Counting Euler circuits through labelled trees where $v_1$ and $v_2$ have distance two

Question

Let $T_n$ be the set of all labelled trees with $n$ vertices. For any $T \in T_n$ let $D(T)$ be the 'doubled tree', where each edge of $T$ is replaced by one directed edge in each direction. $D(T)$ is now an Eulerian directed graph with $2(n-1)$ edges and by the B.E.S.T theorem it has $\prod\limits_{i=1}^n (\deg_{D(T)}(v_i)-1)!$ many Euler circuits.

Let $A_n \subset T_n$ be the set of all labelled trees with $n$ vertices, with the property that vertices $v_1$ and $v_2$ have a distance of $2$, i.e. there is no connection between $v_1$ and $v_2$, but there is a $t \in \{3,...,n\}$ such that there are edges $(v_1,v_t)$ and $(v_2,v_t)$.

Question: Is there a simpler formula for the expression $$ \sum\limits_{T \in A_n} \prod\limits_{i=1}^n (\deg_{D(T)}(v_i)-1)! \ , $$ which counts the number of Euler circuits through trees from $A_n$.

If the distance is prescribed as $1$ instead of $2$, then by using the fact that there are ${n-2 \choose d_1-1,...,d_{n}-1}$ many $T \in T_n$ with $d_i = \deg_{T}(v_i) = \deg_{D(T)}(v_i)$ the formula is found to be $2 \frac{(2n-3)!}{n!}$.

Another way of formulating the question would be: For given $d_1,...,d_{n} \in \{1,...,n-1\}^n$ with $d_1+...+d_n = 2n-2$, how many of the ${n-2 \choose d_1-1,...,d_{n}-1}$ many $T \in T_n$ with $d_i = \deg_{T}(v_i) = \deg_{D(T)}(v_i)$ satisfy the condition that $v_1$ and $v_2$ have distance two?

Any help is much apprechiated.

Answer 1

Wlog we can use $v_{n-2}$ and $v_{n-1}$ instead of $v_1$ and $v_2$. Then if we let $B_n \subset T_n$ be the set of labelled trees with edges $(v_{n-2}, v_n)$ and $(v_{n-1}, v_n)$, the count for $A_n$ is just $(n-2)$ times the count for $B_n$ (there are $n-2$ choices for $v_t$; swap $v_t$ for $v_n$ if they're different).

The advantage to numbering things this way is that when constructing the Prüfer codes for the trees in $B_n$ the three labelled vertices are the last survivors. The final label in the Prüfer code will be $n$, and the penultimate one (if $n > 3$) will be in $\{n-2, n-1, n\}$. This appears empirically also be to sufficient, although a proof eludes me at present.

If so then the number of trees with given vertex degrees $d_i$ (where each $d_i \ge 1$ and $d_n \ge 2$) having subtree $v_{n-1} - v_n - v_{n-2}$ is a sum of three multinomial coefficients which simplifies to $$\binom{n-2}{d_1-1, \ldots, d_n-1} \frac{(d_n - 1)(d_{n-2} + d_{n-1} + d_n - 4)}{(n-3)(n-2)}$$

Then taking the alternative formulation

For given $d_1,...,d_{n} \in \{1,...,n-1\}^n$ with $d_1+...+d_n = 2n-2$, how many of the ${n-2 \choose d_1-1,...,d_{n}-1}$ many $T \in T_n$ with $d_i = \deg_{T}(v_i) = \deg_{D(T)}(v_i)$ satisfy the condition that $v_1$ and $v_2$ have distance two?

the answer is given by relabelling to put the special vertices at $v_1$, $v_2$ and summing over all possible intermediate vertices. $$\sum_{i=3}^n [d_i \ge 2] \binom{n-2}{d_1-1, \ldots, d_n-1} \frac{(d_i - 1)(d_1 + d_2 + d_i - 4)}{(n-3)(n-2)}$$ simplifies to $$\frac{1}{(n-3)(n-2)} \binom{n-2}{d_1-1, \ldots, d_n-1} \left((n - d_1 - d_2)(d_1 + d_2 - 5) - (n-2) + \sum_{i=3}^n d_i^2 \right)$$