I have a possible interesting question about the Collatz Conjecture, that I think it is "answerable".
Let me state my question as follows.
We are only interested in odd natural numbers under the Collatz function/process and if the natural number goes to $1$ under the Collatz process $C(n)$, then we will consider these numbers as "Collatz numbers" in general.
Consider a function that generates the following odd "Collatz numbers" :
$$f(n)=\frac {4^n-1}{3}$$
This function generates all $1-$odd step odd Collatz numbers.
Here, the odd step exactly means under the Collatz function $C(n)$, we consider the number of "jumping" from one odd number to the next odd number in the iterated Collatz sequence. Similarly, the even step means under the Collatz function we count the number of "jumping" from even number to the next even number in the iterated Collatz sequence.
Then consider the following function :
$$f(n)=\frac{2^{8n-3}-2^{2n-1}-3}{9}$$
Note that, the above function does not generate all $2-$odd step odd collatz numbers, but only produces possible infinitely many subset of $2-$odd step odd collatz numbers, under the iterated collatz function/process.
My question :
In terms of $m$ and $n$, does there exist an effective explicit algebraic formula “like above” such that, the explicit algebraic formula $f_m(n)$ yields exactly $\color{red}{\text{m - odd step}}$ odd Collatz numbers ? Here, the existence of the formula $f_m(n)$ does not imply $f_m(n)$ covers all of the odd Collatz numbers.
Note that, the above formulas does not depend on the variable $m$. In above formulas $m=1$ or $m=2$ are fixed numbers, which are the specific cases. Concretely, we are looking for a possible $2-$ variable function or formula, which depends on the variables $m$ and $n$ and generates infinitely many $m-$odd step odd Collatz numbers under the iterated Collatz function. Obviously, we know that the single function $f_m(n)$ can never cover all of the $m-$ odd step odd collatz numbers.
So, the question is very short : Does there exist any possibility to derive such a formula $f_m(n)$ ?
Note that, the question does not necessarily imply $m\neq n$. The construction of $f(n)$ , which exactly produces $n-$ odd step odd collatz numbers, under the iterated collatz function does perfectly answer the original question. For instance, $f(5)$ generates infinitely many $5$ odd step odd collatz numbers, $f(100)$ generates infinitely many $100$ odd step odd collatz numbers.
What I am looking for, the formula $f_m(n)$ includes only algebraic operations, e.g. $+,-,÷,\times, a^b, \sqrt {\cdot}$ and involve only elementary algebraic functions e.g. $\lfloor x \rfloor,\,\lceil x \rceil$ or possible modular restrictions which purely interchangeable with floor/ceiling functions.
Definitions :
What are the $\color{red}{\text{m - even step}}$ collatz numbers ?
For instance, take $f(n)=2^n$ generates exactly $m=n$ even step collatz numbers.
What are the $\color{red}{\text{m - odd step}}$ collatz numbers ?
The algebraic function $f_m(n)$ produces exactly$m-$ odd steps odd collatz numbers in terms of $m$ and $n$, which is equivalent to my original question.
Comment :
We are not looking for an explicit algebraic formula that covers all odd step Collatz numbers. My question is only about my curiosity. Like me, everyone can also observe that, the above question has nothing to do with to prove the conjecture.
I have just convinced myself that looking for a formula $n-$ odd step function, similar to $n-$even step function $f(n)=2^n$ would be pretty interesting and unexpected.