Vertex PI Topological Index of Titania Carbon Nanotubes

Let G be a connected graph with vertex and edge sets V(G) and E(G), respectively. As usual, the distance between the vertices u and v of G is the number of edges in a minimal path connecting them, denoted as d(u,v). Define N_G(u) to be the set of all vertices adjacent to u. The diameter is the greatest distance between two vertices of G, denoted as diam(G).

Let e=uv be an edge of the graph G. The number of vertices of G whose distance to the vertex u is smaller than the distance to the vertex v is denoted by n_u(e). Analogously, n_v(e) is the number of vertices of G whose distance to the vertex v is smaller than the distance to the vertex u.

Suppose X denoted the class of all graphs. A map Top from X into real numbers is called a topological index if G ≌ H implies that Top(G) = Top(H). Obviously, the maps Top_vand Top_e defined as the number of edges and vertices respectively, are topological indices.

The Wiener index was the first reported topological index based on graph distances [1]. This index is defined as the sum of all distances between vertices of the graph under consideration. The Wiener index is applicable to acyclic graphs only. For cyclic compounds a novel molecular graph based descriptor, referred to as the Szeged index [2]. This is considered as the modification of Wiener index to cyclic graph. The Szeged index is defined as

$S z (G) = \sum_{e = u v \in E (G)} n_{u} \cdot n_{v}$ $$\begin{array}{} \displaystyle Sz(G)=\sum _{e=uv\in E(G)}n_{u} \cdot n_{v} \end{array}$$

Note that vertices equidistant to u and v are not counted.

For acyclic graphs the Szeged and Wiener indices coincide. As a consequence, Padmaker and Ivan introduced another index called Padmakar-Ivan index (PI_v) [3, 4]. PI of a graph G is defined as

$P I_{v} (G) = \sum_{e = u v \in E (G)} (n_{u} + n_{v}) .$ $$\begin{array}{} \displaystyle PI_{v} (G)=\sum_{e=uv\in E(G)}(n_{u} +n_{v} ). \end{array}$$

Note that vertices equidistant to u and v are not counted. Many methods for the calculation of these indices of some systems are considered in [5–11].

As a well-known semiconductor with numerous technological applications, Titania nanotubes are comprehensively studied in material sciences. Titania nanotubes were systematically synthesized during the last 10-15 years using different methods and carefully studied as prospective technological materials. Since the growth mechanism for TiO₂nanotubes is still not well defined, their comprehensive theoretical studies attract enhanced attention. The TiO₂ sheets with a thickness of a few atomic layers were found to be remarkably stable.

In this paper, we compute the vertex PI index of the Titania nanotubes. For further results we refer [12–14].

Main Results

The 2-dimensional graph of the Titania nanotube, TiO₂[m,n], is shown in Figure 1, where m and n denotes the number of octagons in a column and the number of octagons in a row of the Titania nanotube. This graph has 2(3n+2)(m+1) vertices and 10mn + 6m + 8n + 4 edges.

2-dimensional model to Titania nanotube TiO₂[m,n].

By using the orthogonal cuts and cut method of the Titania nanotubes, we can determine all edge cuts of the Titania nanotubes. The edge cut C(e) is an orthogonal cut, such that the set of all edges f∈E(G) are strongly co-distant to e. For further research and study of the cut method and orthogonal cuts in some classes of chemical graphs see [8, 9, 11, 15].

By using the cut method and finding orthogonal cuts, we can compute the quantities of n_u(e|TiO₂[m,n]) and n_v(e|TiO₂[m,n]), ∀e∈E(TiO₂[m,n]), which are the number of vertices in two sub-graphs TiO₂[m,n]-C(e). In case the Titania Nanotubes TiO₂[m,n] ∀e=uv∈E(TiO₂[m,n]), we denote n_u(e|TiO₂[m,n]) as the number of vertices in the left component of TiO₂[m,n]-C(e) and alternatively n_v(e|TiO₂[m,n]) as the number of vertices in the right component of TiO₂[m,n]-C(e), since all edges in TiO₂[m,n] Nanotubes sheets are oblique or horizontal.

