Uncomputability of a function based on the Busy Beaver function

Question

Let $\log _b^ac$ denote an iterated base-$b$ logarithm function. For example, $$\log _2^3({2^{65536}}) = {\log _2}({\log _2}({\log _2}({2^{65536}}))) = 4.$$

Pick an arbitrary model M of Turing machines, assuming that a machine operates with the two-symbol alphabet: $0$ as the blank symbol and $1$ as the non-blank symbol. We will call such machines “M-machines.”

Let $f(n)$ denote the maximum number of non-blank symbols which can occur at the tape when a particular M-machine halts, assuming that all machines start with the blank input and the table of instructions contains at most $n$ operational states.

Then the function $F(n)$ is defined as follows: $${F}(n) = \left\{ \begin{array}{l} 0\quad {\text{if}}\;{x_n}\;{\text{is even}}{\text{,}}\\ 1\quad {\text{if}}\;{x_n}\;{\text{is odd}}{\text{,}} \end{array} \right.$$ where $x_n$ is the smallest natural number such that $$\log _{{f}(n) + 2}^{x_n}({f}(n + 1) + 3) < 2.$$

Question: is the function $F(n)$ uncomputable for any model M (that is, there does not exist an M-machine which can compute the $F(n)$ function, no matter which M we choose)? If yes (or no), is it possible to prove this?

EDIT
Suppose that we pick a model described in the "The game" section in the "Busy Beaver" Wikipedia article. Is $F$ uncomputable by such machines? I am also interested in how to prove this.

Answer 1

The nested logarithm doesn't really do much here. It is a computable function with a computable inverse, and thus the functions $n \mapsto x_n$ and $f$ are not only Turing equivalent, but related via computable rescaling.

As such, the answer to this question applies here, too.

We quite certainly could set up our Turing machines in a way that they never write a total number of non-blank symbols falling into a range that would make $x_n$ odd. We'd need to do some side computations to figure out acceptable ranges, but given the size of the ranges I expect that the symbols we need to determine the next interval we can use will eventually not bring us out of the interval.

On the other hand, for a natural setup I see no reason why $F$ would ever be computable. But actually going proving this for a particular setup would require thinking much more about the details of a particular Turing machine model than I would want to do.