I was wondering if we could limit the divisions in the Collatz function and still have an analogous conjecture. If we limit it to divisions by "$2, 4, 8$", then we have divergence. But if we limit it to divisions by "$2, 4, 8, 16$", then it appears that it still holds except for one additional cycle.
That is, define the "limited" Collatz function as:
$$ f(n)= \begin{cases} 3n+1, & \text{if } n \text{ is odd}\\ n/2, & \text{if } n/2 \text{ is odd}\\ n/4, & \text{if } n/4 \text{ is odd}\\ n/8, & \text{if } n/8 \text{ is odd}\\ n/16, & \text{if } n/16 \text{ is odd}\\ 3(n/16)+1, & \text{} \text{otherwise}\\ \end{cases} $$
The one extra cycle appears because in the normal Collatz we have: $$15\to 23\to 35\to 53\to 5\to \dots$$
But in the "limited version", we have a "redirection" (due to "otherwise" case): $$15\to 23\to 35\to 53\color{red}{\to 31\to \dots}$$
But the sequence for $31$ already contains $23$, so the loop is formed.
My question is, is this the only new loop?
So far, I have checked odd numbers up to $10^9$.
Edit: I have simplified the question from $c$ in general to $c=4$ (smallest interesting case). The original (before edit) question was asking about $c$ in general (see down below).
In general, we can examine the "$c$ limited" Collatz conjecture.
The Collatz conjecture defines sequence (where $v_n$ is 2-adic valuation of $a_n$) $$a_{n+1}=3\left(\frac{a_{n}}{2^{v_n}}\right)+1.$$
The conjecture is that for every $a_0\in\mathbb N$, the sequence $a_n$ will reach $4$.
Now, let $w_n=\min\{u_n,c\},c\in\mathbb N$ and define a "limited" Collatz sequence $$b_{n+1}=3\left(\frac{b_{n}}{2^{w_n}}\right)+1.$$
Where $u_n$ is 2-adic valuation of $b_n$.
That is, we are allowing at most $c$ consecutive divisions by $2$.
Given a fixed $c$, what is the smallest $b_0$ that fails the "limited" conjecture?
Was a similar question discussed before? I'm looking for references.
For example, if $c\in\{1,2,3\}$, then many numbers diverge to infinity.
For example, if $c=4$, then $b_0=15$ reaches a cycle that does not contain $4$.
But if $c=5$, the "limited" conjecture still holds for all $b_0\le 10^8$, so far.
Is the original conjecture as hard as the "limited" conjecture?