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}

0quad {text{if}};{x_n};{text{is even}}{text{,}}\

1quad {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?