Firstly let's count the number of coins (spots, discs etc..) needed to make the outline of a hexagon

The sequence is 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66 ...

The nth term is given by 6n. This is easy to see if you start at a corner and count the spots on a side up to but excluding the next corner. There are six sides like this. Each number in the sequence is 6 more than the previous number.


The ancient Greeks studied 'polygonal' numbers. Here is how the triangular and square numbers are formed:

The hexagonal numbers are formed thus:

The sequence is 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231 ...

The nth term is given by n.(2n-1)

Every hexagonal number is also a triangular number


The hexagonal centred numbers result from counting the number of spots making up a full hexagon:

The sequence is 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331 ...

We can find the formula for the nth term by partitioning the hexagon into 3 parallelograms, plus the central spot.

The nth term is therefore 3n.(n-1) + 1.


The 'star' numbers (so-named by Martin Gardner in 'Time Travel and Other Mathematical Bewilderments') look like this:

The sequence is 1, 13, 37, 73, 121, 181, 253, 337, 433, 541, 661 ...

Again we can partition them, this time into 6 parallelograms plus the central spot

which gives us the formula for the nth term: 6n.(n-1) + 1.


The pyramidal hexagonal numbers are formed by making a pyramid from the sequence of hexagonal numbers. So, for example, the 4th number is a pyramid with 28 spots on the bottom layer, 15 spots on the next layer, 6 spots on the next, and finally 1 spot on top - a total of 50 spots.

The sequence goes 1, 7, 22, 50, 95, 161, 252, 372, 525, 715, 946 ...:

The nth term of the sequence is given by n.(n+1).(4n-1)/6


We can form pyramidal hexagonal centred numbers by making a pyramid from the sequence of centred hexagonal numbers. I am not sure if this is an official name, but it seems to make sense. So, for example, the 4th number is a pyramid with 37 spots on the bottom layer, 19 spots on the next layer, 7 spots on the next, and finally 1 spot on top - a total of 64 spots.

The sequence goes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331 ...

The nth term of the sequence is given by n^3 - very simple and surprising!

Proof: if the nth term is n^3, then the (n+1)th term will be the same plus the number of spots in the new base layer of the pyramid, ie
n^3 + [3(n+1).((n+1)-1) +1] (substituting n+1 for n in the 3n(n-1)+1 formula for hexagon centred numbers)
= n^3 + 3(n+1).n +1
= n^3 + 3n^2 + 3n +1
= n^3 + 2n^2 + n + n^2 + 2n + 1
= (n+1).(n^2 + 2n + 1)
= (n+1).(n+1)^2
= (n+1)^3

Now the first (and second) terms are indeed cubes, and so therefore are all others. QED

Note that, from this, the hexagonal centred numbers themselves can be represented as the difference between two adjacent cubes.


page date: 23Oct05.      I enjoy correspondence stimulated by this site. You can contact me here.