chemia - Studia

ALI REZA ASHRAFI, HOSSEIN SHABANI, MIRCEA V. DIUDEA
Let xi
∈ Vi
, 0 ≤ i ≤ k . In each orbit d ( v)
= d(
xi
) when v is a fixed
element of V
i
. Define:
0
1
2
s−1
α ( t,
s)
= (1 + t).3
+ (2 + t).3
+ (3 + t).3
+ L+
( s + t)3
.
s
Then obviously α ( t,
s)
= 1 4[1 − 2t
+ (2s
+ 2t
−1).3
] . Therefore,
140
d( x ) = α ( i,
k)
+ α(0,
k − i)
+ [ i − j + 2α
( i − j,
k − j)]
i
i
∑
j = 0
To simplify above equation, we compute d ( x i
) . We claim that
k + 1
5 k
k 3 −i
d(
xi
) = 1+
(2k
− ).3 + (2×
3 ). i + ( ). 3 .
2
2
We now compute the Wiener and Balaban indices of D[k]. The
Wiener index of a graph G is half of the summation of all d (v)
over all
vertices of G . From the orbits of the action of Aut ( D[
k])
on V ( D[
k])
, one
k
1 ⎛
i−1
⎞
can see that W ( G)
= ⎜∑[4×
3 d(
xi
)] + d(
x0)
⎟ . So the Wiener index of
2 ⎝ i=
1
⎠
D[k] is given by the following formula:
1
1
2
W ( D[
k])
= .
.(4 + 16k
+ (144k
+ 16k)3
2
2k
k
4 (32k
−12k
+ 1)3 + (24k
− 4)3 + 4
+ (416k
3
−16k
2
− 40k
+ 5)3
2k
+ (384k
4
− 80k
3
− 44k
2
+ 14k
−1)3
Corollary 2. The orbits of the action of Aut ( D[
k])
on E ( D[
k])
are
E = {1,2,3,4}
1 , … , { 1 2 (3 k −1
1), ,2 (3 k
E = + × − K × − 1)}
.
k
Since D[k] is a tree, μ ( D[
k])
= 0 and next,
k
i−1
k 4×
3
J ( D[
k])
= 2×
(3 −1)
∑
i= 1 d(
xi
−1)
d(
xi
)
To simplify above equation, we first compute d( xi
− 1)
d(
xi
) . We have:
k k 45 2k
2k
2 2k
d(
xi
−1)
d(
xi
) = (1 − 7.3 + 4k3
+ .3 −14k3
+ 4k
3 ) +
4
k 2k
2k
2k
2 k 2k
2k
−i
(4.3 −14.3
+ 8k3
) i + (4.3 ) i + (6.3 −18.3
+ 12k3
)3
2k
−i
27 2k
−2i
+ (12.3 ) i3
+ ( .3 )3
4
2 −i
−i
−2i
Define: f ( i,
k)
= d(
x ) d(
x ) = α + βi
+ λi
+ γ 3 + δi3
μ , where,
i− 1 i
+ 3
3k
)
k