There are lower bounds worked out for the length of nontrivial Collatz-cycles. How can *upper bounds for the disproof* be determined?

Question

There have been lower bounds estimated for the length $N$ of (odd) steps of a nontrivial cycle in the collatz-problem. Such estimates have been based on knowledge of upper bounds $\chi$ for any number $a_1$, where it has been empirically determined that they decrease to $1$ and enter the so-called "trivial cycle" by the iterated Collatz-transformation. One well known estimate for such length $N$ has been based on the value $\chi_{\Tiny \text{TOdS}} = 5 \cdot 2^{60}$ (due to Tomas Oliveira da Silva, for reference see wikipedia).

Recently a new estimate has been published based on $\chi_{\Tiny \text{yoyo}}= 87 \cdot 2^{60} $ such that all $a_1 \lt \chi_{\Tiny \text{yoyo}}$ run into the trivial cycle.

Inspired by another question here (but which seem to have really meant to ask for the highest lower bound for $N$ instead) I've got the somehow reciprocal question:

Q1: How can we estimate an upper bound for the length $N$, for which we can disprove a nontrivial "general cycle" , based on the knowledge of convergence of all $a_1 \lt \chi_{\Tiny \text{yoyo}}= 87 \cdot 2^{60} $

Remark: Question Q1 includes the question for the most meaningful/rigorous method to arrive at a large number for $N$. There are more and less restrictive methods available, but like require accordingly more or less computational effort.

Q2: What would be that "champion" number $N_\max$?

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Remark: The concept of "m-cycles" as defined by R. Steiner, J. Simons and B. de Weger is relevant here. That concept describes somehow the "hilliness" of a trajectory in an assumed cycle, such that an "1-cycle" has one local peak, a "2-cycle" has two local peaks and so on. It has been proven by the named authors that for "m-cycles" with few peaks ($m<72$) there are no nontrivial cycles at all and so any length $N \gt 1$ is already disproved (and this question is answered for such types of cycles).
Thus I have introduced in the above the qualifier "general cycle" to emphasize that I look for a estimate for an upper bound for $N$ without the respect to the solution for the "m-cycle" with small $m$.

For being able to simply translate the notions and notations of the Simons/deWeger-paper into mine here (as requested by @reuns) I extract from [pg 51]:

1. "m-cycle" is the same
2. They use the $(3x+1)/2$ and $x/2$ convention for one step;I use ${3a+1\over 2^A}$ (="Syracuse"-) convention.
3. The number of odd steps in S/dW-text is $K$, in my text is $N$, meaning $K=N$.
4. The number of even steps in S/dW-text is $L$, in my text is $S=\text{sum-of-exponents } A_k$ ; the discrepancy seeming obvious by the definitions of the steps such that $K+L=S$ .
5. Their criterion $\Lambda = (K+L) \log(2) - K \log (3)$ use I unnamed so far as $ S \log (2) - N \log(3)$
6. In the S/dW-text they use $x_\min$ where I use $a_1$ as minimal/ initial element of a cycle/1-cycle/m-cycle.
7. They use $x_\min>X_0=301 \cdot 2^{50}$ where I use the newer value $\chi_{\Tiny \text{yoyo}}=87 \cdot 2^{60}$
8. We apply the same idea for using upper bounds/lower bounds estimates in general. However in the current or forthcoming answers on Estimation-method 1 and Estimation-method 2 I don't use more knowledge about the structure of a trajectory than that there must be at least one element $a_1$ smaller than some sort of average of all elements. In S/dW text for analyzing the possibility of "1-cycle" and "2-cycle" they use the knowledge of the structure of the trajectory having roughly geometric growthrate leading to the disproof of the "1-cycle", then of the "2-cycle" and then of the "m-cycle" for $m\le72$.

_{(Final remark: the reason for the difference of notations is that I'd developed my ideas initially not being aware of the Steiner/Simons/deWeger work. Accidentally my line of thoughts has been essentially the same, but I feel my choices for notation being more coherent and fluent and so kept it. A translation of the famous "1-cycle" disproof into my language/notation can moreover be found here. This might help to understand my arguing here even better)}

Answer 1

... See updates for larger $N$ according to higher $\chi$ at the end ...

Let's for shortness denote the current upper limit for $a_1$ as found by "yoyo@home" $\chi_{\Tiny \text{yoyo}} = 87 \times 2^{60} \approx 1.003042 \times 10^{20}$ by a single letter $\chi$ in the following.
By heuristic we seem to know, that
$\qquad \qquad$ no number $1 \lt a_1 \lt \chi$ can be element of a nontrivial cycle.
Call this condition 1.

