A question about mathematical reasoning regarding observations in Collatz conjecture 1-cycles

collatz conjectureelementary-number-theoryrecreational-mathematics

This is a question about the use of empirical observations to guide strategies in constructing mathematical proofs. It uses prior discussions of 1-cycles in the Collatz conjecture as a model.

In a response to this post regarding the Collatz conjecture, Collag3n helpfully provided the following in summarizing Steiner's proof that there are no 1-cycles:

To have a 1-cycle you start at an odd number $a=b2n−1$, you climb up
to $b3n−1$ and then divide by $2^{m−n}$ to reach a again:

$$b2^n – 1=\frac{b3^n – 1}{2^{m-n}} \tag 1$$

The explanation also included the following: $$\frac{2^m}{3^n} – 1 = \frac{2^{m-n} -1}{b3^n} \tag 2$$

For simplicity, refer to the last equation in the quote as equation (1), and the subsequent equation as equation (2).
The point of interest is the parameter "$b$". It cannot be just any integer, it must be odd for a given $n$, and from (2) we see that it is equal to $$ b=\frac{2^{m-n}-1}{2^m-3^n} \tag {2.1}$$

Conditions(3): We can also show that $b$ is related to $m$ and $n$, for $p = 0,1,2,3,\ldots$ as follows:

If $b = 6*p + 1$ then $n$ must be odd and $m$ must be even;
If $b = 6*p + 3$ then, if $n$ is odd, $m$ must be odd and if $n$ is even, $m$ must be even;
If $b = 6*p + 5$ then $n$ must be even and $m$ must be odd.

The value $b$ is also constrained by allowable values for "$a$" (using the notation in the quote from Collag3n):

"$a$" cannot be a multiple of $3$;
it cannot be of the form $2^{s+1} – 2^s -1 $ for any $s$ less than $n$ as it appears in the definition of $a$. For example,
$\qquad$ it cannot be of the form $4*p + 1$ for any $n$,
$\qquad$ $8*p+3$ for $n > 2$,
$\qquad$ $16*p + 7$ for $n > 3$,
$\qquad$ etc.

Let us assume for purposes of the main question that all of these relationships are true, and call them collectively (3).

Now, there are two more observations relevant to this question. The first is that, for 1-cycles, there is a 1-to-1 correspondence between $m$, $n$, and $a$. In particular, $$m = [n*\ln_2 3] +1 \tag 4$$ (where the quantity in brackets is the integer part of the product). Call this relationship equation (4).

The second observation is that using the definition of "$a$" from the quote, and the conditions of a 1-cycle, i.e. $$2^n*b-1 = \frac{3^n*b-1}{2^{m-n}}$$

we can write $$2^m = \frac{2^n(3^n*b-1)}{2^n*b-1}$$

The value $m$ can be calculated by $$m=\left[\frac{\ln(\frac{2^n(3^n*b-1}{2^n*b-1})}{\ln(2)}\right] + 1 \tag 5$$

where, again, the brackets indicate taking the integer portion of the quantity within them. Call this latter equation (5).

Finally, here is the question: must the $m$ calculated using equation (5) match the $m$ calculated using equation (4) for all allowable values of $b$? Empirically this seems to be the case, but is there a principle of mathematical reasoning by which we can use the simpler formula, or the constraints labeled as (3) above to show that the two equations give the same result?

How do we decide if we happen to calculate an $m$ that disagrees between the two equations which to discard? Are the constraints (3) irrelevant?

As an aside, here why this is of interest. For a given $n$, we can substitute various values of b, starting at $b = 1$ into equation (5). We can modify our calculation of $m$, dispensing with taking the integer part and adding $1$. Call this the "raw $m$." This value seems to start, for $b = 1$, at a value below the value of $m$ calculated using equation (4). "Raw $m$" then decreases monotonically with increasing $b$ toward the value $n*\ln_2 3$. Now, if the answer to the main question of this post is "yes," then it would seem that, if there are any integer values of $m$ that satisfy the conditions of a 1-cycle, then they can only occur if $b = 1$ (since $m$ decreases monotonically with increasing $b$). This implies that the only permissible starting values "$a$" (Steiner of course proved that there are none), would have to be of the form $2^n-1$, $n$ would have to be odd and $m$ would have to be even.

Another empiric observation is that if we calculate $b$ for allowable integer values of $m$ and $n$ using equation (5), $b$ approaches $\frac{1}{2^n}$ as $n$ increases.

Edited to correct the relationships in (3).

