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.
 A: 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.
A: 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
$$
\eqalign{
  & \left( {{{1 - \sqrt {1 - 4z} } \over 2}} \right)^{\,r}  =
 z^{\,r} {\cal B}_{\,2} (z)^{\,r}  =   \cr 
  &  = z^{\,r} \sum\limits_{0\, \le \,k} {{r \over {2\,k + r}}
\left( \matrix{  2\,k + r \cr   k \cr}  \right)z^{\,k} }  \cr} 
$$
A: We seek
$$[z^k] \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
where $k\ge r$ and we get zero otherwise since $\frac{1-\sqrt{1-4z}}{2}
= z + \cdots.$
Using the residue operator this is
$$\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{k+1}} \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
Now put $\frac{1-\sqrt{1-4z}}{2} = w$ so that $z = w(1-w)$ and $dz\; =
(1-2w) \; dw$ to get
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k+1} (1-w)^{k+1}} w^r (1-2w)
= \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k-r+1} (1-w)^{k+1}} (1-2w) 
\\ = {2k-r\choose k} - 2 {2k-r-1\choose k}
= {2k-r\choose k} - 2 \frac{k-r}{2k-r} {2k-r\choose k}
\\ = \frac{r}{2k-r} {2k-r\choose k}.$$
Following the other post we may write
$$\sum_{k\ge r} \frac{r}{2k-r} {2k-r\choose k} z^k
= z^r \sum_{k\ge 0} \frac{r}{2k+r} {2k+r\choose k} z^k.$$
The residue operator returns zero when $k\lt r.$
