Is it possible to discover a "Universal formula" that generates and generalizes all odd Collatz numbers?

I am just an ordinary student, and I have never had the chance to ask this question to mathematicians. Perhaps this is the first opportunity I have to ask it here.

When we refer to "Collatz numbers", we mean all natural numbers that eventually reach $1$ under the Collatz process. By odd-step Collatz numbers, we consider only the odd numbers that appear at each step of the sequence, skipping all even numbers. For example, for $n=4$, one possible sequence of odd-step Collatz numbers is: $$11 \rightarrow 17 \rightarrow 13 \rightarrow 5 \rightarrow 1 $$

Similarly, for $n = 6$, another example sequence could be:

$$ 19 \rightarrow 29 \rightarrow 11 \rightarrow 17 \rightarrow 13 \rightarrow 5 \rightarrow 1 $$

Now, my question is: Is it possible to find a "Universal formula" (or something "magic formula") that generates all sequences of odd numbers with exactly $n$ odd-steps in the Collatz process?

For instance, consider a function like $f(n) = 2^n$. While this generates a sequence, it does not satisfy the criterion of producing a complete sequence of odd-step Collatz numbers, as it results in only one odd step. I use $f(n) = 2^n$ merely as an example to illustrate the kind of formula I am seeking.

To summarize, for a given $n$, is there a general formula — similar to $f(n) = 2^n$ — that universally produces all sequences of $n$-step odd Collatz numbers? Has such a formula been discovered or studied by mathematicians?

Thank you so much .

I see that I needed to phrase the question more clearly. Let me rewrite what I mean by "Universal Formula". For example, when $n = 1$ is a finite number, we have a formula like $\dfrac {4^n-1}{3}$. We can go backwards and construct such a formulas for $n = 2$.

But, it would be a great experience for me to learn whether a "formula" that directly depends on $n$ exists or - if it is already known - to understand it .

The set $S$ of integers $s$ such that the Collatz iteration starting from $s$ ends in 1 after $n$ odd terms is semi-decidable (aka computably enumerable or recursively enumerable). This means that there exists an algorithm that takes $s$ as input, and returns true if $s$ belongs to the set, while it returns false or never finishes if $s$ does not belong to the set. That is simply because you can algorithmically compute the Collatz sequence starting from $s$, and return true if you reach 1 after you have seen $n$ odd numbers.

Therefore, as a consequence of the MRDP theorem (Matiyasevich-Robinson-Davis-Putnam), there exists a polynomial $P$ in $k+1$ variables and with integer coefficients such that $S = \{s \ | \ ∃ x_1, …, x_k ∈ ℤ, P(s, x_1, …, x_k) = 0\}$, i.e., $S$ is the set of values $s$ such that the equation $P(s, x_1, …, x_k)$ of unknowns $x_1, …, x_k ∈ ℤ$ has a solution. For many reasonable meanings of “formula”, this is a formula that generates $S$.

Unfortunately, this does not help much with the Collatz problem. It is a popular misconception that finding a formula for something will somehow magically make understanding that thing easy. For example, contrary to a common belief among non-mathematicians, there are many formulas for prime numbers (the MRDP theorem can also be used, since primes are a semi-decidable set as well), yet this does nothing to alleviate the infamous difficulty of understanding the behavior of prime numbers. In fact the MRDP theorem is so general that it gives you formulas for, say, twin primes, or counterexamples to Goldbach's conjecture, yet we still don't know if there are infinitely many twin primes (because we can find an equation with a parameter $n$ that has a solution iff $n$ and $n+2$ are twin primes, but we don't know how to prove that this equation has a solution for infinitely many values of $n$), nor whether there are any counterexamples to Goldbach's conjecture (likewise, there's an equation that has a solution iff $n$ is a counterexample, but how do you prove it never has a solution?).