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May 17, 2020 at 4:40 comment added Quotable @user36212 Yes, you are right. I actually came across an even more extreme example when investigating this question: Suppose $|V(G)| = n$ is divisible by 3 (for simplicity). Then consider the Turan graph $T(n, n/3)$. Then this is a complete multipartite graph with $n$ stable sets of size 3. Attaching a vertex $v$ adjacent to all vertices in $T(n, n/3)$ then makes $v$ have a largest star of size 3, but who belongs to $3^{n/3}$ maximal cliques. This is in fact as bad as it possibly can get (see the paper Moon and Moser, "On Cliques in Graphs").
May 17, 2020 at 4:34 comment added Quotable @WlodAA I agree with the sentiment of what you say; however, when I asked this question, I did not yet have an answer. When I cam upon an answer, I thought I would share the results of this with others who may be interested in the answer
May 16, 2020 at 14:33 comment added user36212 Can be much worse. Take a complete graph on mn+1 vertices for your favourite m,n, let v be one vertex and remove the edges of a collection of m disjoint n-cliques avoiding v. The largest induced star centred at v has n leaves, but v is in n^m maximal cliques.
May 16, 2020 at 3:26 comment added Wlod AA To ask a question when you already have an answer is against the spirit of this Q&A group. Thus, I stilled "plused-1" you here but not the question. Please, next time avoid this kind of combination.
May 16, 2020 at 3:21 comment added Wlod AA Perhaps $\ 2\cdot n\,$?
Apr 16, 2020 at 2:01 history answered Quotable CC BY-SA 4.0