COMPUTING THE SECOND- AND THIRD- CONNECTIVITY INDEX OF AN INFINITE CLASS OF DENDRIMER NANOSTARS
HONGZHUAN WANG, HONGBO HUA♦
Faculty of Mathematics and Physics, Huaiyin Institute of Technology, Huai’an City, Jiangsu 223003, P. R. China
The m-connectivety index of a graph G is defined to be
1 2 1 1 2
1
...
) 1 (
... + +
∑
=
m m i i
iv v i i i
v m
d d G d
χ
.where runs over all paths of length m in G and di is the degree of vertex νi. In this paper, we give explicit formulas for the second- and third- order connectivity index of an infinite class of dendrimer nanostars.
1 2
1 ... +
im
i
iv v
v
(Received May 8, 2010; accepted May 14, 2010)) Keywords: Connectivity index, Dendrimer nanostars
1. Introduction
Let G be a connected simple graph. The m-connectivity index of G is defined as
1 2
1 1 2
1
...
) 1 (
... + +
∑
=
m m i i
iv v i i i
v m
d d G d
χ
where runs over all paths of length m in G and is the degree of
1 2
1 ... +
im
i
iv v
v
d
ivertex vi . In particular, 2-connectivity and 3-connectivity index are defined as
3 2 3 1
2 1
) 1
2
(
i i v i
v
v
d d d
G
i i i
∑
χ =
and
4 3 2 4 1
3 2 1
) 1
3
(
i i i v i
v v
v
d d d d
G
i ii i i
∑
χ =
,respectively.
During the past several decades, there are many papers dealing with the connectivity index.
♦Corresponding author (E-mail address:[email protected]).
The reader may consult [1-11 ] and references cited therein.
In this paper, we shall give explicit computing formulas for 2- and 3- connectivty of a type of dendrimer nanostars.
2. Main results
Let NS[n] denote a kind of dendrimer nanostars with n growth stages, see for example, Fig.s 1 and 2.
hn
−1
hn
hn
h0 h1
h1 h2
h2
h2
h2
Fig. 1. The dendrimer nanostar NS[n]
Fig. 2. The dendrimer nanostar NS[1]
In the following, we shall compute the second- and third- order connectivity index for the
We first give an exact formula of the second-order connectivity index for this dendrimer nanostar.
Theorem 1. Let NS[n] be the dendrimer nanostar as shown in Fig. 1. Then
2 5 3 3 1
2χ( N S [ n ] ) = 1 2 2 + 3 + ( 6 2 + 2 3 ) ( 2n-1- 1 ) Proof. Let d
i j k
denote the number of 2 paths whose three consecutive vertices are of degree
, respectively. Also, we use
, ,
i j k
( )sd i j k to mean
d
in the stage. Obviously,ijk
s th
( )s ( )s k j i d i j k = d .
Firstly, we compute the value of
2 ( χ N S [1] )
. It is easily seen that(1) 6 , (1) 1 2 , (1) 6 , (1) 2 4 , (1) 3, (1) 1 2
2 2 2 2 2 3 2 3 2 2 3 3 3 2 3 3 3 3
d = d = d = d = d = d = .
So we have
6 1 2 6 2 4
2 ( [1])
2 2 2 2 2 3 2 3 2 2 3 3
3 1 2
3 2 3 3 3 3
3 2 2 6 3 3 4 2 2 2 4 3 3
2 5 3 1 2 2
3 .
χ N S = + + +
× × × × × × × ×
× × + × ×
= + + + + +
= +
+
) )
Now, we are ready to deduce the relation between
2 ( χ N S s [ ]
and2 ( χ N S s [ − 1]
for2.
s ≥
( ) ( 1) 3 2 3 2 1 ( 1) 3 2 1 .
2 2 2 2 2 2 2 2 2
( ) ( 1) 2 2 2 1 ( 1) 2 .
2 2 3 2 2 3 2 2 3
( ) ( 1) 2 2 2 1 ( 1) 2 1 .
2 3 2 2 3 2 2 3 2
( ) ( 1) 2 2 2 1 ( 1) 2 2 .
2 3 3 2 3 3 2 3 2
( ) ( 1) 3 2
3 2 3 3 2 3
2
4
s s s s s s
d d d
s s s s s s
d d d
s s s s s s
d d d
s s s s s s
d d d
s s s
d d
− − − −
= + ⋅ − ⋅ = + ⋅
− + ⋅ ⋅ − −
− − − +
= + + ⋅ = +
− ⋅ ⋅ − − +
− ⋅
= − = +
= + + = +
= + − 1.
Obviously, for any
( i j k , , ) ≠ (2, 2, 2 ), (2, 2, 3),(2, 3, 2 ), (2, 3, 3),(3, 2, 3),
we have ( )s 0d ijk = or ( )s (s 1) (1)
di j k di j k di j k
= − =L = for
s = 2, L , . n
Thus,
1 1
3 2 2 2 2
2 ( [ ]) 2 ( [ 1])
2 2 2 2 2 3 2 3 2 2 3 3
3 2 1 3 2 3
2 1 2 1
3 2 2 2 2 2
2 ( [ 1])
2 3 3 3 2 2
2 ( [ 1]) (2 3 31 2 6 )2 2 .
n n n n
N S n N S n
n
n n n n n
N S n N S n n
χ χ
χ χ
− + + 2
= − + ⋅ + + + +
× × × × × × × ×
⋅ −
× ×
− − + −
= − + ⋅ + + + +
= − + + −
By the above recursive formula for
2 ( χ N S n [ ] )
, we obtain3 1 2
2 ( [ ]) 2 ( [ 1]) (2 3 )2 2
6 3 1 2
2 ( [ 2 ]) (2 3 )(2 2 2 )
6
3 1 2
2 ( [1]) (2 3 )(2 2 2 1)
6
2 5 3 3 1 2 1
1 2 2 3 (2 3 6 )(2 1).
