Well this theory is structured as a plausible solution to complex hyper operators. This is a proof sketch and it requires a bit more work here and there; requires an evaluation of a limit; but there is an over all sense of plausibility.

We start by defining a sequence of analytic functions that obey the following rules:

\( \vartheta_n : \mathbb{C} \to \mathbb{C} \)

\( \vartheta_n(n) = 1 \) and \( k \in \mathbb{N}_0\,\,\,\vartheta_n(k) = 0 \,\,\Leftrightarrow\,\, k \neq n \)

Necessarily; these functions can be arbitrary and they pose the true uniqueness criterion. Nonetheless we suggest these functions and give a plausible solution.

\( {\bf I}:\,\,\,\,\vartheta_n(s) = \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\,\,\,;\,\,\,n \ge 1 \)

\( \vartheta_0(s) = e^{e^s-1} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{-\frac{s}{k}} \)

Reason for this is it has fast convergence and satisfies the required conditions.

Now we perform a trick. Considering hyper operators; which are written by the following:

\( x\,\,\bigtriangleup_{s}\,\,(x \,\,\bigtriangleup_{s+1}\,\,y) = x \,\,\bigtriangleup_{s+1}\,\,(y+1) \)

\( x \,\,\bigtriangleup_0\,\,y = x + y \)

and the identity of \( \bigtriangleup_n \) for any natural \( n \) is \( 1 \).

We get the usual sequence where zero is addition, one is multiplication, two is exponentiation, etc... We then write that complex operators are products of natural operators with complex exponents. Here; \( \vartheta_n \) is as before.

\( {\bf II}:\,\,\,\,x \,\,\bigtriangleup_s\,\, y = \prod_{n=0}^{\infty}(x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} = (x + y)^{\vartheta_0(s)} \cdot (x\cdot y)^{\vartheta_1(s)} \cdot (x^y) ^{\vartheta_2(s)} \cdot (^y x)^{\vartheta_3(s)} \cdot ... \)

It's easy to identify that at natural arguments the product converges to the natural hyperoperator it indexes.

This product; however; converges if and only if the series:

\( \sum_{n=0}^{\infty} \ln(x\,\,\bigtriangleup_n\,\,y)\vartheta_n(s) \) converges.

With this; we perform the ratio test.

\( L = \lim_{n \to \infty} |\frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) \vartheta_{n+1}(s)}{\ln(x \,\,\bigtriangleup_{n}\,\, y) \vartheta_{n}(s)}| \)

It is clear that

\( \lim_{n\to\infty} |\frac{\vartheta_{n+1}(s)}{\vartheta_{n}(s)}| = 0 \)

Without proof; we can safely assume when x and y aren't 2 or one of them 1:

\( \lim_{n \to \infty}| \frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) }| = \infty \)

The question is if they cancel each other out. Or if one takes over the other. I'm hoping it's zero and the result is entire--the \( s \) disappears in \( L \); so the equation either diverges for all \( s \) or converges for all \( s \) or inconclusive. I do not believe that \( \frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) } \) grows too too fast as n varies. So maybe \( {\bf I} \) will induce convergence. If it doesn't; we'll use a function for \( \vartheta \) that decreases to zero faster than the definition as it is now. Essentially, the true difficulty is finding a solution for \( \vartheta \) that allows for the convergence of \( {\bf II} \).

The recursive pattern becomes much simpler to find a solution for because we only have to account for instances where \( x \,\,\bigtriangleup_s\,\,y = z\,\,\in \mathbb{N}_0 \). Where of course we must have \( x\,\,\bigtriangleup_{s-1}\,\,z = x\,\,\bigtriangleup_s\,\,y+1 \). However. Most of the returned values will not be natural and for these most of our hyper operators are undefined. So the question becomes much simpler. It's almost as if we can extend the hyper operators beyond natural numbers. If you cannot get what I mean; consider:

\( x\,\,\bigtriangleup_{1.5}\,\, 2 = z \,\, \not \in \mathbb{N}_0 \)

however:

\( x\,\,\bigtriangleup_{0.5}\,\,z = x\,\,\bigtriangleup_{1.5}\,\, 3 \)

This allows us to extend operators to some real or complex numbers. Most definitely probably not all.

A brief note about the operators. They are definitely not commutative for almost all values. I haven't proved this yet however; so in the fractal chaos there may emerge a commutative operator. They are not associative for almost all values. Again; haven't proved this yet; but I think this one is pretty obvious. There may or may not exist identities for the operators. This one may be difficult to prove.

I'm mostly working on trying to prove convergence at the moment. One step at a time. Once that is proven we'll have an entire function that passes through every hyper operator at every two natural arguments. Then we tweak \( \vartheta \) to be recursive. Maybe; if possible; tweak it to have a right hand identity.

