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 w₂(n)=2[S₂(n)–A₂(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 A₂(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 A₂(n) ≥ S₂(n)–(3n+48)/5 for n≥40.

A2(1)=0

A2(2)=1

A2(3)=5

A2(4)=10

A2(5)=21

A2(6)=41

A2(7)≥65

A2(8)≥98 (David Cantrell)

A2(9)≥138

A2(10)≥187

A2(11)≥247

A2(12)≥319 (Maurizio Morandi)

A2(13)≥403 (David Bevan)

A2(14)≥503 (Maurizio Morandi)

A2(15)≥615 (Maurizio Morandi)

A2(16)≥744 (Maurizio Morandi)

A2(40)≥11058 (Maurizio Morandi)

A3(1)=0

A3(2)=0

A3(3)=1

A3(4)=5

A3(5)=10

A3(6)=17

A3(7)=30

A3(8)≥52

A3(9)≥74

A3(10)≥106

A3(11)≥146

A4(1)=0

A4(2)=0

A4(3)=0

A4(4)=1

A4(5)=5

A4(6)=10

A4(7)=17

A4(8)=26

A4(9)=41

A4(10)≥65

A4(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(n²) ≤ n² + (n+1)² + . . . + (n+k–1)² = kn(n+k–1) + k(k–1)(2k–1)/6.

C2(1)=4

C2(2)=6 Berend van der Zwaag

C2(3)≤9

C2(4)≤11 Berend van der Zwaag

C2(5)≤13 David Cantrell

C2(6)≤15

C2(7)≤17 Berend van der Zwaag

C2(8)≤20 David Cantrell

C2(9)≤22

C2(10)≤24

C2(11)≤26

C2(12)≤28

C3(1)≤10 David Cantrell

C3(2)≤13 David Cantrell

C3(3)≤19 Berend van der Zwaag

C3(4)≤21 Berend van der Zwaag

C3(5)≤26 David Cantrell

C3(6)≤29 David Cantrell

C3(7)≤31

C4(1)≤20 Berend van der Zwaag

C4(2)≤24 Berend van der Zwaag

C4(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 T₂(n)=2n+1 for all n?

Best Known Upper Bounds for T_k(n)

k \ n	1	2	3	4	5	6
2	3	5	7	9	11	13
3	6	9	12	15
4	10	14

T2(1)=3

T2(2)=5

T2(3)=7

T2(4)=9

T2(5)=11

T2(6)=13

T3(1)=6

T3(2)=9

T3(3)=12 David Cantrell

T3(4)=15 David Cantrell

T4(1)=10 David Cantrell

T4(2)=14 David Cantrell

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If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 3/25/23.