This blog entry poses a genuine question from me and, I hope, a puzzle that might interest many. It is inspired by an entry in the “Short Dictionary of Heuristic” found in How to Solve it: A New Aspect of Mathematical Method.

First, a quick description of Pascal’s Triangle.

This is the triangle of integers – with 1 at the apex – where each number is the sum of the numbers immediately above it – hence it begins (apologies for the slightly unorthodox presentation – a consequence of wordpress.com’s insistence on suppressing leading spaces on lines):

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

…and so on.

As is so often the case with maths, this is more than just a pleasing pattern. The numbers also, correspond, for instance, to the coefficients of the expansion of the binomial eg:

and so on…

But it is the relationship of the triangle to geometry I am interested in.

The simplest (the only) null dimensional shape: a point, has 1 vertex: 1

The simplest 1 dimensional shape – a line: has 2 vertices (null dimensions) and 1 one-dimensional shape: 2 1

The simplest 2 dimensional shape – a triangle: 3 vertices, 3 one dimensional elements (bounding lines) and 1 two-dimensional space: 3 3 1

The simplest 3 dimensional shape – a tetrahedron: 4 vertices, 6 bounding lines, 4 two-dimensional elements (faces) and contains 1 three dimensionsal space: 4 6 4 1

So we can see Pascal’s triangle emerge there (and thus propose by induction that the simplest four dimensional space has 5 vertices, 10 bounding lines, 10 faces, contains 5 three dimensional spaces and 1 four dimensional space). This induction is correct – the shape is the pentachoron.

What puzzles me is whether we can draw any analogies for other shapes.

So for the families of squares and cubes:

We could have the same first two members – a point, 1

A line: 2, 1

Then a square: 4 vertices, 4 bounding lines, 1 space

A cube: 8 vertices, 12 bounding lines, 6 faces, 1 space

A hypercube (tesseract): 16 vertices, 32 bounding lines, 24 faces, 8 three dimensional spaces (cells) and 1 four dimensional space.

So we have a triangle like this:

`1`

2 1

4 4 1

8 12 6 1

16 32 24 8 1

My question is: is there some fundamental relation between the rows of this triangle and there is for Pascal’s triangle? I cannot see one, but there *looks* like there should be one!

###### Related articles

- Quasiperiodic designs from waves and higher dimensional space (geometricolor.wordpress.com)
- Visualizing Pascal’s triangle remainders (mathlesstraveled.com)
- pascals triangle in c with gmp (stackoverflow.com)
- A Compilation of Theory and Ideas about the HYPERCUBE! (darkknight47.wordpress.com)
- Interview Question: Pascal’s Triangle (bobhancock.org)
- Pascal’s Triangle Profiled (bobhancock.org)
- Data, Dimensions and Geometry oh my ! (geomblog.blogspot.com)
- Program to Calculate Large Binomial Coefficients (codecompiler.wordpress.com)
- 10. Famous Thinkers – Mathematician Is a Creationist 2 (biblescienceguy.wordpress.com)
- Mathematics: An Illustrated History of Numbers (Ponderables) (bergersbookreviews.com)

## 3 responses to “Pascal’s triangle and hypercubes”

Yes, there is: given two entries a and b, to get the entry below them compute a + 2b. For example, 8 + 2*12 = 32. Note also that these are the coefficients in the expansion of .

See http://oeis.org/A038207 .

Quite simple when you know 🙂 Thanks

[…] Pascal’s triangle and hypercubes (cartesianproduct.wordpress.com) […]