Start with the sequence of positive integers:

1, 2, 3, 4, …

Now take partial sums, the *n*th term of the new series being the sum of the first n terms of the previous series. This gives us the triangular numbers, so called because they count the number of coins at each level of a triangular arrangement:

1, 3, 6, 10, …

If we repeat this again we get the tetrahedral numbers, the number of balls on each level of a tetrahedral stack of balls.

1, 4, 10, 20, …

We can repeat this process and general define *T*_{n, d}, the *n*th tetrahedral number of dimension *d*, recursively. We define *T*_{n, 1} = *n* and for *d* > 1,

This is just a formalization of the discussion above.

It turns out there’s a simple expression for tetrahedral number of all dimensions:

Here’s a quick proof by induction. The theorem clearly holds when *n* = 1 or *d* = 1. Now assume it hold for all *n* < *m* and *d* < *m*.

The last line follows from the binomial addition identity

It also turns out that *T*_{n, d} is the number of ways to select *d* things from a set of *n* with replacement.

**Related posts**:

- When is a triangle a square?
- Sums of consecutive powers
- Special numbers

Afaik, the generalization of triangle, tetrahedron to arbitrary dimension is called a “simplex”.

Yes, “simplex numbers” would make sense, but the name “tetrahedral numbers” is conventional.

Of course this is the same as n multichoose d, which you wrote about not too long ago (https://www.johndcook.com/blog/2018/05/22/combinatorics/). If you express it in this form, you can make the proof even simpler: You assign to the i-th “floor” of the simplex the d-element multisets of {1, 2, …, n} whose largest element is i. How many of those are there? You have to choose the other d-1 elements to be at most i, so (inducting on d) there are T(i, d – 1) such multisets.

Typo: “Not take partial sums” should be “Now take partial sums”