Timeline for Given k, what is the minimum n such that n choose n/2 is greater than k?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2019 at 16:56 | vote | accept | David White | ||
| Sep 30, 2015 at 11:47 | comment | added | David White | Okay, I wrote out an argument using Newton's method. I was going to post it as an answer, but I guess people felt the question was too elementary. I have to say: it was not too elementary to me, and I'm a pretty active mathematician. In fact, I think this is exactly what MO is supposed to be for. Anyway, thanks Igor for your answer and the pointer to Newton's method, and thanks GH for suggesting the target to approximate. | |
| Sep 29, 2015 at 20:42 | comment | added | Igor Rivin | The way you usually do such things is by Newton's method, with the first step ($n=\log_2 k$) being fairly obvious... | |
| Sep 29, 2015 at 20:18 | comment | added | David White | Thanks for the answer. I just skimmed through that paper and while it seems to have great asymptotics for ${2n \choose n}$, I could not seem to find much about the inverse of this function (i.e. a bound on $k$ above). Can you see how to use these asymptotics to derive a bound like the one in GH's comment above? Thanks again. | |
| Sep 29, 2015 at 19:33 | history | answered | Igor Rivin | CC BY-SA 3.0 |