Restatement of the question

Given the above, it might be useful to more explicitly ask my question. The question, again is about the use of empirical observations and mathematical logic.

Here is an explicit description of the reasoning behind the question:

Using the relationship $a=\frac{3^n-2^n}{2^m-3^n}$, along with the fact that "$a$" must be $> 1$, I am pretty sure we can show, for a 1-cycle, that $$n*\ln_2 3 <m < n*\ln_2 3 +1 \tag 6$$ In other words, there is a 1-to-1 relationship between $n$ and $m$, such that $m$ is the first integer greater than $n*\ln_2 3$.

This introduces a difficulty: the relationship between $m$ and $n$ is not smooth. When we increment $n$ by $1$, depending on what it is, $m$ will sometimes increase by $1$ and sometimes by $2$. So, for example, if $n = 7$ then $m = 12$, and if we increment $n$ by $1$, $m$ also increases by one; e.g. $n = 8$ and $m = 13$. However, if we start with $n = 11$, $m = 18$, but when $n = 12$, $m$ increases to $20$. Although the relationship between $n$ and $m$ is not smooth, it is bounded; $m$ only increases by $1$ or $2$ when $n$ increases by $1$. This boundedness in the relation between $n$ and $m$ suggests that we might find other bounds that can be used to either prove or disprove hypotheses related to 1-cycles. This appears to be what Steiner did in his proof of the non-existence of 1-cycles, as Collag3n explained in his response to the post linked above.

So the question is how to go about finding candidate bounds using empiric observation and mathematical reasoning?

I think it can be shown fairly easily that, in order for there to be a 1-cycle, $2^{m-n} – 1$ must be divisible by $2^m – 3^n$, or in other words, $b$ must be a positive integer. Now if we empirically calculate a few values for $b$ using the relationship $$b = \frac{2^{m-n} – 1}{2^m – 3^n} \tag 7$$ we notice that the values for $b$ are all less than $1$, and usually significantly less than $1$. Do the permissible values of $b$ for all $n$ have an upper bound? If they do, and that bound is less than one, then there can be no 1-cycles in the Collatz conjecture. (I think Collag3n explicitly showed that this bound exists using the same reasoning Steiner used in his proof, but I am looking for additional ways to validate bounds). We would have found an additional bound, in addition to the one used by Steiner to show that 1-cycles cannot exist. A brief empirical inquiry seems to suggest this, but this is not mathematically rigorous. We need to use mathematical logic to demonstrate that the hypothesis is true.

Likewise, if we instead substitute candidate values for $b$ into equation (5) above, we seem to consistently find that $$\frac{2^n(3^*b-1)}{(2^n*b-1)} \text{ is less than } 2^m$$. This suggests another bound: the permissible values for $m$ as a function of integer values of $b$ is less than the $m$ calculated using equation (4). In other words, $$\frac{2^n*(3^n*b-1)}{(2^n*b -1)} \tag 8$$ would fall between $n* \ln_2 3$ and the first integer greater than $n*\ln_2 3$ for all $n$ and $b$. We would have found another bound that might be used to show the non-existence of 1-cycles. Again however, all we have is a sample of empirical observations, not a mathematical rigorous proof. What principles of mathematical reasoning can we use to go from one to the other?

We also observe that calculated values of $m$, using equation (5) and a given $n$, increase as $b$ decreases. That implies that, unless a calculated value for $m$ using $b=1$ is greater than the first integer greater than $n*\ln_2 3$, the first $m$ that satisfies the conditions for a 1-cycle will occur when $b = 1.$ If that is the case, then the smallest element in the loop will be of the form $2^n – 1$. We would have found another bound, not one that disproves the existence of 1-cycles, but one which constrains their possible values.

The question is not whether anyone can prove or disprove any of the hypothesis obtained from empirical observation, although that would be awesome, it is what principles and insights might suggest ways to approach such proofs. Can proof by induction be used even though the relationship between $m$ and $n$ is not smooth? Do the empiric observations, if proven, imply bounds on $2^{m-n}-1$ or $2^m-3^n$?

Continuing on, we find other contingent hypotheses. For example, if $b =3$ then $m-n$ must be an even number, which means that $$2^{m-n}-1$$ can be written $$(2^\frac{m-n}{2} +1)(2^\frac{m-n}{2}-1) \tag 9$$ I think this implies that one factor or the other must be divisible by $2^m-3^n$. This would be another bound, but I could be wrong.

