# Pairs of a tree and a nontree graph with the same status sequence

## TL;DR

This paper disproves two conjectures by constructing specific pairs of trees and unicyclic graphs with identical status sequences, showing that trees and nontree graphs can share status sequences and that status injective trees are not necessarily unique.

## Contribution

It provides counterexamples to longstanding conjectures by explicitly constructing pairs of graphs with the same status sequence, including status injective trees that are not unique.

## Key findings

- Constructed pairs of trees and unicyclic graphs with identical status sequences for all n ≥ 10.
- Disproved that trees and nontree graphs cannot share status sequences.
- Showed that status injective trees are not always unique in their status sequences.

## Abstract

The status of a vertex $x$ in a graph is the sum of the distances between $x$ and all other vertices. Let $G$ be a connected graph. The status sequence of $G$ is the list of the statuses of all vertices arranged in nondecreasing order. $G$ is called status injective if all the statuses of its vertices are distinct. Let $G$ be a member of a family of graphs $\mathscr{F}$ and let the status sequence of $G$ be $s.$ $G$ is said to be status unique in $\mathscr{F}$ if $G$ is the unique graph in $\mathscr{F}$ whose status sequence is $s.$ In 2011, J.L. Shang and C. Lin posed the following two conjectures. Conjecture 1: A tree and a nontree graph cannot have the same status sequence. Conjecture 2: Any status injective tree is status unique in all connected graphs. We settle these two conjectures negatively. For every integer $n\ge 10,$ we construct a tree $T_n$ and a unicyclic graph $U_n,$ both of order $n,$ with the following two properties: (1) $T_n$ and $U_n$ have the same status sequence; (2) for $n\ge 15,$ if $n$ is congruent to $3$ modulo $4$ then $T_n$ is status injective and among any four consecutive even orders, there is at least one order $n$ such that $T_n$ is status injective.