The binomial coefficients T(n)=n(n+1)/2 are called triangular numbers because T(n) points can be arranged into a triangular array with n points on a side. For example, T(4)=10 points can be arranged in the familiar pattern of bowling pins. Call this triangular array with n points on a side Array(n).
If we count the number of triples of points in Array(n) which are vertices of equilateral triangles with sides parallel to those of the whole array, we get the binomial coefficient (n+1)n(n1)/6: a nice formula in closed form (i.e., no sum of stuff).
What surprised me when I dared look at it is that if we count the number E(n) of ALL triples of points in Array(n) which are vertices of equilateral triangles, we also get a nice formula.
What is that nice formula for E(n)?
(In reply to
re(2): Not a nice formula by Bob Smith)
Brute force can get the numbers I gave you. However, your formula says E(2)=5, which, of course, is wrong. Whatever formula you come up with, I'd like to see the proof. Otherwise, how do we know the formula works for all n?

Posted by McWorter
on 20050811 15:10:37 