Timeline for Growth Rate of the Square-Free Part
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| May 1, 2018 at 23:03 | history | edited | Richard Voepel | CC BY-SA 3.0 |
Final revisions to comments.
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| May 1, 2018 at 22:42 | history | edited | Richard Voepel | CC BY-SA 3.0 |
This problem is now solved.
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| May 1, 2018 at 22:39 | comment | added | Richard Voepel | @GerryMyerson Oh my! I'm sorry, I did not stumble across that before posting. My apologies. | |
| May 1, 2018 at 22:38 | comment | added | Gerry Myerson | Previous question on this topic: mathoverflow.net/questions/149511/… See also mathoverflow.net/questions/203502/… although that one is just about $2^p-1$, $p$ prime. | |
| May 1, 2018 at 22:37 | history | edited | Richard Voepel | CC BY-SA 3.0 |
Adding notes on conditional proofs of the problem.
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| May 1, 2018 at 22:29 | history | edited | Richard Voepel | CC BY-SA 3.0 |
Small typo.
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| May 1, 2018 at 22:18 | history | edited | Richard Voepel | CC BY-SA 3.0 |
Major revisions for clarity.
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| May 1, 2018 at 22:12 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Incorporated image with a bang.
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| May 1, 2018 at 22:07 | comment | added | Pasten | $rad(n)^2\le sqf(n).n$ | |
| May 1, 2018 at 22:01 | comment | added | Wojowu | Yes, that sounds like a good idea - I've thought squarefree part and radical are the same... either way, abc should let one show that primes with exponent 1 should be prevalent, but that is less straightforward. I'll try to give more details tomorrow. | |
| May 1, 2018 at 22:01 | history | edited | Richard Voepel | CC BY-SA 3.0 |
Clarifying terminology.
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| May 1, 2018 at 21:59 | comment | added | Richard Voepel | @Wojowu Right, but won't that pick up all of the prime divisors? Not just the primes which divide an odd number of times? I'm afraid I will have to edit my post to make clear what I mean by the square-free part... | |
| May 1, 2018 at 21:57 | comment | added | Wojowu | The crux is that in this case, the radical of $abc$ is just twice the radical of $a$. | |
| May 1, 2018 at 21:56 | comment | added | Richard Voepel | @Wojowu I'm not sure how to use the conclusion of the $abc$ conjecture to get at the lower bounds you are describing; can you elaborate on how the radical of $abc$ (which is really just the radical of $a$) can tell us about the square-free part of $a$? | |
| May 1, 2018 at 21:47 | comment | added | Wojowu | This is very closely related to the $abc$ conjecture (for $a=2^n-1,b=1,c=2^n$). This gives very strong conditional lower bounds. I don't know much about unconditional bounds, but just showing that the squarefree parts go to infinity might have been proven. | |
| May 1, 2018 at 21:40 | comment | added | user44191 | $2^n-1$ will be squarefree whenever $n$ is not divisible by $\text{ord}_{p^2}(2)$ for $p$ an odd prime. I would guess you can do some approximations with replacing $2$ with an arbitrary unit mod $p^2$ which is not $1$ mod $p$. | |
| May 1, 2018 at 21:05 | history | asked | Richard Voepel | CC BY-SA 3.0 |