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I've only done a bit of research on the current findings, not sure if anyone here can answer this.

Q1: I just haven't been able to find,

  • has it been shown yet that a it is impossible for a loop to exist,
  • and for a counter example to exist does it have to grow to infinity?

Q2: I've seen that all numbers up to 2^60 have been checked, and using this information I'm able to show that no loops exist of length 101 or less (as a lower bound).

Has something like this been shown before? I just can't seem to find much information on it.

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    You're obviously not very good at researching - https://en.wikipedia.org/wiki/Collatz_conjecture#Cycles – Peter Foreman Jun 18 '19 at 13:15
  • @PeterForeman I'm not honestly, thanks for finding this though! Do you know where the best place is for me to make a post about how to show that no cycles up to 101-cycles exist, and how to extend this once more numbers are checked. – Cian Flint Jun 18 '19 at 13:58
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    @CianFlint it's known cycles would have to be massive. If you exhaust Wikipedia, try Lagarias. – it's a hire car baby Jun 18 '19 at 16:22
  • $2^{60}$ is not a very big number. Imagine a line of $60$ pixels that is filled with either black or white color. Think about all possible permutations those b/w colors then you've covered all of the initial numbers of that region. If you watch movie that has $1024$ pixels wide resolution, you've only covered $5.8$ percent of that width. Even if this example is totally relative since you can do $2048$ or $4096$ wide bitstrings and get a smaller percentage, its difficult to make a point about why it is so difficult to know where there could be a potential counter-example or non-trivial cycle. –  Jun 19 '19 at 01:28
  • Cian Flint - I edited the title, hoping I understand your question correctly. Note that for researchers (professional or amateur) using MSE it is very helpful when the titles refer to the core content of the question and this is viewable when browsing through the ocean of questions....(as I did just now...) If you have an even better reframing for the title please feel free to improve even more... . – Gottfried Helms Aug 19 '19 at 08:32

3 Answers3

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It has neither been shown that a loop other than the trivial $\{4,2,1\}$ does not exist, nor that there exists a trajectory that grows to infinity.

Klangen
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On Q1: @Klangen has already answered.
On Q2: That has been done, at least since R.Crandall's study in 1978.

There is an easy formula how you can disprove loops of the length of N (odd numbers) from the knowledge, that no number $a_1 < \chi$ can be member of a nontrivial loop (since they've all checked up to some limit $\chi$, which by the time seem to be $\chi_{\small 2019}=87 \cdot 2^{60}$ ). That process is very simple and can be done for instance with Pari/GP in few lines of code.

Here the first few $N$ can be used to determine, that for loop-length $N$ (odd numbers) the minimal element $a_1$ must be smaller than some $\chi$. If $\chi$ is our upper limit of all consecutive numbers $a$ checked to be convergent, a loop of this length $N$ cannot occur (the number $S$ in the table is the number of divisions by $2$):

for(N=2,30,S=ceil(N*ld3);a_mean=(2^(S/N)-3);if(a_mean>1e-1,next()); print("N=",N," S="S,," a_1 <", 1/a_mean))

N=5 S=8 a_1 <31.81356435
N=8 S=13 a_1 <11.84530260
N=10 S=16 a_1 <31.81356435
N=13 S=21 a_1 <15.64144437
N=15 S=24 a_1 <31.81356435
N=16 S=26 a_1 <11.84530260
N=17 S=27 a_1 <146.7715891
N=18 S=29 a_1 <18.22480824
N=19 S=31 a_1 <10.15029702
N=20 S=32 a_1 <31.81356435
N=21 S=34 a_1 <13.94273766
N=22 S=35 a_1 <80.70304398
N=23 S=37 a_1 <20.09651637
N=24 S=39 a_1 <11.84530260
N=25 S=40 a_1 <31.81356435
N=26 S=42 a_1 <15.64144437
N=27 S=43 a_1 <62.86002764
N=28 S=45 a_1 <21.51503268
N=29 S=46 a_1 <386.2846256
N=30 S=48 a_1 <31.81356435

I stopped the output here, but for $N \le 100$ the upper bound for $a_1$ is very small, much smaller than the current known $\chi_{2019}$.
Here we proceed simply to $2 \le N \le 3000$ and document only that $N$ where the upper bound for $a_1$ is at least $1e5$

for(N=2,3000,S=ceil(N*ld3);a_mean=(2^(S/N)-3);if(a_mean>1e-5,next()); print("N=",N," S="S,," a_1 < ", 1/a_mean, " ~ 2^", floor(-log(a_mean)/l2)))