Addendum

I overlooked an obvious bound: the maximum allowable "$a$" for a given $n$. Since $a=\frac{3^n-2^n}{2^m-3^n}$ and $m$ is a bounded function of $n$, there is an upper limit on "$a$" given $n$.

Response to Gottfried:

Thank you for the very thoughtful and helpful answer. I had been dabbling with lower bounds for $mln2-nln3$ and came up with $$mln2-nln3>(\frac{n}{n + 2ln(n)})^{nln2}$$ I guess making the exponent nln2-1 or nln2-(some other number) would give a tighter bound, but I can't think of another justification for doing so.

The reason I tried to find a tight lower bound is because of a thought that was stimulated by your responses to a couple of my other questions. I was unaware of continued fractions until you mentioned them in response to my question about periodicity in the permissible values of $X_1$, the smallest element in a 1-cycle. I wondered why the progressions of these values seemed so well-behaved, i.e., why they appeared to have an orderly progression and not contain outliers. So, I read a little on continued fractions. Then I tried to better understand Collag3n's explanation of Steiner's proof and it occurred to me that, if $mln2-nln3$ has a lower bound, the progression of the terms of the continued fraction must have an upper bound. My reasoning went something like this:

Let $a_i$, i=0,1,2,3,4…. be the terms of the continued fraction expansion of $ln_23$, and $|\frac{p_i}{q_i}-\frac{ln3}{ln2}|$ be the absolute value of the error associated with the i^th convergent of the continued fraction expansion. Further, let $f(n)$ be some function such that $|mln2-nln3|>f(n)$ for all n (e.g., $f(n)=\frac{1}{2^n}$ using Rhin's bound and $\frac{1}{10nln(n)}$ using your tighter bound described in your answer). Also, at the convergents of the continued fraction expansion of $ln_23$, $m=p_i$ and $n=q_i$.

Now, $\frac{1}{q_iq_{i+1}}>|\frac{p_i}{q_i}-\frac{ln3}{ln2}|$, or $$\frac{1}{nq_{i+1}}>|\frac{m}{n}-\frac{ln3}{ln2}|$$

Multiplying both sides of this last inequality by $nln2$, we get $$\frac{nln2}{nq_{i+1}}>|mln2-nln3|$$ so we can write $$\frac{ln2}{q_{i+1}}>|mln2-nln3|>f(n)$$ or $$\frac{ln2}{q_{i+1}}>f(n)$$

By definition, $q_{i+1} = a_{i+1}q_i+q_{i-1}$ where $a_i$ is the i^th term in the continued fraction representation of $ln_23$.

Again, letting $q_i=n$, we can write $$\frac{ln2}{(a_{i+1}n+q_{i-1})}>f(n)$$ or$$\frac{ln2-f(n)q_{n-1}}{nf(n)}>a_{n+1}$$Using your formula for f(n), and understanding that $n_{i-1}=q_{i-1}$ is the n associated with the i-1 convergent we get$$10ln(n)ln2-\frac{n_{i-1}}{n}>a_{i+1}$$It would thus seem that there is a limit for how large an odd convergent can be relative to the previous odd convergent, and if we graph computed values of $X_1$ at the odd convergents, we get a reasonably well-behaved curve.

Thank you again for your patient help.

Addendum to reply to Gottfried:

After staring at your explanation in (1) below (wherein you demonstrate that b cannot be 1), it occurs to me that b cannot be 3, 9, 27 or $3^i$, i=1,2,3,…n either. The reason is that, for a loop defined by $$X_1=2^n-1$$ the second element is $$\frac{3(X_1)+1}{2}=\frac{3(2^n-1)+1}{2}\tag{10}$$ This simplifies to $$\frac{3(2^n)-2}{2}=2^{n-1}(3)-1$$ So, $2^{n-1}(3)-1$, does not give the value of smallest element of a supposed loop having n-1 elements, it gives the value of the second element of a supposed loop of n elements having $2^n-1$ as the smallest element.

So, $2^{n-1}(3)-1$ and $2^n-1$ describe the same (theoretical) loop. Similarly, $$2^{n-2}(3^2)-1$$ gives the third element in a 1-cycle loop with $$X_1=2^n-1$$and defines the same loop. Therefore, if there are no non-trivial loops with b=1, there are no such non-trivial loops with b=3 (or 9, etc). So, loops with b=1 and those with b=3 are not independent, which makes me wonder if there are any other pairs of ($b_1,b_2$) that describe the same loop.

