Timeline for Relaxed Collatz 3x+1 conjecture
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2022 at 16:01 | answer | added | Terry Tao | timeline score: 8 | |
| Sep 22, 2022 at 4:02 | answer | added | Terry Tao | timeline score: 16 | |
| Dec 27, 2017 at 5:27 | comment | added | Alexander Burstein | At the risk of sounding completely clueless, would it be any easier or harder to prove or disprove a generalized version of the conjecture (this one or the original) with the operations $bq\mapsto q$ and $bq+r\mapsto (b+1)(bq+r)+(b-r)$ for any other $b\in\mathbb{N}$ and all $q,r\in\mathbb{N}$ such that $0<r<b$? | |
| Dec 15, 2017 at 10:12 | answer | added | Dan Brumleve | timeline score: 11 | |
| Dec 15, 2017 at 3:42 | comment | added | Dan Brumleve | Expanding the tree generated by the two operations gives a set with polynomial density, but I think we won't be able to prove it that way, since apparently the best result on the density of counterexamples to the Collatz conjecture is in the wrong direction. terrytao.wordpress.com/2011/08/25/… If there were fewer than $n^c$ counterexamples less than $n$, then we could jump out. | |
| S Dec 14, 2017 at 20:01 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
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| Dec 14, 2017 at 18:06 | review | Suggested edits | |||
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| Nov 28, 2017 at 15:53 | comment | added | Robert Frost | @GerhardPaseman "The Threshold Might Be Two" Paseman. There's an interesting connection between this problem and a special property the number $2$ has, as the exponent in discrete calculus. The weakened Collatz Conjecture (no nontrivial cycles) is equivalent to a statement in discrete calculus which looks a bit like a Hensel lift in base 2. Robert "$\Delta_n(x)$ has to be precisely $2x+2^{v_2(x)}$" Frost | |
| Jul 5, 2017 at 10:28 | comment | added | Robert Frost | This describes every $x$ as is the set of all strings in base $3$, and the conjecture is that for every such string there's some length of string of $1$'s which can follow it, to make it a power of $2$. Some measure of orthogonality of the function $3x+1$ to the integers in the $2$-adic metric space away from the vicinity of $1$ would seem to do it. | |
| Jul 5, 2017 at 10:13 | comment | added | Robert Frost | The simplest answer in the affirmative would follow if for every integer $x$ there exists some $n$ such that $3^nx+\frac{3^n-1}{2}\in\{2^m:m\in\mathbb{N}\}$. In other words we can iterate $3x+1$ and always eventually arrive at some power of $2$. | |
| Sep 4, 2015 at 13:00 | comment | added | Suvrit | Perhaps the Markov Chain view may be interesting to look at here... | |
| Sep 4, 2015 at 3:19 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Sep 4, 2015 at 2:48 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Sep 4, 2015 at 0:19 | comment | added | Gerhard Paseman | @YaakovBaruch, I am surprised. I must have been thinking about something different from what I said. Thank you for the links. Gerhard "And It Wasn't Floor Either" Paseman, 2015.09.03 | |
| Sep 3, 2015 at 23:48 | comment | added | Max Alekseyev | @wythagoras: In fact, 27 can take just 23 steps (and that's the minimum): 27 -> 82 -> 41 -> 124 -> 373 -> 1120 -> 560 -> 280 -> 140 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. | |
| Sep 3, 2015 at 22:17 | comment | added | Yaakov Baruch | @GerhardPaseman: the idea fails for $\alpha=\frac{\sqrt(5)+1}{2}$ (and 7 as starting point) and more generally for salem numbers. See item 29 in the Lagarias's survey on the $3x+1$ problem: arxiv.org/pdf/math/0309224.pdf; for more detail: math.grinnell.edu/~chamberl/papers/3x_survey_eng.pdf | |
| Sep 3, 2015 at 18:47 | comment | added | Gerhard Paseman | If instead of 3x+1 you had ceil(\alpha x + 1) with \alpha smaller than 2, you could show convergence pretty easily. My feeling is that there is a threshold between 2 and 3 where for \alpha below the threshold, it will be easy to prove convergence, and for \alpha above it, it will be hard. Gerhard "The Threshold Might Be Two" Paseman, 2015.09.03 | |
| Sep 3, 2015 at 18:24 | comment | added | wythagoras | It can certainly reduce the time for reducing it to zero by a good amount. For example 27 takes at most 48 steps (out of possibilities considered) instead of 112. | |
| Sep 3, 2015 at 17:05 | history | asked | Max Alekseyev | CC BY-SA 3.0 |