On ABS Estrada index of trees

Let G be a graph with n vertices, and \({d_{i}}\) $d_i$ be the degree of its i-th vertex. The ABS matrix of G is the square matrix of order n whose (i, j)-entry is equal to \({\sqrt{({d_{i}+d_{j}-2})/({d_{i}+d_{j}})}}\) $\sqrt{(d_i+d_j-2)/(d_i+d_j)}$ if the i-th vertex and the j-th vertex of G are adjacent, and 0 otherwise. Let \(\rho _1\ge \rho _2\ge \cdots \ge \rho _{n}\) ${\rho }_1\ge {\rho }_2\ge \cdots \ge {\rho }_n$ be the eigenvalues of the ABS matrix of G. Then the ABS Estrada index of G, denoted by \(E_{ABS}(G)\) $E_{\mathit{ABS}}(G)$ , is defined as \(E_{ABS}(G)=\sum _{i=1}^{n}e^{\rho _i}\) $E_{\mathit{ABS}}(G)={\sum }_{i=1}^n{e}^{{\rho }_i}$ . In this paper, the chemical importance of the ABS Estrada index is investigated and it is shown that the predictive ability of ABS Estrada index is stronger than other connectivity Estrada indices (including Randić Estrada index, harmonic Estrada index and ABC Estrada index) and ABS index for octane isomers. Since chemical graphs of octane isomers are trees, we study the extremal problem of ABS Estrada index of trees, and prove that for any tree \(T_n\) $T_n$ with \(n\ge 3\) $n\ge 3$ vertices, \(\begin{aligned} E_{ABS}(P_n)\le E_{ABS}(T_n)\le E_{ABS}(K_{1,n-1}) \end{aligned}\) $$\begin{array}{c}\hfill E_{\mathit{ABS}}(P_n)\le E_{\mathit{ABS}}(T_n)\le E_{\mathit{ABS}}(K_{1,n-1})\end{array}$$ with equality in the left (resp., right) inequality if and only if \(T_n\) $T_n$ is isomorphic to the path \(P_n\) $P_n$ (resp., the star \(K_{1,n-1}\) $K_{1,n-1}$ ).