We start by defining a sequence of analytic functions that obey the following rules:

\( \vartheta_n : \mathbb{C} \to \mathbb{C} \)

\( \vartheta_n(n) = 1 \) and \( k \in \mathbb{N}_0\,\,\,\vartheta_n(k) = 0 \,\,\Leftrightarrow\,\, k \neq n \)

Necessarily; these functions can be arbitrary and they pose the true uniqueness criterion. Nonetheless we suggest these functions and give a plausible solution.

\( {\bf I}:\,\,\,\,\vartheta_n(s) = \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\,\,\,;\,\,\,n \ge 1 \)

\( \vartheta_0(s) = e^{e^s-1} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{-\frac{s}{k}} \)

Reason for this is it has fast convergence and satisfies the required conditions.

Now we perform a trick. Considering hyper operators; which are written by the following:

\( x\,\,\bigtriangleup_{s}\,\,(x \,\,\bigtriangleup_{s+1}\,\,y) = x \,\,\bigtriangleup_{s+1}\,\,(y+1) \)

\( x \,\,\bigtriangleup_0\,\,y = x + y \)

and the identity of \( \bigtriangleup_n \) for any natural \( n \) is \( 1 \).

We get the usual sequence where zero is addition, one is multiplication, two is exponentiation, etc... We then write that complex operators are products of natural operators with complex exponents. Here; \( \vartheta_n \) is as before.

\( {\bf II}:\,\,\,\,x \,\,\bigtriangleup_s\,\, y = \prod_{n=0}^{\infty}(x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} = (x + y)^{\vartheta_0(s)} \cdot (x\cdot y)^{\vartheta_1(s)} \cdot (x^y) ^{\vartheta_2(s)} \cdot (^y x)^{\vartheta_3(s)} \cdot ... \)

It's easy to identify that at natural arguments the product converges to the natural hyperoperator it indexes.

This product; however; converges if and only if the series:

\( \sum_{n=0}^{\infty} \ln(x\,\,\bigtriangleup_n\,\,y)\vartheta_n(s) \) converges.

With this; we perform the ratio test.

\( L = \lim_{n \to \infty} |\frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) \vartheta_{n+1}(s)}{\ln(x \,\,\bigtriangleup_{n}\,\, y) \vartheta_{n}(s)}| \)

It is clear that

\( \lim_{n\to\infty} |\frac{\vartheta_{n+1}(s)}{\vartheta_{n}(s)}| = 0 \)

Without proof; we can safely assume when x and y aren't 2 or one of them 1:

\( \lim_{n \to \infty}| \frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) }| = \infty \)

The question is if they cancel each other out. Or if one takes over the other. I'm hoping it's zero and the result is entire--the \( s \) disappears in \( L \); so the equation either diverges for all \( s \) or converges for all \( s \) or inconclusive. I do not believe that \( \frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) } \) grows too too fast as n varies. So maybe \( {\bf I} \) will induce convergence. If it doesn't; we'll use a function for \( \vartheta \) that decreases to zero faster than the definition as it is now. Essentially, the true difficulty is finding a solution for \( \vartheta \) that allows for the convergence of \( {\bf II} \).

The recursive pattern becomes much simpler to find a solution for because we only have to account for instances where \( x \,\,\bigtriangleup_s\,\,y = z\,\,\in \mathbb{N}_0 \). Where of course we must have \( x\,\,\bigtriangleup_{s-1}\,\,z = x\,\,\bigtriangleup_s\,\,y+1 \). However. Most of the returned values will not be natural and for these most of our hyper operators are undefined. So the question becomes much simpler. It's almost as if we can extend the hyper operators beyond natural numbers. If you cannot get what I mean; consider:

\( x\,\,\bigtriangleup_{1.5}\,\, 2 = z \,\, \not \in \mathbb{N}_0 \)

however:

\( x\,\,\bigtriangleup_{0.5}\,\,z = x\,\,\bigtriangleup_{1.5}\,\, 3 \)

This allows us to extend operators to some real or complex numbers. Most definitely probably not all.

A brief note about the operators. They are definitely not commutative for almost all values. I haven't proved this yet however; so in the fractal chaos there may emerge a commutative operator. They are not associative for almost all values. Again; haven't proved this yet; but I think this one is pretty obvious. There may or may not exist identities for the operators. This one may be difficult to prove.

I'm mostly working on trying to prove convergence at the moment. One step at a time. Once that is proven we'll have an entire function that passes through every hyper operator at every two natural arguments. Then we tweak \( \vartheta \) to be recursive. Maybe; if possible; tweak it to have a right hand identity.