On further consideration:

I seems that if $2^n(B)-1$ gives the value of the first element in a 1-cycle with n elements, then $2^{n-1}(3B)-1$ gives the value of the second element in that loop, and therefore, b=B and b=3B describe the same (theoretical) loops.

If $X_1=2^n(B)-1$ then the second element of this loop will be:$$X_2=\frac{3(2^n(B)-1)+1}{2}=\frac{2^n(3B)-2}{2}=2^{n-1}(3B)-1$$Therefore, regardless of the value of B, $2^n(B)-1$ and $2^{n-1}(3B)-1$ describe the same loop.

Or, going the other direction, if $2^s(3N)-1$ defines a loop, the same loop is described by $2^{s+1}N-1$ and has s+1, rather than s elements.

***HOWEVER!***I over-interpreted this fact. It does not demonstrate that there can be no loops with b=3, only that if there is such a loop, it can also be characterized by an equation of the form $2^{n+1}-1$. This latter equation does not identify a loop, it gives a value that is related to the loop, namely a number that when multiplied by 3, add 1 and dived the result by 2 gives the starting number for a loop consisting of n-1 elements having. Gottfried's sense was correct.

The reason that the two equations identify the same loop, without characterizing the loop, is that the powers of 2 are not the same.

I do think, however, I can demonstrate that there are no 1-cycles when b = 5 or b = 7 (or 13,15,21,23…)

Unless I made a mistake somewhere.

Best Answer

At the beginning this is a comment, but too long for the comment function, and an answer to your last (more general) question is appended

At eq (5) and (7) I'd like to add two comments/informations.

(1) If we write (using $A=m-n$) $$ 2^A = {3^nb-1\over2^nb -1} \qquad \text{from eq (5)} \tag 1 $$ then we can use a very old result (of the medieval N.Oresme) to know, that $b \not = 1$.

(1a) Assume $b=1$ and rewrite $$ 2^A = {3^n-1\over2^n -1} \tag {1a} $$ then for the numerator to have the primefactor $2$ to a larger power than $1$ we need to have $n$ to be even. But if $n$ is even, then the denominator has the primefactor $3$, then -because the numerator cannot be divisible by $3$- the rhs is fractional.
Conclusion:

if $b=1$ then $n$ must be odd, then $A=1$ then $m-n=1$, then $(n,m)=(1,2)$ This defines uniquely the trivial cycle and no nontrivial 1-cycle is possible.

(1b) Moreover, in eq (1) $b$ cannot be even, because then the numerator in the rhs would be odd, and cannot equal $2^A$ with $A>0$.

(1c) Looking at whether $b$ can grow arbitrarily we would rewrite $$ 2^A \overset?= {3^n-1/b\over2^n -1/b} \tag {1c.1} $$ and we see, that there must be an upper bound for $b$ from where on the rhs approximates $(3/2)^N$ below its next above integer $$ \lim_{b \to \infty} {3^n-1/b\over2^n -1/b} = (3/2)^n < \lceil(3/2)^n \rceil \implies \ne 2^A \tag {1c.2}$$ $ \implies$ There is an upper bound for $b$ from where and above eq (1) has no integer solution.

(Of course, there is another upper bound for $b$, which can immediately be derived from your eq. (2.1): $ b= (2^A-1)/(2^m-3^n)$. Here the denominator is -for $(n,m)>(1,2)$ - equal or larger $5$ (the differences $\{1,2,3,4\}$ can be trivially be excluded) so $b \le (2^A-1)/5$)

(1.d) There is the obvious inequality
$$ \underset{\text{lo}}{\underbrace{ { 3^n+1/b \over 2^n+1/b } } }< \underset{\text{mid}}{\underbrace{ 3^n \over 2^n }} <\underset{\text{hi}}{\underbrace{{ 3^n -1/b \over 2^n -1/b }}} \tag {1d.1} $$ and it can be derived (but I don't have my notes at hand) that using $1/b= 1/2 \cdot (4/3)^n$ we get that the interval from lo to hi is smaller than $1$ (and tends to $1$ with $n \to \infty$). That means, there is at most one integer between them. Unfortunately this does not mean that mid and hi have the same "ceiling" integer value. (I think those bounds are an interesting item, but I didn't proceed in that investigations when I found this)

(2) Connection to the Waring-problem (unsolved) The idea in (1c) problematizing the next integer above $ (3/2)^n$ reminds of the Waring-problem; and if the Waring-problem could be resolved, then this would include the nonexistence of an integer solution in (1). The conjecture from the Waring-problem is

