Dominance number on cyclooctane chains
Abstract
Introduction/purpose: Chemical structures are conveniently represented by graphs where atoms are nodes (vertices) and chemical bonds are branches (lines) in the graph. A graphical representation of a molecule provides a lot of useful information about the chemical properties of the molecule. It is known that numerous physical and chemical properties of molecules are highly correlated with theoretical invariants of graphs, which we call topological indices. One such theoretical invariant is the dominance number. The aim of this research is to determine the k-dominance number for cyclooctane chains COn1, COn2, COn3 and COn4, for k ∈ {1,2,3}, n ∈N.
Methods: The cyclooctane chain is a chain of octagons connected by a single line. The vertices of the octagon are treated as nodes of the graph, and the sides and the line connecting them, as branches in the graph. Using mathematical methods, k-dominance was determined on one octagon, k∈{1,2,3}. Then, by representing the cyclooctane chains COn1, COn2, COn3 and COn4, in a convenient, isomorphic way, we determined their k-dominance number, k∈{ 1,2,3}.
Results: Determining k-dominance, k∈{1,2,3}, for 4 cyclooctane chains COn1, COn2, COn3 and COn4, we obtained 12 different formulas to calculate their k-dominance number. All formulas are composed of several alternative algebraic expressions, the selection of which is conditioned by the divisibility of the number n by the number 2, 3 or 4, depending on the type of cyclooctane chain and k-dominance to be determined. The results of the research are fully presented in the paper through mathematically proven theorems and graphical representations.
Conclusion: The results show that the k-dominance numbers, k∈{1,2,3}, on cyclooctane chains COn1, COn2, COn3 and COn4, are determined and explicitly expressed by mathematical expressions. They also indicate the possibility of their application in molecular graphs of cyclooctane rings, in computational chemistry, chemical and biological industry.
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