Some definitions:
$\qquad$ Let's define a single step in the Collatz-transformation by $$ a_{k+1} = {3a_k+1\over 2^{A_k}} \tag 1$$ $\qquad \qquad \qquad$ where all $a_k$ are odd. (This is also what is sometimes called "odd-step")

$\qquad$ A cycle of $N$ (odd) steps is then the occurence $$a_2 = {3a_1+1\over 2^{A_1}} {\Large\;\mid\;} a_3 = {3a_2+1\over 2^{A_2}} {\Large\;\mid\;} \cdots {\Large\;\mid\;} a_1 = {3a_N+1\over 2^{A_N}} \tag 2$$
$\qquad $ and we use the letter $S$ for the sum of all $A_k$ (this is thus the number of even steps) $$ S = A_1 + A_2 + \cdots + A_N \tag 3$$

$\qquad$ We assume that in the cycle $a_1$ is the minimal element.
$\qquad \qquad $ This does not reduce the generality of the following reasoning.

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There is a small formula which allows *for a given, specific, N* to test whether such a cycle-length can exist, dependend on *cond. 1*. That is, to put all elements of an assumed cycle $a_k$ into a trivial equation having the product-formulas $$ a_1 \cdot a_2 \cdot a_3 \cdots a_N = ({3a_1+1 \over 2^{A_1}})({3a_2+1 \over 2^{A_2}})\cdots ({3a_N+1 \over 2^{A_N}}) $$ This equation can be reformulated into the much non-trival equation introducing $S=A_1+A_2+...+A_N$ ($S$ meaning the sum of all $x/2$-steps): $$ 2^S = (3+\frac1{a_1})(3+\frac1{a_2})...(3+\frac1{a_N}) \tag 4$$ where $S$ is the number of $x/2$-steps and $N$ is the number of $3x+1$-steps. The $S$ to a certain $N$ can simply be computed by $S=\lceil N \cdot \log_2(3) \rceil$ and with a software like Pari/GP this can be done to fairly large $N$ with thousands of digits.

Estimation method 1:

So if, in eq (4) we assume some average value $\alpha$ for the $a_k$ then the formula reduces to $$ 2^S = (3+\frac1{\alpha})^N \tag 5 $$ and then $$ \alpha = {1 \over 2^{S/N} -3 } \tag 6 $$ Of course, since $\alpha$ is some average value, some of the $a_k$ must be smaller and some must be larger. But if the $\alpha$, that we get for some specific $N$, is smaller than our heuristical limit $\chi$, such that $ \alpha < \chi$ then that cycle can only exist if some $a_k$ are as well smaller than $\chi$ --- but we know (condition 1), that none of such small numbers can be in a cycle, hence a cycle of that specific length $N$ has been disproven.

_{Usually we look for the lower bound for $N$ such that $\alpha \gt \chi$ and thus that we cannot exclude that $a_1$ of the cycle is larger than $\chi$ and this would be the smallest $N$ that a nontrivial cycle can not been ruled out. A similar computation for such a lower bound has been done already by R. Crandall in 1978 introducing the convergents of continued fraction of $\log_2(3)$ as tools to find the smallest $N$ where a cycle cannot been ruled out by this method.
But here we look out for an upper bound}

Let us now try some random $N$ in the near of $N \approx 10^{20}$

Here are some $N$ for which the existence of a general cycle is disproven by this simple formula:

The Pari/GP-program:

ld3 = log(3)/log(2)
chi = 87*2^60 \\ ~ 1.003042 * 10^20          
{for(k=1,20,
       N=10^20+random(320000);
       S=ceil(ld3*N);
       alpha = 1/(2^(S/N)-3);       
       if( alpha < chi , \\ the cycle of that N is impossible
             print([ N ; S ; alpha ]);
         )
      )}
           

gave this certificates/disproves:

  N                          S                  alpha <1.003 e20
[100000000000000097907; 158496250072115773325; 6.8450652 E19]
[100000000000000063958; 158496250072115719517; 8.0893339 E19]
[100000000000000296617; 158496250072116088273; 5.9811055 E19]
[100000000000000088735; 158496250072115758788; 4.9141255 E19]
[100000000000000053362; 158496250072115702723; 5.6104855 E19]
[100000000000000022312; 158496250072115653510; 5.1008029 E19]
[100000000000000241880; 158496250072116001517; 5.3645875 E19]
[100000000000000093691; 158496250072115766643; 5.3170221 E19]
[100000000000000031371; 158496250072115667868; 6.2658134 E19]
[100000000000000300515; 158496250072116094451; 7.7539063 E19]
[100000000000000305361; 158496250072116102132; 5.3917032 E19]

The listing means, that for $11$ out of $20$ randomly chosen $N$ in that region the existence of a cycle is thus already disproved.

Of course, the referred number $N$ for the disproved cycle-length $9,283,867,937$ is much smaller than the last documented $N$ (number of odd steps $3x+1$) as well as $S$ (number of even steps $x/2$)

      N                         S                   alpha
[100000000000000305361; 158496250072116102132; 5.3917032 E19]
            9283867937             9283867937

So this method gives some values for $N$ for which we can now claim in all open, that a cycle with that number of odd steps cannot exist. Those are much larger $N$ than that brought into the play from the wikipedia-reference!

