Timeline for Divisibility of a binomial sequence
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2019 at 3:48 | vote | accept | T. Amdeberhan | ||
| Mar 18, 2017 at 19:26 | answer | added | GH from MO | timeline score: 3 | |
| Mar 16, 2017 at 11:20 | comment | added | Ofir Gorodetsky | @juan In a recent preprint, Sun's conjecture was proven by Mao: arxiv.org/pdf/1511.06221.pdf (see Theorem 1.1). | |
| Mar 15, 2017 at 20:08 | comment | added | juan | This appears as (part of) Conjecture 5.6 in Zhi-Wei Sun "Two new kinds of numbers and related divisibility results", arXiv:1408.5381 v8. In Remark 5.3 Zhi-Wei Sun asserts it is divisible by n+1. | |
| Mar 15, 2017 at 16:06 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 4 characters in body
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| Mar 15, 2017 at 4:28 | comment | added | D. Ror. | @darijgrinberg ...and divisible by $(n+1)$ but not $(n+1)^2$ for odd exponents greater than $1$. | |
| Mar 14, 2017 at 22:37 | comment | added | T. Amdeberhan | How is a "possible duplicate"? Please check carefully. | |
| Mar 14, 2017 at 22:00 | comment | added | zeraoulia rafik | possible duplicate with :mathoverflow.net/q/11335/51189 | |
| Mar 14, 2017 at 19:29 | comment | added | darij grinberg | Also empirically, $\sum_{k=0}^n\sum_{j=0}^k\binom{k}j^2\binom{2j}j(2j+1)^{2p}$ also seems to be divisible by $\left(n+1\right)^2$ for any positive integer $p$. | |
| Mar 14, 2017 at 4:59 | comment | added | T. Amdeberhan | @darijgrinberg: It's intentional because the identity is true only if you hold on to $(2j+1)$ on the LHS, not true with $(2j+1)^2$. | |
| Mar 14, 2017 at 4:50 | comment | added | darij grinberg | @DannyRorabaugh: Is the lack of a square at the end of the LHS intentional? | |
| Mar 14, 2017 at 3:25 | comment | added | D. Ror. | Empirically, $\sum_{k=0}^n \sum_{j=0}^k {k\ \choose j}^2 {2j \choose j} (2j+1) = (n+1)^2 \sum_{i = 0}^n {n \choose i}^2 C_i$, where $C_i$ is the $i$-th Catalan number. | |
| Mar 14, 2017 at 2:37 | comment | added | T. Amdeberhan | That, too, is not clear. If proved, it might shade some light into the divisibility by $(n+1)^2$. | |
| Mar 14, 2017 at 2:37 | comment | added | Noam D. Elkies | Is it clearly divisible by $n+1$? | |
| Mar 14, 2017 at 2:22 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |