Problem #1:
Assume we have squares with sides 1, 2, 3, . . . n of fixed thickness 1. By stacking them flat, what is the largest connected area A_{k}(n) of height k that we can produce? Each square must either be on the lowest level or be fully supported by one or more squares below. What are the values of A_{k}(n) for large k and n?
Problem #2:
Assume we have a bunch of cubes of side 1. We want to stack them so that for every i, the boundary of level i+1 is contained within the interior of the boundary of level i. What is the fewest number of cubes C_{k}(n) that we need to have n cubes on level k? What are the values of C_{k}(n) for large k and n?
Since the combined area of squares of sides 1 through n is n(n+1)(2n+1)/6, an upper bound is A_{k}(n) ≤ S_{k}(n) = n(n+1)(2n+1)/6k.
David Bevan defined the waste w_{2}(n)=2[S_{2}(n)–A_{2}(n)], the area of the lower layer not covered by the upper layer. He showed how to build a cell, 8 consecutive squares placed so that the waste is exactly 16. From this he gets the lower bound A_{2}(n) ≥ S(n)–n+4 for n≥8.
He improves this by showing how to build a supercell, 40 consecutive squares made from 5 cells with a waste of exactly 48. From this he gets a better lower bound A_{2}(n) ≥ S_{2}(n)–(3n+48)/5 for n≥40.
k \ n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2 | 0 | 1 | 5 | 10 | 21 | 41 | 65 | 98 | 138 | 187 | 247 | 318 | 403 | 500 | 611 |
3 | 0 | 0 | 1 | 5 | 10 | 17 | 30 | 52 | 74 | 106 | 146 | ||||
4 | 0 | 0 | 0 | 1 | 5 | 10 | 17 | 26 | 41 | 65 | 89 |
A_{2}(1)=0 | A_{2}(2)=1 | A_{2}(3)=5 | A_{2}(4)=10 | A_{2}(5)=21 | A_{2}(6)=41 |
A_{2}(7)≥65 | A_{2}(8)≥98 (David Cantrell) | A_{2}(9)≥138 |
A_{2}(10)≥187 | A_{2}(11)≥247 |
A_{2}(12)≥318 (David Bevan) | A_{2}(13)≥403 (David Bevan) |
A_{2}(14)≥500 (David Bevan) |
A_{2}(15)≥611 (David Bevan) |
A_{2}(40)≥22116 (David Bevan) |
A_{3}(1)=0 | A_{3}(2)=0 | A_{3}(3)=1 | A_{3}(4)=5 | A_{3}(5)=10 | A_{3}(6)=17 | A_{3}(7)=30 |
A_{3}(8)≥52 | A_{3}(9)≥74 | A_{3}(10)≥106 |
A_{3}(11)≥146 |
A_{4}(1)=0 | A_{4}(2)=0 | A_{4}(3)=0 | A_{4}(4)=1 | A_{4}(5)=5 | A_{4}(6)=10 | A_{4}(7)=17 | A_{4}(8)=26 |
A_{4}(9)=41 | A_{4}(10)≥65 | A_{4}(11)≥89 |
Problem #2:
David Cantrell and Berend van der Zwaag sent several solutions.
By building pyramids in which every level is a square, we get the upper bound C_{k}(n^{2}) ≤ n^{2} + (n+1)^{2} + . . . + (n+k-1)^{2} = kn(n+k-1) + k(k–1)(2k–1)/6.
k \ n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|---|---|
2 | 4 | 6 | 9 | 11 | 13 | 15 | 17 | 20 | 22 | 24 | 26 | 28 |
3 | 10 | 13 | 19 | 21 | 26 | 29 | 31 | |||||
4 | 20 | 24 | 33 | 35 |
C_{2}(1)=4 | C_{2}(2)=6 Berend van der Zwaag | C_{2}(3)≤9 | C_{2}(4)≤11 Berend van der Zwaag | C_{2}(5)≤13 David Cantrell |
C_{2}(6)≤15 | C_{2}(7)≤17 Berend van der Zwaag | C_{2}(8)≤20 David Cantrell | C_{2}(9)≤22 |
C_{2}(10)≤24 | C_{2}(11)≤26 | C_{2}(12)≤28 |
C_{3}(1)≤10 David Cantrell | C_{3}(2)≤13 David Cantrell | C_{3}(3)≤19 Berend van der Zwaag | C_{3}(4)≤21 Berend van der Zwaag |
C_{3}(5)≤26 David Cantrell | C_{3}(6)≤29 David Cantrell | C_{3}(7)≤31 |
C_{4}(1)≤20 Berend van der Zwaag | C_{4}(2)≤24 Berend van der Zwaag | C_{4}(4)≤35 |
Claudio Baiocchi thought this problem would be interesting for equilateral triangles, tans, hexagons, and other shapes that tile the plane. Here are the best known solutions for tans T_{k}(n). Is it possible that T_{2}(n)=2n+1 for all n?
k \ n | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|
2 | 3 | 5 | 7 | 9 | 11 | 13 |
3 | 6 | 9 | 12 | 15 | ||
4 | 10 | 14 |
T_{2}(1)=3 | T_{2}(2)=5 | T_{2}(3)=7 | T_{2}(4)=9 | T_{2}(5)=11 | T_{2}(6)=13 |
T_{3}(1)=6 | T_{3}(2)=9 | T_{3}(3)=12 David Cantrell | T_{3}(4)=15 David Cantrell |
T_{4}(1)=10 David Cantrell | T_{4}(2)=14 David Cantrell |
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 8/30/07.