(conj. from Waringproblem) for all $n>7$ we have that $$(3/2)^n + (3/4)^n < \lceil (3/2)^n \rceil \tag {2.1} $$ To see how this includes the disproof of any nontrivial 1-cycle we rewrite the rhs (1) as expansion into a geometric series: $$ \begin{array}{ll} \large {3^nb-1\over2^nb -1} & =\large {3^nb\over 2^nb} + {3^nb\over 4^nb^2} + {3^nb\over 8^nb^3}+ \cdots - {1\over 2^nb} - {1\over 4^nb^2} - {1\over 8^nb^3} - \cdots \\ &= (3/2)^n + \underset{\lt (3/4)^n}{\large \underbrace{ \frac1b{3^n\over 4^n} - {4^n-3^n/b^2\over 8^nb}- {4^n-3^n/b^2\over 16^nb^2} - \cdots }} \\& < (3/2)^n + (3/4)^n \\& \qquad \text{ if Waring conjecture is true and } n \gt 7 \implies \\ & < \lceil (3/2)^n \rceil \\ \implies 2^A & = \large{3^nb-1\over2^nb -1} \not \in \mathbb N \qquad \text{for } (A,b)>(1,1) \end{array}$$

(3) Non-elementary approaches (an answer to your question on "which methods")

Non-elementary approaches, namely use of results from theory of transcendental numbers, here from A. Bakers "linear logarithmic forms", allow to establish rigorous upper bounds for $b$, effectively providing disproofs of the nontrivial 1-cycle.

I show one standard and one conjectural example.

For both examples, a reformulation of the expression $2^m / 3^n$ as logarithmic difference is essential: $$ \text{(some_lower_bound)} < \log( 2^m / 3^n) = m \ln2 - n \ln3 \tag {3.1} $$ $ \qquad \qquad $ where in the following we use the short notation $ \Lambda = m \ln2 - n \ln3 $.

Your equation (7) $ b = (2^A-1)/(2^m-3^n) $ can be reformulated to have the $\Lambda$-expression on the lhs: $$ \begin{array} {} b &= (2^A-1)/(2^m-3^n) \\ 2^m-3^n &= (2^A-1)/b \\ {2^m\over 3^n} &= 1 + (2^A-1)/b/3^n \\ m \ln2 - n \ln3 &= \log(1 + (2^A-1)/b/3^n) \\ \Lambda &= \log(1 + (2^A-1)/b/3^n) \\ \end{array} \tag {3.2}$$

(3a) The "Rhin"-bound
According to J. Simons we have due to Georges Rhin the lower bound for $\Lambda$: $$ {1\over 453 n^{13.3}} < \Lambda \tag {3a.1}$$ So we compare $$ {1\over 453 n^{13.3}} < \log(1 + (2^A-1)/b/3^n) \tag{3a.2} $$ While this can be exact by exponentiation and rearrangement we show this by approximation to the $\log(1+x)$-expression which is - already for not too large $n$ - precise enough for our goal: $$ {1\over 453 n^{13.3}} < (2^A-1)/b/3^n \\ b < 453\cdot n^{13.3}\cdot (2^A-1)/3^n \tag{3a.3} $$ Because in the denominator the term $3^n$ grows much more than the numerator, there shall be one value $n$, from where the rhs begins to decrease monotonuously. It will be enough for us to find the value $n$ from where the rhs goes below $1$ (even $5$ would suffice). This can be found in the following table for $n=2..100$ where from $n=97$ the rhs is smaller than $1$: $$ \small \begin{array} {rrr|rl} n & m & A & b<(value) \\ \hline 2 & 4 & 2 & 1522915.79015 \\ 3 & 5 & 2 & 111575381.470 \\ 4 & 7 & 3 & 3982071995.90 \\ 5 & 8 & 3 & 25816123268.5 \\ 6 & 10 & 4 & 2.08388410882E11 \\ 7 & 12 & 5 & 1.11535994763E12 \\ 8 & 13 & 5 & 2.19576918045E12 \\ 9 & 15 & 6 & 7.12480257148E12 \\ 10 & 16 & 6 & 9.64331169146E12 \\ \vdots & \vdots & \vdots & \vdots \\ 80 & 127 & 47 & 8828.85628812 \\ 81 & 129 & 48 & 6943.31890371 \\ 82 & 130 & 48 & 2724.70408950 \\ 83 & 132 & 49 & 2134.23617282 \\ 84 & 134 & 50 & 1668.50339003 \\ 85 & 135 & 50 & 650.973105014 \\ 86 & 137 & 51 & 507.025120020 \\ 87 & 138 & 51 & 197.099208421 \\ 88 & 140 & 52 & 152.970293799 \\ 89 & 142 & 53 & 118.517747955 \\ 90 & 143 & 53 & 45.8353363847 \\ 91 & 145 & 54 & 35.3943797566 \\ 92 & 146 & 54 & 13.6439662782 \\ 93 & 148 & 55 & 10.5025435332 \\ 94 & 149 & 55 & 4.03599363595 \\ 95 & 151 & 56 & 3.09729599650 \\ 96 & 153 & 57 & 2.37342165864 \\ 97 & 154 & 57 & 0.908051150051 & (***)\\ 98 & 156 & 58 & 0.693843932287 \\ 99 & 157 & 58 & 0.264716993321 \\ 100 & 159 & 59 & 0.201716987763 \end{array} $$ From $n=97$ on we observe, that (due to the Rhin-bound) $b<1$ . Because it is also decreasing with increasing $n$ (monotonicity of (3a.3)) we have a proof, that a nontrivial 1-cycle with $n>97$ is impossible. Of course, we can test all smaller $n$ computationally, whether a 1-cycle does exist with this parameter, and find no nontrivial 1-cycle.
So other than by the consideration of the Waring-conjecture we even have a proof here, that no nontrivial 1-cycle exists.

