Timeline for How $a+b$ can grow when $a!b! \mid n!$
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 2, 2013 at 19:02 | vote | accept | CommunityBot | ||
| Aug 12, 2013 at 7:07 | comment | added | GH from MO | @Jash: On the one hand, the exponent of 2 in the prime factorization of $\frac{b!}{(n-a)!}$ is at least $\frac{b-(n-a)-1}{2}$, because there are at least that many even numbers in the sequence $n-a+1,\dots,b$. On the other hand, the exponent of 2 in the prime factorization of $\binom{n}{a}$ is at most $\log_2(n)$ by Kummer's theorem (see the link in my original post). These two facts imply the inequality you ask about. | |
| Aug 12, 2013 at 6:58 | comment | added | user38122 | Would you explain more on how did you drive $\frac{b-(n-a)-1}{2} \leq \log_2(n)$ ? | |
| Jul 27, 2013 at 16:34 | history | answered | GH from MO | CC BY-SA 3.0 |