As a warm-up we might consider allowing only 1-by-1 and 2-by-2 tiles. In that case, minimizing the number of tiles is equivalent to maximizing the number of 2-by-2 tiles; that is, we're trying to disjointly pack as many 2-by-2 squares as possible into an n-by-n square. It's obvious that a(2k) = k^2. Can anyone prove that a(2k+1) = k^2+4k+1? Equivalently, can anyone prove that you can't pack more than k^2 2-by-2 squares inside a (2k+1)-by-(2k+1) square? Jim Propp On Friday, August 18, 2017, Allan Wechsler <acwacw@gmail.com> wrote:
Suppose you are asked to tile an integer square of side n with smaller squares, but the tiles are all of side 1, 2, or 3. What is the smallest number of tiles, a(n), that you can get away with?
For n = 1, 2, 3, 4, 5, 6 the answers are fairly easily seen to be 1, 1, 1, 4, 8, 4; I'm not quite as sure that my value a(7) = 13 is correct.
If these values are right, then a(n) is not in OEIS.
It's obvious that a(3k) = k^2. I would expect that big squares can be optimally tiled by filling most of the space with 3's, with a "fringe" of some sort around two sides if n is not a multiple of 3. But I maintain a small hope that there will be "weird" solutions that don't look like this. Can anybody provide more values? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun