Yes, sureley. In Computablilty Theory you have the famous Ackermann-function[1]. It is actually an operator-extension, like you just described. It is important, because it grows overexponentially, but is still computable (unlike e.g. the busy-beaver-function).
Given that Graham's Number is an (over-)extension of the Ackermann-function that means that it's still computable, correct?
Given the series leading up to Graham's Number, G = g_64 (g_1, g_2, ...), BB(n) will outgrow g_n, right? If so what's the smallest n such that BB(n) > g_n?
Yes, sureley. In Computablilty Theory you have the famous Ackermann-function[1]. It is actually an operator-extension, like you just described. It is important, because it grows overexponentially, but is still computable (unlike e.g. the busy-beaver-function).
[1]: https://en.wikipedia.org/wiki/Ackermann_function