | edited | Vincent Granville | CC BY-SA 4.0 |
Added some source code (which is correct) in case there are typos left in my text
|
| May 22, 2021 at 23:46 | comment | added | Vincent Granville | @Steven: $L(n)$ and $L^*(n)$ can be different, because $L(n)$ is chaotically oscillating (as a function of $n$) while $L^*(n)$ is very smoothly decreasing. My expectation is that $\lim\sup (|\log L(n)|/\log n) = \lim\sup (|\log L^*(n)|/\log n)$. Probably easy (not requiring RH) to prove the latter one is $1/2$. Not sure how difficult to prove that the two are identical (because $L$ and $L^*$ are close enough) without using RH. If it can be done without RH (based on matrix algebra for matrices that are close enough) then RH would result. I am not optimistic though. More on this later. | |
| May 22, 2021 at 21:26 | comment | added | Steven Clark | For $n=22500$ and $w_n=20$, I get $L=-86$, $L*=-99.0133$, $C=298.523$ ,and $D=-397.536$ which are fairly close to your results, but I believe $a_{w_n+1}=a_{22500-22479}=a_{21}=-0.000177809$ whereas $a_{n-(w_n+1)}=a_{22500-21}=a_{22479}=-0.000177332$. I'm not sure what your approach is to selecting $w_n$. For $n=15810$ and $w_n=20$, I get $L=-150$, $L*=-83.3228$, $C=249.43$ ,and $D=-332.753$ which is a much larger error for a much smaller value of $n$ when using the same value of $w_n$. | |
| May 22, 2021 at 19:36 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 42 characters in body
|
| May 22, 2021 at 7:31 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 65 characters in body
|
| May 22, 2021 at 5:33 | comment | added | Vincent Granville | @Steven (continued): Thus if we can prove that $L(n)$ and $L^*(n)$ have basically the same order of magnitude (when you take the $\log$) then you would have proved RH. But I have no doubt that this is probably nearly as difficult as proving RH, though at this point it's a (difficult) problem of matrix algebra. | |
| May 22, 2021 at 5:33 | comment | added | Vincent Granville | @Steven: Could you confirm this? I have a sense based on my computations so far that this may be correct. Regarding $L^*(n)= C(n)+D(n)$, you have $C(n)\approx 2\sqrt{n}$ and $a_{w_n +1}\approx a_1\approx -4/n$. Thus it turns out (without assuming RH, looks not very difficult to prove or maybe I am wrong) that $L^*(n)\sim-\frac{2}{3}\sqrt{n}=-0.666\dots \times\sqrt{n}$. Interestingly for $L_0(n)$ we have the following order of magnitude, assuming RH: $\sqrt{n}/\zeta(\frac{1}{2})=-0.684\dots \times \sqrt{n}$. Almost identical to $L(n)$. | |
| May 22, 2021 at 3:42 | comment | added | Steven Clark | With respect to your approach, it seems like you're trying to find a shortcut to computing $L(x)=\sum\limits_{n=1}^x\mu(n)\left\lfloor\sqrt{\frac{x}{n}}\right\rfloor$ which I believe is the full and exact relationship. | |
| May 22, 2021 at 3:22 | vote | accept | Vincent Granville | ||
| May 22, 2021 at 3:22 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 1 character in body
|
| May 22, 2021 at 0:44 | history | edited | Vincent Granville | CC BY-SA 4.0 |
I added Update 1 and 2 at the bottom. I think update 2 is the most interesting one.
|
| May 21, 2021 at 5:11 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 12 characters in body
|
| May 21, 2021 at 0:21 | history | edited | Vincent Granville | CC BY-SA 4.0 |
Added a section called "Note"
|
| May 21, 2021 at 0:11 | comment | added | Steven Clark | @OfirGorodetsky I believe the explicit formula for the Mertens function $M(x)$ is $M_o(x)=\sum\limits_\rho\frac{x^{\rho}}{\rho\,\zeta'(\rho)}-2+\sum\limits_n\frac{x^{-2\,n}}{(-2\,n)\,\zeta'(-2\,n)}$ where $\rho$ indicates a non-trivial zero of the Riemann zeta function $\zeta(s)$. See londmathsoc.onlinelibrary.wiley.com/doi/10.1112/plms/s3-1.1.86 for the explicit formula for the summatory Liouville function $L(x)$ which leads to the asymptotic $L(x)\approx\frac{\sqrt{x}}{\zeta\left(\frac{1}{2}\right)}$. | |
| May 20, 2021 at 23:38 | comment | added | Ofir Gorodetsky | On further thought, the matrix you are studying resembles the Redheffer matrix to some extent. It is well known that the Redheffer's determinant is the summatory Mobius function, so it attracted a lot of attention and its eigenvalues are understood to some extent. You might want to look into it. | |
| May 20, 2021 at 22:51 | answer | added | Steven Clark | timeline score: 7 | |
| May 20, 2021 at 22:20 | history | edited | Vincent Granville | CC BY-SA 4.0 |
added 3 characters in body
|
| May 20, 2021 at 21:15 | history | asked | Vincent Granville | CC BY-SA 4.0 |