TITLE: Number of labeled trees with n vertices such that all vertices have degree 1 or 3
QUESTION [6 upvotes]: How many labeled trees with $n$ vertices exist such that their degree is $1$ or $3$?
I succeeded to get a range but not a particular answer.
EDIT
the range I found:
$$\frac{n-2}{2^n}\lt\text{ number of trees with }n\text{ vertices }\lt n-2$$

REPLY [8 votes]: We      present     two      answers,      one     using      Pruefer
codes   and   one
using combinatorial classes.  Recall that the Pruefer code  for a tree
on $n$  nodes is a list of  $n-2$ elements taking its  values from the
$n$ nodes, for a total of $n^{n-2}$ trees. The degree of a node in the
resulting tree is one more than  the number of times it appears in the
list. Therefore with  degrees one and three a  node either appears not
at all or it appears twice.  We see immediately that there are no such
trees on an odd number of nodes because $n-2$ must be even.
Now for the closed formula  we must choose the $(n-2)/2$ nodes that
are  present in the  code and  distribute these  pairs into  the $n-2$
slots. We obtain
$${n\choose (n-2)/2} {n-2\choose 2,2,\ldots,2}
= {n\choose (n-2)/2} \frac{(n-2)!}{2^{(n-2)/2}}$$
for $n$ even and zero otherwise. This yields the sequence
$$1, 0, 4, 0, 90, 0, 5040, 0, 529200, 
\\ 0, 89812800, 0, 22475653200,\ldots$$
For  the  alternate proof  introduce  labeled  combinatorial class  of
rooted trees with outdegree two.  These have total degree three or one
at all  nodes except  for the  root, which has  degree two.  The class
equation is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} = \mathcal{Z} + 
\mathcal{Z}\times\textsc{SET}_{=2}(\mathcal{T}).$$
We thus obtain the functional equation
$$T(z) = z\left(1+\frac{1}{2} T(z)^2\right).$$
Now  observe that  an element  of  the target  class $\mathcal{V}$  of
unrooted labeled  trees with node degree  one or three is  obtained by
connecting two  elements of $\mathcal{T}$  by an edge between  the two
roots.  In this  way we obtain every element  of $\mathcal{V}$ exactly
$n-1$ times, so we are after the following coefficient
$$\frac{n!}{2(n-1)} [z^n] T(z)^2.$$
To  compute   the  desired  value   we  use  a  variant   of  Lagrange
inversion. With $n\ge 2$ we have
$$\frac{n!}{2(n-1)} [z^n] T(z)^2 =
\frac{n!}{2(n-1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)^2 dz.$$
We have $$z = \frac{2T(z)}{2+T(z)^2}$$ so putting $T(z) = w$
we otain
$$z = \frac{2w}{2+w^2} \quad\text{and}\quad
dz = \left( \frac{2}{2+w^2}
- \frac{2w}{(2+w^2)^2} 2w \right) dw.$$ 
This yields for the integral
$$\frac{n!}{2(n-1)} \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(2+w^2)^{n+1}}{2^{n+1} w^{n+1}}
\frac{2(2-w^2)}{(2+w^2)^2} w^2 \; dw
\\ = \frac{n!}{2^{n+1}(n-1)} \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(2+w^2)^{n-1}}{w^{n-1}}
(2-w^2) \; dw.$$
This is clearly zero when $n-2$ is odd. When $n$ is even we obtain
$$\frac{n!}{2^{n+1}(n-1)}
[w^{n-2}] (2+w^2)^{n-1} (2-w^2).$$
Now when $n=2$ this evaluates to
$$\frac{2}{2^3\times 1} \times {1\choose 0} 2^{n-1} \times 2
= 1$$
which is the correct value. Continuing with $n\ge 4$ we obtain
$$\frac{n!}{2^{n+1}(n-1)}
\left({n-1\choose (n-2)/2} 2^{n-1-(n-2)/2} \times 2
- {n-1\choose (n-4)/2} 2^{n-1-(n-4)/2}\right)
\\ = n\times(n-2)!\times
\left({n-1\choose (n-2)/2} 2^{-1-(n-2)/2}
- {n-1\choose (n-4)/2} 2^{-2-(n-4)/2}\right)
\\ = \frac{(n-2)!}{2^{n/2}} \times n \times
\left({n-1\choose (n-2)/2}
- {n-1\choose (n-4)/2} \right)
\\ = \frac{(n-2)!}{2^{n/2}} \times 
\left(n {n-1\choose (n-2)/2}
- n \frac{(n-2)/2}{n} {n\choose (n-2)/2} \right)
\\ = \frac{(n-2)!}{2^{n/2}} \times 
\left((n+2)/2 - (n-2)/2 \right) {n\choose (n-2)/2}
\\ = \frac{(n-2)!}{2^{(n-2)/2}} {n\choose (n-2)/2}.$$
This is the claim.
