TITLE: Generating function for planted planar trees
QUESTION [5 upvotes]: I need your help to solve this problem :


Give a generating function for planted planar trees with all degrees odd.
  Show that the number of such trees with $2k+1$ non-root vertices is 
  $$\displaystyle \frac{\binom{3k}{k}}{(2k+1)}$$


Please help !

REPLY [4 votes]: Start by  computing the generating  function for the odd  total degree
trees  without the  degree-one  root node.  There is an even number of children plus a link to the parent for a total odd degree. This  has the  functional
equation 
$$T(z) = z + z \frac{T(z)^2}{1-T(z)^2}.$$
Now recall the Lagrange Inversion Formula as presented e.g. at this
MathWorld link
(consulted May 26 2014.)
Using their variables we have $w=z$, $\alpha=z$ and
$$\phi(w) = \frac{w^2}{1-w^2}.$$
Setting $F(v) = v$ in the above we obtain
$$T(z) = z + \frac{z}{1} \frac{z^2}{1-z^2}
+ \frac{z^2}{2!} \frac{\partial}{\partial z} 
\left(\frac{z^2}{1-z^2}\right)^2
+ \frac{z^3}{3!} \frac{\partial^2}{\partial z^2} 
\left(\frac{z^2}{1-z^2}\right)^3
+ \cdots.$$
Recall that
$$\left(\frac{z}{1-z}\right)^q
= \sum_{n\ge q} {n-1\choose q-1} z^n$$
so that
$$\left(\frac{z^2}{1-z^2}\right)^q
= \sum_{n\ge q} {n-1\choose q-1} z^{2n}.$$
This implies that
$$\frac{\partial^{q-1}}{\partial z^{q-1}} 
\left(\frac{z^2}{1-z^2}\right)^q
= \sum_{n\ge q} (2n)^{\underline{q-1}} 
{n-1\choose q-1} z^{2n-(q-1)}.$$
This in turn yields 
$$[z^m] \frac{z^q}{q!}\frac{\partial^{q-1}}{\partial z^{q-1}} 
\left(\frac{z^2}{1-z^2}\right)^q
\\= [z^m] \frac{z^q}{q!}
\sum_{n\ge q} (2n)^{\underline{q-1}} {n-1\choose q-1} z^{2n+1-q}
\\ = \frac{1}{q!} [z^{m-q}]
\sum_{n\ge q} (2n)^{\underline{q-1}} {n-1\choose q-1} z^{2n+1-q}.$$
If $m$ is odd this gives  $2n+1=m$ and we must have $(m-1)/2\ge q$ for
a contribution of
$$\frac{1}{q!} (m-1)^{\underline{q-1}}{m/2-3/2\choose q-1}.$$
Collecting everything into one formula  we obtain for $m\ge 3$ and $m$
odd the closed form 
$$1 + \sum_{q=2}^{m/2-1/2} \frac{1}{q!}
(m-1)^{\underline{q-1}}{m/2-3/2\choose q-1}$$
which is
$$1 + \sum_{q=2}^{m/2-1/2} \frac{1}{q!}
\frac{(m-1)!}{(m-q)!}
{m/2-3/2\choose q-1}$$
or equivalently
$$1 + \frac{1}{m}
\sum_{q=2}^{m/2-1/2} {m\choose q} {m/2-3/2\choose q-1}.$$
Simplifying this to
$$\frac{1}{m}{3m/2-3/2\choose m/2-1/2}$$
is best done with a CAS and Zeilberger's algorithm /
Sister Celine's method.
This is the following sequence
$$ 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, \ldots$$
which is OEIS A001764 where 
a variety of relevant information can be found.
Addendum Tue May 27 22:12:28 CEST 2014.
Here is some additional material  to document how to obtain the simple
closed form.

