Generating function for planted planar trees

Question

I need your help to solve this problem :

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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 $$\displaystyle \frac{\binom{3k}{k}}{(2k+1)}$$

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Please help !

Answer 1

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.