Number of vertices of degree n not connected to vertices of degree 1 Imagine a tree made up only of vertices of degree 1 and vertices of degree $n$.
Let $m$ be the number of vertices of degree 1 and $\frac{m-2}{n-2}$ be the number of $n$-vertices of the tree.
How can I find the number of vertices not connected to a vertex of degree 1? I.e. the number of $n$-vertices connected only to other $n$-vertices. 
 A: You can't, at least not from the degree sequence alone. Consider a tree $T_1$ where node A is the root its neighbours are B,C,and D, and in turn B,C, and D have 2 leaf children each. Then $T_1$ has 4 vertices of degree 3--namely $A,B,C$ and $D$--and 6 vertices of degree 1, and exactly one vertex NOT connected to a vertex of degree 1, namely A.
Now consider another tree $T_2$ where nodes $A,B,C$,and $D$ form a path, where nodes A is adjacent to nodes $E$ and $F$, $B$ is adjacent to $G$, $C$ to $H$, and $D$ to $I$ and $J$. [So in $T_2$ nodes $A,B,C,D$ all have degree exactly 3 while nodes $E$ through $J$ have degree exactly 1.] Then $T_2$ also has  4 vertices of degree 3 and 6 vertices of degree 1, but no vertex not connected to a vertex of degree 1
T1:  L--B--A--D--L     T2:   E--A--B--C--D--J
       /   |   \                |  |  |  |
      L    C    L               F  G  H  I
          / \
         L   L

A: A tree with $k$ vertices has $k-1$ edges. 
On the other side, the number of edges in a simple graph with $m$ vertices of degree $1$ and $\frac{m-2}{n-2}$ vertices of degree $n$ is $\frac{1}{2}(m+n(\frac{m-2}{n-2}-m)).$
Solve $\frac{m-2}{n-2}-1=\frac{1}{2}(m+n(\frac{m-2}{n-2}-m)),$ we obtain $m=0$ or $n=2.$ None of them is convenient and thus such a tree does not exist. 
COMMENT ADDED AFTER YOUR EDIT:
With $\frac{m-2}{n-2}$ vertices of degree $n$, you will solve $\frac{m-2}{n-2}-1=\frac{1}{2}(m+n(\frac{m-2}{n-2})),$ which leads to $m=0$ again.
