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 $N_max$?

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
$$ a_{k+1} = {3a_k+1over 2^{A_k}} tag 1$$
$qquad qquad qquad$ 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
$$a_2 = {3a_1+1over 2^{A_1}} {Large;mid;} a_3 = {3a_2+1over 2^{A_2}} {Large;mid;} cdots {Large;mid;} a_1 = {3a_N+1over 2^{A_N}} tag 2$$
$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.

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.

<sub>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</sub>

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:
$$text{a cycle with } N_8=208,576,659,753,891,832,997 [ gt 2.085text{ E}20 ] text{ odd steps cannot exist} $$
$qquad qquad$ 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)

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
$$ alpha = {1over 2^{S/N}-3} le chi qquad text{to have a cycle of length $N$ been disproved} tag 7$$
To speed up computing time I rewrote that criterion. Denote $beta = log_2(3)$
$$ {1overalpha} = 2^{S/N}-3 ge {1over chi} \
{1over 3alpha} = 2^{S/N-beta}-1 ge {1over 3 chi} \
1+{1over 3alpha} = 2^{S/N-beta} ge 1+ {1over 3 chi} \
log_2(1+{1over 3alpha}) = S/N-beta ge log_2(1+ {1over 3 chi}) tag 8 $$
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
$$S/N-beta= { lceil N cdot betarceil over N } - beta
={ 1 + N cdot beta - {N cdot beta} over N } - beta
= { 1 - {N cdot beta} over N } tag 9$$
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
$$ N_chi le {1over chi^star} tag {11} $$

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):