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

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.