# What is the smallest (in terms of vertices) 5-chromatic $K_5$-free graph?

I am looking for graphs that can be vertex-colored using at least 5 colors, but does not contain $K_5$ (a clique of size 5) as a sub-graph. The question is what is the smallest number of vertices a graph need to have in order to meet the desired properties?

By coloring I mean assignment of colors (numbers from 1 to 5) to vertices such that no adjacent vertices are assigned the same color.

The closest result I was able to find is "The size of a minimum five-chromatic K4-free graph" by Denis Hanson, Gary MacGillivray and Dale Youngs:

https://www.sciencedirect.com/science/article/pii/0012365X9390309H

We can certainly accomplish this with $$7$$ vertices: start with a $$5$$-cycle (a $$3$$-chromatic graph with no $$K_3$$) and add two more vertices adjacent to all others (and to each other), getting this graph.
The two added vertices must use a color not present anywhere else in the graph, and then we need $$3$$ more colors to color the $$C_5$$, so we need $$5$$ colors total. And there is no $$K_5$$, because a $$K_5$$ would need to contain three vertices of the cycle, but the cycle does not contain a $$3$$-clique.
It's clear that $$5$$ vertices are not enough: the only graph on $$5$$ vertices with chromatic number $$5$$ is $$K_5$$ itself.
If we had a $$6$$-vertex example, then it would need to have minimum degree $$4$$. Otherwise, we could delete a vertex with degree $$3$$, getting a $$5$$-vertex graph that's not $$K_5$$, color that with $$4$$ colors, and give the deleted vertex a color not used by its neighbors.
A $$6$$-vertex graph with minimum degree $$4$$ is a $$K_6$$ with a matching (not necessarily a perfect one) removed. If we remove just one edge, the result still contains a $$K_5$$; if we remove two independent edges, the result is already $$4$$-colorable, so we do not get a smaller example here either.