Laplacian coefficients of trees

Let G be a simple and undirected graph with Laplacian polynomial ψ(G, λ) = Σk=0n (−1)n-kck(G)λk. In this paper, exact formulas for the coefficient cn−4 and the number of 4-matchings with respect to the Zagreb indices of a given tree are presented. The chemical trees with first through the fifteenth greatest cn−4-values are also determined

On Symmetry of Some Non-transitive Chemical Graphs

The automorphism group of a chemical graph has to be generated for computer-aided structure elucidation. A Euclidean graph associated with a molecule is defined by a weighted graph with adjacency matrix M = [dij], where for i ≠ j, dij is the Euclidean distance between the nuclei i and j. In this matrix, dii can be taken as zero if all the nuclei are equivalent. Otherwise, one may introduce different weights for distinct nuclei. A. T. Balaban introduced some monster graphs and then M. Randić computed complexity indices for them. This paper describes a simple method, by means of which it is possible to calculate the automorphism group of weighted graphs

ON CAPABLE GROUPS OF ORDER p4

A group $H$ is said to be capable, if there exists another group$G$ such that $\frac{G}{Z(G)}~\cong~H$, where $Z(G)$ denotes thecenter of $G$. In a recent paper \cite{2}, the authorsconsidered the problem of capability of five   non-abelian $p-$groups of order $p^4$ into account. In this paper, we continue this paper by considering three other groups of order $p^4$.  It is proved that the group $$H_6=\langle x, y, z \mid x^{p^2}=y^p=z^p= 1, yx=x^{p+1}y, zx=xyz, yz=zy\rangle$$ is not capable. Moreover, if $p > 3$ is  prime and $d \not\equiv 0, 1 \ (mod \ p)$ then the following groups are not capable:\\{\tiny $H_7^1=\langle x, y, z \mid x^{9} = y^3 = 1, z^3 = x^{3}, yx = x^{4}y, zx = xyz, zy = yz \rangle$,\\$H_7^2= \langle x, y, z \mid x^{p^2} = y^p = z^p = 1, yx = x^{p+1}y, zx = x^{p+1}yz, zy = x^pyz \rangle,$ \\$H_8^1=\langle x, y, z \mid x^{9} = y^3 = 1, z^3 = x^{-3}, yx = x^{4}y, zx = xyz, zy = yz \rangle$,\\$H_8^2=\langle x, y, z \mid x^{p^2} = y^p = z^p = 1, yx = x^{p+1}y, zx = x^{dp+1}yz, zy = x^{dp}yz \rangle$.