Simulation of collatz tapes by k-state 2-symbol Turing Machines

General
1

Consider an empty tape with all unmarked cells, such that the reading head (standing initially in the middle of the tape) applies the collatz function to a starting n:

f(n) = \begin{cases} n/2 & \text{if} \quad n \equiv 0 \quad (\text{mod}\, 2) \\ 3n + 1 & \text{if} \quad n \equiv 1 \quad (\text{mod}\, 2) \\ \end{cases}

flipping the state of the cell it currently stands in, and then moving left if n is odd, and right if n is even. It will do this until n = 1 is reached. Consider the example of n = 5:

collatz_tape5

and the corresponding tape development over time (\downarrow):

developed_collatz_tape5
developed_collatz_tape53059×239 468 Bytes

What is the smallest k-state 2-symbol Turing machine ™ which simulates such tape development? It is the following 3-state machine 1LB0LA_1RC0LA_1LH0RB (in corresponding string format and according to standard TM notation - as for instance described in this blog post).

Powers of 2, for instance, will be simulated by k-state TMs, according to the relationship k(n) = 2^{k} if (n & (n - 1)) == 0 for n > 0, with & being the bitwise AND operator (e.g. 2 - 1LH0LA; 4 - 1RB0LA_1LH0LA; 8 - 1RB0LA_1RC0LA_1LH0LA).

But overall, what will be the dependence of k on the corresponding collatz numbers n?