at most $k_E(i,j)$. We have $$k_E(i,j)=\min\left\{ \min( s_{i,k}, s_{j,k}): s_{i,k}\neq s_{j,k} \right\},$$ where $s_{i,k}$ is the integer such that $$M_i[k]=\sum_{l=k+1}^{s_{i,k}}{M_i[l]}.$$ Thus, if $M_1$ and $M_2$ are two distinct multiplicity sequences, we will define the compatibility between $M_1$ and $M_2$ as the integer $\textrm{Comp}(M_1,M_2)=k_E(1,2) $ where $E=\left\{M_1,M_2 \right\}$. While if $M_1=M_2$ we set by definition $\textrm{Comp}(M_1,M_2)=+\infty$. A tree $T$ is untwisted if two non-consecutive branches are glued at level $l$ if and only if all the consecutive branches between them are glued at a level greater or equal to $l$. We will call twisted a tree that it is not untwisted. From the definition it follows that the matrix of an untwisted tree $T \in \tau(E)$ is such that: $$p_{i,j}=\min\left\{ p_{i,i+1},\ldots,p_{j-1,j}\right\} \textrm{ for all } i<j.$$ So an untwisted tree can be completely described by the second diagonal of its matrix. Thus in the following we will indicate an untwisted tree by a vector $T_E=(p_1,\ldots,p_{r-1})$ where $p_i=p_{i,i+1}$.

\[remark0\] It is easy to see that a twisted tree can be converted to an untwisted one by accordingly permuting its branches (the corresponding Arf semigroups are therefore isomorphic). Thus in the following we can focus only on the properties of the untwisted trees, 