You are given a multigraph $G$ with $n$ vertices as follows:

$V := (v_1, v_2, \dots ,v_n)$

$C := \{c_1, c_2, \dots\}$, be an infinite set of colors.

$f: V \rightarrow \mathbb{P}_{\le m}(C) $, a function that maps each vertex to a subset of colors of size $\le m$.

We define the set of edges of the graph as:
$E := \big\lbrace \small\{ \normalsize v_i, v_j, c_k \small\} \normalsize \mid c_k \in f(v_i) \cap f(v_j) \big\rbrace$, i.e. there is an edge colored with a given color between every pair of vertices that share that color. Note that a pair of vertices may have more than one edge if they share multiple colors.

It is guaranteed that every pair of vertices has at least one edge in common. i.e. $\forall v_i, v_j \in V$ there exists $c_k \in C$ such that $\{v_i, v_j, c_k\} \in E$.

Define $F(n,m) = s$ such that for any graph with $n$ vertices, each vertex colored with $\le m$ colors as above, there is a clique of order $s$ where each edge is colored with the same color. Another way of phrasing this is that there is a subset of size $s$, $S \subseteq V$, $|S| = s$, all vertices of which share a color. i.e. $\exists c_k \in C$ such that $\forall v_i \in S, c_k \in f(v_i)$.

So a few simple examples: $F(n,1)=n$, clearly because every vertex has to have the same color. It is easy to show that, $F(n,2)=n−1$ and $F(n,n-1)=2$.

One can show that $F(9,3)=5$ by applying the pigeonhole principle repeatedly.

What method can be used to solve this for general $n$ and $k$? Or how can this problem be represented as a known problem in graph-theory/combinatorics.

*Note:* This function can be represented as an inverse to resemble the Ramsey numbers as follows:
$G(s, m) = n$, where $n$ is the smallest number where any graph with $n$ vertices, each vertex colored with at most $m$ colors as above contains a clique of order $s$. But that is harder since $F$ only gives us bounds on $G$.