Some basic thoughts
If we want to summarize our operation sequence [n] we can perhaps write:
)
, where
 = b+x)
)
, 
So we have a certain operator
that assigns to a given function 
the function =t\to f^{\circ t}(1))
, this operator may be based on the natural Abel method, the diagonalization method, or the regular Abel method (with restrictions of 
) and we can write
)
, moreover
 = E^{\circ n}(x\mapsto b+x))
(x))
.
Note: I original found it more appropriate to start with the addition as 0th instead of the 1st operation. However I adapted to the already established nomenclature. If we would stick to my original counting (addition as 0th operation) we had the better looking formula(x))
As we have some methods for real functions to switch from
to non-natural iterations 
maybe there are also methods for the operator to compute 
which then gives definition for real and complex iteration ranks:
(x))
.
In the moment however it is even unclear how to generally express such operators which map powerseries.
At least we can determine that
is the inverse operator, i.e. )
is a function such that =f(g(x)))
, i.e. (x)=g(g^{-1}(x)+1))
. So is unique (independent of the method of and independent on the initial condition).
If we compute
from this view point we get also:
=E^{-1}(f_{1,b})(x)=f_{1,b}(f_{1,b}^{-1}(x)+1)=b+(x-b+1)=x+1)
and
=E^{-1}(f_{0,b})(x)=f_{0,b}(f_{0,b}^{-1}(x)+1)=(x-1+1)+1=x+1)
...
If we want to summarize our operation sequence [n] we can perhaps write:
So we have a certain operator
Note: I original found it more appropriate to start with the addition as 0th instead of the 1st operation. However I adapted to the already established nomenclature. If we would stick to my original counting (addition as 0th operation) we had the better looking formula
As we have some methods for real functions to switch from
In the moment however it is even unclear how to generally express such operators which map powerseries.
At least we can determine that
If we compute
...
