This is an indirect follow-up of the previous post I did on the Collatz conjecture. After a few responses, we managed to get to the fact that if we have $n\in\mathbb N$ and cyclic $(e_n)$ such that $e_0\equiv1~(\textrm{mod}~2)$, $e_{n+1}=e_{min}=e_0$ and that for all $k\in\mathbb N$ we have the relation $e_{k+1}=\frac{3e_k+1}{2^{\nu_2(3e_k+1)}}$, then
$$\sum_{k=0}^{n-1}\nu_2(3e_k+1)>n\log_23$$
and I found out that for $n\le108$, this inequation was violated. I asked if this inequation was violated for all $n$ such that for all $k < n \implies e_0< e_k$, which would henceforth imply the inexistence of cyclic sequences. However, I know realize this might be a bit out of reach, so we could rather interpret this result as a way to say that all non-trivial cycles have length $>108$, as it's basically the number of steps to get from $e_0$ to $e_n$ in $(a_n)$ such that $a_n=\left\{\begin{array}{cc}(3a_n+1)/2&a_n~\text{odd}\\a_n/2&\rm otheriwse\end{array}\right.$. However, I manage to get way higher recently, and I don't really know what the current record holder is (I only know Lagaria's $301\;994$…), so I'll put this here anyway to know whether it is a correct lower bound or if someone points out an error… ^^' Anyway, first off, we know that since $e_0$ is minimal, then for all $n\in\mathbb N^*$ we'd have
$$\sum_{k=0}^{n-1}\nu_2(3e_k+1) < n\log_2\left(3+\frac1{e_0}\right)$$
However, since LHS is an integer, we would have
$$\sum_{k=0}^{n-1}\nu_2(3e_k+1) < n\log_23$$
as long there no integer $m$ between $n\log_23$ and $n\log_2(3+1/e_0)$… Hence, we have
$$\sum_{k=0}^{n-1}\nu_2(3e_k+1) \ge n\log_23$$
$$\iff\exists~m\in\mathbb N^*:n\log_23 < m < n\log_2\left(3+\frac1{e_0}\right)$$
$$\iff\exists~m\in\mathbb N^*:2^{m/n} < 3+\frac1{e_0}$$
We can get the minimal value of $m$ with $m=\lceil n\log_23\rceil$. Hence,
$$\iff 2^{\lceil n\log_23\rceil/n} < 3+\frac1{e_0}$$
$$\iff 2^{\lceil n\log_23\rceil/n}e_0 < 3e_0+1$$
So what I did is, I made an algorithmic which computes for larger and larger values of $n$ if $\lfloor2^{\lceil n\log_23\rceil/n}e_0\rfloor < 3e_0+1$ because the programming language I used works better with integer values. Despite all of this, I only managed to get up to $n=225\;640\;000$ until it crashed (it's numerically correct though, since I have set precision to 50 digits after the decimal point, despite the amount needed being only 21). Using Wolfram Alpha, the first instance of $\lfloor n\log_2(3e_n+1)\rfloor\ne\lfloor n\log_23\rfloor$ I managed to check manually was $n=10^{18}+283$ (by descending from $10^{20}$), so all I can say is that there exists a lower bound of $(e_n)$ (and definitely $(a_n)$ as well) cycle lengths between $225\;640\;000$ and $10^{18}+283$ (or between $357\;630\;939$ $(10^{18}+283)\log_23$ in $(a_n)$ dynamics if I'm correct). It would take me 20 million years with the algorithm I did to get to $10^{18}+283$ because my programming skills are kind of weak… Anyway, if there is no higher lower bound yet, mine is gonna have to be $225\;640\;000$, I guess ! And if there's higher, then… I'm going to have to do it in another programming language or something to get more efficient results, I don't know…
New record for for lower bound of non-trivial cycle lengths of Collatz sequences
collatz conjecturesequences-and-series
Related Solutions
$$\frac{3e_0+1}{2^{\nu_2(3e_0+1)}}=e_1$$ can be rewritten as $$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$$ Now you have
$(3+\frac{1}{e_0})=2^{\nu_2(3e_0+1)}\frac{e_1}{e_0}$
$(3+\frac{1}{e_1})=2^{\nu_2(3e_1+1)}\frac{e_2}{e_1}$
...
