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

re(3): Not a nice formula by McWorter)

You're right, brute force can get the numbers, but I still missed a few
of the triangles and was therefore was heading down the wrong
path. Luckily you corrected that.

Then I misread my binomial coefficient table/Pascal's triangle and
grabbed the wrong diagonal for my last answer. Luckily, you now
corrected that as well. Thank you again for the help.

C(n,k) = n!/(k!(n-k)!)

So T(n) is one of the diagonals of the Pascal's triangle, namely C(n+1,n-1) which simplifies as

(n+1)n[(n-1)!]/[(n-1)!(n+1-(n-1))!]

= (n+1)n/2!

= (n+1)n/2

Likewise the second formula, which I also mistated in the last posting
is the next lower diagonal in the Pascal's triangle, or C(n+1,n-2)
which simplifies as

(n+1)n(n-1)(n-2)!/[(n-2)!(n+1-(n-2))!]

= (n+1)n(n-1)/3!

= (n+1)n(n-1)/6

The final series, 1, 5, 15, 35, 70, etc. is the next lower diagonal on the triangle, or C(n+2,n-2)

(n+2)(n+1)n(n-1)(n-2)!/(n-2)![n+2-(n-2)]!

= (n+2)(n+1)n(n-1)/4!

= (n+2)(n+1)n(n-1)/24

which I believe is finally the correct answer. No proof yet, however.

*He cowers, awaiting the Wrath of McWorter* :)

*Edited on ***August 12, 2005, 2:01 am**