Theorem 1

Let TiO₂[m,n] be the Titania Nanotubes, where m,n∈N. Then vertex PI index of TiO₂[m,n] is:

$P I_{v} (T i O_{2} (m, n)) = 2 (m + 1) (3 n + 2) (n + 1) (11 m + 9)$ $$\begin{array}{} \displaystyle PI_{v} (TiO_{2} \left(m,n\right))=2(m+1)(3n+2)(n+1)(11m+9) \end{array}$$

Proof. According to the structure of the Titania Nanotubes TiO₂[m,n] for all integer numbers m,n>1, we have following results:

For the edge e₁=u₁v₁ that belong to the first square of TiO₂[m,n] Nanotubes (in the first column and row), we see that

$\begin{matrix} n_{u 1} (e_{1} | T i O_{2} [m, n]) = 2 (m + 1) \\ n_{v 1} (e_{1} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - 2 (m + 1) = 6 m n + 2 m + 6 n + 2. \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_{u1}}({e_1}|Ti{O_2}[m,n]) = 2(m + 1)} \\ {{n_{v1}}({e_1}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - 2(m + 1) = 6mn + 2m + 6n + 2.} \\ \end{array} \end{array}$$

For the edge e₂=u₂v_2:

$n_{u 2} (e_{2} | T i O_{2} [m, n]) = 3 \times 2 (m + 1) + 1 \times 2 (m + 1) = 8 (m + 1)$ $$\begin{array}{} \displaystyle {n_{u2}}({e_2}|Ti{O_2}[m,n]) = 3 \times 2(m + 1) + 1 \times 2(m + 1) = 8(m + 1) \end{array}$$

and

$n_{v 2} (e_{2} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - 8 (m + 1) = 6 m n - 4 m + 6 n - 4.$ $$\begin{array}{} \displaystyle {n_{v2}}({e_2}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - 8(m + 1) = 6mn - 4m + 6n - 4. \end{array}$$

For the edge e_n+1=u_n+1v_n+1:

$n_{u}_{n + 1} (e_{n + 1} | T i O_{2} [m, n]) = (3 n + 1) \times 2 (m + 1)$ $$\begin{array}{} \displaystyle {n_u}_{n + 1}({e_{n + 1}}|Ti{O_2}[m,n]) = (3n + 1) \times 2(m + 1) \end{array}$$

and

$n_{v}_{n + 1} (e_{n + 1} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - (6 m n + 6 n + 2 m + 2) = 2 (m + 1) .$ $$\begin{array}{} \displaystyle {n_v}_{n + 1}({e_{n + 1}}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - (6mn + 6n + 2m + 2) = 2(m + 1). \end{array}$$

Thus, by a simple induction for i=1,2,. . . ,n; we can see that for the edge e_i=u_iv_i:

$n_{u}_{i} (e_{i} | T i O_{2} [m, n]) = 2 (m + 1) \times (3 (i - 1) + 1)$ $$\begin{array}{} \displaystyle {n_u}_i({e_i}|Ti{O_2}[m,n]) = 2(m + 1) \times (3(i - 1) + 1) \end{array}$$

and

$\begin{matrix} n_{v}_{i} (e_{i} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - (6 m i + 6 i - 4 m - 4) \\ = 6 m (n - i) + 6 (n - i) + 8 (m + 1) \\ = 6 (m + 1) (n - i) + 8 (m + 1) \\ = 2 (m + 1) (3 (n - i) + 4) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_v}_i({e_i}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - (6mi + 6i - 4m - 4)}\\ { = 6m(n - i) + 6(n - i) + 8(m + 1)}\\ { = 6(m + 1)(n - i) + 8(m + 1)}\\ { = 2(m + 1)(3(n - i) + 4).} \end{array} \end{array}$$

