Tetrahedral numbers

Start with the sequence of positive integers:

1, 2, 3, 4, …

Now take partial sums, the nth 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 Tn, d, the nth tetrahedral number of dimension d, recursively. We define Tn, 1 = n and for d > 1,

T(n, d) = \sum_{i=1}^n T(i, 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:

T(n, d) = \binom{n+d-1}{d}

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.

\begin{eqnarray*} T(n, m) &=& \sum_{i=1}^n T(n, m-1) \\ &=& T(n, m-1) + \sum_{i=1}^{n-1} T(n, m-1) \\ &=& T(n, m-1) + T(n-1, m) \\ &=& \binom{n+m-2}{m-1} + \binom{n+m-2}{m} \\ &=& \binom{n+m-1}{m} \end{eqnarray*}

The last line follows from the binomial addition identity

\binom{a}{b} = \binom{a-1}{b} + \binom{a-1}{b-1}

It also turns out that Tn, d is the number of ways to select d things from a set of n with replacement.

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4 thoughts on “Tetrahedral numbers

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

  2. 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.

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