I need you help to solve this problem:
Consider a planar tree with $n$ non-root vertices.
- Give a generating function for vertices distance $d$ from the root.
- Proof that the total number is $$\displaystyle \binom{2n}{n-d}\frac{2d+1}{(n+d+1)}$$
We are supposed to have an exponential generating function then use Lagrange Inversion Theorem...
What we want here is ordinary generating functions since we are working with unlabeled trees. Start by computing the generating function $T_n(z)$ for ordered planar trees of at most a given height $n$ classified according to the nodes of a distance $d$ from the root (variable $u_d$) and the total number of nodes (variable $z$), where the singleton node has height zero, to get
$$T_0(z) = u_0 z \quad\text{and}\quad T_n(z) = u_0 z + u_0 z \times \left.\frac{T_{n-1}(z)}{1-T_{n-1}(z)} \right|_{u_0=u_1, u_1=u_2,\ldots}.$$
It then follows that the generating function $G_d(z)$ of the number of nodes of a distance $d$ from the root in all ordered planar graphs on $n$ nodes is given by $$G_d(z) = \left.\frac{\partial}{\partial v} T_d(z)(u_0=1, u_1=1, \ldots u_{d-1}=1, u_d = v\times Q(z)/z) \right|_{v=1}.$$
Here $Q(z)$ generates ordered trees on $n$ nodes and has the functional equation $$Q(z) = z + z\frac{Q(z)}{1 - Q(z)} = z \frac{1}{1-Q(z)}.$$ Re-write this as $$Q(z) = z + Q(z)^2.$$
We use $Q(z)/z$ in the substitution because when we attach a tree at a node of depth $d$ we get two overlapping nodes, one of which we remove. Working with this we discover that it encodes too much information. There is a closed form of $T_n(z)$ but many of its properties can only be conjectured, requiring considerable algebra to prove. We see that we can in fact skip $T_d(z)$ and work directly with a corresponding generating function $F_d(z)$ where only leaves of a distance $d$ are marked and we do not need to shift markers from $u_f$ to $u_{f+1}$ in the recursive step. We thus obtain from first principles the recurrence
$$F_0(z) = vz \quad\text{and}\quad F_d(z) = z + z \frac{F_{d-1}(z)}{1-F_{d-1}(z)} = z \frac{1}{1-F_{d-1}(z)}.$$
We now claim that for $d\ge 2$ we have
$$F_d(z) = z\frac{(-1)^d P_{d-1}(z) + vz P_{d-2}(z)} {P_d(z) - (-1)^d vz P_{d-1}(z)}$$
where (these are closely related to Fibonacci polynomials)
$$P_d(z) = \begin{cases} 1 & \quad\text{if}\quad d=0 \\ -1 & \quad\text{if}\quad d=1 \\ (-1)^d P_{d-1}(z) + z P_{d-2}(z) & \quad\text{if}\quad d\ge 2. \end{cases}$$
The proof is by induction. We have by the combinatorial recurrence $F_1(z) = \frac{z}{1-vz}$ and hence $F_2(z) = z\frac{vz-1}{z-1+vz}.$ The formula in terms of the $P_d(z)$ yields
$$F_2(z) = z \frac{P_1(z)+vzP_0(z)}{P_2(z)-vzP_1(z)} = z \frac{-1+vz}{z-1-vz\times -1}$$
and the base case holds. For the induction step we find
$$z \frac{1}{1-F_{d}(z)} = z \frac{1}{1-z\frac{(-1)^d P_{d-1}(z) + vz P_{d-2}(z)} {P_d(z) - (-1)^d vz P_{d-1}(z)}} \\ = z \frac{P_d(z) - (-1)^d vz P_{d-1}(z)} {P_d(z) - (-1)^d vz P_{d-1}(z)-z(-1)^d P_{d-1}(z) - vz^2 P_{d-2}(z)} \\ = z \frac{P_d(z) - (-1)^d vz P_{d-1}(z)} {P_d(z) - vzP_d(z) -z(-1)^d P_{d-1}(z)} \\ = z \frac{(-1)^{d+1} P_d(z) + vz P_{d-1}(z)} {(-1)^{d+1} P_d(z) - (-1)^{d+1} vzP_d(z) + z P_{d-1}(z)} \\ = z \frac{(-1)^{d+1} P_d(z) + vz P_{d-1}(z)} {P_{d+1}(z) - (-1)^{d+1} vzP_d(z)}.$$
This is the formula for $F_{d+1}(z)$ and the induction is complete. We attach $Q(z)/z$ at $v$ as explained earlier to obtain
$$z\frac{(-1)^d P_{d-1}(z) + v Q(z) P_{d-2}(z)} {P_d(z) - (-1)^d v Q(z) P_{d-1}(z)}.$$
Differentiating with respect to $v$ we get for the denominator
$$(P_d(z) - (-1)^d v Q(z) P_{d-1}(z))^2$$
and for the numerator (scalar $z$ will be restored at end)
$$Q(z) P_{d-2}(z) (P_d(z) - (-1)^d v Q(z) P_{d-1}(z)) \\ - ((-1)^d P_{d-1}(z) + v Q(z) P_{d-2}(z)) (-1)^{d+1} Q(z) P_{d-1}(z) \\ = Q(z) P_d(z) P_{d-2}(z) + Q(z) P_{d-1}(z)^2.$$
We claim that
$$P_d(z) P_{d-2}(z) + P_{d-1}(z)^2 = z^{d-1}.$$
For an induction proof we get with $d=2$
$$(z-1) \times 1 + (-1)^2 = z$$
