Consider the following problem:
For $k \le n$, what is the smallest number of edges in the graph with $n$ vertices so that the maximum independent set has size at most $k$?
I.e., given an empty graph on $n$ vertices, I want to add the smallest number of edges so that the maximum independent set has size $k$. Let's denote it as $A(n,k)$. For example, $A(8,3)=7$, with edges $1-2-3-1$, $4-5-6-4$ and $7-8$ (two triangles and an edge).
I know that $A(n,k) \le T(n-1, k)$ where $T$ is https://oeis.org/A134546. In particular, I know that $A(n,n) = 0$ and $A(n + 1,k) \le A(n, k) + \lfloor n / k \rfloor$.
The particular construction is to maintain $k$ cliques so that the independent set can pick only vertex per clique. For example, for $A(8,3)$, the solution $1-2-3-1$, $4-5-6-4$ and $7-8$ consists of three cliques $(1,2,3)$, $(4,5,6)$, and $(7,8)$. When we add another vertex, we have $A(9,3) \le A(8,3) + 2$, since we can add edges $8-9$ and $7-9$ to have cliques $(1,2,3)$, $(4,5,6)$, and $(7,8,9)$.
Q: How to prove that we have equality, i.e. $A(n + 1,k) = A(n, k) + \lfloor n / k \rfloor$? My experiments show that it indeed holds, and, intuitively, it should hold.
I suspect there is some relation to Clique Cover, but I couldn't find it.
