Just want to throw in - because Leo.W seems to make quite some stretches to get a formula for the coefficients -
there is a well-known formula for the coefficients of the t-iteration of a function/formal powerseries with fixed point multiplier 1 by Jabotinsky:
\[ {f^{\circ t}}_1=1, \quad {f^{\circ t}}_n=\sum_{m=0}^{n-1} \binom{t}{m} \sum_{k=0}^m \binom{m}{k} (-1)^{m-k}
{f^{\circ k}}_{n} \]
Just plug in t=1/2 and f(x) = e^x - 1 (as powerseries of course)
There is also a recurrence that you can directly achieve from comparing the formal powerseries \(f\circ f^{\circ t} = f^{\circ t}\circ f\), which I don't have handy though and which typically does not help much either
(but is faster in terms computation).
there is a well-known formula for the coefficients of the t-iteration of a function/formal powerseries with fixed point multiplier 1 by Jabotinsky:
\[ {f^{\circ t}}_1=1, \quad {f^{\circ t}}_n=\sum_{m=0}^{n-1} \binom{t}{m} \sum_{k=0}^m \binom{m}{k} (-1)^{m-k}
{f^{\circ k}}_{n} \]
Just plug in t=1/2 and f(x) = e^x - 1 (as powerseries of course)
There is also a recurrence that you can directly achieve from comparing the formal powerseries \(f\circ f^{\circ t} = f^{\circ t}\circ f\), which I don't have handy though and which typically does not help much either
(but is faster in terms computation).
