I've really only made functions/notations that have a growth rate of only \(\omega ^ {3} + 1\), so I really, really hope that this goes higher than that.

RandomFunction(n) = The largest number that any F(n) defined within the rules stated below can return. Rules: You have to define the largest F(x) such that it can only use the lvl function (defined on my first blog post), recursion, you could also define other functions and use them in F(x), you can use inverses to lvl(x, y, z) like subtraction, logarithms, roots, etc., you can only have one parameter of F(x), and only whole numbers are allowed. For example, define F(x) = x. This would be 3 symbols because of the F(), the =, and the x. But why not the x inside of F? Because, when you use a variable by itself, it counts as a symbol, but if it is a parameter, it does not count as a symbol, however, a function will count as a symbol. Recursion is also allowed. It could go like this. F(x) = F(Subtraction(x, 1)) F(0) = x. This would have 6 symbols because of the F(), the =, the other F(), the Subtraction(), the other F(), and the isolated x. Also, lvl(x, y, z) is allowed, so we can take that into an advantage. F(x) = lvl(x, x, x) which means this would only have 3 symbols and still have a growth rate of f_w(n)! There are many more listed 4 "paragraphs" above, but I think you know how the symbols work because of the many examples I gave you. When you use this in RandomFunction(n), the highest whole number that can be in the equation is n, so F(x) = lvl(5, x, x) would only be valid if x was equal or greater than 5. Also, for every function F you can define with the rules above, RandomFunction(n) = the greatest F(n), which means you would have to plug in n in F(). Also, there can only be at the maximum n symbols for F(x).

What is the growth rate of RandomFunction(n) and how can I improve it? Is this even well-defined?