The Abel equation for the slog was =\text{slog}_b(x)+1)
.
Our original recurring for the super exponential (which I will call similarly) sexp is=b^{\text{sexp}_b(x)})
.
So if we develop it at 0, say=\sum_{n=0}^\infty \rho_n x^n)
satisfying the inverted Abel equation:
^n = \exp\left(\log(b)\sum_{n=0}^\infty \rho_n x^n\right))
we get
 x^k = 1 + \sum_{n=0}^\infty B(\log(b)\rho_1,\dots,\log(b)\rho_n) \frac{x^n}{n!})
where)
is the complete Bell polynomial.
The left side develops to)
and so we have the infinite equation system
 = B(\log(b)\rho_1,\dots,\log(b)\rho_k) \frac{1}{k!})
And I wonder if we solve it the natural way whether we get exactly the inverse of the slog (which I assume). Unfortunately there is no complete Bell polynomial in Maple (at least I didnt find it) and I am too lazy in the moment to program it myself
And yes, it is not a linear equation system. Perhaps it is despite solvable, who knows ...
Our original recurring for the super exponential (which I will call similarly) sexp is
So if we develop it at 0, say
we get
where
The left side develops to
And I wonder if we solve it the natural way whether we get exactly the inverse of the slog (which I assume). Unfortunately there is no complete Bell polynomial in Maple (at least I didnt find it) and I am too lazy in the moment to program it myself
And yes, it is not a linear equation system. Perhaps it is despite solvable, who knows ...
