Many, many octahedra

Geometry Level 3

This shelf is displaying several Sierpinski tetrahedra:

The linear dimensions of the five big ones from left to right are 1, 2, 4, 8, and 16.

You can fill the void in the second one perfectly with one octahedron.
The third one can be filled with five octahedra, a big one and four smaller ones.

How many octahedra (of different sizes) would it take to fill all the voids in a Sierpinski tetrahedra of linear dimension 1024?

Image credit: http://www.fractalnature.com/

The answer is 349525.

2 solutions

Geoff Pilling
Nov 16, 2016

Every time you double the linear dimension, the number of octahedra needed is multiplied by four (one for each of the four pyramids replicated) plus one for the big hole in the middle.

So, if $N_n =$ The number of octahedra required to fill a Sierpinski Tetrahedra with linear dimension $2^n$ , then:

$N_0 = 0$
$N_n = 4\cdot N_{n-1} + 1$

From this recursion relation:

$N_{10} = \boxed{349525}$

Nice explanation! Could you turn the recursion relation into the explicit formula (1/3)(4^n) - (1/3)?

Jason Dyer Staff - 4 years, 6 months ago

Lets see, I think my recursion relation is equivalent to the explicit formula:

$N_n = \sum_{i = 0}^{n-1}4^i$

It looks like thats equivalent to $(\frac{1}{3} \cdot 4^n - \frac{1}{3})$ ?

Geoff Pilling - 4 years, 6 months ago

Ah... Interesting... Lemme think about it for a bit...

Geoff Pilling - 4 years, 6 months ago

Sure. You also know in this circumstance the format is a*4^n+b and you can just set a system of two equations where n = 1 and n = 2.

Jason Dyer Staff - 4 years, 6 months ago

Christopher Williams
Dec 26, 2016

Just listing this out in case it's easier for someone:

Sierpinski Tetrahedra

Linear Dimension	Missing Octahedra
1	0
2	1
4	5
8	21
16	85
32	341
64	1365
128	5461
256	21845
512	87381
1024	349525

Many, many octahedra

The answer is 349525.

2 solutions

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