| The problem is to find a formula for the number of distinct MacMahon
triangles using n colours.
We could start with a few small numbers to get a feel for what is
going on.
| # Colours |
# Triangles |
| 0 |
0 |
| 1 |
1 |
| 2 |
4 |
| 3 |
17 |
| etc |
|
We can also investigate if there is an easy transition between n
colours and n+1 colours.
When a new colour is added to the arrangements with n colours, the
number of new triangles that have that colour in one position is n2,
the number that have that colours in two positions is n and there is one
triangle that has that colour in all three positions.
Thus the number of new triangles made when one colour is added to n
existing ones, is
n2 + n + 1
This means that, if Tn is the number of triangles formed
with n colours then:
Tn+1 = Tn + n2 + n + 1
Another implication is that Tn is a cubic in n, and we can
use the data gathered in the table of values to find out that
Tn = n(n2 + 2)/3
|
