Let $f(n) = \begin{cases} n/2, & \text{if $n$ is even} \\ 3n+1, & \text{if $n$ is odd} \end{cases}$
For an arbitrary prime $p$ are there some start value $x_0$ such that $p = x_k$ for some $k > 0$ in the sequence defined by $x_{n+1} = f(x_n)$?
I was confused. What I really wondered, turns out to be if all primes $> 3$ could be written as $3n + 1$ with $n$ odd, which is also trivially true. Or? No! I'll try again without Collatz.
Define $f(p_1^{n_1} \cdots p_k^{n_k}) = p_2^{n_2}\cdots p_k^{n_k}$, where $p_k$ is the $k$th prime. Is any prime $> 3$ in the image of the function $g:\mathbb N\to\mathbb N$, defined by $g(n) = f(3f(n) + 1)$.
I see now that all odd numbers not divisible by $3$ are in the image.
Yes, just take $x_0=2p$ and then $x_1=p$.