Let the edge f₁=u₁v₁∈E(TiO₂[m,n]) be the first oblique edge in the first square of TiO₂[m,n] Nanotubes (in the first column and row), we see that

$n_{u 1} (f_{1} | T i O_{2} [m, n]) = m + 1$ $$\begin{array}{} \displaystyle {n_{u1}}({f_1}|Ti{O_2}[m,n]) = m + 1 \end{array}$$

and

$\begin{matrix} n_{v 1} (f_{1} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - (m + 1) \\ = 6 m n + 3 m + 6 n + 3 \\ = 6 n (m + 1) + 3 (m + 1) \\ = (6 n + 3) (m + 1) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_{v1}}({f_1}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - (m + 1)} \\ { = 6mn + 3m + 6n + 3} \\ { = 6n(m + 1) + 3(m + 1)} \\ { = (6n + 3)(m + 1).} \\ \end{array} \end{array}$$

For the edge f₂=u₂v_2:

$n_{u 2} (f_{2} | T i O_{2} [m, n]) = (m + 1) + 3 \times 2 (m + 1) = 7 (m + 1)$ $$\begin{array}{} \displaystyle {n_{u2}}({f_2}|Ti{O_2}[m,n]) = (m + 1) + 3 \times 2(m + 1) = 7(m + 1) \end{array}$$

and

$n_{v 2} (f_{2} | T i O_{2} [m, n]) = 6 m n + 4 m + 6 n + 4 - 7 (m + 1) = 6 n (m + 1) - 3 (m + 1) = (6 n - 3) (m + 1) .$ $$\begin{array}{} \displaystyle {n_{v2}}({f_2}|Ti{O_2}[m,n]) = 6mn + 4m + 6n + 4 - 7(m + 1) = 6n(m + 1) - 3(m + 1) = (6n - 3)(m + 1). \end{array}$$

For the edge f_(n+1)=uv_:

$n_{u}_{(n + 1)} (f_{(n + 1)} | T i O_{2} [m, n]) = (m + 1) + 3 n \times 2 (m + 1) = (6 n + 1) (m + 1)$ $$\begin{array}{} \displaystyle {n_u}_{(n + 1)}({f_{(n + 1)}}|Ti{O_2}[m,n]) = (m + 1) + 3n \times 2(m + 1) = (6n + 1)(m + 1) \end{array}$$

and

$n_{v}_{(n + 1)} (f_{(n + 1)} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - (6 n + 1) (m + 1) = 3 (m + 1) .$ $$\begin{array}{} \displaystyle {n_v}_{(n + 1)}({f_{(n + 1)}}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - (6n + 1)(m + 1) = 3(m + 1). \end{array}$$

Therefore, by a simple induction for j=1,2,. . . ,n+1; we can proof the previous statement.

For the edge f_j=u_jv_j:

$n_{u}_{j} (f_{j} | T j O_{2} (m, n)) = 3 (j - 1) \times 2 (m + 1) + (m + 1) = (m + 1) (6 j - 5)$ $$\begin{array}{} \displaystyle {n_u}_j({f_j}|Tj{O_2}(m,n)) = 3(j - 1) \times 2(m + 1) + (m + 1) = (m + 1)(6j - 5) \end{array}$$

and

$\begin{matrix} n_{v}_{j} (f_{j} | T j O_{2} (m, n)) = (6 n + 4) (m + 1) - (m + 1) (6 j - 5) \\ = (m + 1) (6 n - 6 j + 9) \\ = 3 (m + 1) (2 n + 3 - 2 j) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_v}_j({f_j}|Tj{O_2}(m,n)) = (6n + 4)(m + 1) - (m + 1)(6j - 5)} \\ { = (m + 1)(6n - 6j + 9)} \\ { = 3(m + 1)(2n + 3 - 2j).} \\ \end{array} \end{array}$$

