Recall the factorial function \(n!\)

and the binomial coefficients \(B(n, k)\):

With the binomial coefficients, you can define the \(n\)-th \(k\)-simplex number \(S_{k}(n)\):

When \(k = 1\), you have the line numbers:

When \(k = 2\), you have the triangle numbers:

When \(k = 3\), you have the tetrahedron numbers:

When \(k = 4\), you have the pentatope numbers:

Besides the \(k\)-simplex numbers, you also have the \(n\)-th \(k\)-cube number \(C_{k}(n)\):

The cases \(k = 2\), \(k = 3\), and \(k = 4\) are, respectively, the square, cube, and tesseract numbers. The \(n\)-th \(k\)-cube number corresponds to the number of entries in a \(k\)-array of size \(n\). For example, a 2-array of size \(n\) is an \(n \times n\) square matrix. I have been thinking about \(k\)-simplex and \(k\)-cube numbers lately because I have been thinking about square matrices and cubic hypermatrices.

The \(k\)-volume of a \(k\)-cube and the \(k\)-volume of a \(k\)-simplex are related via:

Trivially,

That is, a 1-cube is also a 1-simplex. As a first nontrivial example, take \(k = 2\). A square can be divided into two triangles with the same area. You can translate this statement into a property of square matrices. An \(n \times n\) square matrix has \(n^{2}\) entries (its "area"). These \(n^{2}\) entries can be divided into two identical triangular domains with \(S_{2}(n - 1)\) entries, and one linear domain between the two triangles with \(S_{1}(n)\) entries. Thus,

Equivalently, you can absorb the interface between the two triangles into one of them, and arrive at the result that the sum of two consecutive triangle numbers is a square number:

But here the size of each triangular domain is different.

Similarly, a cube can be divided into six tetrahedra with the same volume. You can translate this statement into a property of \(n \times n \times n\) hypermatrices. An \(n \times n \times n\) hypermatrix has \(n^{3}\) entries (its "volume"). These entries can be divided into six identical tetrahedral domains with \(S_{3}(n-2)\) entries, six identical triangular domains with \(S_{2}(n-1)\) entries, and one linear domain with \(S_{1}(n)\) entries. Thus,

Since

you can absorb the interface triangular domains into the tetrahedral domains.

I tried to find the analogous result in four dimensions. Since \(4! = 24\), a tesseract can be divided into 24 identical pentatopes. Thus, you can imagine that

It took me some effort to find the full result:

That is, you can divide a tesseract array into one linear array, 14 triangular arrays, 36 tetrahedral arrays, and 24 pentatopic arrays. I found this by first writing

and then solving the equation \(C_{4}(n) - n^{4} = 0\). Using this technique you can also show that

That is, you can divide a 5-cube into a one 1-simplex array, 30 2-simplex arrays, 150 3-simplex arrays, 240 4-simplex arrays, and 120 5-simplex arrays.

It seems that in general, you have

For a long time I wondered if there is a neat expression for \(N_{k}(l)\). I have to confessed of being lazy and not actively trying to find \(N_{k}(l)\) based on what I know of \(C_{k}(n)\) and \(S_{k}(n)\). In the end, I rediscovered the Online Encyclopedia of Integer Sequences and searched "24, 36, 14, 1" there. The first hit is the sequence A090582. After searching for "1, 14, 36, 24" I found the sequence A019538. Both of these sequences are related to Stirling numbers of the second kind.