## 2 Topological Indices of Line Graph of the Subdivision Graph of Rooted Product of Graphs
### Topological Indices of Line Graph of the Subdivision Graph of \(C_{n}\{P_{k}\}\)  
Let \(P_{n}\) and \(C_{n}\) denotes the path and cycle with order \(n\). Let \(C_{n}\{P_{k+1}\}\) be the cluster product of cycle of length \(n\) with a path of length \(k\). The graph of \(C_{n}\{P_{k+1}\}\) is shown in Figure 1. It clearly shows that the number of vertices and number of edges in \(C_{n}\{P_{k+1}\}\) are both \(n+nk\). In the next theorem we will compute the general Zagreb index of the line graph of subdivision graph of \(C_{n}\{P_{k+1}\}\). **Theorem 1**. : _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{P_{k+1}\}\). Then \(M_{\alpha}(G)=n+(nk-n)2^{\alpha+1}+n\cdot(3)^{\alpha+1}\)._  
Proof. : The structure of \(G\) is shown in Figure 1. There are total \(2n+2 nk\) vertices among which the number of vertices of degree 1, 2, and 3 are \(n\), \(2 nk-2n\), and \(3n\), respectively. Thus, \(M_{\alpha}(G)\) is obtained in terms of Equation (2). **Theorem 2**. : _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{P_{k+1}\}\). Then_  
1. \(R_{\alpha}(G)=\left\{ \begin{array}{ll}n\cdot 3^{\alpha}+4n\cdot 3^{2\alpha},&\text{if}\ \ k=1\ ;\\ n\cdot 2^{\alpha}+(2 nk-3n)4^{\alpha}+n\cdot 6^{\alpha}+4n\cdot 3^{2\alpha},& \text{if}\ \ k>1\.\end{array} \right.\)__
2. \(\chi_{\alpha}(G)=\left\{ \begin{array}{ll}n\cdot 4^{\alpha}+4n\cdot 6^{ \alpha},&\text{if}\ \ k=1\ ;\\ n\cdot 3^{\alpha}+(2 nk-3n)4^{\alpha}+n\cdot 5^{\alpha}+4n\cdot 6^{\alpha},& \text{if}\ \ k>1\.\end{array} \right.\)__
3. \(ABC(G)=\left\{ \begin{array}{ll}\left(\frac{1}{3}\sqrt{6}+\frac{8}{3}\right)n,&\text{if}\ \ k=1\ ;\\ \left(-\frac{\sqrt{2}}{2}+\frac{8}{3}\right)n+nk\sqrt{2},&\text{if}\ \ k>1\.\end{array} \right.\)__
4. \(GA(G)=\left\{ \begin{array}{ll}\left(\frac{1}{2}\sqrt{3}+4\right)n,&\text{if} \ \ k=1\ ;\\ \left(\frac{2}{3}\sqrt{2}+1+\frac{2}{5}\sqrt{6}\right)n+2\,nk,&\text{if}\ \ k>1\.\end{array} \right.\)__  
Proof. : It is easily seen that \(|E(G)|=3n+2 kn\). The edge partitions, based on the degrees of the vertices for the cases \(k=1\) and \(k>1\), are shown in Tables 1 and 2, respectively. In view of (1), (3), (4), and (6), we infer the required conclusions. Figure 1: The graph of \(C_{n}\{P_{k}\}\); Line graph of the subdivision graph of \(C_{n}\{P_{k}\}\). **Theorem 3**: _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{P_{k+1}\}\). Then 1. \(ABC_{4}(G)=\)_  
\[\left\{\begin{array}{ll}\left(\frac{2}{21}\sqrt{42}+\frac{2}{3}\sqrt{2}+\frac{ 8}{9}\right)n,&\text{if}\ \ k=1\ ;\\ \left(\frac{1}{2}\sqrt{2}+\frac{1}{5}\sqrt{1  
**Theorem 4**. : _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{S_{m+1}\}\). Then \(M_{\alpha}(G)=mn+n\cdot 3^{\alpha+1}+n(m+1)^{\alpha+1}\)._  
Proof. : The structure of \(G\) is manifested in Figure 2. There is a total \(2 mn+4n\) vertices among which the number of vertices with degree 1, 3, and \(m+1\) are \(mn\), \(3n\), and \(mn+n\), respectively. Thus, the expression of \(M_{\alpha}(G)\) is deduced by means of (2). **Theorem 5**. : _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{S_{m+1}\}\). Then_  
1. \(R_{\alpha}(G)=mn(m+1)^{\alpha}+n(3m+3)^{\alpha}+4n\cdot 9^{\alpha}+\frac{mn}{2}(m+1) ^{2\alpha+1}\)_._
2. \(\chi_{\alpha}(G)=mn(m+2)^{\alpha}+n(m+4)^{\alpha}+4n\cdot 6^{\alpha}+\frac{mn(m+1)}{ 2}(2m+2)^{\alpha}\)_._
3. \(ABC(G)=mn\sqrt{\frac{m}{m+1}}+\frac{n}{\sqrt{3}}\sqrt{\frac{2+m}{m+1}}+\frac{8 n}{3}+\frac{mn\sqrt{m}}{\sqrt{2}}\)_._
4. \(GA(G)=\frac{2 mn\sqrt{m+1}}{2+m}+\frac{2n\sqrt{3+3}}{4+m}+4n+\frac{mn(m+1)}{ 2}\)_._  
Proof. : It is noted that there are \(\frac{1}{2}(m^{2}n+3 mn+10n)\) edges in \(G\). Its edge partition, follow from the degrees of the vertices are depicted in Table 6. From (1), (3), (4), and (6), we get the required results. **Theorem 6**. : _Let \(G\) be a graph isomorphic to the line graph of the subdivision graph of \(C_{n}\{S_{m+1}\}\) and let \(\beta=m^{2}+m+1\). Then_  
1.