doing a little bit of algebra, it seems "meta superfunctions" are expressible by nested iteration:
\( f^{\diamond n}(x) = (f^{\diamond n-1} \circ f^{\diamond n})(x-1) \)
which is obvious but it gives the relation:
\( f^{\diamond 2}(2) = (f^{\diamond 1} \circ f^{\diamond 1}) ( C ) \)
which, when plugged in with the iteration formula
\( f^{\diamond n}(x) = (f^{\diamond n-1})^{\circ x} ( C ) \)
gives
\( f^{\diamond 2}(2) = f^{\circ f^{\diamond 1} ( C )} ( C ) = f^{\circ f^{\circ C}( C )} ( C ) \)
which means
\( f^{\diamond 2}(x) = f^{\circ f^{\diamond 2}(x-1)} ( C ) \)
which to me resembles something like "compositional"/"iterational" tetration.
\( f^{\diamond 2}(x) = f^{\circ f^{\circ ...\,\, ^{\circ f ( C )}}} ( C ) \)
where there are x amount of nested iterations.
actually, the generalized formula to give nested iterations is:
\( f^{\diamond n}(x) = (f^{\diamond n-2})^{\circ f^{\diamond n}(x-1)}( C ) \)
I think I just need a bit more time to make sense of this "meta compositional" operator \( \diamond \) such that it gives a solution
\( (f^{\diamond \frac{1}{2}} \diamond f^{\diamond \frac{1}{2}})(x) = f^{\circ x}( C ) \)
which gives a recurrence relation which allows us to solve half -operators.
\( f^{\diamond n}(x) = (f^{\diamond n-1} \circ f^{\diamond n})(x-1) \)
which is obvious but it gives the relation:
\( f^{\diamond 2}(2) = (f^{\diamond 1} \circ f^{\diamond 1}) ( C ) \)
which, when plugged in with the iteration formula
\( f^{\diamond n}(x) = (f^{\diamond n-1})^{\circ x} ( C ) \)
gives
\( f^{\diamond 2}(2) = f^{\circ f^{\diamond 1} ( C )} ( C ) = f^{\circ f^{\circ C}( C )} ( C ) \)
which means
\( f^{\diamond 2}(x) = f^{\circ f^{\diamond 2}(x-1)} ( C ) \)
which to me resembles something like "compositional"/"iterational" tetration.
\( f^{\diamond 2}(x) = f^{\circ f^{\circ ...\,\, ^{\circ f ( C )}}} ( C ) \)
where there are x amount of nested iterations.
actually, the generalized formula to give nested iterations is:
\( f^{\diamond n}(x) = (f^{\diamond n-2})^{\circ f^{\diamond n}(x-1)}( C ) \)
I think I just need a bit more time to make sense of this "meta compositional" operator \( \diamond \) such that it gives a solution
\( (f^{\diamond \frac{1}{2}} \diamond f^{\diamond \frac{1}{2}})(x) = f^{\circ x}( C ) \)
which gives a recurrence relation which allows us to solve half -operators.