Timeline for Divisibility of (finite) power sum of integers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2021 at 13:38 | vote | accept | T. Amdeberhan | ||
| Jul 10, 2021 at 17:46 | comment | added | T. Amdeberhan | Yes, you did say that. :-) | |
| Jul 10, 2021 at 17:43 | comment | added | darij grinberg | That's why I'm saying "suspect"... | |
| Jul 10, 2021 at 17:15 | comment | added | T. Amdeberhan | @darijgrinberg: interesting. How do you ensure that $m$ and $r$ are not divisible by $3$? | |
| Jul 10, 2021 at 3:09 | answer | added | Max Alekseyev | timeline score: 6 | |
| Jul 9, 2021 at 22:02 | comment | added | darij grinberg | I suspect that your conjecture is indeed true, and in fact that $S_a\left(b\right)$ is a polynomial in $a$ of the form $\dfrac{1}{m} \left(2a-1\right) \left(3 q\left(a\right) + r\right)$, where $q$ is an integral polynomial and where $m$ and $r$ are two integers not divisible by $3$. This should really be a property of Faulhaber polynomials, but I am too faul to translate it into that language :) (I have checked it for all $b \leq 8$.) | |
| Jul 9, 2021 at 21:59 | comment | added | darij grinberg | Questions 1 and 2 are clearly equivalent, since $S_a\left(1\right) = 1^2 + 2^2 + \cdots + \left(3a-2\right)^2 = \dfrac12 \left(2a-1\right) \left(3a-1\right) \left(3a-2\right)$ has the same $3$-adic valuation as $2a-1$. | |
| Jul 9, 2021 at 18:10 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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| Jul 9, 2021 at 16:29 | history | asked | T. Amdeberhan | CC BY-SA 4.0 |