The usual Collatz map is $C:n \mapsto n/2$ if $n$ even, $(3n+1)/2$ if $n$ odd. Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$.

We suspect that for every fixed $n>0$, the sequence $C^{\circ r}(n)$ never diverges to infinity (and more strongly, always converges to $1$), because (together with experimental checking) *heuristically* the application of $C$ to a random number multiplies it *in geometric average* by $(\frac{1}{2} \cdot \frac{3}{2})^{1/2} = \frac{\sqrt{3}}{2} < 1$.

**The (borderline) Collatz-like problems:**

A map $f: \mathbb{N} \to \mathbb{N}$ will be called a *Collatz-like* map if $$ 0 \neq \lim_{n \to \infty} \left( \prod_{r=1}^n \frac{f(r)}{r} \right)^{1/n} \le 1 \ \ \ \ \ (*) $$ If the inequality $(*)$ is an equality then the map $f$ will be called a **borderline** Collatz-like map.

For each (borderline) Collatz-like map $f$, we have the *(borderline) Collatz-like problem* asking whether its iterations diverges nowhere to infinity, i.e. $$\forall n>0, \ \exists m,r>0 \text{ with } f^{\circ (m+r)}(n) = f^{\circ m}(n).$$

If the answer is yes, then let us call $f$ an **acceptable** (borderline) Collatz-like map.

This post focus on a specific family of borderline Collatz-like problems:

For any given $\alpha >0$, let us consider the following map $$ f_{\alpha}: n \mapsto \left\{ \begin{array}{ll} \left \lfloor{n\alpha} \right \rfloor & \text{ if } n \text{ even,} \\ \left \lfloor{n/\alpha} \right \rfloor & \text{ if } n \text{ odd.} \end{array} \right. $$

The map $f_{\alpha}$ is borderline Collatz-like. Let $S$ be the set of $\alpha>0$ for which $f_{\alpha}$ is acceptable.

Question: What is the set $S$, explicitly?

It is obvious to find some $\alpha$ in $S$, and some other out, for example $S \cap \mathbb{Z}_{>0} = \{ 1 \}$.

*Proposition*: $3/2 \in S$.

*Proof*: If $n$ is even then $n=2^ra$ with $a$ odd and $r>0$. Then $f_{3/2}(n)=2^{r-1}3a$, $f^{\circ r}_{3/2}(n)=3^ra$ and $f^{\circ(r+1)}_{3/2}(n)=3^{r-1}2a = f^{\circ(r-1)}_{3/2}(n)$. Next, if $n$ is odd, then $n=6k+i$ with $i \in \{1,3,5\}$, and $\left \lfloor{2n/3} \right \rfloor = \left \lfloor{2(6k+i)/3} \right \rfloor = 6k + \left \lfloor{2i/3} \right \rfloor$, but $\left \lfloor{2i/3} \right \rfloor = 0,2,3$ for $i=1,3,5$. So the cases $i=1,3$ reduce to the even case, next, if $i=5$ then $\left \lfloor{2n/3} \right \rfloor = 6k+3$. $\square$

*Remark*: See this comment of user35593 which proves that $\alpha = 2/3$ is also in $S$.

Now what about $\alpha$ irractional? This post (and its comments) provides a family of quadratic integers not in $S$ (one of which being the golden ratio $\phi$).

Below are the pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$, of the plot of the map which for every fixed $n$ gives the minimal $m$ such that $f_{\alpha}^{\circ (m+r)}(n) = f_{\alpha}^{\circ m}(n)$ for some $r$.

**Subquestion 1**: Is it true that $\{\pi, 1/\pi,\sqrt{2} \} \subset S$?

Then, we *could* expect that $\alpha = 1/\sqrt{2}$ is also in $S$, but in fact it seems not! The following picture shows the value of $\log_{10}(f_{\alpha}^{\circ r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.

**Subquestion 2**: Is it true that $2^{-1/2} \not \in S$?

[for a focus on this specific problem, see this post]

With $n=15$ we get the following loop of length $33$
$$15,
21,
29,
41,
57,
80,
56,
39,
55,
77,
108,
76,
53,
74,
52,
36,
25,
35,
49,
69,
97,
137,
193,
272,
192,
135,
190,
134,
94,
66,
46,
32,
22,
15, \dots
$$The smallest $n$ for which the iterations of $f_{2^{-1/2}}$ *seems* to diverge to infinity is $73$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, \dots , 102868753471, 145478386303, \dots$$

The difference of shape of the pictures for $\pi$ and $\sqrt{2}$ is surprising. Below is the picture for $\alpha = \sqrt{2}^{\sqrt{2}}$, it is also unsimilar to those above:

Moreover, the fact that $\sqrt{2}, \pi$ seem to be in $S$ and $\sqrt{2}/2, \phi$ not, is also surprising because then the answer would be very irregular on the irrational numbers.

All these surprises lead to feel that the answer to the main question could be reachable.

**Extended family of (borderline) Collatz-like maps**

We can extend the above family as follows: let the tuple $A=(\alpha_1, \alpha_2, \dots , \alpha_r)$ such that $\alpha_i>0$ for all $i$ and $\prod_i \alpha_i \le 1 $. Consider the map $f_A: n \mapsto \left \lfloor{n\alpha_i} \right \rfloor \text{ if } n \equiv i \mod (r)$.

*Problem*: What are the tuples $A$ for which the map $f_A$ is acceptable?

**Other borderline Collatz-like maps**

Here are other examples of boderline Collatz-like maps, provided by Tom Crawford in this Numberphile video (which inspired this post). Let $\alpha$ be a normal number in base $10$ (you can think to $\pi$, which is strongly suspected to be normal). Consider the function $g_{\alpha}(n)$ equals to the position of the first occurence of $n$ (written in base $10$) in the decimal digits of $\alpha$. Because $\alpha$ is normal, $g_{\alpha}$ should satisfy $(*)$. Note that $g_{\pi}$ is available in OEIS A014777.

Here is an other Collatz-like map involving the set of prime numbers $\mathbb{P}$:

First consider the usual bijection $b: \mathbb{P} \to \mathbb{N}$, next the map $g: \mathbb{P} \to \mathbb{P}$ with $$g(p):= \text{ the greatest prime factor of } 2p+1.$$ Then $f:= b \circ g \circ b^{-1}$ should also be a (borderline?) Collatz-like map.

For more details on this map and its generailzations, see this post.