Timeline for Extending the discussion on "super Catalan" numbers
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2016 at 16:26 | comment | added | T. Amdeberhan | I don't mean a proof, but an alternative verification of your first line. | |
| Sep 24, 2016 at 16:24 | comment | added | Jan-Christoph Schlage-Puchta | Yes, but being non-negative is not enough. We need it to be large enough to cancel the factor $(x+y+z)$. | |
| Sep 24, 2016 at 16:18 | comment | added | T. Amdeberhan | $\nu_p$ is the $p$-adic valuation. Since $\frac{(3x)!}{x!^3}=\binom{3x}{2x}\binom{2x}{x}\in\mathbb{N}$ and by Legendre's formula $\nu_p(n!)=\frac{n−s_p(n)}{p−1}$, the first quantity you wrote is non-negative. | |
| Sep 24, 2016 at 8:58 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |