TITLE: Counting on rectangles？
QUESTION [8 upvotes]: In an array of dots, how many can be colored red without forming a red rectangle?
We use rectangular arrays of dots, and consider only those rectangles with horizontal and vertical
sides. A dot-rectangle is called “red” if its four corners are red. (Other dots can be of any color.)For whole numbers $m, n,$ define $R(m, n)$ to be the largest number of dots in an $m ×n$ array that can be colored red without making a red rectangle. Can you evaluate (with proof) $R(3, 4), R(3, 5), . . . ?$ How about $R(4, 4)$? Eventually, find a general formula for $R(m,n)$ and provide a rigorous proof.
So I think that this is needed to be solved by Ramsey theory and induction but I am having a hard time to find a place to start. I tried a few cases but I am extremely unconfident if those countings are right. Currently I am stuck here and any help will be appreciated( e.g. how to check if I put the maximum red dots in). 
About the comment of results, I get 7 for  $R(3, 4)$. My reasoning is basically to fill out 3 spots of a rectangle and left the one blank, while trying to cram those spots together as tight as possible.

REPLY [2 votes]: An approximate solution (for at least part of the domain of $n$) for $R(n,n)$ is:
If $n$ is even then a solution of the form: $n + (n/2-m)(n/2+m)$ (where $0\le m<n/2$) is "close" in the cases listed in A07267. For values of $n: 2\le n \le 8$, then $m=0$ works best. For $n = 10$, then $m=1$, and for $n=12$, then $m=2$.
If $n$ is odd then $n+((n+1)/2+m)(n/2-m)$ is also "close". For values of $n: 1\le n \le 9$, then $m=0$ works best. For $n = 11$, then $n=1$ and for $n=13$ then $m=2$.
As $n$ increases, increasing $m$ yields a better estimate. However I don't have enough data points to determine what relationship between $n$ and $m$ works best.