 Alternating count of common-center graphs

For $n\ge 3$, count the empty graph once, all one-edge graphs once, and each larger accepted graph under its unique center. The latter weighted contribution is $n(n-2)$, giving $1-\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n(n-2)$. Its nonzero value proves evasiveness by the alternating-sum criterion for decision-tree evasiveness.