Timeline for Integers solutions of products of truncated Riemann zeta functions
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 8, 2022 at 20:48 | comment | added | Steven Clark | $F_1(n)=\frac{n\ (n+1)}{2} H_n$ and $F_k(n)=H_n^{(-k)}\ H_n^{(k)}$ where $H_n$ is the harmonic number and $H_n^{(k)}$ is the harmonic number of order $k$. | |
| Nov 8, 2022 at 20:26 | comment | added | Conrad | for $k=2$ one needs that $n(n+1)(2n+1)/6$ cancels out all $1/p^2$ for $n/2 <p \le n$ prime, so in particular $n^2/4 \le p^2 \le 2n+1$ for at least one such $p$ by Bertrand since $p$ can divide only one of the three factors, or $n \le 8$ etc; similar arguments work for higher $k$ given that we know that $1^k+2^k+\ldots+n^k$ is a polynomial of degree $k+1$ in $n$ | |
| Nov 8, 2022 at 19:44 | comment | added | Aeryk | It should be noted that $n=m=1$ is also a solution. | |
| Nov 8, 2022 at 18:59 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing (bracket scaling)+ minor grammar improvement
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| Nov 8, 2022 at 17:47 | history | asked | gigi | CC BY-SA 4.0 |