This time my question is related to Ramsey number $R(4\ 4; 3) = 13$.
DEFINITIONS: Functions $c : \binom X3\rightarrow \{0\ 1\}$ are called 2-colorings of triangles in $X$. The $4$-element subsets $A\subseteq X$ are called tetrahedra. Each 2-coloring $c$ of triangles induces $(\alpha\ \beta)$-coloring of each tetrahedron $A$, where $$\beta := \sum_{T\subseteq A,\ |T|=3}\ c(T)\quad\quad\quad \alpha := 4-\beta$$
A 2-coloring of triangles is called lively $\Leftarrow:\Rightarrow$ there are no tetrahedra which have all four walls painted in the same color, i.e. when no tetrahedron has coloring of type $(4\ 0)$ or $(0\ 4)$. Integer $n\ge 5$ is called lively $\Leftarrow:\Rightarrow$ an (hence any) $n$-element set admits a lively 2-coloring.
QUESTION: Let $n\ge 5$ be an arbitrary lively integer. What is the minimal number $m:=m_{22}(n)$ for which there exists a 2-coloring of triangles in an $n$-element set such that no tetrahedron has all four walls of the same color, and there are exactly $m$ tetrahedra with $(2\ 2)$-coloring.
MOTIVATION: I want to estimate anew $R(4\ 4; 3)$, from scratch, ignoring the fact that this Ramsey number is perfectly known--it was obtained after years of effort by some mathematicians and computer scientists assisted by computers. So far, my last partial result listed below helped me to obtain only $R(4\ 4; 3) \le 17$.
A PARTIAL RESULT:
- $m_{22}(5) = 1$
- $m_{22}(6) = 3$
- $m_{22}(n) \ge \lceil\frac 15\cdot\binom n4\rceil$ for every lively $n=5\ 6\ 7\ 8$
- $m_{22}(n) \ge \lceil\frac {13}{63}\cdot\binom n4\rceil$ for every lively $n=9\ 10\ 11\ \ldots$
REMARK Lively integers are simply integers in the range $5\ldots R(4\ 4;3)$--but I pretend that I have no a priori information about this Ramsey number. I am computing it from scratch.
