In this section, we discussed the techniques to find the eccentricity of vertices of the Polycyclic Aromatic Hydrocarbons PAHk and gave the closed formula of third Zagreb index of PAHk. We use the ring cut method [17,18] to divide the vertices in partition sets. From Figure 2 we have the vertex set as $\begin{array}{}
\displaystyle
V(PA{H_k}) = \left\{ {{\alpha _{z,l}},\beta _{z,l}^i,\gamma _{z,j}^i:l = 1,...,k,j \in {Z_i},l \in {Z_{i - 1}}\& z \in {Z_6}} \right\}
\end{array}$, where α,β and γ represents the vertices with degree 1, 3 and 3, respectively, and Zi = {1,2,...,i}.
Theorem 1
Let G be the Polycyclic aromatic hydrocarbons PAHk. Then the third Zagreb index of PAHk is equal to
![]()
$$\begin{array}{}
\displaystyle
{M_3}(PA{H_k}) = 6\mathop \Sigma \limits_{i = 1}^k (12i(i + k) + 2k - 9i + 5).
\end{array}$$
Proof. To obtain the final result we used the ring cut method. The i-th ring cut contains 6(2i − 1) vertices also $\begin{array}{}
\displaystyle
d(\beta _{z,l}^i,\beta _{z,l}^k) = d(\gamma _{z,j}^i,\gamma _{z,j}^k) = 2(k - i)
\end{array}$. So, we have
1. For all vertices αz,j of PAHk ( j ∊ Zk, z Ῥ Z6)
![]()
$$\begin{array}{}
\displaystyle
\varepsilon (\alpha _{z,j}^{}) = \underbrace {d({\alpha _{z,j}},\gamma _{z,j}^k)}_1 + \underbrace {d(\gamma _{z,j}^k,\gamma _{z',j'}^k)}_{4k - 1} + \underbrace {d(\gamma _{z',j'}^k,{\alpha _{z',j'}})}_1 = 4k + 1
\end{array}$$
1. For all vertices $\begin{array}{}
\displaystyle
\beta _{z,j}^i
\end{array}$of PAHk (∀i = 1,···,k;z ∈ Z6, j ∊ Zi−1)
![]()
$$\begin{array}{}
\displaystyle
\textit{$\varepsilon (\beta_{z,j}^{i}
)=\underbrace{d(\beta _{z,j}^{i} ,\beta _{z+3,j}^{i}
)}_{4i-3}+\underbrace{d(\beta _{z+3,j}^{i} ,\gamma _{z+3,j}^{k}
)}_{2(k-i)+1}+\underbrace{d(\gamma _{z+3,j}^{k} ,\alpha _{z+3,j}
)}_{1}=2k+2i-1$.}
\end{array}$$
1. For all vertices $\begin{array}{}
\displaystyle
\gamma _{z,j}^i
\end{array}$ of PAHn (∀i = 1,···,k;z ∊ Z6, j ∊ Zi.
![]()
$$\begin{array}{}
\displaystyle
\textit{$\varepsilon (\gamma _{z,j}^{i}
)=\underbrace{d(\gamma _{z,j}^{i},\gamma _{z+3,j}^{i}
)}_{4i-1}+\underbrace{d(\gamma_{z+3,j}^{i},\gamma _{z+3,j}^{k}
)}_{2(k-i)}+\underbrace{d(\gamma_{z+3,j}^{k},\alpha _{z+3,j}
)}_{1}=2(k+i)$.}
\end{array}$$
Now we apply these results on the definition of third Zagreb index to obtain the required result.
