ECCENTRIC CONNECTIVITY INDEX OF V-PHENYLENIC NANOTUBES
N. PRABHAKARA RAO*, K LAVANYA LAKSHMI
Department of mathematics, Bapatla Engg. College, Bapatla,Guntur (D.t), Andhra Pradesh, 522101, India
If G = (V, E) is a connected graph then the eccentric connectivity index of G is defined as . In this paper we obtain explicit formulas for eccentric connectivity index of phenylenic nanotubes.
(Received December 8, 2010; accepted December 17, 2010)
Keywords: Eccentric connectivity index, Nanotube; Phenylenic nanotubes
1. Introduction
Chemical compounds can be modeled by molecular graphs where the points represent atoms and the edges represents covalent bonds. This model has become an important tool in the prediction of physico-chemical, pharmacological and toxicological properties of a compound directly. The study of eccentric connectivity index gained more importance as the topological models involving this index show a high degree of predictability of pharmaceutical properties.
PI index of V-Phenylenic nanotube is computed by Amir Bahrani [1] in 2008. We compute eccentric connectivity index of nanotubes in this paper.
2. Main results
In this section, the eccentric connectivity index of the molecular graph of the V- Phenylenic nanotube VPHX [p, q] is computed, For simplicity, we denote this nanotube by V = V [p, q]. To compute eccentric connectivity index, we need which is maximum of
The edges in the longest path from x1,1 in V[4,6] and V[3,7] are shown below with dark edges.
* Corresponding author: [email protected], [email protected]
2
1 2 3
X1,1 X1,2 X1,3 X2,1 X2,2
X4p,1 X4p,2 X4p,q
q X1,q X2,2q-1 X2,2q
p
. . . . row 2 . . . . row 3
. . . . row 4p . . . . row 4p‐1
Graph of V [p,q] with p=4, q=6
1 2 3
X1,1 X1,2 X1,3 X1,q
X2,1 X2,2 X2,2q-1 X2,2q
X4p,1 X4p,2 X4p,q
p
q
Graph of V[p,q] with p=3, q=7
Theorem 2.1. If G is a phenylenic nanotube (see above figures) then ξ(G) =
Proof. From the lemmas given below the proof is clear.
Lemma. 2.2. ξ(G) =
Proof. In this case the graph has only 4 rows. The degree of every vertex in first and 4th rows is 2 where as the degree of every vertex in 2nd and 3rd rows is 3. We have
Case (1).
Then ξ(G) = 2[(2)(3)+(2)(3)(2)] = 36.
Case (2).
In this case the eccentricity of every vertex in every row is .
Then ξ (G) = 2 .
Case (3).
In this case the eccentricity of every vertex in every row is .
Then ξ(G) = 2 .
Lemma 2.3. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where .
Hence ξ(G) = = 2 = Lemma 2.4. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where . Hence ξ(G) =
= 2 =
Lemma 2.5. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where .
Hence ξ(G) = = 2
= .
Lemma 2.6. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where .
Now ξ(G) = = 2
= .
Lemma 2.7. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where .
Let be a vertex in th or throw.
The eccentricity of is where .
Hence ξ(G) = = 2
2 =
Lemma 2.8. ξ (G) =
.
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in ith row.
The eccentricity of is where .
Let be a vertex in th or throw.
The eccentricity of is where .
Hence ξ(G) = = 2
2
= .
Lemma 2.9. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in th or throw.
The eccentricity of is where .
Now ξ(G) = = 2
= .
Lemma 2.10. ξ (G) =
Proof. In this case the graph has 4 rows and the eccentricity of a vertex in ith row is same as the eccentricity of a vertex in (4p+1-i)th row where .
Here all the vertices in first and last rows are of degree 2 and the remaining vertices are of degree 3.
Let be a vertex in th or throw.
The eccentricity of is where .
Hence ξ(G) = = 2
= .
References
[1] Amir Bahrami, Javad Yazdani.: Padmakar-Ivan indes of H-Phenylenic nanotubes and nanotori, Digest Journal of Nanomaterials and Biostructures, 3(4), 265 (2008).
[2] Mahboubeh Saheli, Ali Reza Ashrafi.: The eccentric connectivity index of Armchair Polyhex nanotubes, Macedonian journal of chemistry and chemical engineering,
29(1), 71 (2010).
[3] F. Harary., Graph theory, Addison-Wesley.Reading MA. (1969).
[4] V. Sharma, R. Goswami and A.K.Madan.: Eccentric connectivity index,
A novel highly discriminating topological descriptor for structure-property and structure- activity studies, J. Chem. inf. Comput. Sci, 37, 273 (1997).