(3b) The (conjectured) "n_log_n"-bound

By looking at values of $\Lambda$ for $n$ up to $10^{2 000 000}$ I found a much plausible lower bound; it is much tighter than the Rhin-bound, and as well as the Ellison-bound. The form of its graph looks also much plausible, so I conjecture this lower bound ^{heuristically seen for $1<n<10^{2 000 000}$}: $$ {1\over 10\cdot n \cdot \ln n} < \Lambda \tag {3b.1}$$
Using this for the reformulated equation $(3.2)$ we get

$$ {1\over 10 \cdot n \cdot \ln n} < \log(1 + (2^A-1)/b/3^n) \tag{3b.2} $$ Again we do not use the exponentiation but the approximation using the general inequality $\log(1+x)<x$ for small $x$ and get $$ \begin{array}{rl} {1\over 10 \cdot n \cdot \ln n} &< (2^A-1)/b/3^n \\ b &< 10 \cdot n \cdot \ln n \cdot (2^A-1)/3^n \\ \end{array} \tag{3b.3} $$ The heuristic table shows us that already for all $n>7$ the 1-cycle is disproved: $$ \small \begin{array} {rrr|rl} n & m & A & b<(value) \\ \hline 2 & 4 & 2 & 4.62098120373 \\ 3 & 5 & 2 & 3.66204096223 \\ 4 & 7 & 3 & 4.79212865572 \\ 5 & 8 & 3 & 2.31812045001 \\ 6 & 10 & 4 & 2.21204872744 \\ 7 & 12 & 5 & 1.93078418996 \\ 8 & 13 & 5 & 0.786010520251 & *** \\ 9 & 15 & 6 & 0.632945351496 \\ 10 & 16 & 6 & 0.245665228638 \\ 11 & 18 & 7 & 0.189100560331 \\ 12 & 20 & 8 & 0.143079181854 \end{array} $$ and, in fact since we know, that $b$ must be larger or equal than $5$ our lower bound for $b$ indicates already for $n \ge 2$ the disproof of the 1-cycle.

Conclusion
So introducing transcendental numbertheory and linear forms of logarithms is as well - besides of diophantine/factorization properties of the 1-cycle equation - an option for our handling of the 1-cycle problem, and moreover which would allow us not only to disprove certain subsets of lengthes $n$ but to determine an absolute lower bound (or in other cases an upper bound) for the interesting parameters at all (here: length $n$ of the considered 1-cycle).

Related Solutions

[Math] Intuition behind lack of cycles in the Collatz Conjecture

Consider the "compacted" formula (as also used in the "Syracuse"-version of the Collatz), defining one "step" beginning on $a$ going to $b$ (both odd $\ge 1$): $$ b = {3a+1\over 2^A } \tag 1$$ where $A$ contains the number of halving-steps.

Now to have a (very) simple cycle of just one step we must have: $$ a = {3a+1\over 2^A } \tag 2$$ It is simple to show, there is no solution except when $a=1$ only by rearranging: $$ a = {3a+1\over 2^A } \\ 2^A = {3a+1\over a } $$ $$ 2^A = 3+{1\over a } \tag3\\ $$ and having the rhs a perfect power of $2$ requires $a=1$, thus allowing (only) the well known "trivial" cycle $1 \to 1$.