Let the edge g₁=u₁v₁∈E(TiO₂[m,n]) be the second oblique edge in the first square of TiO₂[m,n] Nanotubes (in the first column and row), so we have

$n_{u 1} (g_{1} | T i O_{2} [m, n]) = 2 (m + 1) + (m + 1)$ $$\begin{array}{} \displaystyle {n_{u1}}({g_1}|Ti{O_2}[m,n]) = 2(m + 1) + (m + 1) \end{array}$$

and

$n_{v 1} (g_{1} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - 3 (m + 1) = (6 n + 1) (m + 1)$ $$\begin{array}{} \displaystyle {n_{v1}}({g_1}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - 3(m + 1) = (6n + 1)(m + 1) \end{array}$$

For g₂=u₂v_2:

$n_{u 2} (g_{2} | T i O_{2} [m, n]) = 3 \times 2 (m + 1) + 2 (m + 1) + (m + 1) = 9 (m + 1)$ $$\begin{array}{} \displaystyle {n_{u2}}({g_2}|Ti{O_2}[m,n]) = 3 \times 2(m + 1) + 2(m + 1) + (m + 1) = 9(m + 1) \end{array}$$

and

$n_{v 2} (g_{2} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - 9 (m + 1) = (6 n - 5) (m + 1)$ $$\begin{array}{} \displaystyle {n_{v2}}({g_2}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - 9(m + 1) = (6n - 5)(m + 1) \end{array}$$

For g_n+1=u_n+1v_n+1:

$n_{u}_{n + 1} (g_{n + 1} | T i O_{2} [m, n]) = 3 n \times 2 (m + 1) + 2 (m + 1) + (m + 1) = (6 n + 3) (m + 1)$ $$\begin{array}{} \displaystyle {n_u}_{n + 1}({g_{n + 1}}|Ti{O_2}[m,n]) = 3n \times 2(m + 1) + 2(m + 1) + (m + 1) = (6n + 3)(m + 1) \end{array}$$

and

$n_{v}_{n + 1} (g_{n + 1} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - (6 n + 3) (m + 1) = (m + 1) .$ $$\begin{array}{} \displaystyle {n_v}_{n + 1}({g_{n + 1}}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - (6n + 3)(m + 1) = (m + 1). \end{array}$$

And these imply that ∀_j=1,2,. . . ,n+1

$n_{u}_{j} (g_{j} | T j O_{2} (m, n)) = 3 j \times 2 (m + 1) + 3 (m + 1) = (m + 1) (6 j + 3)$ $$\begin{array}{} \displaystyle {n_u}_j({g_j}|Tj{O_2}(m,n)) = 3j \times 2(m + 1) + 3(m + 1) = (m + 1)(6j + 3) \end{array}$$

and

$n_{v}_{j} (g_{j} | T j O_{2} (m, n)) = (6 n + 4) (m + 1) - (m + 1) (6 j + 3) = (m + 1) (6 n - 6 j + 1) .$ $$\begin{array}{} \displaystyle {n_v}_j({g_j}|Tj{O_2}(m,n)) = (6n + 4)(m + 1) - (m + 1)(6j + 3) = (m + 1)(6n - 6j + 1). \end{array}$$

Finally, let h₁=u₁v₁ & l₂=u₂v₂∈E(TiO₂[m,n]) be the first and second oblique edges in the second square of the first row (or the first square in the second column) of TiO₂[m,n] Nanotubes, then

and

$\begin{matrix} n_{u 1} (h_{1} T i O_{2} [m, n]) = 2 \times 2 (m + 1) \\ n_{u 2} (l_{2} T i O_{2} [m, n]) = 3 \times 2 (m + 1) \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_{u1}}({h_1}Ti{O_2}[m,n]) = \it2 \times \it2(m + 1)}\\ {\qquad \qquad \;\:{n_{u2}}({l_2}Ti{O_2}[m,n]) = \it3 \times \it2(m + 1)} \end{array} \end{array}$$

