@Gottfried:
That's interesting, your matrix seems to be related to the iteration of the decremented exponential.
I'm thinking that perhaps that would be a better line of attack here. The regular Schroder function for the fixed point at 0 of a function with such a fixed point, i.e.
\( f(z) = \sum_{n=1}^{\infty} f_n z^n \)
is given by
\( \chi(z) = \sum_{n=1}^{\infty} \chi_n z^n \)
where the coefficients \( \chi_n \) satisfy the recurrence
\( \chi_1 = 1 \)
\( \chi_n = \frac{1}{a_1 - a_1^n} \sum_{m=1}^{n-1} \chi_m {f^{\cdot m}}_n \)
where \( {f^{\cdot m}}_n \) is the nth coefficient of the mth exponentiation (NOT iteration!) of \( f \) (for iteration, it would be \( {f^{\circ m}}_n \), and \( {f^m}_n \) is too ambiguous here to be used), i.e. \( {f^{\cdot m}}_n = \frac{1}{n!} \left[\frac{d^n}{dx^n} f(x)^m\right]_{x = 0} \). For \( f(z) = e^{uz} - 1 \), i.e. the decremented exponential, we get, using the binomial theorem (here, \( \left\{{n \atop m}\right\} \) is a Stirling number of the 2nd kind),
\( {f^{\cdot m}}_n = u^n \frac{m!}{n!} \left\{{n \atop m}\right\} \).
So, the recurrence is
\( \chi_n = \frac{u^{n-1}}{1 - u^{n-1}} \sum_{m=1}^{n-1} \frac{m!}{n!} \left\{{n \atop m}\right\} \chi_m \),
which looks to be much more "linear" than the weird Bell-polynomials recurrence. Could it be possible to come up with an explicit formula for the \( \chi_n \)? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function \( \chi^{-1} \), which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired \( UT \) function (although it may be offset from the original, but at least we would have a formula with explicit, non-recursive coefficients).
ADD: I tried some expansions of those \( \chi_n \) coefficients. I noticed that the numerators seemed to contain coefficients that look like the second column of your matrices (the -1, -5, -6, -5 thing). Could there be some simpler way to transform that second column into the last column? If so, then we might be able to extract a simpler formula from the explicit formula for \( \chi_n \) once it is developed.
That's interesting, your matrix seems to be related to the iteration of the decremented exponential.
I'm thinking that perhaps that would be a better line of attack here. The regular Schroder function for the fixed point at 0 of a function with such a fixed point, i.e.
\( f(z) = \sum_{n=1}^{\infty} f_n z^n \)
is given by
\( \chi(z) = \sum_{n=1}^{\infty} \chi_n z^n \)
where the coefficients \( \chi_n \) satisfy the recurrence
\( \chi_1 = 1 \)
\( \chi_n = \frac{1}{a_1 - a_1^n} \sum_{m=1}^{n-1} \chi_m {f^{\cdot m}}_n \)
where \( {f^{\cdot m}}_n \) is the nth coefficient of the mth exponentiation (NOT iteration!) of \( f \) (for iteration, it would be \( {f^{\circ m}}_n \), and \( {f^m}_n \) is too ambiguous here to be used), i.e. \( {f^{\cdot m}}_n = \frac{1}{n!} \left[\frac{d^n}{dx^n} f(x)^m\right]_{x = 0} \). For \( f(z) = e^{uz} - 1 \), i.e. the decremented exponential, we get, using the binomial theorem (here, \( \left\{{n \atop m}\right\} \) is a Stirling number of the 2nd kind),
\( {f^{\cdot m}}_n = u^n \frac{m!}{n!} \left\{{n \atop m}\right\} \).
So, the recurrence is
\( \chi_n = \frac{u^{n-1}}{1 - u^{n-1}} \sum_{m=1}^{n-1} \frac{m!}{n!} \left\{{n \atop m}\right\} \chi_m \),
which looks to be much more "linear" than the weird Bell-polynomials recurrence. Could it be possible to come up with an explicit formula for the \( \chi_n \)? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function \( \chi^{-1} \), which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired \( UT \) function (although it may be offset from the original, but at least we would have a formula with explicit, non-recursive coefficients).
ADD: I tried some expansions of those \( \chi_n \) coefficients. I noticed that the numerators seemed to contain coefficients that look like the second column of your matrices (the -1, -5, -6, -5 thing). Could there be some simpler way to transform that second column into the last column? If so, then we might be able to extract a simpler formula from the explicit formula for \( \chi_n \) once it is developed.
