Timeline for Can one show combinatorially how $\operatorname{lcm}(1, \dotsc, n)$ grows?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2020 at 9:06 | answer | added | Emil Jeřábek | timeline score: 0 | |
| May 21, 2020 at 4:20 | answer | added | Mark Lewko | timeline score: 5 | |
| May 19, 2020 at 17:38 | history | edited | LSpice | CC BY-SA 4.0 |
TeX fix
|
| May 19, 2020 at 17:00 | answer | added | Gerald Edgar | timeline score: 4 | |
| May 19, 2020 at 16:39 | answer | added | Gerhard Paseman | timeline score: 2 | |
| May 19, 2020 at 15:47 | comment | added | Gerhard Paseman | @Sam, right. However the base of the prime power does, and that is enough for the argument. Gerhard "Willing To Tweak The Slick" Paseman, 2020.05.19. | |
| May 19, 2020 at 15:39 | comment | added | Sam Hopkins | @DavidESpeyer: $3^2$ is a prime power between $6$ and $12$, but it doesn't divide $\binom{12}{6}$. | |
| May 19, 2020 at 15:32 | comment | added | Gerhard Paseman | Indeed, @David, I am looking at slick arguments and seeing how they can be tweaked. Your second comment is good enough for an answer in my opinion. When I try to recall the Chebyshev proofs though, combinatorics is not what first leaps to mind. Perhaps I should look at them more closely. Thanks! Gerhard "That's What I'm Talking About!" Paseman, 2020.05.19. | |
| May 19, 2020 at 14:55 | comment | added | David E Speyer | There is a slick combinatorial proof of $M(n) < 4^n$, though. Let $Q$ be the set of all prime powers between $n/2$ and $n$. Then $\mathrm{LCM}(1,2,\ldots,n) \leq \mathrm{LCM}(1,2,\ldots, \lfloor n/2 \rfloor)\cdot \prod_{q \in Q} q$. But $\prod_{q \in Q} q$ divides $\binom{n}{\lfloor n/2 \rfloor} < 2^n$. So $M(n) < M(\lfloor n/2 \rfloor) \cdot 2^n$ and induction gives $M(n) < 4^n$. | |
| May 19, 2020 at 14:49 | comment | added | David E Speyer | In what sense are the standard proofs of Chebyshev's estimates not combinatorial? And we know from Diamond and Erdos mathscinet.ams.org/mathscinet-getitem?mr=610529 that, for any $\delta>0$, similar methods can prove $(1-\delta)^n < M(n) < (1+\delta)^n$ for $n$ sufficiently large. | |
| May 19, 2020 at 14:46 | comment | added | Mark Lewko | If we take a logarithm, then you are asking for upper and lower estimates on the familiar $\psi(x)$ from prime number theory. So you are asking for a combinatorial proof of Chebyshev's prime number estimates. | |
| May 19, 2020 at 14:35 | history | asked | Gerhard Paseman | CC BY-SA 4.0 |