Edit: O.k.. The pattern is that for the n base division triangle
there are a total of
n^2 + [n*(n-1)/2] + [(n-1)*(n-2)/2] + [(n-2)*(n-3)/2} + ..... +
[(n - (n - 3))*(n - (n - 2))/2] + [(n - (n - 2))*(n - (n - 1))/2] =
n^2 + (1/2)*sum(k=1 to n)(k*(k - 1)) =
n^2 + (1/6)*[(n - 1)*n*(n + 1)] =
(1/6)*n*(n^2 + 6n - 1) triangles.
This checks out for n = 1,2,3.
For n = 4 we have (1/6)*4*(16 + 24 - 1) = 26 triangles.
For n = 5 we have (1/6)*5*(25 + 30 - 1) = 45 triangles.
So the general formula looks promising.
For n = 15 we have (1/6)*15*(225 + 90 - 1) = 785 triangles.
I may have lost you a while back but I think I have the
correct general formula. If I have time later I'll write out
a proof for the last sum identity I used above, namely
sum(k=1 to n)(k*(k-1)) = (1/3)*(n-1)*n*(n+1).
Edit #2: sum(k=1 to n)(k^2 - k) =
sum(k=1 to n)(k^2) - sum(k=1 to n)(k) =
(1/6)*n*(n+1)*(2n + 1) - [n*(n+1)/2] =
(1/6)*n*(n+1)*[(2n + 1) - 3] =
(1/6)*n*(n+1)*(2n - 2) = (1/3)*(n-1)*n*(n+1) as promised. :)