User blog comment:Edwin Shade/The Most Powerful Array Notation Yet Devised/@comment-5529393-20170924012309

Well, the least ordinal definable in n symbols or less will presumably be 0, assuming we are using a somewhat normal language to describe ordinals. Assuming you mean "greatest" rather than "least", I should point out that if you want to plug your ordinal into a fast-growing hierarchy, it should be a countable ordinal, since there is no obvious definition for $$F_{\omega_1}(n)$$, and it is conceivable that there is no function that exceeds $$F_\alpha(n)$$ for all countable $$\alpha$$. Also, it is extremely likely that there is no way to explicitly define $$F_\alpha$$ for all countable \alpha, since it appears that to do so we would need some way to specify all countable $$\alpha$$, and this is impossible since there are uncountably many countable ordinals. So we need to stop at some countable ordinal.

You definition of N(a) does not make sense, since it depends on variables that are not inputs of the function. For example, you have $$\alpha = N(\varepsilon_0) = N(\omega^2 2 + \omega 2 + 3)$$, but only when $$\alpha$$ as a subscript in $$F_\alpha(3)$$; it would presumably be different if you had $$F_\alpha(10)$$. This is not how functions are supposed to work. You could change your notation to N(a,n), or N(a)[n], for instance, to indicate which n you are using to reduce your ordinals.

This leads to the other issue: fundamental sequences for ordinals are not uniquely defined, until you define them. This is not too hard to do if you have have a notation for all your ordinals up to the largest ordinal that you are plugging in to the FGH, but you do need to do so.

I don't know why you think $$F_{N(\omega,\omega)}(10^{100})$$ would be bigger than BIG FOOT.