What are the values of m_{n}(r) for non-square values of n? A much harder question is to find M_{n}(r), the smallest value of m so that so that an m x mr rectangle contains n non-overlapping 1 x r rectangles that might be tilted. For example, when r > 1.9762, M_{2}(r) < m_{2}(r):
What are the values of M_{n}(r) for small values of n?
Here are the piecewise definitions for m_{n}(r) for small n:
If | r ≥ 1 | r ≥ √2 | r ≥ 2 |
n=2 | |||
then | m_{2}(r) = 2/r | m_{2}(r) = r | m_{2}(r) = 2 |
If | r ≥ 1 | r ≥ 3/2 | r ≥ √3 | r ≥ 2 |
n=3 | ||||
then | m_{3}(r) = 2 | m_{3}(r) = 3/r | m_{3}(r) = r | m_{3}(r) = 2 |
If | r ≥ 1 | r ≥ (√13-1)/2 | r ≥ (√5+1)/2 | r ≥ 2 | r ≥ √5 | r ≥ 3 |
n=5 | ||||||
then | m_{5}(r) = 3/r | m_{5}(r) = r + 1 | m_{5}(r) = 2 + 1/r | m_{5}(r) = 5/r | m_{5}(r) = r | m_{5}(r) = 3 |
If | r ≥ 1 | r ≥ √(3/2) | r ≥ (√3+1)/2 | r ≥ 3/2 | r ≥ (√17-1)/2 | r ≥ 2 | r ≥ √6 | r ≥ 3 |
n=6 | ||||||||
then | m_{6}(r) = 3/r | m_{6}(r) = 2r | m_{6}(r) = 2 + 1/r | m_{6}(r) = 4/r | m_{6}(r) = r + 1 | m_{6}(r) = 6/r | m_{6}(r) = r | m_{6}(r) = 3 |
If | r ≥ 1 | r ≥ 4/3 | r ≥ √2 | r ≥ 3/2 | r ≥ √3 | r ≥ 2 | r ≥ 7/3 | r ≥ √7 | r ≥ 3 |
n=7 | |||||||||
then | m_{7}(r) = 3 | m_{7}(r) = 4/r | m_{7}(r) = 2r | m_{7}(r) = 1 + 3/r | m_{7}(r) = r + 1 | m_{7}(r) = 3 | m_{7}(r) = 7/r | m_{7}(r) = r | m_{7}(r) = 3 |
If | r ≥ 1 | r ≥ 4/3 | r ≥ √2 | r ≥ 3/2 | r ≥ 8/3 | r ≥ √8 | r ≥ 3 |
n=8 | |||||||
then | m_{8}(r) = 3 | m_{8}(r) = 4/r | m_{8}(r) = 2r | m_{8}(r) = 3 | m_{8}(r) = 8/r | m_{8}(r) = r | m_{8}(r) = 3 |
If | r ≥ 1 | r ≥ √5-1 | r ≥ √2 | r ≥ 3/2 | r ≥ √(5/2) | r ≥ (√17+3)/4 |
n=10 | ||||||
m_{10}(r) = 4/r | m_{10}(r) = 2 + r | m_{10}(r) = 2 + 2/r | m_{10}(r) = 5/r | m_{10}(r) = 2r | m_{10}(r) = 3 + 1/r | |
r ≥ 2 | r ≥ (√13+1)/2 | r ≥ 1+√2 | r ≥ 3 | r ≥ √10 | r ≥ 4 | |
then | m_{10}(r) = 2 + 3/r | m_{10}(r) = 1 + r | m_{10}(r) = 3 + 1/r | m_{10}(r) = 10/r | m_{10}(r) = r | m_{10}(r) = 4 |
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 6/7/04.