Question regarding 1-cycles in the Collatz conjecture

Question

The question is whether there is a simple explanation for the following observation, or whether the observation is an incorrect inference from limited cases. The context assumes 1-cycle loops in the Collatz conjecture (I know that Steiner proved that there are no 1-cycles; I am thinking about an old question from a distinguished reader regarding any approaches that do not require sophisticated relations among transcendental numbers or advanced theorems).

Assumptions:

(1) A loop of n elements, {$X_1, X_2,...X_n$} can be characterized by the equation $$\frac{3^n-2^n}{2^m-3^n}=X_1$$

(2) m is not free but is limited to integers that satisfy $nLog_2{3}<m<nLog_2{3}+1$.

(3) Every $X_i$ in the loop is associated with a number $b_i$ where $$\frac {b_i}{2^m - 3^n} = X_i$$

(4) For a 1-cycle, $b_1$ = $3^n - 2^n$

(5) For a 1-cycle, the $b_i$ are related by $$b_{i+1} = \frac {3}{2} * b_i + \frac {2^m - 3^n}{2}$$

Inferences:

(6) From (3) every $b_i$ must be divisible by $2^m - 3^n$.

(7) If two numbers a and b are divisible by d, then a-b is divisible by d. Since the differences are divisible by d then the differences of the differences must be divisible by d, etc.

Observation:

(8) If we compute the appropriate values $b_i$ for a given n (and consequently by assumption (2), a pair {m,n}), then take the differences between consecutive $b_i$, then the differences of the differences, etc., we can construct tables such as the following:

For n=5 and m=8:

$b_i$	1st diff	2nd diff	3rd diff	4th diff
211
323	112
491	168	56
743	252	84	28
1121	378	126	42	14

Following the same procedure for n=8, m=13 and we get:

$b_i$	1st diff	2nd diff	3rd diff	4th diff	5th diff	6th diff	7th diff
6305
10273	3968
16225	5952	1984
25153	8928	2976	992
38545	13392	4464	1488	496
58633	20088	6696	2232	744	248
88765	30132	10044	3348	1116	372	124
133963	45198	15066	5022	1674	558	186	62

For n = 9 and m = 15 the last element is 126.

(9) The last element in the table is equal to $2^{m-n+1} - 2$

(10) The diagonal starting from the last element of the table and working backward proceeds in multiples of 2.

(11) The last row, starting from the last value, proceeds backward in multiples of 3.

(12) Stating from the last element we can fill out the table up to the first column. We can do this, for the ith column, 1 < i < n, by either by adding terms from the $i+1$ column to the diagonal elements that vary by factors of 2, or by subtracting terms from the $i+1$ column from the last row of the table, which vary by factors of 3. If we know the first element of the first column, we can fill out the rest by adding the terms in the second column.

(13) For the pair n=7, m=12 the difference is 5, the same for the pair n=8, m=13. The last element in the tables for each is 62.

(14) It would seem that any function in which the elements of the first column are related to each other in an orderly way will produce an orderly table; for example, if we start with the Fibonacci sequence, even if we start somewhere in the middle, the table will reproduce the earlier terms in the sequence:

21
34	13
55	21	8
89	34	13	5
144	55	21	8	3

Questions:

(14) Is it true that the last term in the 1-cycle related tables should be $2^{m-n+1} - 2$ for all permissible {n,m} pairs? By permissible I mean those which satisfy (2).

(15) If the answer to (14) is true, is there a straight-forward explanation for this?

(15) By (6) and (7) it would seem that every entry a given table would have to be divisible by $2^m - 3^n$ in order for there to be a 1-cycle. (Obviously 211 in the table for n=5 is a prime number, so there are no 5 element 1-cycles.) Is this true?

(16) Since all $b_i$ are odd numbers, and $2^{m-n+1} - 2$ is an even number, does this imply that $2^m - 3^n$ should also divide $2^{m-n} -1$, assuming that that the other assumptions regarding 1-cycles are satisfied?

(17) And I guess I should throw in, are any of the assumptions wrong as applied to 1-cycles in the Collatz conjecture?

(18) Any suggestions as to how to go about searching for answers to these on the internet?

ADDENDUM:

Thank you to Collag3n, Eric Shumard and Gottfried Helms. You are all very helpful. I didn't quite follow all of the math, but I did notice this in Mr. Shumard's response:

"from which $(2^{Pn}−3^n)|(2^{Pn−n}−1)$ follows", and from Collag3n's response:

"so in the end $2^m-3^n\nmid2^{m-n}-1$"

(19) So is this the kind of contradiction that demonstrates that there can be no 1-cycles?

Final Observations

(20) The equation given in (1) completely characterizes a 1-cycle for a given n; that is, given n we can compute all of the $X_i$ in the loop. This is because there is a one-to-one relationship between m and n for 1-cycles. This is not necessarily the case for any k-cycles other than 1. However, equations for the first 2 $b_i$ do completely describe these cycles, because the last term of the equation for $b_2$ is $2^{m-1}$.

