Is it possible to find $ψ(m_1,m_2,\ldots,m_k)$? Inspired by the question 
Is $\lim_{k \to \infty}\left[ \lim_{p \to \infty} \frac{M}{1+3+5+7+\cdots+ [2^{p(k-1)}-2^{p(k-2)}-2^{p(k-3)}-\cdots-1]}\right]=1$? which that, I asked before.
I researched the pdf books before asking this question at MSE. When I could not find an answer anywhere, I decided to ask. Even if the question is absurd.
For this function:
$$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ n+1 & \text{if } n\equiv 1 \pmod 2. \end{cases}$$
We know that for any positive number, there is such a number $\text{“} k \text{''}$, which that $f^k(n)=1$
For function $f(n)$ go "backward" from number $1$ for only odd numbers sequence:
Let step number is $k$
 $$[2^{\sum_{z=1}^k m_z}-2^{\sum_{z=2}^k m_z}-2^{\sum_{z=3}^k m_z}-\cdots-1]\stackrel{k\to \infty}{\longleftarrow}\mathbf{\cdots} \stackrel{k=5}{\longleftarrow} \mathbf{[2^{m_4+m_3+m_2+m_1}-2^{m_4+m_3+m_2}-2^{m_4+m_3}-2^{m_4}-1]}\stackrel{k=4}{\longleftarrow} \mathbf{[{2^{m_1+m_3+m_2}-2^{m_2+m_3}-2^{m_3}-1}]}\stackrel{k=3}{\longleftarrow} \mathbf{[{2^{m_1+m_2}-2^{m_2}-1}]}\stackrel{k=2}{\longleftarrow} \mathbf{[{2^{m_1}-1}]}\stackrel{k=1}{\longleftarrow} \mathbf1$$
I used this formula:
$$φ({m_1,m_2,\ldots,m_k})=φ({m_1,m_2,\ldots,m_{k-1}})×2^{m_k}-1$$
and  we can find general distrubition function: $φ({m_1,m_2,\ldots,m_k})$
$$φ({m_1,m_2,\ldots,m_{k}})=[2^{\sum_{z=1}^k m_z}-2^{\sum_{z=2}^k m_z} - 2^{\sum_{z=3}^k m_z}-\ldots-1]$$
$$f^k(φ(m_1,m_2,\ldots,m_k))=f^k([2^{\sum_{z=1}^k m_z}-2^{\sum_{z=2}^k m_z} - 2^{\sum_{z=3}^k m_z}-\cdots-1])=1$$
Example: 
$$f^2(φ(m_1,m_2))=f^2({2^{m_1+m_2}-2^{m_2}-1})=\frac{\frac{{2^{m_1+m_2}-2^{m_2}-1} + 1}{2^{m_2}}+1}{2^{m_1}}=1$$
Then, look at this function:
$$g(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod 2 \\ 3n+1 & \text{if } n \equiv 1 \pmod 2. \end{cases}$$
For function $g(n)$ go "backward" from number $1$ for only odd numbers sequence, again:
$$ψ(m_1,m_2,\ldots,m_k)\stackrel{k\to \infty}{\longleftarrow}\mathbf{\cdots} \stackrel{k=2}{\longleftarrow} \mathbf{[\frac{4^{m_1}-1}{3}]}\stackrel{k=1}{\longleftarrow} \mathbf1$$
$$ψ(m_1)=\frac{4^{m_1}-1}{3}$$
Because, $g^1(ψ(m_1))=1$
The problem starts here. I found $ψ(m_1)$ for $k=1$. But, I dont know, how can I find $ψ(m_1,m_2)$ or $ψ(m_1,m_2,m_3)$.I can not continue here.
Anyway.My question's "meat" is this: 
Is it possible to make a general formula by going back from $1$?

How can we find and is it possible $ψ(m_1,m_2),ψ(m_1,m_2,m_3)...ψ(m_1,m_2,\ldots,m_k)$?
If we find, general dispersion $ψ(m_1,m_2),ψ(m_1,m_2,m_3),...,ψ(m_1,m_2,\ldots,m_k)$ can we answer that question: Why, is there such a number $k$ for function $f(n)$ always $f^k(n)=1$ and $g^k(n)=1$ or $g^k(n)≠1$(counterexample)?

The question is open to any editing, because I know, there are flaws in question and formulas.
 A: For $k=3$ it seems a bit more complicated. The third exponent must be something like $2o + (l-n)mod$ 3. I'll try to find out later.
I think this needs a more systematic way.
let's say $n_{(k-1)}=\frac{3n_k+1}{2^{m_k}}$, with $n_0=1$.
If you want to climb up the tree, you can't pass through multiple of 3 (the $n_{(k-1)}$ cannot be multiple of 3, this is well known and obvious when you replace in the above formula). Here are the valid $m_k$ for different $n_{(k-1)}$ to avoid multiple of 3:
$n_{(k-1)}\equiv 1 \pmod {18}$ => $m_k$ must be of the form $6x+2$ or $6x+4$
$n_{(k-1)}\equiv 11 \pmod {18}$ => $m_k$ must be of the form $6x+1$ or $6x+3$
$n_{(k-1)}\equiv 13 \pmod {18}$ => $m_k$ must be of the form $6x+2$ or $6x+6$
$n_{(k-1)}\equiv 5 \pmod {18}$ => $m_k$ must be of the form $6x+3$ or $6x+5$
$n_{(k-1)}\equiv 7 \pmod {18}$ => $m_k$ must be of the form $6x+4$ or $6x+6$
$n_{(k-1)}\equiv 17 \pmod {18}$ => $m_k$ must be of the form $6x+1$ or $6x+5$
Perhaps if we use the general formula with the desired $k$ level: 
$\frac{2^{(m_k+m_{k-1}+...+m_3+m_2+m_1)}}{3^k} - \frac{2^{(m_k+m_{k-1}+...+m_3+m_2)}}{3^k} - \frac{2^{(m_k+m_{k-1}+...+m_3)}}{3^{k-1}} - \frac{2^{(m_k+m_{k-1}+...+m_4)}}{3^{k-2}} -  ....  -  .... - \frac{2^{(m_k+m_{k-1})}}{3^3} - \frac{2^{(m_k)}}{3^2} - \frac{2^0}{3^1}$
and by replacing the $m_k$ by valid forms ($6x+y$ from the modulo list above), like i did in the previous answer, we must be able to construct the formulas.
Note: for the highest $m_k$ you don't need to avoid multiple of 3 (they are valid values), so you can use the $2l$ or $2l+1$ forms instead, like i did in the previous answer too.
I have not much time now, i'll take a look later, but i still think that having a unique formula or working with the $m_k$ alone is almost impossible.
EDIT:
Ok, one of the formulas for $k=3$ is
$\frac{2^{m_3+m_2+m_1}}{3^3}-\frac{2^{m_3+m_2}}{3^3}-\frac{2^{m_3}}{3^2}-\frac{1}{3^1}$ with $m_1=18n+2$, $m_2=18l+2$ and $m_3=2o$ which gives
$\frac{16\cdot4^o\cdot262144^{n+l}-4\cdot4^o\cdot262144^l-3\cdot4^o-9}{27}$
I used $18x+2$ instead of $6x+2$ because the patern repeats every 3 times (e.g. the multiple of 3 is in first position for exponent $6\cdot3\cdot x+2$)
So i think you can split to avoid using modulos like i thought. Anyway, this gives a lot of possible formulas to cover $k=3$, and even more for $k=4$... which is not practicable...unless we can find some sort of systematic way to do it.
