For my introduction to combinatorics class, we are being asked to compute the Ramsey number for $R(P_4,C_7)$ where $R(P_4,C_7)=k$ is the minimum number of vertices needed such that the 2-coloring of $K_k$ contains either a monochromatic $P_4$ or a monochromatic $C_7$.
To do this, I used a lemma/note from class stating that the connected components of a (maximal) $P_4$ graph are:
,
,
,
and the star on $n$ vertices (here is photo example of it on 7 vertices:
).
Using this info, I found (I think) all the maximal graphs on 7 vertices that do not contain a $P_4$. So if we were coloring $K_7$ red and blue, let the edges shown in the graphs below be the red coloring and the edges in the complement of the graphs be the blue coloring. I just wanted to ensure that these were the only ones as then obviously in their complements there will be a blue $P_4$ which should prove that $R(P_4,C_7)=7$. Here are those five graphs in question:
First:
Second:
Third:
Fourth:
Fifth:
All of which obviously have $P_4$ in it's complement.
My concern stems from this paper which on page 10 Theorem 9, it states:
$$R(P_n,C_m)= \begin{cases} 2n − 1 & \text{for $3 \leq m \leq n$, $m$ odd} \\ n − 1 + \frac m 2 & \text{for $4 \leq m \leq n$, $m$ even} \\ \max\{ m − 1 + \lfloor \frac n 2\rfloor, 2n − 1 \} & \text{for $2 \leq n \leq m$, $m$ odd} \\ m − 1 + \lfloor \frac n 2\rfloor & \text{for $2 \leq n \leq m$, $m$ even} \end{cases} $$
But that would mean $R(P_4,C_7)=\max\{ 7 − 1 + \lfloor \frac 4 2\rfloor, 2\cdot 4 − 1 \}=\max\{ 8, 7 \}=8\neq 7$. I most certainly did something wrong but it's been an hour now and I cannot find the counterexample. At least I believe there is a counterexample I am not seeing.
Also, apologizes if the formatting around the photos is strange/makes the post enlongated. I tried doing the thing where you add s to the end of the MathSE link to the photo but that created weirder inconsistencies and photo quality issues (at least in the preview).
Thanks in advanced for the help and cheers!