N S n N S n n
n n
N S n
n n
N S
n
χ χ
χ
χ
= − + + −
− −
= − + + +
=
− −
= + + + +
= + + + − −
L
L
3
3 +
Next, we shall give an exact formula of the third-order connectivity index for the dendrimer nanostar as shown in Fig. 1.
Theorem 2. Let NS[n] be the dendrimer nanostar as shown in Fig. 1. Then
1 5 5 4 5 6 4 6
3χ( N S [ n ] ) = 1 8 + 1 8 + ( 5 + 3 ) ( 2n-1-1 ) .
Proof. Let denote the number of 3 paths whose four consecutive vertices are of degree
, respectively. Also, we use
d i j k l
, l , ,
i j k ( ) s
d ijkl
to meand
in the stage. Obviously,ijkl
st h( )s ( )s l k j i d i j k l = d .
Firstly, we compute the value of
3 ( χ N S [1] )
.=
= It is easy to obtain the following:
(1 ) 4 , (1 ) 1 0 , (1 ) 6 , (1 ) 1 8 , (1 ) 1 2 ,
2 2 2 2 2 2 2 3 2 2 3 2 2 2 3 3 2 3 3 2
d = d = d = d = d
(1 ) 1 7 , (1 ) 1, (1 ) 6 , (1 ) 4 , (1 ) 1 3 .
2 3 3 3 3 2 2 3 3 2 3 2 3 2 3 3 3 3 3 3
d = d = d = d = d
4 10 6 18 3 ( [1])
2 2 2 2 2 2 2 3 2 2 3 2 2 2 3 3
12 17 1 6
2 3 3 2 2 3 3 3 3 2 2 3 3 2 3 2
4 13
3 2 3 3 3 3 3 3 155 45 6 .
18 18
χ N S = + + +
× × × × × × × × × × × ×
+ + + +
× × × × × × × × × × × ×
× × × + × × ×
= +
+
)
)
Now, we are ready to deduce the relation between 3 (
χ
N S s[ ] and3 χ ( N S s [ − 1]
for2.
s ≥
( ) ( 1) 2 2 2 1 ( 1) 2 .
2 2 2 2 2 2 2 2 2 2 2 2
( ) ( 1) 2 2 2 1 ( 1) 2 .
2 2 2 3 2 2 2 3 2 2 2 3
( ) ( 1) 2 2 2 1 ( 1) 2 .
2 2 3 2 2 2 3 2 2 2 3 2
2 2 2
s s s s s s
d d d
s s s s s s
d d d
s s s s s s
d d d
− − −
= + ⋅ ⋅ = +
− + ⋅ ⋅ − −
− ⋅ ⋅ − −
−
= − =
= + − = +
+
( ) ( 1 ) 2 2 2 1 ( 1 ) 2 .
2 2 3 3s 2 2 3 3s s 2 s 2 2 3 3s s
d = d − + ⋅ − ⋅ − = d − +
( ) ( 1) 6 2 1 ( 1) 3 2 .
2 3 2 3 2 3 2 3 2 3 2 3
( ) ( 1) 4 2 4 2 1 ( 1) 2 1 .
2 3 3 2 2 3 3 2 2 3 3 2
( ) ( 1) 6 2 1 ( 1) 3 2 .
3 2 3 3 3 2 3 3 3 2 3 3
s s s s s
d d d
s s s s s s
d d d
s s s s s
d d d
− − −
= + ⋅ = + ⋅
− − − +
= + ⋅ − ⋅ = +
− − −
= + ⋅ = + ⋅
Obviously, for any
we have
( , , , ) i j k l ≠ ( 2, 2, 2, 2 ), ( 2, 2, 2, 3),( 2, 2, 3, 2 ), (2, 2, 3, 3),( 2, 3, 2 ,3 ),
( 2 , 3 , 3 , 2 ) ,( 3 , 2 , 2 , 3 ) , ( )s 0
d ijk l = or d ( )s d (s 1) (1) i j k l = i j k l− = L d i j k l
2, , .
= for
s = L n
Thus,
2 2 2 3 ( [ ]) 3 ( [ 1])
2 2 2 2 2 2 2 3 2 2 3 2
2 3 2 2 1 3 2
2 2 3 3 2 3 2 3 2 3 3 2 3 2 3 3
4 2 1 1
3 ( [ 1]) 2 2 ( ) 2 1 ( ) 2
3 6 3 6
3 ( [ 1]) (5 4 6 3 )2 2 .
n n n
NS n NS n
n n n n
n n n
NS n NS n n
χ χ
χ χ
= − + + +
× × × × × × × × ×
⋅ + ⋅
+ + +
× × × × × × × × × × × ×
− −
= − + + + ⋅ + + ⋅
= − + + −
+
)
By the above recursive formula for
3 ( χ N S n [ ]
, we obtain3 ( [ ]) 3 ( [ 1]) (5 4 6 3 ) 2 2
3 ( [ 2]) (5 4 6 3 )(2 2 2 N S n N S n n
n n
N S n
χ χ
χ
= − + + ⋅ −
− − 3 )
= − + + +
3 ( [1]) (5 4 6 )(2 2 2 1
3 4 5 6
1 5 5 (5 4 6 )(2 1 1).
1 8 1 8 3
n n
N S
n χ
=
− −
= + + + +
= + + + − −
L
3
L+ )
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