[Enrique]

I think the answer is (n+2) choose 3.

I did some cases and it looked like it was the sum of triangular numbers.
Here is why. The triangles of length 1 that are face upwards are: (1+2+3+...+n-1).
The triangles of length 2 are (1+2+3+...+(n-1)).
And you continue until you get 1 triangle of length n.
This sum comes out to (n)*(n+1)*(n+2) / 6.