and

$\begin{matrix} n_{v 1} (h_{1} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - 2 (m + 1) = (6 n + 2) (m + 1) \\ n_{v 2} (l_{2} | T i O_{2} [m, n]) = (6 n + 1) (m + 1) \end{matrix}$ $$\begin{array}{} \displaystyle % MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! \begin{array}{*{20}{c}} {{n_{v1}}({h_1}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - 2(m + 1) = (6n + 2)(m + 1)}\\ {{n_{v2}}({l_2}|Ti{O_2}[m,n]) = (6n + 1)(m + 1)} \end{array}% MathType!End!2!1! \end{array}$$

And by a simple induction on ∀i=1,2,. . . ,n; for the edges h_i=u_iv_i and l_i=a_ib_i we have

$n_{u}_{i} (h_{i} | T i O_{2} [m, n]) = 3 (i - 1) \times 2 (m + 1) + 2 \times 2 (m + 1) = 2 (m + 1) (3 i - 1)$ $$\begin{array}{} \displaystyle {n_u}_i({h_i}|Ti{O_2}[m,n]) = 3(i - 1) \times 2(m + 1) + 2 \times 2(m + 1) = 2(m + 1)(3i - 1) \end{array}$$

and

$\begin{matrix} n_{v}_{i} (h_{i} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - 2 (m + 1) (3 i - 1) \\ = (m + 1) (6 n - 6 i + 6) \\ = 6 (m + 1) (n + 1 - i) . \\ n_{a}_{i} (l_{i} | T i O_{2} [m, n]) = 3 (i - 1) \times 2 (m + 1) + 3 \times 2 (m + 1) = 6 i (m + 1) \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_v}_i({h_i}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - 2(m + 1)(3i - 1)}\\ { = (m + 1)(6n - 6i + 6)}\\ { = 6(m + 1)(n + 1 - i).}\\ {{n_a}_i({l_i}|Ti{O_2}[m,n]) = 3(i - 1) \times 2(m + 1) + 3 \times 2(m + 1) = 6i(m + 1)} \end{array} \end{array}$$

and

$\begin{matrix} n_{b}_{i} (l_{i} | T i O_{2} [m, n]) = (6 n + 4) (m + 1) - 6 i (m + 1) \\ = (m + 1) (6 n - 6 i + 4) \\ = 3 (m + 1) (3 n + 2 - 2 i) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{n_b}_i({l_i}|Ti{O_2}[m,n]) = (6n + 4)(m + 1) - 6i(m + 1)}\\ { = (m + 1)(6n - 6i + 4)}\\ { = 3(m + 1)(3n + 2 - 2i).} \end{array} \end{array}$$

On the other hands, by according to Figure 2, we can see that the size of all orthogonal cuts for these edge categories in the Titania Nanotubes TiO₂[m,n] are equal to (∀i=1,2,. . . ,n+1):

Categories for edges of the Titania Carbon Nanotubes.

$| C (e_{i}) | = | C (h_{i}) | = | C (l_{i}) | = 2 (m + 1)$ $$\begin{array}{} \displaystyle |C({e_i})| = |C({h_i})| = |C({l_i})| = 2(m + 1) \end{array}$$

and

$| C (f_{i}) | = | C (g_{i}) | = 2 m + 1.$ $$\begin{array}{} \displaystyle |C({f_i})| = |C({g_i})| = 2m + 1. \end{array}$$

From the above calculations and Figure 2, we can compute the vertex PI index of Titania Nanotubes TiO₂[m,n].