Note that we can absorb the term in front into the sum, turning
$$1 + \frac{1}{m}
\sum_{q=2}^{m/2-1/2} {m\choose q} {m/2-3/2\choose q-1}$$
into
$$\frac{1}{m}
\sum_{q=1}^{m/2-1/2} {m\choose q} {m/2-3/2\choose q-1}.$$
Now since $m$ is odd put $m=2k+1$ to get
$$\frac{1}{2k+1}
\sum_{q=1}^k {2k+1\choose q} {k-1\choose q-1}.$$
It turns out that Maple has an implementation of creative telescoping,
the method we cited above. In the present case it yields the following
data:

> restart;
> with(SumTools[DefiniteSum]):
> CreativeTelescoping(binomial(2*k+1,q)*binomial(k-1,q-1),q=1..k);
                       1/2   k
                      3    27  GAMMA(k + 1/3) GAMMA(k + 2/3)
                  1/2 --------------------------------------
                                Pi GAMMA(2 k + 1)

Now using the multiplication theorem of the
Gamma function
we have that
$$\Gamma(k+1/3)\Gamma(k+2/3)
=\frac{2\pi\times 3^{1/2-3k}}{\Gamma(k)} \Gamma(3k).$$
Substituting this into the closed form for the sum yields
$$\frac{1}{2\pi}
\frac{3^{1/2+3k}}{\Gamma(2k+1)}
\frac{2\pi\times 3^{1/2-3k}}{\Gamma(k)} \Gamma(3k)
= \frac{3k\Gamma(3k)}{k\Gamma(k)\Gamma(2k+1)}.$$
It follows that the closed form for the number of trees is
$$\frac{1}{2k+1}
\frac{3k\Gamma(3k)}{k\Gamma(k)\Gamma(2k+1)}
= \frac{(3k)!}{(k!)\times (2k+1)!}
= \frac{1}{2k+1} \frac{(3k)!}{(k!)\times (2k)!}
= \frac{1}{2k+1} {3k\choose k}.$$
Note that the  method of creative telescoping provides  sound proof of
the closed forms  of the sums that it is applied to as explained e.g. by
Wilf in his book generatingfunctionology.
Addendum Tue May 15 2018. Starting from the functional equation
$$T(z) = z + z \frac{T(z)^2}{1-T(z)}$$
we may solve for $z$ to get
$$z = \frac{T(z)}{1/(1-T(z)^2)}$$
and we recognize that Lagrange-Burmann applies with
$$[z^m] T(z) = \frac{1}{m} [w^{m-1}] \frac{1}{(1-w^2)^m}.$$
With $m=2k+1$ this becomes
$$\frac{1}{2k+1} [w^{2k}] \frac{1}{(1-w^2)^{2k+1}}
= \frac{1}{2k+1} [w^{k}] \frac{1}{(1-w)^{2k+1}}
\\ = \frac{1}{2k+1} {2k+k\choose 2k} = \frac{1}{2k+1} {3k\choose 2k}.$$
This is the claim. 
Concerning the evaluation of the sum form we find
that
$$\frac{1}{2k+1} \sum_{q=1}^k {2k+1\choose q} {k-1\choose k-q}
\\ = \frac{1}{2k+1} \sum_{q=1}^k {2k+1\choose q}
[z^{k-q}] (1+z)^{k-1}
\\ = \frac{1}{2k+1} [z^k] (1+z)^{k-1}
\sum_{q=1}^k {2k+1\choose q} z^q.$$
Now with  $k\ge 1$ we  may add  in the term  for $q=0$ with  no change
since $[z^k] (1+z)^{k-1} = 0.$ We may extend the upper limit to $2k+1$
as there  is no contribution  to the coefficient extractor  when $q\gt
k,$ getting
$$\frac{1}{2k+1} [z^k] (1+z)^{k-1}
\sum_{q=0}^{2k+1} {2k+1\choose q} z^q
= \frac{1}{2k+1} [z^k] (1+z)^{k-1} (1+z)^{2k+1}
= \frac{1}{2k+1} [z^k] (1+z)^{3k}
= \frac{1}{2k+1} {3k\choose k}.$$
This is the claim as before.