Now extending this to a two-step cycle we need to have $$\begin{array}{} b = {3a+1\over 2^A } & a = {3b+1\over 2^B }\\ \end{array} \tag 4$$ with some so far undetermined $A$ and $B$.
To see better the space of possible solutions we can take the product of the lhs's $a \cdot b$ and which must equal the product of the rhs's: $$\begin{array}{} a \cdot b = {3a+1\over 2^A } \cdot {3b+1\over 2^B } \end{array}$$ resulting in the required equality

$$\begin{array}{} 2^{A+B} = (3+{1\over a}) \cdot (3+{1\over b }) \end{array} \tag 5$$ This shows, we can only have a solution if the rhs becomes at least integer, which is difficult enough , but actually must even equal a perfect power of 2, larger than $3^2=9$ and namely must equal $16$. Now what values must $a,b$ have such that the rhs equals $16$? Both must contain $a=b=1$ and that is the already known trivial cycle, and no other solution in possible.

Now you see the pattern, how this generalizes to the disproof of the 3-step-cycle, 4-step-cycle, and so on for the N-step-cycle.

Unfortunately, for several $N$ the possibility for small $a,b,c,...$ exists "theoretically", which means, that the rhs-product can reach a perfect power of 2 by some $1 \lt (a_1 \ne a_2 \ne a_3 \cdots \ne a_N)$ - one might just try some $N$ by hand, remembering the conditions that all members of an assumed cycle should be greater than $1$, should be odd, should not be divisible by $3$ and that all involved $a_k$ should be different from each other, to get a better intuition.

Now to proceed more we introduce the knowledge, that by simple heuristics we already know, that $a_k \gt 1000$ (can be done manually with a programming language) or $a_k \gt 1 000 000 $ and even $a_k \gt 2^{60} $ (the latter by an extensive numerical search by de Oliviera and by Rosendaal).
If we introduce that knowledge and assume the minimal $a_1$ being, say $a_1=1+2^{60} $ , $a_2 = {3a_1+1\over 2^{A_1}} $ and so on , we find that for all manually reachable $N$ the rhs is disappointingly near to the value of $3^N$ and no perfect power of $2$ is only in any realistic distance from that. One can find, using the continued fraction of $\beta = \log(3)/\log(2)$, we need $N \gt 150 000 $ (or even much more don't have the actual value at hand).

Well, this shall just give an intuition as to why cycles are much unlikely.
If you like to do more, allow negative numbers for $a_1,a_2,...$ and see how and which solutions for small $N$ you can get.

Or proceed and compare the $5x+1$ problem with this: we find actually two or three possible cycles for small $N$ and $a_1 \lt 100 $ but after that the above method can be used to say much more about the nonexistence of certain $N$-step cycles. I've even found two cycles for the $181 x+1$-version having $N=2$ and small $a_1 \lt 100$ but after that, the formula comparing the rhs-product for $N$ steps to perfect powers of $2$ indicates the same "difficulty" for cycles to exist.

