Timeline for How $a+b$ can grow when $a!b! \mid n!$
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2013 at 4:04 | comment | added | Will Sawin | Yes. I think $a=v$, $b= 2^{(v!)_2 -1}$, $c=2^{(v!)_2}$ gives the best possible $2$-adic solutions to $a+b-n=v$ for any $v$, which your formula is the best closed-form approximation to. It's not clear how much more one can gain from including the other primes. | |
| Jul 29, 2013 at 3:58 | comment | added | S. Carnahan♦ | I think the $p$-valuation of $x$ consecutive integers is at least $\lfloor \frac{x}{p-1} \rfloor - \lfloor \log_p(x+1) \rfloor$, so $a+b-n$ is at most $(p-1)\lfloor \log_p(n) \rfloor + (p-1) \lfloor \log_p \log_p(n) \rfloor +2p-2$. You can use $n=2^k$, $a = k+ \lceil \log_2(k) \rceil$, $b = 2^k-1$ to get close to a 2-adic solution, when $k+\lceil \log_2(k) \rceil$ has the form $2^m-1$. | |
| Jul 28, 2013 at 19:32 | history | answered | Will Sawin | CC BY-SA 3.0 |