When generalizing addition, multiplication, and exponentiation, there exists a certain hierarchy upon which these operators can be placed: the hyperoperator hierarchy. Starting with succession, each hyperoperator is an iterated version of the previous one.
A hyperoperator function $f(x, y, n)$ can be defined as follows: $$ f(x, y, n) = \left\{\\y + 1, n = 0\\x + y, n = 1\\x * y, n = 2\\x^y, n = 3\\1, n > 3, y = 0\\f(x,f(x, y-1, n), n-1), n > 3, y\neq1\right\} $$
The domain of this function is fairly limited. The value of $n$ is always restricted to non-negative integers. For $n = 0, 1, 2, 3$, the values of $x$ and $y$ can be any complex numbers, provided $n\neq0$; in that case, $y$ must be a non-negative real number. For $n > 3$, however, while $x$ can be any complex number except 0, $y$ must be a non-negative integer. Overall, the domain of $f(x, y, n)$ is as follows: $$ \left\{n\in \mathbb{N^0}\\x\in \mathbb{C}, n = 0, 1, 2, 3\\x\in \left\{\mathbb{C}|x\neq 0\right\}, n > 3\\y\in \mathbb{C}, n = 0, 1, 2\\y\in \mathbb{C}, n = 3, x\neq 0\\y\in \mathbb{R^+}, n = 3, x = 0\\y\in \mathbb{N^0}, n > 3\right\} $$
How could the domain of $f$ be extended? Specifically, how could the domain of $n$ be extended to $\mathbb{C}$? How could the domain of $y$ be extended when $x = 0$ and $n = 3$? How could the domain of $y$ be extended to $\mathbb{C}$ when $n > 3$?
In trying to accomplish the first task, I tried first extending the domain of $n$ to all integers $\mathbb{Z}$. By the definition of $f$: $$ f(x, y, 0) = f(x, f(x, y - 1, 0), -1)\\ y + 1 = f(x, y, -1)\\ $$
In other words, when $n = -1$, $f$ simply outputs $y + 1$, the same result as when $n = 0$. This process can be repeatedly applied, and it can be deduced that when $n$ is any non-positive real integer, $f$ always outputs $y + 1$.
I tried extending this result to non-integer values of $n$, but I could not find a generalized recursive definition of $f$ that is defined in terms of $n - k$, where $k$ is any complex number. ($f$ is currently defined in terms of $n - 1$.)
In trying to accomplish the second task, I defined $0^0 = 1$; this value is useful in Taylor Series expansions of polynomials, and it is the value Python uses. Trying to define $0^x$, where x is a negative number proved to be impossible because such would involve dividing by zero. Similarly, trying to define $0^i$ proved to be impossible because such would involve computing $\cos(\ln(0)) + i\sin(\ln(0))$, which is impossible.
In trying to accomplish the third task, I tried finding patterns involving tetration (the function $f(x, y, 4)$). In exponentiation (the function of $f(x, y, 3)$), it is known that $x^y * x^z = x^{y+z}$, which is how the domain of the power in exponentiation can be extended to all real numbers. I could not find such patterns for tetration, however. Because I could not find patterns for tetration, I could not find patterns for pentation, hexation, etc....
By my treatment of the hyperoperation function $f$, an extended function $f^{ext}(x, y, n)$ can be defined as follows: $$ f^{ext}(x, y, n) = \left\{\\y + 1, n < 1\\x + y, n = 1\\x * y, n = 2\\x^y, n = 3\\1, n > 2, y = 0\\f^{ext}(x,f^{ext}(x, y-1, n), n-1), n > 3, y\neq1\right\} $$
The domain of $f^{ext}$ is now as follows: $$ \left\{n\in \mathbb{Z}\\x\in \mathbb{C}\\y\in \mathbb{C}, n < 3\\y\in \mathbb{C}, n = 3, x\neq 0\\y\in \mathbb{R}, n = 3, x = 0\\y\in \mathbb{N^0}, n > 3\right\} $$
Are there ways to extend the domain of this function further than what I have done? Could this be extended like the factorial function could be extended to the gamma function while preserving its core property? For the factorial function, the core property that had to be preserved was $x! = x*(x-1)!$; for this hyperoperator function, the core property that must be preserved is $f(x, y, n) = f(x,f(x, y-1, n), n-1)$. Is this possible, and how would this be accomplished?