Extract Coefficients From A Function So I am trying to determine the average number of nodes with an even amount of children in a plane planted tree with n nodes. I created the generating function, did some manipulation, then applied LIFT (Lagrange Implicit Function Theorem) which gave me the following: $A = 2^{n-1}[u^{n-1}](\frac{1}{1-u})^n$, where $[u^{n-1}]$ denotes the coefficient of $u^{n-1}$ in the function above. So my question is... where do I go from here? Typically, these functions have just been binomial-like, so extracting the coefficient has been easy. However, I have no clue how to extract it in this case. Could anyone show me how?
I should also add that once I have this coefficient and obtain the value of A, in order to calculate the "average value", I will need to divide it by the total number of plane planted trees with n nodes, which I also have as $T= \frac{1}{n}\binom{2n-2}{n-1}$
Thanks!
 A: It’s a standard generating function:
$$\frac1{(1-x)^n}=\sum_{k\ge 0}\binom{n+k-1}kx^k\;.$$
You can prove this by induction on $n$:
$$\begin{align*}
\frac1{(1-x)^{n+1}}&=\frac1{1-x}\sum_{k\ge 0}\binom{n+k-1}kx^k\\
&=\sum_{k\ge 0}x^k\sum_{k\ge 0}\binom{n+k-1}kx^k\\
&=\sum_{k\ge 0}\sum_{i=0}^k\binom{n+i-1}ix^k\\
&=\sum_{k\ge 0}\binom{n+k}kx^k\;.
\end{align*}$$
In particular, you have
$$A = 2^{n-1}[u^{n-1}]\left(\frac{1}{1-u}\right)^n=2^{n-1}\binom{2n-2}{n-1}\;.$$
A: If you want some details, the generalized binomial theorem says that for $|x|<1$, $\alpha\in\Bbb R$
$$(1+x)^\alpha=\sum_{k=0}^\infty {\alpha\choose k}x^k$$
where $${\alpha\choose k}:=\frac{1}{k!}\prod_{m=0}^{k-1}(\alpha-m)$$
If $\alpha=-n$ we have
$$(1-x)^{-n}=\sum_{k=0}^\infty {-n\choose k}(-1)^kx^k$$ so 
$$\begin{align}
  {\left( { - 1} \right)^k}{-n\choose k} &= {\left( { - 1} \right)^k}\frac{1}{{k!}}\prod\limits_{m = 0}^{k - 1} {( - n - m)}  \cr 
   &= \frac{1}{{k!}}\prod\limits_{m = 0}^{k - 1} {\left( { - 1} \right)( - n - m)}  \cr 
   &= \frac{1}{{k!}}\prod\limits_{m = 0}^{k - 1} {(n + m)}  \cr 
   &= \frac{1}{{k!}}\prod\limits_{m = 0}^{k - 1} {(n + k - 1 - m)}  ={n+k-1\choose k}\end{align} $$
whence
$$\frac1{(1-x)^n}=\sum_{k= 0}^\infty\binom{n+k-1}kx^k\;.$$
A: Here is some enrichment  material to complete this calculation.  First
note that  these planted plane trees correspond to ordinary plane trees
with an extra node attached at the root.
The  species equation  for  these enumerating  by  the internal  nodes
(i.e. excluding the node where the tree is planted) is
$$\mathcal{T} = \mathcal{Z}\times \mathfrak{S}(\mathcal{T}).$$
This gives the functional equation
$$T(z) = z\frac{1}{1-T(z)}
\quad\text{or}\quad
z = T(z)(1-T(z)).$$
Now to extract coefficients from this by Lagrange inversion use
$$[z^n] T(z)
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)\;dz$$
and put $w=T(z)$ where $dz = 1-2w\; dw$ to obtain
$$\frac{1}{2\pi i} 
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
w \times (1-2w) \;dw$$
which is
$$\frac{1}{2\pi i} 
\int_{|w|=\epsilon} 
\left(\frac{1}{w^n (1-w)^{n+1}}
- 2\frac{1}{w^{n-1} (1-w)^{n+1}} \right) \; dw.$$
This yields
$${n-1+n\choose n} - 2{n-2+n\choose n}
= {2n-1\choose n} - 2{2n-2\choose n}$$
which is
$$\frac{2n-1}{n}{2n-2\choose n-1} 
-2\frac{n-1}{n}{2n-2\choose n-1} 
= \frac{1}{n} {2n-2\choose n-1}.$$
These are of course the Catalan numbers.
The species equation for these trees with the even outdegree marked is
$$\mathcal{Q} = 
\mathcal{Z}\times 
\mathcal{U}\mathfrak{S}_\mathrm{even}(\mathcal{Q})
+ \mathcal{Z}\times 
\mathfrak{S}_\mathrm{odd}(\mathcal{Q}).$$
This gives the functional equation
$$Q(z) = uz\frac{1}{1-Q(z)^2}
+ z\frac{Q(z)}{1-Q(z)^2}$$
or
$$Q(z)(1-Q(z)^2) = uz + z Q(z).$$
To compute the total number of even degree nodes introduce
$$G(z) = \left.\frac{\partial}{\partial u} Q(z)\right|_{u=1}.$$
Differentiate the functional equation and put $u=1$ to get
$$G(z)(1-T(z)^2) + T(z) (-2T(z) G(z))
= z + z G(z).$$
This yields
$$G(z) = \frac{z}{1-z-3T(z)^2}.$$
To extract coefficients from $G(z)$ use
$$[z^n] G(z)
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\frac{z}{1-z-3T(z)^2}\;dz.$$
Using the same substitution as before we obtain
$$\frac{1}{2\pi i} 
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
\frac{w(1-w)}{1-w(1-w)-3w^2} \times (1-2w) \;dw$$
which is
$$\frac{1}{2\pi i} 
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
\frac{w(1-w)}{(1-2w)(1+w)} \times (1-2w) \;dw
\\ = \frac{1}{2\pi i} 
\int_{|w|=\epsilon} \frac{1}{w^n (1-w)^n}
\frac{1}{1+w}\;dw
\\ = \frac{1}{2\pi i} 
\int_{|w|=\epsilon} \frac{1}{w^n (1-w)^n}
\sum_{q\ge 0} (-1)^q w^q \; dw.$$
Extracting coefficients from this we obtain
$$\sum_{q=0}^{n-1} {q+n-1\choose n-1} (-1)^{n-1-q}.$$
This gives OEIS A026641 which is
$$1, 1, 4, 13, 46, 166, 610, 2269, 8518, 32206,\ldots$$
where we find the above workings confirmed.

It follows that the average  number of even outdegree
nodes in a random rooted plane tree is given by the formula
$$ n \times  {2n-2\choose n-1}^{-1} \times
\sum_{q=0}^{n-1} {q+n-1\choose n-1} (-1)^{n-1-q}.$$
