1.
The Wiener index of weighted stars
I studied the Wiener index of linear chains
which arcs were weighted by their index. Similarly can be indexed stars. Their
distance matrix of weighted stars has the form shown for S(5)
|
Index |
1 |
2 |
3 |
4 |
5 |
Upper sum |
|
|
0 |
1 |
2 |
3 |
4 |
10 |
|
|
1 |
0 |
3 |
4 |
5 |
12 |
|
|
2 |
3 |
0 |
5 |
6 |
11 |
|
|
3 |
4 |
5 |
0 |
7 |
7 |
|
|
4 |
5 |
6 |
7 |
0 |
|
|
Column sums |
|
1 |
5 |
12 |
22 |
|
|
Wiener index |
|
1 |
6 |
18 |
40 |
|
The sum of opposite pairs of distances on
side diagonals is always 2(n 1), the odd values are equal to (n -1) which is
therefore the mean distance. When we eliminate the zero main diagonal value, we
get (n2 - n) distances which are counted in the matrix twice, once
in the upper triangle, once in the lower triangle. The Wiener index of stars
weighted by the index is thus
W(Sn) = n(n
1)2/2.
More complicated proof is possible. We see
that the consecutive columns (rows) start with (n 1) value and end with (2n
3) value. The sums thus equal to the difference of two binomial coefficients
(2n
-2)(2n 3)/2 (n 1)(n 2)/2 = (3n
The the equation is then verified by the
full induction. The weighted distance matrices of stars have rather interesting
eigenvalues:
|
n |
|
|
|
|
|
|
|
2 |
1 |
-1 |
|
|
|
|
|
3 |
4.1115 |
-3.2019 |
-0.9112 |
|
|
|
|
4 |
9.1415 |
-5.3569 |
-3.1470 |
-0.9076 |
|
|
|
5 |
16.8737 |
-7.4972 |
-5.3254 |
-3.1448 |
-0.9063 |
|
|
6 |
16.4947 |
-9.6414 |
-7.4853 |
-5.3253 |
-3.1426 |
-0.9000 |
Starting from the third column, the
eigenvalues on side diagonals decrease slowly.