Timeline for Counting natural numbers in a set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$, for every $n$
Current License: CC BY-SA 4.0
9 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Sep 15, 2018 at 2:39 | comment | added | Gerhard Paseman | This (mathoverflow.net/q/253783) just might be related. I'd appreciate someone's opinion on this. Gerhard "Gotta Cover The Possible Connections" Paseman, 2018.09.14. | |
| Sep 15, 2018 at 0:47 | comment | added | Alexander Kalmynin | I think that all the analytic results that I used can be replaced by results of the form "there are primes in $((1-\varepsilon)x,(1+\varepsilon)x)$" for some small enough $\varepsilon$. And these facts are known to have a purely combinatorial proof. | |
| Sep 15, 2018 at 0:46 | comment | added | Right | That is, with minimal amount of analysis. | |
| Sep 15, 2018 at 0:40 | comment | added | Right | I am thinking about some purely combinatorial/number-theoretic approach without any use of analysis. | |
| Sep 15, 2018 at 0:27 | comment | added | Alexander Kalmynin | In fact, I think one can use this sort of the argument (even without theorems on primes in short intervals: just classical formulas for $\pi(x)$ with nice enough remainder terms) to show that $f(n) \gg \frac{n}{\log^2 n}$. | |
| Sep 15, 2018 at 0:21 | history | edited | Alexander Kalmynin | CC BY-SA 4.0 |
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| Sep 15, 2018 at 0:19 | comment | added | Alexander Kalmynin | So, basically we just choose $p$ and $q$ to be primes with $q/2<p<q$ and both $q$ and $2p$ rather close to $\sqrt{n}$. | |
| Sep 15, 2018 at 0:15 | history | answered | Alexander Kalmynin | CC BY-SA 4.0 |