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.

At page 200 of the the renowned book
"Concrete Mathematics: a foundation for computer science" R. L. Graham - D.E. Knuth - O. Patshnik
the authors analyze the properties of what they call Generalized Binomial Series
(but it seems not to be a widely accepted denomination) defined as
$$
\eqalign{
  & {\cal B}_{\,t} (z) = \sum\limits_{0\, \le \,k}
 {{{\left( {t\,k} \right)^{\,\underline {\,k - 1\,} } } \over {k!}}z^{\,k} }
  = \sum\limits_{0\, \le \,k} {{1 \over {\left( {tk - k + 1} \right)}}
\left( \matrix{  t\,k \cr   k \cr}  \right)z^{\,k} }  =   \cr 
  &  = 1 + z\sum\limits_{0\, \le \,k} {{1 \over {k + 1}}
\left( \matrix{  t\,k + t \cr   k \cr}  \right)z^{\,k} }  \cr} 
$$
These satisfy the interesting functional identity
$$
{\cal B}_{\,t} (z)^{\,1 - t}  - {\cal B}_{\,t} (z)^{\, - t}  = z\quad 
 \Leftrightarrow \quad {\cal B}_{\,t} (z) - z{\cal B}_{\,t} (z)^{\,t}  = 1
$$
and their power have a simple expression
$$
\eqalign{
  & {\cal B}_{\,t} (z)^{\,r}  = \sum\limits_{0\, \le \,k} {{r \over {t\,k + r}}
\left( \matrix{  t\,k + r \cr   k \cr}  \right)z^{\,k} } 
 = \sum\limits_{0\, \le \,k} {{{r\left( {t\,k + r - 1} \right)^{\,\underline {\,k - 1\,} } }
 \over {k!}}z^{\,k} }  =   \cr 
  &  = 1 + z\sum\limits_{0\, \le \,k} {{r \over {k + 1}}\left( \matrix{  \,t\,k + t + r - 1 \cr 
  k \cr}  \right)z^{\,k} }  = 1 + r\,z\,\sum\limits_{0\, \le \,k}
 {{{\left( {\,t\,\left( {k + 1} \right) + r - 1} \right)^{\;\underline {\,k\,} } } 
\over {\left( {k + 1} \right)!}}z^{\,k} }  \cr} 
$$
valid for all real $r$ and $t$ ( with some care in interpreting the case $tk+r=0$).
Putting $t=2$
$$
{\cal B}_{\,2} (z) - z{\cal B}_{\,2} (z)^{\,2}  = 1\quad 
\Leftrightarrow \quad {\cal B}_{\,2} (z) = {{1 - \sqrt {1 - 4z} } \over {2z}}
$$
where we have taken only the solution that is limited for $z \to 0$.
Then the coefficients you are looking for are
$$
\eqalign{
  & \left( {{{1 - \sqrt {1 - 4z} } \over 2}} \right)^{\,r}  =
 z^{\,r} {\cal B}_{\,2} (z)^{\,r}  =   \cr 
  &  = z^{\,r} \sum\limits_{0\, \le \,k} {{r \over {2\,k + r}}
\left( \matrix{  2\,k + r \cr   k \cr}  \right)z^{\,k} }  \cr} 
$$
