The result follows easily by applying the method of differences to
f(n,k+1) – f(n-1,k+1) = n(n – 1)…(n – k) – (n – 1)(n – 2)…(n – k – 1).
= (k + 1)(n – 1)…(n – k).
= (k + 1) f(n-1,k).

I think there is a small error in the last two formulae of the advanced proof. Should they be like this (http://bit.ly/hzjQCf) or I’m missing something?


1
1 1
1 2 1
1 3 3 1
1 4 6 4 1


A “preview” function for these comments would be useful.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

If you sum down along any diagonal from (say) left to right, and then take one step down and left, you find the sum. For instance, start at the top and take three steps down-right — 1+1+1 — and then one step down-left to find 3.

Or start at the first “1″ on the second row and sum down-right — 1+2+3 — then take one step down-left to find 6.

This works for all diagonals in the triangle, and it is easy to show by induction, given the fact that that every element is the sum of the two directly above it.

But these sums are exactly your k-dimensional triangle numbers. So T(n,k) = (n+k-1)-choose-k = (n+k-1)! / ((n-1)!k!)

P.S. You could have used T(n,0) = 1 for your definition instead of T(n,1) = n

I was asking myself the same question the other day. My wife was reading a book to our children that “hid” the items from the 12 days of Christmas on each page. The process of finding all of the gifts was tiresome, and drove me to scribbling out a computation before she completed the book.

I didn’t think about tetrahedrons, that’s really helpful. I recognized that each day is a triangular number. Knowing that triangular numbers are a diagonal of Pascal’s triangle, it follows that the tetrahedral numbers, being sums of triangular numbers, would be an adjacent diagonal.

One quickly observes that the desired diagonal is the numbers of the form n choose 3 for some n. Since 3 choose 3 is one it must be that the nth tetrahedral number is n+2 choose 3.

This lead me to conclude that there are 14 choose 3 gifts. I then erroneously computed that to be 1092.

Anyway, the combination of the tetrahedrons and Pascal’s Triangle is fascinating.

Last thought: The number of gifts is also the number of triangles formed placing 14 points on a circle and connecting each point to all of the others. (Not so useful, but there you are.)

As for the tetrahedron being “a sort of 3-dimensional triangle”, if you want to get a little more formal, a tetrahedron is a dimension 3 simplex, just as the triangle is a dimension 2 one. Guess that makes T(n,k) the n-th k dimensional simplicial number…