Remark. Lagrange Inversion is more machinery than we need here,
we can also use the quadratic formula to obtain
$$T(z) = \frac{1-\sqrt{1-2z^2}}{z}
\quad\text{so that}\quad T(z)^2 = \frac{2-2z^2}{z^2}
-2\frac{\sqrt{1-2z^2}}{z^2}.$$
We get for $n\ge 2$ even $$[z^n] \sqrt{1-2z^2}
= {1/2\choose n/2} (-1)^{n/2} 2^{n/2}
\\ = (-1)^{n/2} 2^{n/2} \times 
\frac{1}{2} \frac{1}{(n/2)!} \prod_{q=0}^{n/2-2} (-1/2-q)
= - \frac{1}{(n/2)!} \prod_{q=0}^{(n-4)/2} (1+2q)
\\ = - \frac{(n-3)!}{(n/2)!} \frac{1}{2^{(n-4)/2}} 
\frac{1}{((n-4)/2)!}.$$
We then get for $n\ge 4$
$$[z^n]\left(-\frac{2}{z^2}\right) \sqrt{1-2z^2}
= -2 [z^{n+2}] \sqrt{1-2z^2}
= 2 \frac{(n-1)!}{((n+2)/2)!} \frac{1}{2^{(n-2)/2}} 
\frac{1}{((n-2)/2)!}.$$
Collecting everything we finally have
$$2 \frac{n!}{2(n-1)} \frac{(n-1)!}{((n+2)/2)!} \frac{1}{2^{(n-2)/2}} 
\frac{1}{((n-2)/2)!}
\\= n! \frac{(n-2)!}{((n+2)/2)!} \frac{1}{2^{(n-2)/2}} 
\frac{1}{((n-2)/2)!}
\\ = \frac{(n-2)!}{2^{(n-2)/2}} {n\choose (n-2)/2}$$
again as claimed.

Addendum.  We now  show  how  to compute  the  number of  unrooted
labeled trees with  node degree at most three  i.e. including nodes of
degree two.   This means we have  pairs and singletons  present in the
Pruefer code. Suppose we have $q$ pairs. We must choose these from the
$n$  available  nodes and  fill  the rest  of  the  Pruefer code  with
singletons. This yields
$$\sum_{q=0}^{\lfloor (n-2)/2\rfloor}
{n\choose q} {n-q\choose n-2-2q} 
{n-2\choose 2,2,\ldots, 1,1,\ldots 1}
\\ = \sum_{q=0}^{\lfloor (n-2)/2\rfloor}
{n\choose q} {n-q\choose n-2-2q}  \frac{(n-2)!}{2^{q}}.$$
There  are  several  possibilities  to  simplify  this,  we  will  use
combinatorial classes.  The class equation here now becomes
$$\mathcal{T} = \mathcal{Z} + 
\mathcal{Z}\times\textsc{SET}_{=1}(\mathcal{T}) + 
\mathcal{Z}\times\textsc{SET}_{=2}(\mathcal{T}).$$
We thus obtain the functional equation
$$T(z) = z\left(1+ T(z) + \frac{1}{2} T(z)^2\right).$$
The construction goes through as before and we seek
$$\frac{n!}{2(n-1)} [z^n] T(z)^2.$$
To  compute   the  desired  value   we  use  a  variant   of  Lagrange
inversion. With $n\ge 2$ we have
$$\frac{n!}{2(n-1)} [z^n] T(z)^2 =
\frac{n!}{2(n-1)} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)^2 dz.$$
We have $$z = \frac{2T(z)}{2+ 2T(z) + T(z)^2}$$ so putting $T(z) = w$
we otain
$$z = \frac{2w}{2+2w+w^2} \quad\text{and}\quad
dz = \left( \frac{2}{2+2w+w^2}
- \frac{2w \times (2w+2)}{(2+2w+w^2)^2}  \right) dw.$$ 
This yields for the integral
$$\frac{n!}{2(n-1)} \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(2+2w+w^2)^{n+1}}{2^{n+1} w^{n+1}}
\frac{2(2-w^2)}{(2+2w+w^2)^2} w^2 \; dw
\\ = \frac{n!}{2^{n+1}(n-1)} \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{(2+2w+w^2)^{n-1}}{w^{n-1}}
(2-w^2) \; dw.$$
We write
$$(2+2w+w^2)^{n-1} = (1+(w+1)^2)^{n-1}
= \sum_{q=0}^{n-1} {n-1\choose q} (w+1)^{2q}$$
Extracting the residue we obtain the formula
$$\frac{n!}{2^{n+1}(n-1)}
\sum_{q=0}^{n-1} {n-1\choose q} 
\left(2\times {2q\choose n-2} - {2q\choose n-4}\right).$$
This yields the sequence
$$1, 3, 16, 120, 1170, 14070, 201600, 3356640, 63730800, 
\\ 1359666000, 32212857600, 839350512000, 23860289653200,
\\  734964075846000, \ldots $$
which is OEIS A003692.
We introduce a binomial coefficient  integral in an effort to get a
better closed form. Put
$${n-q\choose n-2-2q} =
\frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{1}{v^{n-1-2q}} (1+v)^{n-q}\; dv.$$
Observe that this vanishes  when $\lfloor(n-2)/2\rfloor \lt q\le n$ so
we may extend the upper limit of the sum to $n,$ obtaining
$$\frac{(n-2)!}{2\pi i}
\int_{|v|=\epsilon} \frac{1}{v^{n-1}} (1+v)^{n}
\sum_{q=0}^n {n\choose q}  \frac{v^{2q}}{2^q(1+v)^q}
\; dv
\\ = \frac{(n-2)!}{2\pi i}
\int_{|v|=\epsilon} \frac{1}{v^{n-1}} (1+v)^{n}
\left(1+\frac{v^2}{2(1+v)}\right)^n
\; dv
\\ = \frac{(n-2)!}{2^n\times 2\pi i}
\int_{|v|=\epsilon} \frac{1}{v^{n-1}} 
(2+2v+v^2)^n\; dv.$$
Extracting the residue we get a very slight improvement which yields
$$\frac{(n-2)!}{2^n}
\sum_{q=0}^{n} {n\choose q} {2q\choose n-2}.$$