Using symbolic method to calculate number of particular plane trees I was asked to use the symbolic method for calculate the number of plane rooted trees with $n$ vertices such that each inner vertex have exactly three leaves or a sequence of at least two trees.
So I stated $$\mathcal{T}=\mathcal{Z}\times (\mathcal{Z}^3+\mathcal{T}^2\times SEQ(\mathcal{T})),$$
where $\mathcal{Z}$ is the atomic class. The problem was that, solving it, I got $$T(x)=\frac{x^4+1\pm\sqrt{(x^4+1)^2-4x^4(x+1)}}{x+1}$$
which seems impossible to solve. I think I did an important misstake in the equation for $\mathcal{T}$, but I can't see it. 
 A: I would start with the following combinatorial class:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{Z} \times \mathcal{U} +
\mathcal{Z} \times \textsc{SEQ}_{\ge 2}(\mathcal{T})$$
which gives the functional equation
$$T(z) = uz + z \frac{T(z)^2}{1-T(z)}$$
so that
$$z = \frac{T(z) (1-T(z))}{T^2(z) - uT(z) + u}$$
We then have
$$n \times T_n = [z^{n-1}] T'(z)$$
and from the Cauchy Coefficient Formula
$$[z^{n-1}] T'(z) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} T'(z) \; dz.$$
Now put $T(z) = w$ so that $T'(z) \; dz = dw$ and we obtain
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(w^2-uw+u)^n}{w^n (1-w)^n} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(w^2+u(1-w))^n}{w^n (1-w)^n} \; dw.$$
Next for trees having $k$ leaves of type $\mathcal{U}$ we get
(we must evidently have $0\le k\le n$)
$${n\choose k} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{w^{2n-2k}(1-w)^k}{w^n (1-w)^n} \; dw
\\ = {n\choose k} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{2k-n} (1-w)^{n-k}} \; dw.$$
Observe what happens when $k=n,$ we find
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n}} \; dw
= [[n=1]].$$
This means we have one tree consisting of a node of type $\mathcal{Z}$
with a single child node of  type $\mathcal{U}.$ When $n\gt 1$ we must
have $k\le n-1,$ a refinement. Continuing,
$${n\choose k} {2k-n-1+n-k-1\choose n-k-1}
= {n\choose k} {k-2\choose n-k-1}.$$
The total contribution  where a node of type  $\mathcal{U}$ has weight
three  and of  type $\mathcal{Z}$  has weight  one is  $n+3k.$ Letting
$m=n+3k$ be the  total number of nodes by weight,  the condition $k\le
n-1$ then implies $k\le m-3k-1$ or $4k\le m-1.$ For the residue not to
vanish we must also  have $2k\gt n$ or $5k\gt m.$  Hence the number of
trees with $m$ nodes is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=\lfloor m/5 \rfloor+1}^{\lfloor (m-1)/4\rfloor}
\frac{1}{m-3k}
{m-3k\choose k} {k-2\choose m-4k-1}.}$$
This holds when $n\ge 2$ or  at least two nodes of type $\mathcal{Z}.$
We  get a  single tree  on four  nodes total  when $n=1.$  Due to  the
sequence constraint $\textsc{SEQ}_{\ge 2}$ the  next tree to appear is
when $n=3$ with $m=9.$
