Timeline for Is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 10, 2020 at 8:42 | comment | added | Gerhard Paseman | I hope to avoid a misunderstanding. If by first paragraph of my answer you mean of my post, that appeared a day before your post. (I leave the ramifications of this unstated.) If you instead mean first paragraph of the edit, indeed some of it was inspired by the first paragraph of my answer and by the first line of your answer, where n/\phi(n) appears. However your simple argument giving a fractional exponent bound does not appear as simple to me as the one that gives a logarithmic (in m) bound on the multiplier. Gerhard "Please Add Clarity To Comment" Paseman, 2020.07.10. | |
| Jul 10, 2020 at 6:47 | comment | added | Emil Jeřábek | @Gerhard The argument in my answer is extremely simple, hence it is no doubt “similar” to any other argument estimating $\varphi(n)$, such as the one in your answer. Nevertheless, the first paragraph you added basically copies what I already wrote, yielding the same $O(m\log\log m)$ bound, and I find this rather uncool. And yes, one of the points of my answer is that even though the bound on the product looks generous, it leads to the same asymptotic estimate as if it were bounded in a more complicated way, which is why I did the simple way. | |
| Jul 9, 2020 at 22:25 | comment | added | Gerhard Paseman | If you want credit for writing n/\phi(n), I am willing to recognize you as a recent source of inspiration. Gerhard "Will That Help This Situation?" Paseman, 2020.07.09. | |
| Jul 9, 2020 at 22:20 | comment | added | Gerhard Paseman | I think this could blow up into a major misunderstanding if we aren't careful. It is not clear to me what part for which you want credit. However, the edit I made to my post has an argument similar to the one in my original post. If you need recognition, I need to know more precisely for what. I thought the product over primes had too generous a bound in the first part of your post. Gerhard "Not In It For Points" Paseman, 2020.07.09. | |
| Jul 9, 2020 at 19:43 | comment | added | Emil Jeřábek | @GerhardPaseman I have no idea what all this talk about lilies is about, but did I just see you include parts of my answer in your answer without as much as a thank you? | |
| Jul 9, 2020 at 17:17 | comment | added | Gerhard Paseman | Um, the argument I present above can be adapted to show that phi(n)=m implies n is at most (k+1)m, where k is the exponent of 2 that divides m. When m has many prime factors, this can be changed to ( Clog k) m using work of Mertens or simpler approximations. Gerhard "Not Quite Gilding The Lily" Paseman, 2020.07.09. | |
| Jul 9, 2020 at 16:54 | comment | added | zeraoulia rafik | The nice idea you have used is the compositional inverse of Euler totiont function | |
| Jul 9, 2020 at 16:50 | history | answered | Emil Jeřábek | CC BY-SA 4.0 |