$P I_{v} (T i O_{2} [m, n]) = \sum_{e = u v \in E (G)} (n_{u} (e | T i O_{2} (m, n) + n_{v} (e | T i O_{2} (m, n))$ $$\begin{array}{} \displaystyle P{I_v}(Ti{O_2}[m,n]) = \sum\nolimits_{e = uv \in E(G)} {({n_u}(e|Ti{O_2}(m,n) + {n_v}(e|Ti{O_2}(m,n))} \end{array}$$

$\begin{matrix} = \sum_{\begin{matrix} e_{i} = u v \in E (T i O_{2} (m, n)) \\ \forall i = 1, 2, \dots, n + 1 \end{matrix}} | C (e_{i}) | (n_{u} (e_{i} | T i O_{2} (m, n)) + n_{v} (e_{i} | T i O_{2} (m, n))) \\ + \sum_{_{\begin{matrix} f_{i} = u v \in E (T i O_{2} (m, n)) \\ \forall i = 1, 2, \dots, n + 1 \end{matrix}}} | C (f_{i}) | [n_{u} (f_{i} | T i O_{2} (m, n)) + n_{v} (f_{i} | T i O_{2} (m, n))] \\ + \sum_{\begin{matrix} g_{i} = u v \in E (T i O_{2} (m, n)) \\ \forall i = 1, 2, \dots, n + 1 \end{matrix}} | C (g_{i}) | [n_{u} (g_{i} | T i O_{2} (m, n)) + n_{v} (g_{i} | T i O_{2} (m, n))] \\ + \sum_{\begin{matrix} h_{i} = u v \in E (T i O_{2} (m, n)) \\ \forall i = 1, 2, \dots, n \end{matrix}} | C (h_{i}) | [n_{u} (h_{i} | T i O_{2} (m, n)) + n_{v} (h_{i} | T i O_{2} (m, n))] \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} { = \sum\limits_{\begin{array}{*{20}{c}} {{e_i} = uv \in E(Ti{O_2}(m,n))} \hfill \\ {\forall i = 1,2, \ldots ,n + 1} \hfill \\ \end{array}} {|C({e_i})|({n_u}({e_i}|Ti{O_2}(m,n)) + {n_v}({e_i}|Ti{O_2}(m,n)))} } \hfill \\ { + \sum\limits_{_{\begin{array}{*{20}{c}} {{f_i} = uv \in E(Ti{O_2}(m,n))} \hfill \\ {\forall i = 1,2, \ldots ,n + 1} \hfill \\ \end{array}}} {|C({f_i})|[{n_u}({f_i}|Ti{O_2}(m,n)) + {n_v}({f_i}|Ti{O_2}(m,n))]} } \hfill \\ { + \sum\limits_{\begin{array}{*{20}{c}} {{g_i} = uv \in E(Ti{O_2}(m,n))} \hfill \\ {\forall i = 1,2, \ldots ,n + 1} \hfill \\ \end{array}} {|C({g_i})|[{n_u}({g_i}|Ti{O_2}(m,n)) + {n_v}({g_i}|Ti{O_2}(m,n))]} } \hfill \\ { + \sum\limits_{\begin{array}{*{20}{c}} {{h_i} = uv \in E(Ti{O_2}(m,n))} \hfill \\ {\forall i = 1,2, \ldots ,n} \hfill \\ \end{array}} {|C({h_i})|[{n_u}({h_i}|Ti{O_2}(m,n)) + {n_v}({h_i}|Ti{O_2}(m,n))]} } \hfill \\ \end{array} \end{array}$$

