TITLE: Why does this cycle of 44 show up in the Collatz Conjecture?
QUESTION [6 upvotes]: Consider this function:
$$f\left(x\right)=\frac{x-b^{\left(\operatorname{floor}\left(\log_{b}x\right)\right)}}{b^{\left(\operatorname{floor}\left(\log_{b}x\right)\ +\ 1\right)}-b^{\left(\operatorname{floor}\left(\log_{b}x\right)\right)}}$$
Here is a StackExchange post with more information about the function. But basically, you could describe the function as:

A relative measure of how far away a number is from the smallest number with the same number of digits.

or

A relative measure of how far a number 'x' is from the biggest power of 'b' that is still less than or equal to 'x'

So anyways, if we set b=6, then apply f(x) to elements of Collatz orbits that have a length > 44, we see a cycle. It's easy to see this when you graph the points (i, f(x)) on a polar graph.
i = the index of the element in the sequence 
x = the value at COLLATZ[i] where COLLATZ is an array created from the elements of the Collatz orbit starting at n and terminating at 1.
Here is the graph for n = 27. The points are labeled (i, COLLATZ[i])

It clearly cycles at 44.
To test this hypothesis further, I tried graphing n = 670617279, which has 949 steps. Here is the result.

I thought it may be simply because I was using polar coordinates, but I see similar symmetry when using cartesian coordinates.
I think I'm on the verge of understanding why, but I just wanted to ask the community for any insight. I find it odd that 44 shows up.
By the way, here is a medium article where I go over the use of this function in more detail as it applies to Collatz.

REPLY [2 votes]: $f(x)$ for $b=6$ (bellow in Cartesian coordinates for $n=27$) is almost equivalent to the fractional part of $log_6(x)$:  $g(x)=\{log_6(x)\}$ (second graph bellow) which also measure the distance to the next power of $6$

In red in the graph above (SageMathCell) is the Collatz sequence for $n=27$ with $f(x)$, and in blue, also with $f(x)$, is a random Collatz sequence (randomly apply $3n+1$ or $n/2$) starting with $n=27$ for 111 steps, witch I generated with this little Pari/GP code (didn't took the time to make it in SageMathCell):
n=27;for(g=1,111,print1(floor(n),", ");if(random%2==0,n=3*n+1,n=n/2))
You can see that as long as we apply either $3n+1$ or $n/2$, you will get the same type of graph whatever you do (with a small shift here and there depending on the starting value, the "+1" and which way you go from step to step).
This is because the fractional part of $log_6(x)$ will be nearly the same on step $i$ since you either multiply by $3$ or divide by $2$ from the previous step which land on the same spot (both results being separated by a factor of 6, so having the same fractional part)
Same result for $g(x)$ bellow (can be found in the same SageMathCell linked above)

So a 111 steps Collatz sequence, whatever the path, will always look like this in Polar coordinates (each group of 3 dots are the dots we find horizontally separated by 44 on the Cartesian graph, at different "heights" from 0 to 1) :

Now, if we look at 1 of the 3 dots group (e.g the 3 most outer values in OP first graph, which are the nearest to a power of 6, or near y=1.0 on the Cartesian plot):
$[214, 1276, 35]$
We can see that $$1276=\frac{3^{18}}{2^{26}}\cdot 214+\frac{2722925818}{2^{26}}$$
$$35=\frac{3^{15}}{2^{29}}\cdot 1276+\frac{481276588}{2^{29}}$$
To have almost the same value for the fractional part of lets say $\{log_6(214)\}\approx\{log_6(1276)\}$, we must have $\{log_6(\frac{3^n}{2^{m-n}})\}=\{log_6(\frac{1}{2^m})\}\approx 0$ or $1$, which is the case for $m=44$, but also for
$\{13, 18, 31, 44, 49, 62, 75, 88, 93\}$ with precision $<.05$, which is mostly the same list as Dan proposed.