[Update] Just to have an example, how a (possibly infinite, don't know at the moment) set of $N$ can be ruled out for a solution to exist.
Assume for example $N=12$. So we have on the rhs 12 parentheses of the form $(3+1/a_k)$, whose product should equal $2^S$ (where I use in general the letter $S$ for $S=A_1+A_2+...A_{12}$, just the sum of exponents, which is also the number of even-steps). Then what is the next possible perfect power of 2 above $3^{12}$? we get, using $\beta=\log(3)/ \log(2)$: $S=\lceil N \cdot \beta \rceil = 20$,thus we have the condition $$2^{20} =(3+ {1\over a_1})(3+ {1\over a_2})...(3+ {1\over a_{12}}) \tag 6$$and the solution for smallest $1 \ne (a_1 \ne a_2 ... \ne a_{12})$ is $$2^{20} \overset?=(3+ {1\over 5})(3+ {1\over 7})(3+ {1\over 11})...(3+ {1\over 37}) \tag 7$$ Of course we could simply compute the values of the LHS and the RHS getting $$ 2^{20}= 1048576 \gt 697035.738776 $$ and because increasing the values for the $a_k$ would even decrease the value of the rhs there is no solution and thus no $N=12$-step cycle. But for the sake of generality we proceed differently.
Instead we do a rough estimate for the mean-value of the $a_k$. Assume all $a_k$ are equal to their "mean" $a_m$, then we can rewrite the equation $$ 2^S = (3+{1\over a_m})^N \\ 2^{S/N} = (3+{1\over a_m})\\ 2^{S/N} -3 = {1\over a_m}\\ {1 \over 2^{S/N} -3} = a_m \tag 7 $$ $$ a_m = {1 \over 2^{20/12}-3}\approx 5.72075494219... \tag 8 $$ and because $a_m$ is somehow a rough mean, some $a_k$ must be smaller and some must be larger. But there is only one possible value for any $a_k \lt a_m$ namely $a_1=5$. After that, rearranging one parenthese with that value assumed to the lhs in eq(6) we get $$ 2^S/(3+1/5) = (3+{1\over a_m})^{N-1} \\ 2^{20} \cdot 5/16 = (3+{1\over a_m})^{11} \\ a_m \approx 5.79638745091$$ and we find, that there is no way that $a_m$ can be a rough mean of the remaining $11$ $a_k$ since there is no more odd integer $a_k$ in $1 \ne 3 \ne 5 \lt a_k $ so a $N=12$-step-cycle cannot exist.
We see nicely, that for some $N$ we can exclude the possibility of a cycle just based on the basic assumptions on the form of the members of a possible cycle $a_k$ by the formula (6) and for such $N$ do not need to recurse to the $a_k \gt 2^{60}$ found by de Oliviera and Rosendaal.

However, and this leads to the observation that we need for the general solution of the cycle-problem some deeper thinking, there are some $N$ for which $2^S$ is comfortably near to $3^N$ so we can allow a set of small $a_k$ such that the rhs can approach the lhs. The continued fraction of $\beta$ gives us pairs of $N$ and $S$ where $2^S$ are especially near to $3^N$ and for which a cycle cannot be excluded by this method alone.

[update 2]
I've not done this before explicitely, but trying the continued fraction-convergents and filling into $a_k$ the set of consecutive smallest possible integers ($5,7,11,13,...)$ we get the following small table

N           S      lhs=2^S            rhs = (3+1/a_k)*()...    lhs/rhs        a_m
    1,      2,             4       ,          3.2        ,      1.25               1    ~ 2^0
    5,      8,           256       ,        292.571428571,      0.875             31.81 ~ 2^5
   41,     65, 3.68934881474 E   19, 5.44736223436 E   19,      0.677           1192.08 ~ 2^10
  306,    485, 9.98959536101 E  145, 5.57867460455 E  146,      0.179          99780.79 ~ 2^16
15601,  24727, 3.70427126979 E 7443, 1.06756786898 E 7444,      0.346      285817586.21 ~ 2^28
79335, 125743, 2.59863196329 E37852, 8.97264176433 E37852,      0.289     7216102492.69 ~ 2^33

and we see, that the RHS can reach the LHS for that specific $N$ and thus the assumption of the smallest possible values for the $a_k$ does not suffice to exclude the possibility of a cycle. The "mean" $a_m$, estimated by the geometric mean of all parentheses as proposed above, are in the last column; they increase with the size of $N$ and we get an impression as at which cyclelength $N$ this allows values of $a_1 >2^{60}$ and thus this method has its heuristical limit: the last entry in the last row means $a_m \approx 2^{33}$ and this means, that the knowledge that $a_1>2^{60}$ suffices to have disproved all cycles of steps $N \le 79335$.

But it might anyway be interesting to see, what explicitely smallest value for $a_1$ (which we can assume to be the smallest of the cycle) and the following sequence of consecutive possible members would suffice to have the LHS smaller than the RHS. That would surely be a nice exercise ...

Question regarding 1-cycles in the Collatz conjecture

Yes, you should have $2^m-3^n|2^{m-n}-1$ since $X_1+1=\frac{2^m-3^n+3^n-2^n}{2^m-3^n}=\frac{2^m-2^n}{2^m-3^n}$ and the denominator is odd so you can ignore the power of 2 in the numerator for the divisibility.

Notice that since $m<2n$ you can see that $2^{m-n}-1<2^n$, but we know (with Rhin's bound and for $n>5$) that $2^m-3^n>2^n$ so in the end $2^m-3^n\nmid2^{m-n}-1$

Also, I didn't try it myself, but it made me think of this Mathlologer video. Perhaps something you would want to explore: https://youtu.be/4AuV93LOPcE?t=2042

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