zero coordinate is in the $i$-th position. Now, from the properties of the integers $d(i)$, it follows that

-   $\textbf{n}_i^{d(i)}=e_i$ for all $i=1,\ldots,r$;

-   $\textbf{n}_i^{d(i)-1}\neq e_i$ for all $i=1,\ldots,r$.

We consider the subtree $T'$ of $T$ such that $N(T')=\left\{\textbf{n}_i^{j(i)}: i=1,\ldots,r; j(i)=1,\ldots,d(i)-1 \right\}$. Then we have

-   $T'$ is rooted in $\textbf{n}_1^{1}$ (it corresponds to an element of the associated Arf good semigroup);

-   $e_i \notin N(T')$ for all $i=1,\ldots,r$;

-   If $T''$ is such that $T' \subseteq T'' \subseteq T$ then $N(T'') \setminus N(T') $ consists only of nodes of the type $e_i$.

From the previous properties it is clear that the element corresponding to the subtree $T'$ must be the conductor of the Arf semigroup associated to $T$. It is also trivial that the sum of all the elements of $N(T')$ is equal to $\textbf{c}=(c[1],\ldots,c[n])$ where $$c[i]=\sum_{k=1}^{d(i)-1}{M_i[k]}  \textrm{ for }  i=1,\ldots,n.$$

With the notations of the previous proposition, given a tree $T_{E}=(p_1,\ldots,p_{r-1})$ where $E=\left\{M_1,\ldots,M_r\right\}$, it is easy to show that $$d(i)-1=\max(l(M_i),p_i,p_{i-1}) \textrm{ for  } i=2,\ldots,r-1$$ $$d(1)-1=\max(l(M_1),p_1) \textrm { and } d(r)-1=\max(l(M_r),p_{r-1}).$$

Now, we focus on the case $r=2$ and we determine a procedure to compute $\textrm{Cond}(\textbf{c})$ where $\textbf{c}$ is a fixed arbitrary vector $(c[1],c[2])$. Suppose that $T_{E}=(p) \in \textrm{Cond}(\textbf{c})$ 