N=971 S=1539 a_1 < 330749.4970 ~ 2^18
N=1277 S=2024 a_1 < 212745.4992 ~ 2^17
N=1583 S=2509 a_1 < 174546.8464 ~ 2^17
N=1636 S=2593 a_1 < 583287.1405 ~ 2^19
N=1889 S=2994 a_1 < 155653.6267 ~ 2^17
N=1942 S=3078 a_1 < 330749.4970 ~ 2^18
N=2195 S=3479 a_1 < 144382.7969 ~ 2^17
N=2248 S=3563 a_1 < 251503.8361 ~ 2^17
N=2301 S=3647 a_1 < 860563.0163 ~ 2^19
N=2501 S=3964 a_1 < 136895.8416 ~ 2^17
N=2554 S=4048 a_1 < 212745.4992 ~ 2^17
N=2607 S=4132 a_1 < 454137.6774 ~ 2^18
N=2807 S=4449 a_1 < 131561.1466 ~ 2^17
N=2860 S=4533 a_1 < 189759.9235 ~ 2^17
N=2913 S=4617 a_1 < 330749.4970 ~ 2^18
N=2966 S=4701 a_1 < 1166399.316 ~ 2^20

The largest upper bounds relative to $N$ occur at convergents of the continued fraction of the $\log_2(3)$ . Here is a table only for $N$ from this set of first 30 convergents. At the somehow famous $N=397573379$ the upper bound for $a_1$ is approx $2^{60}$ so up to until recent years a cycle with that length $N$ could not be excluded with this simple formula alone. The more recently considered length $N=6586818670$ with $a_1 < 2.168901553 E20 \approx 2^{67}$ has been excluded by the yoyo@home-project in 2017 (if I've that correctly in mind):

for(k=2,30,N=cvgts[2,k];S=ceil(N*ld3);a_mean=(2^(S/N)-3);if(a_mean>1e-5,next()); print("N=",N," S="S,," a_1 < ", 1/a_mean, " ~ 2^", floor(-log(a_mean)/l2)))

N=15601 S=24727 a_1 < 285817586.2 ~ 2^28
N=79335 S=125743 a_1 < 7216102493. ~ 2^32
N=190537 S=301994 a_1 < 9.845728110 E11 ~ 2^39
N=10590737 S=16785922 a_1 < 5093068.134 ~ 2^22
N=10781274 S=17087915 a_1 < 2.944018243 E14 ~ 2^48
N=53715833 S=85137582 a_1 < 25831855.26 ~ 2^24
N=171928773 S=272500658 a_1 < 3.203055713 E16 ~ 2^54
N=225644606 S=357638240 a_1 < 108512118.1 ~ 2^26
N=397573379 S=630138897 a_1 < 1.252079477 E18 ~ 2^60   ****************
N=6189245291 S=9809721695 a_1 < 2976397830. ~ 2^31
N=6586818670 S=10439860591 a_1 < 2.168901553 E20 ~ 2^67  ***************
N=65470613321 S=103768467014 a_1 < 3.148470972 E10 ~ 2^34
N=137528045312 S=217976794617 a_1 < 5.101255583 E22 ~ 2^75
N=753110839881 S=1193652440099 a_1 < 3.621697580 E11 ~ 2^38
N=5409303924479 S=8573543875303 a_1 < 2.734923048 E25 ~ 2^84
N=6162414764360 S=9767196315402 a_1 < 2.963495073 E12 ~ 2^41
N=11571718688839 S=18340740190704 a_1 < 2.990877298 E26 ~ 2^87
N=52449289519716 S=83130157078218 a_1 < 2.522277663 E13 ~ 2^44
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Reformulate the Collatz iteration in a compact form as

(1) $v_n = z_n + 2^{2n+1-r}m_r$

for any input $u \in 2\mathbb{N}+1 \not \equiv 0\pmod{3}$ such that $u=3m_r+r$ where $r=1$ if $u \equiv 1\pmod{3}$ or $r=2$ if $u \equiv 2\pmod{3}$ and $z_n = \sum_{i=1}^{n} 2^{i-1}$ for $n \in \mathbb{N}$. Iterating (1) by one step for each $n \in \mathbb{N}$ yields an infinite set $\{v_n\}_{n \in \mathbb{N}}$, termed sibling set, arising from parent $u$. Suppose we consider $u$ and $v_n$ as vertices connected by edge $(u, v_n)$ from $u$ to $v_n$ for every $n \in \mathbb{N}$. Then starting with $u_0=1$, repeatedly iterating (1) constructs an arborescence graph $G=(V,E)$ that is the union of infinitely many sibling sets, with root at $r_0$, vertex set $V$ and edge set $E$, as shown here. Every path of any length from the root in $G$ represents an odd Collatz sequence in reverse. It can be shown that every vertex in $G$ is unique and that $V=2\mathbb{N}+1$, i.e. $V$ exactly covers the set of odd natural numbers. Since $G$ is weakly connected, then we infer that: (i) only the trivial loop exists (but is necessarily ignored in $G$) and (ii) every odd Collatz sequence always converges to $1$ (hence, cannot diverge). These results, subject to further scrutiny, answer both questions under Q1 above.