structural:summarization, information:extraction, stylistic:style_transfer, quality:corruption
Number of elements of hypercubes

A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2), instead of (x + 1). There are a couple ways to do this. The simpler is to begin with row 0 = 1 and row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:

That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:

The other way of producing this triangle is to start with Pascal's triangle and multiply each entry by 2, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2 = 6 × 2 = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.

To understand why this pattern exists, first recognize that the construction of an n -cube from an (n − 1) -cube is done by simply duplicating the original figure and displacing it some distance (for a regular n -cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n -cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n -cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n -cube.

In this triangle, the sum of the elements of row m is equal to 3. Again, to use the elements of row 4 as an example: 1 + 8 + 24 + 32 + 16 = 81, which is equal to 3 4 = 81 {\displaystyle 3^{4}=81}.