<s>[INST] Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\begin{equation}
\sqrt{1 - 4z} = \sum_{n \geq 0}\binom{1/2}{n}(-4 x)^n.
\end{equation}
and therefore
\begin{align*}
  f(z) = \frac{1 - \sqrt{1 - 4z}}{2} &= \frac{\sum_{n \geq 1}\binom{1/2}{n}(-4)^n z^n}{2} \\
  &= \sum_{n \geq 1}\binom{1/2}{n}(-1)^n 2^{n} z^n \\
  &= \sum_{n \geq 1}\binom{1/2}{n}(-2)^n z^n \\
  &= \sum_{n \geq 0}a_n z^n
\end{align*}
with coefficients $a_0 = 0$ and $a_n = \binom{1/2}{n}(-2)^n$.
I wanted to use cauchys integral formula for
$g(z) = f(z)^r$ to extract the $n$ coefficient
but then I get
\begin{align}
  \left( \frac{1 - \sqrt{1 - 4z}}{2}\right)^r &= \frac{1}{2^r} \left(1 - \sqrt{ 1 - 4z} \right)^r \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sqrt{1 - 4z}^m \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m (1 - 4z)^{m/2} \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sum_{k \geq 0}\binom{m/2}{k}(-1)^k 4^k z^k \\
  &= \frac{1}{2^r}\sum^{r}_{m=0} \sum_{k \geq 0}\binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k.
\end{align}
Therefore I should have
\begin{align*}
  [z^n] (f(z))^r &= [z^n] \sum^{r}_{m=0} \sum_{k \geq 0}\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k \\
  &= \sum^{r}_{m=0}[z^n] \sum_{k \geq 0}{\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k} \\
  &= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^{m + n} 4^n \\
  &= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
  &= \frac{1}{2^r} \sum^{r}_{m=0} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
  &= \frac{1}{2^r} \sum^{r}_{m=2 n} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n 
\end{align*}
For $r \geq 2n$ otherwise the coefficient vanishes.
I would be thankful if anyone can give me a hint.
 [/INST] Too long for a comment, but not a full flesh answer.
$f(z)$ can be considered as the root of this quadratic equation:
$$f(z)^2-f(z)+z=0,\tag{1}$$
Let us multiply (1) by $f(z)$, giving:
$$f(z)^3-f(z)^2+zf(z)=0\tag{2}$$
Adding (1) and (2) gives an expression for $f(z)^3$.
More generally, iterating $r-2$ times the process of multiplication by $f(z)$ the previous equation gives by successive cancellations, and using the formula for the sum of a finite geometric series, the following recurrence formula:
$$f(z)^r=f(z)-z\dfrac{1-f(z)^{r-1}}{1-f(z)},$$
which can be transformed, due to relationship $1-f(z)=\dfrac{4z}{1-\sqrt{1-4z}}$ into
$$f(z)^r=f(z)-(1-\sqrt{1-4z})\frac14 (1-f(z)^{r-1}),$$
a formula that looks tractable for a recursive computation of the coefficients.
</s>