Timeline for A mutation of the Collatz disease
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 28, 2023 at 11:31 | answer | added | Gottfried Helms | timeline score: -1 | |
| Jun 30, 2023 at 8:30 | vote | accept | Roland Bacher | ||
| Jun 30, 2023 at 6:28 | answer | added | HenrikRüping | timeline score: 7 | |
| Jun 29, 2023 at 20:13 | comment | added | Roland Bacher | @RodrigodeAzevedo No, the 2 is not a typo: $(3x+3^k)$ is always even for odd $x$, it is therefore natural to divide it by $2$ immediately. | |
| Jun 29, 2023 at 15:06 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
added 7 characters in body
|
| Jun 29, 2023 at 13:22 | comment | added | Rodrigo de Azevedo | Is the $2$ in $\dfrac{3x+3^k}{2}$ a typo? | |
| Jun 29, 2023 at 13:21 | comment | added | Roland Bacher | @HenrikRüping Thanks for your pertinent comments. I should have read the Wikipedia article before posting. | |
| Jun 29, 2023 at 12:30 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
added 61 characters in body
|
| Jun 29, 2023 at 11:42 | comment | added | HenrikRüping | Sorry even that is not true. Maybe the denominator is not $\pm 1$, but the numerator is divisible by it. Then we would still get an integral cycle and these are not covered by my previous comment. | |
| Jun 29, 2023 at 11:39 | comment | added | HenrikRüping | EDIT: oops I missed that the denominator $2^n-3^m$ can be $\pm 1$. Are there only finitely many solutions for $m,n$ for this equation. If so, there should be only finitely many Collatz-cycles in the integers and the section in wikipedia should give a way to list them all. Maybe the 1,4,2 -cycle is the only positive one. | |
| Jun 29, 2023 at 11:25 | comment | added | HenrikRüping | At least the section about parity cycles seem to imply that the Collatz sequence for numbers whose denominator is a power of three either is unbounded or, if it becomes periodic, it always ends up in the trivial cycle above. | |
| Jun 29, 2023 at 11:18 | comment | added | HenrikRüping | This is essentially the usual collatz conjecture for rationals whose denominator is a power of three. I dont know the answer to this question, but I suggest checking out the references in en.wikipedia.org/wiki/…. | |
| Jun 29, 2023 at 10:57 | history | asked | Roland Bacher | CC BY-SA 4.0 |