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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

i

vertex 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]).

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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

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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

( )s

d 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 [ ]

and

2 ( χ N S s [ − 1]

for

2.

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 0

d ijk = or ( )s (s 1) (1)

di j k di j k di j k

= − =L = for

s = 2, L , . n

Thus,

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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 obtain

3 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 mean

d

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

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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[ ] and

3 χ ( N S s [ − 1]

for

2.

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,

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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 obtain

3 ( [ ]) 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

+ )

References

[1] A. R. Ashrafi, P. Nikzad, Digest Journal of Nanomaterials and Biostructures, 4 (2), 269(2009).

[2] X. Li, I. Gutman, Mathematical Chemistry Monographs, 1, 330 (2006).

[3] S. Yousefi, A. R. Ashrafi, MATCH Commun. Math. Comput. Chem., 56, 169 (2006).

[4] M.B.Ahmadi, M. Sadeghimehr, Digest Journal of Nanomaterials and Biostructures, 4 (4), 639(2009).

[5] H. Yousefi-Azari, A. R. Ashrafi, M. H. Khalifeh, Digest Journal of Nanomaterials and Biostructures, 3(4), 315 (2008).

[6] H. Yousefi-Azari, A. R. Ashrafi, M. H. Khalife, Digest Journal of Nanomaterials and Biostructures, 3(4), 251 (2008).

[7] M. H. Khalifeh, H. Yousefi-Azari, A. R. Ashrafi, Digest Journal of Nanomaterials and Biostructures, 4(1), 63 (2009).

[8] A. R. Ashrafi, S. Yousefi, Nanoscale Res Lett., 2, 202 (2007).

[9] M. V. Diudea, A. Graovac, MATCH Commun. Math. Comput. Chem., 44, 93 (2001).

[10] M. V. Diudea, John P. E., MATCH Commun. Math. Comput. Chem., 44, 103 (2001).

[11] M. V. Diudea, MATCH Commun. Math. Comput. Chem., 45, 109 (2002).

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