<s>[INST] Some details about 'Collatz Conjecture'? Yes, there is no one who doesn't know this problem.My question is only about curiosity.

$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$

On this problem, I caught something like this.I'm sure, We all realized that.
For example, $n=19$, we have $6$ odd steps.
We know that, even steps are not important, because each even number is converted to an odd number.

$19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$

Then, for $n=77$, We have also $6$ odd steps.

$77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$

For $n=9$ 

$9\Longrightarrow 7 \Longrightarrow   11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$

Again we have $k=6$ odd steps.

I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result  $1$? 

Thank you! 
 [/INST] If you search a single formula for any $k$, here it is:
$$n_k=\frac{2^{l_1+l_2+...+l_k}}{3^k}-\frac{2^{l_2+l_3+...+l_k}}{3^k}-\frac{2^{l_3+l_4+...+l_k}}{3^{k-1}}-\frac{2^{l_4+l_5+...+l_k}}{3^{k-2}}-...-\frac{2^{l_{k-1}+l_k}}{3^3}-\frac{2^{l_k}}{3^2}-\frac{2^0}{3^1}$$
e.g. 
$$19=\frac{2^{4+3+2+1+3+1}}{3^6}-\frac{2^{3+2+1+3+1}}{3^6}-\frac{2^{2+1+3+1}}{3^5}-\frac{2^{1+3+1}}{3^4}-\frac{2^{3+1}}{3^3}-\frac{2^{1}}{3^2}-\frac{2^0}{3^1}$$
The dificulty is to find the $l_k$ for which $n_k$ is an integer.
The $l_k$ are the number of times you divide by 2 to jump from an odd to another odd.
e.g. for $19$, $l_6=1$ because you divide $3*19+1$ only once to get the next odd $29$. $l_5=3$ because you divide $3*29+1$ three times by 2 to get the next odd $11$...
When an $l_k$ is known, any $l_k$ of the same parity will work (e.g. for $19$, $l_6=1$ is odd, so any odd value of $l_6$ will work).
</s>