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 odd number goes to $1$ under the Collatz process, then we will consider these numbers as "Collatz numbers".
Consider a function that generates the following odd "Collatz numbers" :
$$f(n)=\frac {4^n-1}{3}$$
This function generates all $1-$odd step Collatz numbers.
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 collatz numbers, but only produces possible infinitely many $2-$odd step collatz numbers, under the collatz function/process.
My question :
In terms of $n$, does there exist an effective explicit algebraic formula “like above” such that, the explicit algebraic formula $f(n)$ yields exactly $\color{red}{\text{n - odd step}}$ Collatz numbers ? Here, the existence of the formula $f(n)$ does not imply $f(n)$ covers all of the odd Collatz numbers.
What I am looking for, the formula $f(n)$ 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 is $\color{red}{\text{n - even step}}$ collatz numbers ?
For instance, take $f(n)=2^n$ generates exactly $n-$ even step collatz numbers.
What is $\color{red}{\text{n - odd step}}$ collatz numbers ?
The algebraic function $f(n)$ produces $n-$ odd steps collatz numbers in terms of $n$, which is equivalent to my original question.
Comment :
We are not looking for an explicit algebraic formula that covers all odd step Collatz numbers. We are not trying to prove the Collatz conjecture. I have just convinced myself that looking for a formula similar to $n-$even step function $f(n)=2^n$ would be pretty interesting and unexpected.