Let us denote by $n=r(k_1,k_2,\ldots,k_s)$ the minimal number of vertices such that for every coloring of the edges of the complete graph $K_n$ by $s$ different colors, there is some color $1\le i\le s$ such that the $i$-th graph contains a $k_i$ clique. By the $i$-th graph we mean the graph on the given $n$ vertices and this edges of the complete graph that are colored $i$.
let us denote $r_s = r(3,3,3,3,\ldots)$, with $3$ appearing $s$ times.
a) show that $r(k,s)$ is exactly the regular Ramsey number.
b) show that $r_s \le s(r_{s-1}-1)+ 2$.
c) show that $r_s \le s! \cdot e + 1$.
d) show that $r_3 \le 17$.
Now I don't get how to even start solving this, considering (c) for example, I don't see how we can achieve a factorial... Any hints for the solution, or a better explanation of the definition will be really helpful, thank you!
The definition seems clear enough, especially if you've seen Ramsey numbers before. The questions are all about the special case of $r_s$, which is just the smallest $n$ such that every coloring of the edges of $K_n$ with $s$ colors contains a monochromatic triangle, that is, a triangle whose edges are all the same color; e.g., $r_1=3$, $r_2=6$.
a) I hope this follows directly from the definitions.
b) Have you seen a proof of an upper bound for the "regular" Ramsey numbers? This is a lot like that. Pick a point $v$, consider the edges incident with $v$, and use the pigeonhole principle to show that enough of them are of the same color. [*]
c) This is misstated. The upper bound should be $\lfloor s!e\rfloor+1$ where $e=2.718\dots$ is the base of natural logarithms and $\lfloor x\rfloor$ is the floor function; e.g., if $s=2$ then $\lfloor s!e\rfloor+1=6$. Prove by induction on $s$, using b) and the fact that, for $s\ge1$, $$\lfloor s!e\rfloor=\frac{s!}{0!}+\frac{s!}{1!}+\frac{s!}{2!}+\cdots+\frac{s!}{s!}.$$
d) Just set $s=3$ in c) and do the calculation.
[*] Consider an integer $s\gt1$, let $n=s(r_{s-1}-1)+2$, and suppose the edges of $K_n$ are colored with $s$ colors. Choose a vertex $v$ and consider the $n-1=s(r_{s-1}-1)+1$ edges incident with $v$. By the pigeonhole principle, at least $r_{s-1}$ of those edges have the same color, which I will call "red". That is, there is a set $W\subset V(K_n)$ of size $|W|=r_{s-1}$ such that each vertex in $W$ is joined to $v$ by a red edge. If two vertices $w_1,w_2\in W$ are joined to each other by a red edge, then $v,w_1,w_2$ are the vertices of a red triangle. On the other hand, if no two vertices in $W$ are joined by a red edge, then the edges of the complete graph on $W$ are colored with $s-1$ colors, and the existence of a monochromatic triangle follows from the definition of $r_{s-1}$. This proves that $r_s\le n=s(r_{s-1}-1)+2$.