I have been doing some thinking on the Mondrian Art Problem and think I may have discovered something.
I think I have improved the upper bound for odd numbers from $k$ (for a $k$ by $k$ square) to $(5k+1) / 6$. There may also be a better upper bound, but I haven't heard of one. My idea is to separate the odd numbers into three categories: $6n + 1$, $6n + 3$, $6n + 5$. From there I create upper-bounds for each. For the case of $(6n+1)$ I made this diagram:
You can see that all the sides add up to $6n+1 = k$. The difference between the biggest and smallest is $5n+1$ (which is less than $k$).
I do the same thing for the other $2$ cases:
And...
The equations on the top right are the score (it should have a $k$ instead of $n$ there).
Thanks in advance for taking the time of reading /going through this.
PS: If it is correct where would I post it (or is this it?)
(EDIT: here is the sequence.)