For regular n sided polygons, what's the formula for # of diagonal intersections?

Question

For example, a
4-sided square has 1
5-sided pentagon has 5
6-sided hexagon has 13
etc.

Intersections, not diagonals. Yes, the formula for # of diagonals is 1/2(n)(n-3), but that's not what I asked for.

This question is ONLY for perfect regular polygons, so frequently a number of diagonals can intersect at a single point. Sometimes for large n-polygons, it can look downright beautiful.

Not for "any" n-sided polygons, but for perfect REGULAR polygons, which makes it harder, because of multiplicity of diagonals possible for an intersection.

Answer 1

The sequence appears to be this one:

http://www.research.att.com/~njas/sequences/?q=1+5+13+35&sort=0&fmt=0&language=english

There's also some discussion of it here:

http://mathforum.org/library/drmath/view/55217.html

Answer 2

Copied from the reference supplied by Jim:
0, 0, 0, 1, 5, 13, 35, 49, 126, 161, 330, 301, 715, 757, 1365, 1377, 2380, 1837, 3876, 3841, 5985, 5941, 8855, 7297, 12650, 12481, 17550, 17249, 23751, 16801, 31465, 30913, 40920, 40257, 52360, 46981, 66045, 64981, 82251, 80881, 101270

The first 8 non-trivial numbers can be generated by
-175791 + 187356.5333n - 83855.57778n^2 + 20442.80556n^3 - 2933.486111n^4 + 247.9388889n^5 - 11.43611111n^6 + 0.222222222n^7 =

(-1582119 + (8431044/5)n - (3773501/5)n^2 + (735941/4)n^3 - (211211/8)n^4 + (44629/20)n^5 - (4117/40)n^6 + 2n^7)/9

I have verified that
N = n(n - 1)(n - 2)(n - 3)/24
works for all odd polygons.

Given the listing above, there must be an algorithm for computing the even-sided, as well, but it escapes me.

Answer 3

There are n vertices in an n-gon. From each vertex, you can draw a diagonal to (n - 3) other vertices. [you can't go to the two adjacent vertices nor to itself, therefore (n - 3)] So, there are n(n - 3) diagonal lines, BUT, each diagonal has two vertices as endpoints. The number of diagonals in an n-gon is: [n (n - 3)]/2 QED

Answer 4

n(n-3)/2

Good night!

Hmmm . . . I did a little searching, and I couldn't find a formula for intersections. I did see one for the number of regions formed, but that isn't what you're looking for (it was very ugly anyway). This is a really good question!

OK, I found a discussion of the question here:
http://mathforum.org/library/drmath/view/55217.html

Unfortunately, I don't see a formula anywhere. Just observations. Good luck!