Recently, I have read the article (Eliahou, 1993). The author shows that any nontrivial cycle must contain at least 17 087 915 elements, assuming that the Collatz problem has been verified up to $2^{40}$. To find this out, Eliahou set up the equation $$ \operatorname{Card}\Omega = 301\,994a + 17\,087\,915b + 85\,137\,581c\text{,} $$ where $a, b, c$ are nonnegative integers, $b>0$, and $ac=0$. My first question is: How did he arrive at this equation? And the second question: How can I construct such an equation for the current boundary $662 \times 2^{60}$? I saw a similar equation for the boundary $1.08 \times 2^{60}$, but I don't know how they came up with it.
References
(Eliahou, 1993) Shalom Eliahou. The 3x+1 problem: new lower bounds on nontrivial cycle lengths. Discrete Mathematics, volume 118, issues 1–3, pages 45-56, 1993. https://doi.org/10.1016/0012-365X(93)90052-U
I don't know for Eliahou, but from Simons m-cycles which also use convergents of continued fractions
$K>$ min$\{q_n,\frac{2\cdot x_{min}}{q_n+q_{n+1}}\}$
with $x_{min}=662\cdot2^{60}$ you would get $K>\frac{2\cdot x_{min}}{q_{22}+q_{23}}$ ($q_{22}=65470613321$ $q_{23}=137528045312$)
$K>7.519.596.840$
$\operatorname{Card}\Omega>11.918.279.012$
To find the right $q_n$ just search for the minimum one satisfying $x_{min}<\frac{q_n(q_n+q_{n+1})}{2}$ and verify that $q_{n-1}<\frac{2\cdot x_{min}}{q_n+q_{n+1}}$ (if not, pick $q_{n-1}$)
The $\operatorname{Card}\Omega$ equation is just using properties of consecutives convergents (Farey properties) and use those consecutive $q_n$ to determine a "minimal form" it would have.
(I forgot to mention but as mentioned by Gottfried and Eliahou, these are convergents of $\log_2(3)$)
EDIT:
Eliahou seems to give larger bounds
$\frac{p_{23}}{q_{23}}<\log_2(3+\frac{1}{662\cdot 2^{60}})<\frac{p_{21}} {q_{21}}$
($\frac{p_{21}} {q_{21}}$ being the smallest convergent larger than the center term, or in other word having the largest index $n$ while still being larger than the center term)
It would mean $\operatorname{Card}\Omega\geq p_{23}=217.976.794.617$
Note that odd indexes gives convergents larger than $\log_2(3)$ and even indexes are smaller. $\operatorname{Card}\Omega$ would be constructed with $p_{21}$, $p_{23}$ and $p_{24}$ (but only $p_{23}$ is "guaranteed")
Didn't read that paper yet...I know what to read next now
EDIT2:
I think I should definitely read the paper before posting (which I still didn't do!!). Thanks Gottfried for the tip, indeed $\{\frac{p_{21}}{q_{21}},\frac{p_{23}}{q_{23}}\}$ are not Farey pairs like $\{\frac{p_{13}}{q_{13}},\frac{p_{15}}{q_{15}}\}$ or $\{\frac{p_{19}}{q_{19}},\frac{p_{21}}{q_{21}}\}$ and we can't use the $\frac{p}{q}<\frac{ap+bp'}{aq+bq'}<\frac{p'} {q'}$ property, so we need to look at the intermediate/generalized convergents too, especially what Eliahou call "Upper convergents".
The next "Upper convergent" after $\frac{p_{21}}{q_{21}}$ is not $\frac{p_{23}}{q_{23}}$, but $\frac{p_{21,1}}{q_{21,1}}$ (The one indicated by Gottfried, and in Table 2 from the paper).
So we end up with this inequality (after ensuring that $|p_nq_{n+1}-p_{n+1}q_n|=1$ for consecutive fractions):
$\frac{p_{22}}{q_{22}}<\log_2(3)<\frac{p_{21,1}}{q_{21,1}}<\log_2(3+\frac{1}{662\cdot 2^{60}})<\frac{p_{21}}{q_{21}}$
and with $\log_2(3)<\frac{\operatorname{Card}\Omega}{K}<\log_2(3+\frac{1}{662\cdot 2^{60}})$ we have 3 posibilities:
$\frac{p_{22}}{q_{22}}<\frac{\operatorname{Card}\Omega}{K}<\frac{p_{21,1}} {q_{21,1}}$ or $\frac{\operatorname{Card}\Omega}{K}=\frac{p_{21,1}} {q_{21,1}}$ or $\frac{p_{21,1}}{q_{21,1}}<\frac{\operatorname{Card}\Omega}{K}<\frac{p_{21}} {q_{21}}$
By applying $\frac{p}{q}<\frac{ap+bp'}{aq+bq'}<\frac{p'} {q'}$ we get
$\operatorname{Card}\Omega\geq p_{21,1}=114.208.327.604$
Sorry for the mess...I originally had no intention to comment on Eliahou but couldn't resist to take a peak