$(3+\frac{1}{e_n})=2^{\nu_2(3e_n+1)}\frac{e_{n+1}}{e_n}$
You multiply every LHS/RHS to get
$(3+\frac{1}{e_0})(3+\frac{1}{e_1})...(3+\frac{1}{e_n})=\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}$
From here you get
$$(3+\frac{1}{e_{max}})^{n+1}\leq\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}\leq (3+\frac{1}{e_{min}})^{n+1}$$
But it means that in a cylce where $e_{n+1}=e_0$ you have
$\prod_{k=0}^n2^{\nu_2(3e_k+1)}\gt 3^{n+1}$ or $\begin{array}{|c|}\hline\sum\limits_{k=0}^n\nu_2(3e_k+1)>(n+1)\log_23\\\hline\end{array}$
Unless I messed something in translating to your notations, it does not match what you get.
Since your fraction weights values towards the end of the sequence far more than values earlier on, the decimal value of your sequence up to a certain level of precision will be decided almost entirely by the last few entries in the hailstone sequence.
But this public domain graphic from Wikipedia shows us all of the ways that a hailstone sequence can end (for relatively small $n$):
So the value of your fraction is almost entirely determined by the last five or six elements in the hailstone sequence, and the last five or six elements in the hailstone sequence are almost always the exact same. The values of your fraction are always going to look very close to one another, even for wildly varying hailstone sequences. The condition you're describing about how the fraction tends to always look quite similar is absolutely normal. You're effectively squishing down all of the information that we actually use to differentiate hailstone sequences (the numbers earlier on in the sequence) and over-emphasising the importance of the final few numbers of the sequence which are almost always the exact same. I don't think this actually tells us anything deeper about the Collatz problem.
That said, no, these sequences don't always end up at the same one or two numbers as you conjectured; two hailstone sequences will always give you a different fraction, here. You're just pushing all of the differentiating information way down into the deep levels of precision so it's hard to tell.
Best Answer
Using Pari/GP and hoping I understand you correctly I show programming and results. (Pari/GP is interpreted and easily programmed like python. You can obtain it for free at website)
Now we use the convergents of the continued fraction of $\log_2(3)$ because they give us the numbers $n$ (I use $N$ for the numbers of steps, and $S$ for the sum-of-the-exponents)
Now we use your limit-value from your earlier posting (you should this reminder include in your current OP!):
Now I show, how the formula and the computation of the comparision can be much much improved. You wrote $$ \left \lfloor 2^{\left \lceil n \log_2 3 \right \rceil /n } e_0 \right \rfloor \lt 3 e_0 +1 \tag 1 $$ First improvement: in Pari/GP you don't have the need to transform this into integers. We can stay with $$ 2^{S/N} < 3 + 1/e_0 \qquad \qquad \text{where } S=\lceil N \log_2 3 \rceil \tag 2 $$ but $S$ can also be taken from the first row of the convergents of the cont.frac, however only of each second one. (The other one leads to $2^S< 3^N$ and thus to $e_k$ from the negative numbers).
Now we improve the computation even more. $$ 2^{S/N} < 3 + 1/e_0 \qquad \qquad \text{where } S=\lceil N \log_2 3 \rceil $$ Taking logarithms to base $2$ and improving the rhs: $$ S/N < \log_2(3 + 1/e_0) \\ S /N < \log_2 3 + \log_2(1 + 1/3/e_0) \tag 3 $$ improving the lhs $$ S/N = ( 1 + N \log_2 3 - \{ N \log_2 3 \})/N \\ \qquad = \log_2 3 + ( 1 - \{ N \log_2 3 \} )/N $$ comparing lhs and rhs and reduce then by the equal summand $$( 1 - \{ N \log_2 3 \})/N < \log_2(1 + 1/3/e_0) \tag 4 $$ making constants $\text{ld}_3 = \log_2(3)$ and $\chi^* = \log_2(1 + 1/3/e_0)$ $$( 1 - \{ N \text{ld}_3 \})/N < \chi^* \tag 5 $$
From here on, $N \ge 6586818670 $ , cycles are possible (regarding our type of criteria!).
However, this does not mean, that not for some larger $N$ our criteria still might allow to disprove specific cycles, as we see in the following list.
For instance, the longest cycle whis is disproved by this criteria and where $N$ is from the continued fractions is for $N=127940101513462006853$. Actually, there are some higher $N$ disprovable this way, but not for further $N$ from the convergents (I found some higher $N$ manually searching). But no disproof of this type can be occur for $N>1/\chi^* \approx 208\, 576\, 659\, 774\, 868\, 320\, 450$ (my manually found maximum $N$ that this criteria can disprove was $N=208\, 576\, 659\, 753\, 891\, 832\, 997$ - see here). Here we need stronger criteria than the one given.
The remainder of the list (extended to 46 entries of the convergents):