$\begin{matrix} = \sum_{i = 1}^{n + 1} 2 (m + 1) (2 (m + 1) (3 (i - 1) + 1) + 2 (m + 1) (3 (n - i) + 4)) \\ + \sum_{i = 1}^{n + 1} 2 (m + 1) ((m + 1) (6 i - 5) + 3 (m + 1) (2 n + 3 - 2 i)) \\ + \sum_{i = 1}^{n + 1} 2 (m + 1) ((m + 1) (6 i + 3) + (m + 1) (6 n - 6 i + 1)) \\ + \sum_{i = 1}^{n} (2 m + 1) (2 (m + 1) (3 i - 1) + (m + 1) (n + 1 - i)) \\ + \sum_{i = 1}^{n} (2 m + 1) (6 i (m + 1) + 3 (m + 1) (3 n + 2 - 2 i)) \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} { = \sum\nolimits_{i = 1}^{n + 1} {2(m + 1)(2(m + 1)(3(i - 1) + 1) + 2(m + 1)(3(n - i) + 4))} } \hfill \\ { + \sum\nolimits_{i = 1}^{n + 1} {2(m + 1)((m + 1)(6i - 5) + 3(m + 1)(2n + 3 - 2i))} } \hfill \\ { + \sum\nolimits_{i = 1}^{n + 1} {2(m + 1)((m + 1)(6i + 3) + (m + 1)(6n - 6i + 1))} } \hfill \\ { + \sum\nolimits_{i = 1}^n {(2m + 1)(2(m + 1)(3i - 1) + (m + 1)(n + 1 - i))} } \hfill \\ { + \sum\nolimits_{i = 1}^n {(2m + 1)(6i(m + 1) + 3(m + 1)(3n + 2 - 2i))} } \hfill \\ \end{array} \end{array}$$

$\begin{matrix} \begin{matrix} = 4 {(m + 1)}^{2} \sum_{i = 1}^{n + 1} (3 n + 2) + (2 m + 1) (m + 1) \sum_{i = 1}^{n + 1} (6 n + 4) + (2 m + 1) (m + 1) \sum_{i = 1}^{n + 1} (6 n + 4) + \\ 4 {(m + 1)}^{2} \sum_{i = 1}^{n + 1} (3 n + 2) + 6 {(m + 1)}^{2} \sum_{i = 1}^{n + 1} (3 n + 2) \\ = 4 {(m + 1)}^{2} (3 n + 2) (n + 1) + (2 m + 1) (m + 1) (6 n + 4) (n + 1) + (2 m + 1) (m + 1) (6 n + 4) (n + 1) \\ + 4 {(m + 1)}^{2} (3 n + 2) (n + 1) + 6 {(m + 1)}^{2} (3 n + 2) (n + 1) \\ = 2 (m + 1) (3 n + 2) (n + 1) (11 m + 9) \end{matrix} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} { = 4{{(m + 1)}^2}\sum\limits_{i = 1}^{n + 1} {(3n + 2)} + (2m + 1)(m + 1)\sum\limits_{i = 1}^{n + 1} {(6n + 4)} + (2m + 1)(m + 1)\sum\limits_{i = 1}^{n + 1} {(6n + 4)} + }\\ {4{{(m + 1)}^2}\sum\limits_{i = 1}^{n + 1} {(3n + 2)} + 6{{(m + 1)}^2}\sum\limits_{i = 1}^{n + 1} {(3n + 2)} }\\ { = 4{{(m + 1)}^2}(3n + 2)(n + 1) + (2m + 1)(m + 1)(6n + 4)(n + 1) + (2m + 1)(m + 1)(6n + 4)(n + 1)}\\ { + 4{{(m + 1)}^2}(3n + 2)(n + 1) + 6{{(m + 1)}^2}(3n + 2)(n + 1)}\\ {{{ = 2(m + 1)(3n + 2)(n + 1)(11m + 9)}}} \end{array} \end{array}$$

which is the required result and the proof is over

eISSN:: 2444-8656
Language:: English

Publication timeframe:: Volume Open
Journal Subjects:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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Vertex PI_v Topological Index of Titania Carbon Nanotubes TiO₂(m,n)

Published Online: Feb 04, 2016

Page range: 175 - 182

Received: Oct 18, 2015

Accepted: Jan 31, 2016

DOI: https://doi.org/10.21042/AMNS.2016.1.00013

© 2016 Mohammad Reza Farahani, Muhammad Kamran Jamil, Muhammad Imran, published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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Vertex PIv Topological Index of Titania Carbon Nanotubes TiO2(m,n)