Timeline for A mutation of the Collatz disease
Current License: CC BY-SA 4.0
8 events
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| Jul 6, 2023 at 7:53 | comment | added | Gottfried Helms | I've answered a question in MSE where I show, that the iteration in question is exactly equivalent to the 3x+1 iteration, and that no cycles other than in 3x+1 iteration can occur. I could copy that answer here, or see at math.stackexchange.com/a/4728332/1714 The last two paragraphs focus the question here. | |
| Jun 30, 2023 at 17:29 | comment | added | Robert Frost | Sorry, that was wrong. It should have been $p(x)=2^{\nu_2(x)}$ then $T=\sum_{k=0}^\infty p(c^n(x))$ | |
| Jun 30, 2023 at 17:15 | comment | added | Robert Frost | Sorry a number of ways, all are essentially equivalent to "multiply by two until odd" but the broadest is the extension to $\Bbb Q_2$ you get if you consider the function $c(x)=3x+2^{\nu_2(x)}$ then let $c^n(x)$ indicate the nth composition. Then let $p(x):\Bbb Q_2\to\{0,1\}$ indicate $x\pmod2$. Then let $T(x)=\sum_{n=0}^\infty p(c^n(x))$. In this case $T$ is a 2-adic (isometric) homeomorphism and the Collatz conjecture asks if $T(\Bbb Z[\frac16]^+\setminus\{\frac13\})\subset \frac13\Bbb N$ | |
| Jun 30, 2023 at 16:52 | comment | added | HenrikRüping | sorry but I have no idea how to extend the collatz iteration to rationals with denominator divisible by two. Can you explain what you mean by that ? | |
| Jun 30, 2023 at 13:31 | comment | added | Robert Frost | One can say a little more. All the positive elements of $\Bbb Z[\frac16]$ converge, and in fact one can write an inequality within the negative integers which weighs up the 2-adic and 3-adic values of any given dyadic/ternary rational and tells you whether or not its orbit is eventually positive. | |
| Jun 30, 2023 at 11:30 | history | edited | HenrikRüping | CC BY-SA 4.0 |
deleted 2 characters in body
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| Jun 30, 2023 at 8:30 | vote | accept | Roland Bacher | ||
| Jun 30, 2023 at 6:28 | history | answered | HenrikRüping | CC BY-SA 4.0 |