Timeline for Is there a $c > 1$ such that for all $n \ge 1$ the largest integer $\le c^n$ is prime?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2022 at 5:13 | answer | added | Terry Tao | timeline score: 15 | |
| Dec 21, 2022 at 16:26 | answer | added | user44143 | timeline score: 7 | |
| Dec 21, 2022 at 13:22 | comment | added | David E Speyer | @YCor It could work, but I'm not claiming to know a theorem, I'm saying these are the only cases that I know of where $(b c^n) \bmod 1$ isn't equidistributed. It does seem like it would be worth scouring the literature to see if anyone has proved or conjectured this. | |
| Dec 21, 2022 at 7:19 | comment | added | YCor | @DavidESpeyer anyway does this already show that if $c$ has the required property (or just that $\lfloor c^n\rfloor$ is odd for every $n\ge 1$) then $c$ is rational? | |
| Dec 21, 2022 at 6:54 | answer | added | Sebastien Palcoux | timeline score: 18 | |
| Dec 14, 2022 at 12:01 | comment | added | David E Speyer | Oh, but I am wrong! The Pisot case gives a linear recurrence, but the rational case doesn't, as "so-called friend Don" points out. | |
| Dec 14, 2022 at 11:54 | comment | added | David E Speyer | To explain a bit more about the linear recurrence: Let $a_n$ be a non-periodic sequence of integers obeying a linear recurrence. Then there is an $N$ such that, for any $M>0$, the values of $a_n \bmod M$ are periodic for $n>N$. Choose a prime divisor $p$ of $a_{n_0}$ for some $n_0>N$, then $p$ will divide infinitely many $a_n$. | |
| Dec 14, 2022 at 11:50 | comment | added | David E Speyer | If this is true, then none of the sequences $(c^n/m) \mod 1$ are equidistirbuted, for any $m$. To my knowledge (but I am very far from an expert) the only thing that makes $(b c^n) \mod 1$ not equidistributed (for $c>1$) is when $c$ is either rational or a Pisot number. In those latter cases, $\lfloor c^n \rfloor$ satisfies a linear recurrence relation, and thus can't be infinitely prime. | |
| Dec 14, 2022 at 0:15 | comment | added | so-called friend Don | This is equivalent to asking whether there is a $c>1$ for which $\lfloor c^n\rfloor$ is composite only finitely often. It seems to not even be known there is no such rational number $c$. For the state of the art on that restricted problem, see "Integer parts of powers of rational numbers" by Dubickas and Novikas, Math. Z. 251 (2005), no. 3, 635–648. | |
| Dec 13, 2022 at 23:09 | comment | added | JoshuaZ | In a slightly different direction, mathoverflow.net/questions/286389/… has as a result that there's no constant C such that $\lfloor n^C \rfloor$ is always prime. That's a lot denser than what you want though. | |
| Dec 13, 2022 at 22:56 | history | edited | John Baez | CC BY-SA 4.0 |
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| Dec 13, 2022 at 20:09 | history | asked | John Baez | CC BY-SA 4.0 |