@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.

is given by

where the coefficients satisfy the recurrence

where is the nth coefficient of the mth exponentiation (NOT iteration!) of (for iteration, it would be , and is too ambiguous here to be used), i.e. . For , i.e. the decremented exponential, we get, using the binomial theorem (here, is a Stirling number of the 2nd kind),

.

So, the recurrence is

,

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 ? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function , which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired 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 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 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.

is given by

where the coefficients satisfy the recurrence

where is the nth coefficient of the mth exponentiation (NOT iteration!) of (for iteration, it would be , and is too ambiguous here to be used), i.e. . For , i.e. the decremented exponential, we get, using the binomial theorem (here, is a Stirling number of the 2nd kind),

.

So, the recurrence is

,

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 ? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function , which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired 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 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 once it is developed.