(21) In finding values in the difference table that are not divisible by $2^m-3^n$ we can do consecutive divisions by 2 until we reach an odd number, rather than taking the difference of even numbers. The smallest difference between any two numbers in the difference table will then be an odd number which will have to be divisible by $2^m-3^n.$ This is true for any k-cycle or combination of k-cycles.

(22) For a 2-cycle, if it is possible to show that every difference table produces a number that is not divisible by $2^m - 3^n$, considering only those $b_i$ up to the point where $X_i > X_{i+1}$, i.e., the "break in the ascending sequence of the cycle, then there can be no 3-cycles, nor any other k-cycles, since the element of the difference table that precludes a 2-cycle will also be present in the difference table for higher order cycles. One may get away with including $b_{i+2}$ in computing the difference table.

Answer 1

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

Answer 2

Don't want answer to all question. To question (14) the following might be interesting/instructive. (I take from a sketch that I wrote very recently, so I just use my notation, beg your pardon for this):

• $N$ for the number of odd steps $3x+1$,
• $S$ for the number of even steps $x/2$,
• $a^*_k$ for the $k$'th element of a 1-cycle (let the "**" denote the property of forming a "1-cycle")*.

We have then $$ \begin{bmatrix} \begin{array} {rcrlrrrr} a^*_1 & = &\Large {3^N−2^N\over 2^S−3^N} &=& 3^0 & \cdot 2^N \quad &\cdot \Large {{2^{S−N}−1 \over 2^S − 3^N}}& − 1 \\ \\ a^*_2 &= 1.5 & \cdot(a^*_1 + 1)−1 &=& 3^1 & \cdot 2^{N−1} &\cdot \Large {2^{S−N}−1 \over 2^S − 3^N} & − 1 \\ \\ a^*_3 &= 1.5^2 & \cdot(a^*_1 + 1)−1 &=& 3^2 & \cdot 2^{N−2} &\cdot \Large {2^{S−N}−1 \over 2^S − 3^N}& − 1 \\ \\ \vdots & \qquad \vdots & &= &\; \vdots & \; \vdots & \vdots \qquad\\ \\ a^*_N &= 1.5^{N-1} & \cdot (a^*_1 + 1)−1 &= &3^{N-1} & \cdot 2^1 \quad &\cdot \Large{ 2^{S−N}−1 \over 2^S − 3^N } &− 1 \end{array} \end{bmatrix}$$ The rightmost formulae might shed some light on the "internal machine" of your differences-tables and make the observed property explainable with a little formula.

Answer 3

(2) isn't quite right. The constraint is: $$0 \lt m - n\log_2(3) \le n \log_2(1 + \frac{1}{3x_{min}})$$ where $x_{min}$ is the minimum element of the cycle. For any given $x_{min}$, $n$ can be increased sufficiently so that any integer $m \gt n \log_2(3)$ is allowed. This could mean really big cycles.

Here is a derivation of $(2^m-3^n)|(2^{m-n}-1)$ for a 1-cycle. The odd integers $x_i$ in a sequence are given by: $$x_i = \frac{3x_{i-1}+1}{2^{p_i}}$$ where $p_i$ is the integer such that $x_i$ is an odd integer. Define $$P_i = \sum\limits_{j=1}^i p_j \textrm{ for }i\gt0 \textrm{ and }P_0 = 0$$ This gives the sequence equation: $$x_n = \frac{3^n x_0 + \sum\limits_{i=0}^{n-1}3^i2^{P_{n-i-i}}}{2^{P_n}}$$ Note that $P_n$ = your $m$. This can also be written as: $$2^{P_n}x_n - 3^nx_0 = \sum\limits_{i=0}^{n-1}3^i2^{P_{n-i-i}}$$ This is a linear Diophantine equation with variables $x_0$ and $x_n$ with coprime coefficients which has an infinite number of solutions of the form: $$(x_0,x_n) = (x'_0 + k2^{P_n}, x'_n + k3^n)$$ where $(x'_0,x'_n)$ is any particular solution and $k \in \mathbb{Z}$. We only consider the values of $k$ where $x_0$ and $x_n$ are odd positive integers. The first $n-1$ elements of a 1-cycle have $p_i=1$ hence $P_i=i$ so the Diophantine equation is: $$2^{n-1}x_{n-1} - 3^{n-1}x_0 = \sum\limits_{i=0}^{n-2} 3^i 2^{n-2-i} = 3^{n-1} - 2^{n-1}$$ which has the particular solution $(x'_0,x'_{n-1})=(-1,-1)$ so the solutions are: $$(x_0,x_{n-1}) = (k2^{n-1}-1,k3^{n-1}-1)$$ where $k$ is a positive even integer (so that $x_{n-1}$ is odd). For a 1-cycle $x_n$ is given by: $$x_n = \frac{3x_{n-1} + 1}{2^{p_n}} = x_0$$ where $p_n = P_n-(n-1)$. Substituting in the values for $x_0$ and $x_{n-1}$ gives (after some algebra): $$k(2^{P_n} - 3^n)=2(2^{P_n-n} - 1)$$ from which $(2^{P_n} - 3^n) | (2^{P_n-n} - 1)$ follows.