Well there's an obvious connection between the Ackermann function and superfunctions, but now I've found a way to generalize this to what I am labeling "meta-superfunctions". These have to do with iteration but in a very different complicated light.

consider the definition:

\( f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x \)

\( f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1) \)

or so that

\( f^{\diamond n + 1}(x) \) is the super function of \( f^{\diamond n}(x) \)

and

\( g^{\diamond n + 1}(x) \) is the abel function of \( f^{\diamond n}(x) \)

this gives the relation:

\( f^{\diamond n + 1}(x) = (f^{\diamond n} \circ f^{\diamond n} \circ ...\text{x amount of times}... \circ f^{\diamond n}) ( C ) = (f^{\diamond n})^{\circ x} ( C ) \)

for some arbitrary constant C, that gives no singularities and such.

we'd center the equation at zero, so that:

\( f^{\diamond 0}(x) = f(x) \)

\( g^{\diamond 0}(x) = f^{\circ -1}(x) \)

the question I ask is, how do we find fractional values of n? Or put more linguistically, what function is in between a function and its super function?

The connection to the Ackermann function is simple:

\( \vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b \)

\( L_a^{\sigma}(\vartheta(a,b,\sigma)) = b \)

so that we get

\( \vartheta(a,L_a^{\sigma + 1}(b) + 1, \sigma + 1) = \vartheta(a,b,\sigma) \)

which is a simple exercise in super functions

now to split confusion we'll rewrite this in "meta superfunction" notation, fixing a to some arbitrary constant

\( f^{\diamond \sigma}(b) = \vartheta(a,b,\sigma) \)

\( g^{\diamond \sigma}(b) = L_a^{\sigma}(b) \)

we get the simple result from out previous formula above

\( f^{\diamond \sigma}(b) = f^{\diamond \sigma + 1}(g^{\diamond \sigma + 1}(b) + 1) \)

and this defines the Ackermann function in terms of "meta superfunctions"

Now, the difficulty with defining fractional values of these "meta superfunctions" is that we do not have a recurrence relation for fractional values... The difference between that and regular superfunctions is that we do, namely

\( f^{\circ \frac{1}{2}}(f^{\circ \frac{1}{2}}(x)) = f(x) \)

there is no such property or relation for "meta superfunctions"

the only property we know it must satisfy is:

\( f^{\diamond \frac{1}{2}}(b) = f^{\diamond \frac{3}{2}}(g^{\diamond \frac{3}{2}}(b) + 1) \)

but this is no good because it doesn't give an expression in terms of an even integer superfunction. So my question is, are meta-superfunctions by nature discrete? and inconclusively continuous? Since there is no external property they need satisfy other than maybe

\( R(x) = f^{\diamond x}(b) \)

with R(x) being analytic across x.

So, I believe, before the Ackermann function can be continued analytically we must define a recurrence relation across meta superfunctions that applies to fractional values.

My rather meagre attempt at this recurrence relation comes from defining a new operator. I call it "meta composition". It's rather vague and does not have a concrete definition, but it's the best I can come up with.

\( (f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x) \)

this would mean that the identity of meta composition is the function itself

\( (f \diamond f^{\diamond n})(x) = f^{\diamond n}(x) \)

which implies idempotency and also that \( \diamond \) is defined relatively based on each function. Actually, \( \diamond \) is very much like the composition operator \( \circ \) except it is reformulated with \( f \) as the identity function rather than the normal identity function \( f(x) = x \)

Actually thinking about it, I'm sure I've read somewhere of composition operators redefined to suit different identity functions. If this was the case this could perhaps be an approach to a solution for analytic fractional operators!

So I'm wondering, is this just absurd? It's really the best I can come up with...

any other possible "meta superfunction" recurrence relations would be welcomed with a warm heart!

consider the definition:

\( f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x \)

\( f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1) \)

or so that

\( f^{\diamond n + 1}(x) \) is the super function of \( f^{\diamond n}(x) \)

and

\( g^{\diamond n + 1}(x) \) is the abel function of \( f^{\diamond n}(x) \)

this gives the relation:

\( f^{\diamond n + 1}(x) = (f^{\diamond n} \circ f^{\diamond n} \circ ...\text{x amount of times}... \circ f^{\diamond n}) ( C ) = (f^{\diamond n})^{\circ x} ( C ) \)

for some arbitrary constant C, that gives no singularities and such.

we'd center the equation at zero, so that:

\( f^{\diamond 0}(x) = f(x) \)

\( g^{\diamond 0}(x) = f^{\circ -1}(x) \)

the question I ask is, how do we find fractional values of n? Or put more linguistically, what function is in between a function and its super function?

The connection to the Ackermann function is simple:

\( \vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b \)

\( L_a^{\sigma}(\vartheta(a,b,\sigma)) = b \)

so that we get

\( \vartheta(a,L_a^{\sigma + 1}(b) + 1, \sigma + 1) = \vartheta(a,b,\sigma) \)

which is a simple exercise in super functions

now to split confusion we'll rewrite this in "meta superfunction" notation, fixing a to some arbitrary constant

\( f^{\diamond \sigma}(b) = \vartheta(a,b,\sigma) \)

\( g^{\diamond \sigma}(b) = L_a^{\sigma}(b) \)

we get the simple result from out previous formula above

\( f^{\diamond \sigma}(b) = f^{\diamond \sigma + 1}(g^{\diamond \sigma + 1}(b) + 1) \)

and this defines the Ackermann function in terms of "meta superfunctions"

Now, the difficulty with defining fractional values of these "meta superfunctions" is that we do not have a recurrence relation for fractional values... The difference between that and regular superfunctions is that we do, namely

\( f^{\circ \frac{1}{2}}(f^{\circ \frac{1}{2}}(x)) = f(x) \)

there is no such property or relation for "meta superfunctions"

the only property we know it must satisfy is:

\( f^{\diamond \frac{1}{2}}(b) = f^{\diamond \frac{3}{2}}(g^{\diamond \frac{3}{2}}(b) + 1) \)

but this is no good because it doesn't give an expression in terms of an even integer superfunction. So my question is, are meta-superfunctions by nature discrete? and inconclusively continuous? Since there is no external property they need satisfy other than maybe

\( R(x) = f^{\diamond x}(b) \)

with R(x) being analytic across x.

So, I believe, before the Ackermann function can be continued analytically we must define a recurrence relation across meta superfunctions that applies to fractional values.

My rather meagre attempt at this recurrence relation comes from defining a new operator. I call it "meta composition". It's rather vague and does not have a concrete definition, but it's the best I can come up with.

\( (f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x) \)

this would mean that the identity of meta composition is the function itself

\( (f \diamond f^{\diamond n})(x) = f^{\diamond n}(x) \)

which implies idempotency and also that \( \diamond \) is defined relatively based on each function. Actually, \( \diamond \) is very much like the composition operator \( \circ \) except it is reformulated with \( f \) as the identity function rather than the normal identity function \( f(x) = x \)

Actually thinking about it, I'm sure I've read somewhere of composition operators redefined to suit different identity functions. If this was the case this could perhaps be an approach to a solution for analytic fractional operators!

So I'm wondering, is this just absurd? It's really the best I can come up with...

any other possible "meta superfunction" recurrence relations would be welcomed with a warm heart!