Is it possible to find $ψ(m_1,m_2,\ldots,m_k)$?

Question

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.

Answer 1

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.

Answer 2

$φ({m_1,m_2,\ldots,m_{k}})=\frac{2^{(m_k+m_{k-1}+...+m_3+m_2+m_1)}}{3^k} - \frac{2^{(m_k+m_{k-1}+...+m_3+m_2)}}{3^k} - \frac{2^{(m_k+m_{k-1}+...+m_3)}}{3^{k-1}} - \frac{2^{(m_k+m_{k-1}+...+m_4)}}{3^{k-2}} - .... - .... - \frac{2^{(m_k+m_{k-1})}}{3^3} - \frac{2^{(m_k)}}{3^2} - \frac{2^0}{3^1}$

If you apply successively the condensed formula $g'(n)=\frac{3n+1}{2^{m_i}}$ to it, you'll reach 1.

k being the "distance" to 1, answering to "why is there such a number k" is answering the conjecture.

EDIT: I used your layout so you can understand better:

For function $g(n)$ go "backward" from number $1$ for only odd numbers sequence:

Note: This is how the Collatz tree is contructed

Let step number is $k$ $$[\frac{2^{\sum_{z=1}^k m_z}}{3^k}-\frac{2^{\sum_{z=2}^k m_z}}{3^k}-\frac{2^{\sum_{z=3}^k m_z}}{3^{k-1}}-\cdots-\frac{1}{3}]\stackrel{k\to \infty}{\longleftarrow}\mathbf{\cdots} \stackrel{k=5}{\longleftarrow} \mathbf{[\frac{2^{m_4+m_3+m_2+m_1}}{3^4}-\frac{2^{m_4+m_3+m_2}}{3^4}-\frac{2^{m_4+m_3}}{3^3}-\frac{2^{m_4}}{3^2}-\frac{1}{3}]}\stackrel{k=4}{\longleftarrow} \mathbf{[{\frac{2^{m_3+m_2+m_1}}{3^3}-\frac{2^{m_3+m_2}}{3^3}-\frac{2^{m_3}}{3^2}-\frac{1}{3^1}}]}\stackrel{k=3}{\longleftarrow} \mathbf{[{\frac{2^{m_2+m_1}}{3^2}-\frac{2^{m_2}}{3^2}-\frac{1}{3}}]}\stackrel{k=2}{\longleftarrow} \mathbf{[{\frac{2^{m_1}}{3^1}-\frac{1}{3^1}}]}\stackrel{k=1}{\longleftarrow} \mathbf1$$

I used this formula:

$$φ({m_1,m_2,\ldots,m_k})=φ({m_1,m_2,\ldots,m_{k-1}})×\frac{2^{m_k}-1}{3}$$ (which is the inverse of the Collatz function $\frac{3n+1}{2^{m_k}}$ since we run backward)

and we can find general distrubition function: $φ({m_1,m_2,\ldots,m_k})$

$$φ({m_1,m_2,\ldots,m_{k}})=[\frac{2^{\sum_{z=1}^k m_z}}{3^k}-\frac{2^{\sum_{z=2}^k m_z}}{3^k} -\frac{2^{\sum_{z=3}^k m_z}}{3^{k-1}}-\ldots-\frac{1}{3}]$$

$$g^k(φ(m_1,m_2,\ldots,m_k))=g^k([\frac{2^{\sum_{z=1}^k m_z}}{3^k}-\frac{2^{\sum_{z=2}^k m_z}}{3^k} -\frac{2^{\sum_{z=3}^k m_z}}{3^{k-1}}-\ldots-\frac{1}{3}])=1$$

Example:

$$g^2(φ(m_1,m_2))=g^2(\frac{2^{m_1+m_2}}{3^2}-\frac{2^{m_2}}{3^2}-\frac{1}{3})=\frac{3(\frac{3(\frac{2^{m_1+m_2}}{3^2}-\frac{2^{m_2}}{3^2}-\frac{1}{3})+1}{2^{m_2}})+1}{2^{m_1}}=1$$

Note that $ψ(m_1)=\frac{4^{m_1}-1}{3}$ is correct, but above, it was written as $\frac{2^{m_1}-1}{3}$ because other exponents $m_x$ are not always even.

Also note that my answer [k being the "distance" to 1, answering to "why is there such a number k" is answering the conjecture] refers to the previously unedited question: "Why,is there such a number $k$ for function $g(n)$ always $g^k(n)=1$" which is the conjecture itself and nobody can answer yet: when you run backward, you have to prove that you reach all naturals from 1.

EDIT2: The $m_k$ are totally dependent of the parent value $φ$, so there is no way to have a unique formula (with $m_k$ being any natural). For $k=2$ you could write these two formulas (they give all naturals that are 2 steps from 1): one with $m_1=6n+4$ and $m_2=2l+1$: $\frac{32\cdot64^n\cdot4^l-2\cdot4^l-3}{3^2}$ and one with $m_1=6n+2$ and $m_2=2l$(preferably with l>0, but i think it works with 0): $\frac{4\cdot64^n\cdot4^l-4^l-3}{3^2}$, but it does not help much (with $k=3$ you multiply the number of formulas...). You can have an infinite succession of $m_k=1$ (e.g. for $2^y-1$ there is $y-1$ successives $m_k=1$, so you can't predict $m_{k+1}$ with other $m_k$'s), or other combinations, but they are dependent of the successive values they produce (you can't just replace $m_4=1$ by $m_4=2$ or $3$, it would completly change the possible values of $m_{5+}$. If it was so simple, Collatz would be proved by now.