Timeline for Do we know any bound on $\operatorname{lcm}(2^1-1, 2^2-1,\dots,2^n-1)$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5 at 16:49 | answer | added | Max Alekseyev | timeline score: 7 | |
| Aug 5 at 15:02 | history | edited | Martin Sleziak |
edited tags
|
|
| May 4, 2020 at 13:06 | answer | added | Gerry Myerson | timeline score: 16 | |
| S May 4, 2020 at 0:14 | history | suggested | RobPratt | CC BY-SA 4.0 |
Added operatorname
|
| May 3, 2020 at 23:38 | review | Suggested edits | |||
| S May 4, 2020 at 0:14 | |||||
| S May 3, 2020 at 23:36 | history | edited | LSpice | CC BY-SA 4.0 |
Math mode cleanup
|
| S May 3, 2020 at 23:36 | history | suggested | RobPratt | CC BY-SA 4.0 |
Math mode cleanup
|
| May 3, 2020 at 23:32 | review | Suggested edits | |||
| S May 3, 2020 at 23:36 | |||||
| Apr 9, 2015 at 2:53 | comment | added | Gerry Myerson | I found the Szymiczek reference: On the distribution of prime factors of Mersenne numbers, Prace Mat. 13 (1969) 33–49, MR0252316 (40 #5537). Unfortunately, I don't seem to have the actual paper, nor can I find it online. | |
| Apr 3, 2015 at 6:02 | vote | accept | Amir | ||
| Apr 2, 2015 at 16:27 | comment | added | The Masked Avenger | One can choose prime powers instead of primes in dhy's product and bump up the lower bound by more than 2^n for large n. | |
| Apr 2, 2015 at 11:53 | comment | added | Gerry Myerson | Kazimierz Szymiczek worked this out in a paper around 1970. Sorry I'm away from my office and can't give you a better reference. | |
| Apr 2, 2015 at 10:50 | answer | added | user40023 | timeline score: 25 | |
| Apr 2, 2015 at 7:48 | comment | added | dhy | As $\operatorname{gcd}(2^p-1,2^q-1)=1$ if $p,q$ are distinct primes, it's at least the product of $2^p-1$ where $p$ ranges over the primes at most $n.$ I think this should give you an asymptotic lower bound of $2^{cn^2/\operatorname{log}(n)}$ where $c$ is any constant less than $\frac{1}{2}.$ | |
| Apr 2, 2015 at 7:24 | comment | added | Qiaochu Yuan | It's at least $2^n - 1$ and it's at most $\prod_{i=1}^n (2^i - 1) \le 2^{{n \choose 2}} - 1$. | |
| Apr 2, 2015 at 6:48 | history | asked | Amir | CC BY-SA 3.0 |