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(n-1)/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 closed form solution by Josh70679)

The equilateral triangle formed by a triple can "point" in many directions, not just up and down.  Consider Array(4) and these three points: the second last point in the bottom row, the first point in the first row above the bottom row, and the last point in the second row from the top.  These three points are vertices of an equilateral triangle "pointing" north northeast.

I don't know if your formula is correct, but I know that E(2)=1, E(3)=5, E(4)=15, and E(5)=35.

Posted by McWorter on 2005-08-11 00:11:52