![]()
$$\begin{array}{}
\displaystyle
\begin{array}{*{20}{c}}
{{M_3}\left( {PA{H_k}} \right)} \hfill & = \hfill & {\sum\limits_{uv \in E(G)} {\left( {\varepsilon \left( u \right) + \varepsilon \left( v \right)} \right)} } \hfill \\
{} \hfill & = \hfill & {\left( {{\sum _{\beta _{z,j}^i\gamma _{z,j}^i \in {\rm{E}}\left( {{H_k}} \right)}}\left( {\varepsilon \left( {\beta _{z,j}^i} \right) + \varepsilon \left( {\gamma _{z,j}^i} \right)} \right)} \right) + \left( {{\sum _{\beta _{z,j}^i\gamma _{z,j + 1}^i \in {\rm{E}}\left( {{H_k}} \right)}}\left( {\varepsilon \left( {\beta _{z,j}^i} \right) + \varepsilon \left( {\gamma _{z,j + 1}^i} \right)} \right)} \right)} \hfill \\
{} \hfill & {} \hfill & { + \left( {{\sum _{\beta _{z,j}^i\gamma _{z,j}^{i - 1} \in {\rm{E}}\left( {{H_k}} \right)}}\left( {\varepsilon \left( {\beta _{z,j}^{i + 1}} \right) + \varepsilon \left( {\gamma _{z,j}^{i - 1}} \right)} \right)} \right) + \left( {{\sum _{\gamma _{z,i}^i\gamma _{z + 1,1}^i \in {\rm{E}}\left( {{H_k}} \right)}}\left( {\varepsilon \left( {\gamma _{z,i}^i} \right) + \varepsilon \left( {\gamma _{z + 1,1}^i} \right)} \right)} \right)} \hfill \\
{} \hfill & {} \hfill & { + \left( {{\sum _{{\alpha _{z,j}}\gamma _{z,j}^k}}\left( {\varepsilon \left( {{\alpha _{z,j}}} \right) + \varepsilon \left( {\gamma _{z,j}^k} \right)} \right)} \right)} \hfill \\
{} \hfill & = \hfill & {\sum\limits_{z = 1}^6 {\left( {\sum\limits_{i = 2}^k {\sum\limits_{j = 1}^i {\left( {\varepsilon \left( {\beta _{z,j}^i} \right) + \varepsilon \left( {\gamma _{z,j}^i} \right)} \right)} } } \right)} + \sum\limits_{z = 1}^6 {\left( {\sum\limits_{i = 2}^k {\sum\limits_{j = 1}^i {\left( {\varepsilon \left( {\beta _{z,j}^i} \right) + \varepsilon \left( {\gamma _{z,j + 1}^i} \right)} \right)} } } \right)} } \hfill \\
{} \hfill & {} \hfill & { + \sum\limits_{z = 1}^6 {\left( {\sum\limits_{i = 2}^k {\sum\limits_{j = 1}^i {\left( {\varepsilon \left( {\beta _{z,j}^{i + 1}} \right) + \varepsilon \left( {\gamma _{z,j}^{i - 1}} \right)} \right)} } } \right)} + \sum\limits_{z = 1}^6 {\left( {\sum\limits_{i = 2}^k {\left( {\varepsilon \left( {\gamma _{z,i}^i} \right) + \varepsilon \left( {\gamma _{z + 1,1}^i} \right)} \right)} } \right)} } \hfill \\
{} \hfill & {} \hfill & { + \sum\limits_{z = 1}^6 {\left( {\sum\limits_{i = 1}^k {\left( {\varepsilon \left( {{\alpha _{z,j}}} \right) + \varepsilon \left( {\gamma _{z,j}^k} \right)} \right)} } \right)} } \hfill \\
{} \hfill & = \hfill & {\sum\limits_{i = 2}^k 6 (i - 1)\left[ {\left( {2k + 2i - 1} \right) + \left( {2k + 2i - 2} \right)} \right] + \sum\limits_{i = 2}^k 6 (i - 1)\left[ {\left( {2k + 2i - 1} \right) + \left( {2k + 2i - 2} \right)} \right]} \hfill \\
{} \hfill & {} \hfill & { + \sum\limits_{i = 1}^k 6 i\left[ {\left( {2k + 2i - 1} \right) + \left( {2k + 2i} \right)} \right] + \sum\limits_{i = 1}^k 6 \left[ {2\left( {2k + 2i - 1} \right)} \right]} \hfill \\
{} \hfill & {} \hfill & { + \sum\limits_{i = 1}^k 6 \left[ {\left( {4k + 1} \right) + \left( {2k + 2i} \right)} \right]} \hfill \\
{} \hfill & = \hfill & {\sum\limits_{i = 1}^k {\left( {12\left( {i - 1} \right)\left( {4k + 4i - 3} \right)} \right)} + \sum\limits_{i = 1}^k 6 i\left( {4k + 4i - 1} \right) + \sum\limits_{i = 1}^k 1 2\left( {2k + 2i - 1} \right)} \hfill \\
{} \hfill & {} \hfill & { + \sum\limits_{i = 1}^k 6 \left( {6k + 2i + 1} \right)} \hfill \\
{} \hfill & = \hfill & {6\sum\limits_{i = 1}^k {\left( {12i\left( {i + k} \right) + 2k - 9i + 5} \right)} } \hfill \\
\end{array}
\end{array}$$
which ends the proof.