The largest $N$ as length of a nontrivial cycle to be disproved with the formula above using the value $\alpha \lt \chi_{\Tiny \text{yoyo}} $ I could find so far was after manually searching

    N                         S                   alpha
[127 940 101 513 462006853 ; 202780263237295321100 ; 6.15261833281 E19]
[170 340 101 513 461998797 ; 269982673267872330425 ; 9.99741444442 E19] \\ update
[207 500 101 513 461893633 ; 328879879794670327447 ; 1.00301880212 E20] \\ update 2
[207 500 101 513 471024061 ; 328879879794684798833 ; 9.98536768861 E19] \\ update 3
[208 568 587 248 096949695 ; 330573389616622387583 ; 1.00302673877 E20] \\ update 4
[208 569 494 409 908034699 ; 330574827434075043604 ; 1.00302379880 E20] \\ update 5
[208 576 425 778 542627280 ; 330585813393439547647 ; 1.00304058384 E20] \\ update 6
[208 576 659 701 197283637 ; 330586184152075247118 ; 1.00304170887 E20] \\ update 7
[208 576 659 753 891832997 ; 330586184235594131846 ; 1.00304170901 E20] \\ update 8 (Record?)
 ======

using linear composition of the convergents of the continued fraction of $\log_2(3)$.
This answers now the question when related to "general cycles" (which can be seen as being "m-cyclic" with $m>72$) giving

Theorem 1: $$\text{a cycle with } N_8=208\,576\,659\,753\,891\,832\,997 [ \gt 2.085\text{ E}20 ] \text{ odd steps cannot exist} $$ $\qquad \qquad$ assuming truth of "condition 1".

Conjecture 1:
That $N_8$ in theorem 1 is also the largest possible number of odd steps $N$ for which a cycle can be disproved with the given method (depending on condition 1). (For evidence see image 2)

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A picture showing *alpha*s at some random *N* in the neighbourhood of the heuristically found current largest *N* with valid disproof ($\alpha(N) \lt \chi$): [![image][1]][1]

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An improvement of the formulae and a proof of conjecture 1 ---

An improvement of the search-formula allows to easily give an upper bound for such an $N$ and allowed to confirm conjecture 1.

Our search-criteria was so far $$ \alpha = {1\over 2^{S/N}-3} \le \chi \qquad \text{to have a cycle of length $N$ been disproved} \tag 7$$ To speed up computing time I rewrote that criterion. Denote $\beta = \log_2(3)$ $$ {1\over\alpha} = 2^{S/N}-3 \ge {1\over \chi} \\ {1\over 3\alpha} = 2^{S/N-\beta}-1 \ge {1\over 3 \chi} \\ 1+{1\over 3\alpha} = 2^{S/N-\beta} \ge 1+ {1\over 3 \chi} \\ \log_2(1+{1\over 3\alpha}) = S/N-\beta \ge \log_2(1+ {1\over 3 \chi}) \tag 8 $$ Now I denote the $\log_2()$-expressions shorter as $\alpha^\star$ resp. $\chi^\star$ because the latter is a constant in the Pari/GP-comparisions.
Moreover the middle term can be much simplified $$S/N-\beta= { \lceil N \cdot \beta\rceil \over N } - \beta ={ 1 + N \cdot \beta - \{N \cdot \beta\} \over N } - \beta = { 1 - \{N \cdot \beta\} \over N } \tag 9$$ so that we get the far faster testable expression with that $\chi^\star$ a constant: $$ \alpha^\star = { 1 - \{N \cdot \beta\} \over N } \ge \chi^\star \tag {10}$$ where we need only the fractional part of $N$ multiplied with the constant $\beta$.

An accordingly redesigned image suggests strongly that indeed $N_8=208\,576\,659\,753\,891\,832\,997$ is the largest possible $N$ where this type of $\alpha(N)$-test disproves a general cycle.
Here we find that automatically $\alpha^\star \lt {1 \over N}$ and thus the upper limit for $N$ is $ {1 \over \chi^\star } \ge N_\chi$ or $$ N_\chi \le {1\over \chi^\star} \tag {11} $$

I have empirically/numerically tested that range from $ N_8$ to $N_\chi$ using Pari/GP and indeed

Theorem 2: $N_8$ is the largest $N$ where method 1 disproves the nontrivial cycle depending on the current $\chi_{\tiny \text{yoyo}}=87 \cdot 2^{60}$.

Detail (Zoom):

Updates
By manual search for the largest $N$ for which the existence of a cycle is disproved/disprovable by the described method, I got for the subsequently increasing known values for chi ($\chi$): $$\begin{array}{rrrrrr} \chi_{2210} &= 645 \times 2^{60} &\implies &N_{2210}&=1′546′344′201′699′447′217′164 &(Jan23)\\ \chi_{2301} &= 660 \times 2^{60} &\implies &N_{2301} &=1′582′305′694′802′731′032′874 &(Jul23)\\ \chi_{2307} &= 1 \times 2^{70} &\implies &N_{2307} &=2′454′971′259′794′877′266′896 &(Aug23) \end{array}$$
The given $N_{k}$ are found manually testing linear combinations of values from the generalized continued fractions. Conjectures have been/are, that those $N$ are the largest $N$ for which a cycle can be disproved given a certain $\chi$. $N_{2210}$ has been proved to be maximal by scanning completely the relevant segment of numbers $N$. See first announcements in comments at this question in MSE