and the base case goes through. For the induction step we start with
$$P_d(z) (P_d(z)/z - (-1)^d P_{d-1}(z)/z) + P_{d-1}(z)^2 = z^{d-1}$$
or
$$P_d(z)^2 - (-1)^d P_{d-1}(z) P_d(z) + z P_{d-1}(z)^2 = z^d$$
which is
$$P_d(z)^2 + P_{d-1}(z) ((-1)^{d+1} P_d(z) + z P_{d-1}(z)) = P_d(z)^2 + P_{d-1}(z) P_{d+1}(z)$$
and the induction goes through. We have shown that
$$G_d(z) = \frac{z^{d} Q(z)}{(P_d(z) - (-1)^d Q(z) P_{d-1}(z))^2}.$$
We prepare for Lagrange Inversion to conclude this computation. We have with $w=Q(z)$ that $z=w(1-w)$ and we now claim that
$$(P_d(w(1-w)) - (-1)^d w P_{d-1}(w(1-w)))^2 = (1-w)^{2d}.$$
Proof is by induction one more time. We have with $d=1$ that
$$(-1 - (-1) \times w \times 1)^2 = (-1+w)^2$$
and the base case holds. For the induction step re-write the second inner term on the LHS as
$$(-1)^{d+1} \frac{1}{1-w} (P_{d+1}(w(1-w)) -(-1)^{d+1} P_{d}(w(1-w)))$$
Add the first inner term to get
$$(-1)^{d+1} \frac{1}{1-w} (P_{d+1}(w(1-w)) -(-1)^{d+1} P_{d}(w(1-w)) \\ + (-1)^{d+1} P_d(w(1-w)) (1-w)) \\ = (-1)^{d+1} \frac{1}{1-w} (P_{d+1}(w(1-w)) + (-1)^{d+1} P_d(w(1-w)) (-w)).$$
Square to get
$$\frac{1}{(1-w)^2} (P_{d+1}(w(1-w)) - (-1)^{d+1} w P_d(w(1-w)))^2.$$
This concludes the induction. For Lagrange Inversion integral we thus obtain
$$[z^m] G_d(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} G_d(z) \; dz.$$
We have $dz = (1-2w) \; dw$ and moreover $Q(z) = z + \cdots$ hence we are justified in writing
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1} (1-w)^{m+1}} \frac{w^d (1-w)^d \times w}{(1-w)^{2d}} (1-2w) \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m-d} (1-w)^{m+1+d}} (1-2w) \; dw.$$
This evaluates by inspection to
$${m-d-1+m+d\choose m+d} - 2 {m-d-2+m+d\choose m+d} \\ = {2m-1\choose m+d} - 2 {2m-2\choose m+d}.$$
OP asks for non-root vertices and thus $n=m-1$ and we find
$${2n+1\choose n+1+d} - 2 {2n\choose n+1+d} \\ = \left(\frac{2n+1}{n+1+d} - 2 \frac{n-d}{n+1+d} \right) {2n\choose n+d}.$$
This is indeed
$$\bbox[5px,border:2px solid #00A000]{ \frac{2d+1}{n+1+d} {2n\choose n-d}}$$
as claimed.
The relation to Fibonacci polynomials is readily apparent through the fact that
$$P_d(z) = (-1)^{\lfloor (d+3)/2 \rfloor} \sum_{k=0}^{\lfloor d/2 \rfloor} {d-k\choose k} (-1)^{k+1} z^k$$
The proof is again by induction. We immediately get $P_0(z) = -1 \times - 1 = 1$ and $P_1(z) = 1 \times -1 = -1$ for the base case, which is correct. The induction step is split into a case-by-case analysis modulo four, of which we do one example, namely $d=4q+3.$ We find
$$P_{4q+3}(z) = - P_{4q+2}(z) + z P_{4q+1}(z).$$
This becomes
$$- \sum_{k=0}^{2q+1} {4q+3-k\choose k} (-1)^{k+1} z^k = - \sum_{k=0}^{2q+1} {4q+2-k\choose k} (-1)^{k+1} z^k \\ + z \sum_{k=0}^{2q} {4q+1-k\choose k} (-1)^{k+1} z^k$$
or
$$- \sum_{k=0}^{2q+1} {4q+3-k\choose k} (-1)^{k+1} z^k = - \sum_{k=0}^{2q+1} {4q+2-k\choose k} (-1)^{k+1} z^k \\ + \sum_{k=1}^{2q+1} {4q+2-k\choose k-1} (-1)^{k} z^k.$$
This goes through as claimed since the constant coefficient on $z$ is minus one on both sides and for the coefficients on $[z^k]$ with $1\le k\le 2q+1$ we get (attention to $(-1)^k$)
$$ - {4q+2-k\choose k} - {4q+2-k\choose k-1} = {4q+3-k\choose k} \left( - \frac{4q+3-2k}{4q+3-k} - \frac{k}{4q+3-k} \right) \\ = - {4q+3-k\choose k}.$$
To be rigorous in the induction we have to process $4q+2, 4q+3, 4q$ and $4q+1$ in this order.
Another interesting fact that relates $P_d(z)$ to Fibonacci numbers is that
$$P_d(-1) = (-1)^{\lfloor (d+1)/2 \rfloor} F_{d+1}.$$
Proof is a again by induction with the base case holding by inspection. We then get from the recurrence on the RHS
$$(-1)^d (-1)^{\lfloor d/2 \rfloor} F_{d} - (-1)^{\lfloor (d-1)/2 \rfloor} F_{d-1}.$$
With $d=2f$ the coefficients on the two terms are $(-1)^f = (-1) \times (-1)^{f-1} = (-1)^{\lfloor (2f+1)/2\rfloor}$ and when $d=2f+1$ they are $(-1)\times (-1)^f = (-1) \times (-1)^f = (-1)^{\lfloor (2f+2)/2\rfloor}$ and the claim is proven.
