claims to give series reversion/Lagrange inversion coefficients recursively for:
$$d_n = \frac{d^{n-1}}{dy^{n-1}}\Big( \frac{y}{5/2(e^{-y}-1)+e^{-y}(2y+y^2/2)} \Big)^n\Big|_{y=0} = (n-1)! \, [y^{n-1}] \Big(\frac{5}{2y}(e^{-y}-1)+e^{-y}(2+y/2)\Big)^{-n}$$ It has been written this way because the portion raised to the $-n$ power is now an ordinary power series, and everyone knows (or should know) how to calculate the coefficients recursively. That is, given $$ \Big( \sum_{k=0}^\infty a_k \,x^{k} \Big)^{-n} = \sum_{m=0}^\infty c_m \, x^m, $$ with non-zero $a_0,$ the coefficients $c_m$ are found by $$ c_0=a_0^{-n}$$ $$(*) \quad c_m=\frac{1}{m a_0} \, \sum_{k=1}^m \big( k(1-n)-m\big)a_k c_{m-k} $$ Once the $c_n$ have been calculated, don't forget the multiplication by $(n-1)!$ to get the $d_n.$
which implies $d_n$, or many Lagrange inversion coefficients, can be directly recursively defined using $(*)$. One can expand $f(y)$ as a series and use $(*)$ for a recursion of its series to the $n$th power, but we are trying to instead trying to find series coefficients of $g(y)=f(y)^{-n}$, so it is unclear how $\Big( \sum\limits_{k=0}^\infty a_k \,x^{k} \Big)^{-n} = \sum\limits_{m=0}^\infty c_m \, x^m$ is used. Can someone please complete the derivation in the blockquote/answer to provide a recursion for Lagrange inversion coefficients?
