Stella Octangula

The term "stella octangula" is another name for a "stellated octahedron" such as is shown in Figure 1.

Figure 1: stellated octahedron
Source

I've only made one previous post about stellated polyhedra and that was The Cubohemioctahedron and other Polyhedra back on the 21st July 2019. Here's what Wikipedia had to say about the stellated octahedron:

The stellated octahedron is the only stellation of the octahedron. It is also called the stella octangula (Latin for "eight-pointed star"), a name given to it by Johannes Kepler in 1609, though it was known to earlier geometers. It was depicted in Pacioli's De Divina Proportione, 1509.

It is the simplest of five regular polyhedral compounds, and the only regular compound of two tetrahedra. It is also the least dense of the regular polyhedral compounds, having a density of 2.

It can be seen as a 3D extension of the hexagram: the hexagram is a two-dimensional shape formed from two overlapping equilateral triangles, centrally symmetric to each other, and in the same way the stellated octahedron can be formed from two centrally symmetric overlapping tetrahedra. This can be generalized to any desired amount of higher dimensions; the four-dimensional equivalent construction is the compound of two 5-cells. It can also be seen as one of the stages in the construction of a 3D Koch snowflake, a fractal shape formed by repeated attachment of smaller tetrahedra to each triangular face of a larger figure. The first stage of the construction of the Koch Snowflake is a single central tetrahedron, and the second stage, formed by adding four smaller tetrahedra to the faces of the central tetrahedron, is the stellated octahedron.

Associated with this shape are the stella octangula numbers. These are figurate numbers  of the form \(n(2n^2 − 1) \) and they form OEIS A007588:

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960

To quote from Wikipedia again:

There are only two positive square stella octangula numbers. The first is \(1\) and the other is$$9653449 = 3107^2 = (13 × 239)^2$$corresponding to \(n = 1\) and \(n = 169\) respectively. The elliptic curve describing the square stella octangula numbers is:$$m^2=n(2n^2-1)$$Using Geogebra, this curve is shown in Figure 2.

Figure 2 

The stella octangula numbers arise in a parametric family of instances to the crossed ladders problem in which the lengths and heights of the ladders and the height of their crossing point are all integers. In these instances, the ratio between the heights of the two ladders is a stella octangula number.

The shape and these numbers caught my attention because my diurnal age today is 27624 and this number is a member of the OEIS sequence for the case of \(n=24\). These numbers are quite sparse. The previous number was 24311 and the next will be 31225.

Here is a video which shows how to create a stellated octahedron using origami. It was uploaded on the 13th March 2010 but the technique of course is timeless. Figure 3 shows another representation of the shape.

Figure 3: source

Figure 4 shows how the shape can be inscribed in a cube and also illustrates its connection with the hexagon.

Figure 4: source

The site from which Figures 3 and 4 were taken also contains illustrations of nets can be used to construct the models. This approach is easier than the origami method. Instructions for the construction of wire frame models can also be found there.

Figure 5 shows a screenshot from a site that provides a stellated octahedron calculator. Entering a side length of 53 generates a volume of 26318 to the nearest decimal place. However, the \(a\) shown in Figure 5 is a side length whereas the common formula uses edge length \(b\) where \(b=a/2\). The volume of the shape using edge length of \( b \) is \(b^3 \sqrt {2} \).

Figure 5: source

Centered Platonic Numbers

On the 19th November 2018, I created a post titled Platonic Numbers and at the end of that post I made reference to Centered Platonic Numbers as follows:

Additionally, there are the centered Platonic numbers defined by starting with 1 central dot (for \(n\)=0) and adding regular convex polyhedral layers around the central dot, where the \(n\)th layer, \(n\) ≥ 1, has \(n\)+1 dots per facet ridge (face edge for polyhedrons) including both end vertices. The formulae are very similar to the above and can be explored further here.

Let's look a particular type of centered Platonic number, namely the centered tetrahedral numbers, and the case where n=2. See Figure 1.

Figure 1

This second centered tetrahedral number differs from the equivalent tetrahedral number by 1. The progression of the tetrahedral numbers from n=1 to 4 is shown in Figure 2.

Figure 2

With the tetrahedral numbers, there is a layering process in place as is clearly visible in the progression shown in Figure 2. With the centered tetrahedral numbers, the tetrahedra shown in Figure 2 are being built around the central dot. Each progressive tetrahedron contains all the previous tetrahedra, Russian doll style. 

Thus the centered tetrahedral numbers can be derived by progressively summing the tetrahedral numbers:

1, 1 + 4, 1 + 4 + 10, 1 + 4 + 10 + 20 --> 1, 5, 15, 35 etc.

The formulae for the various centered Platonic numbers can be found at this site and for the centered tetrahedral numbers, the formula is:$$ \frac{ (2n+1)(n^2+n+3)}{3}$$This formula generates the following initial members of OEIS A005894:

1, 5, 15, 35, 69, 121, 195, 295, 425, 589, 791, 1035, 1325, 1665, 2059, 2511, 3025, 3605, 4255, 4979, 5781, 6665, 7635, 8695, 9849, 11101, 12455, 13915, 15485, 17169, 18971, 20895, 22945, 25125, 27439, 29891, 32485, 35225, 38115

The generating function for this series is: $$ \frac{(1+x)(1+x^2)}{(1-x)^4}$$For example, the first few terms of the resulting Taylor series are:$$1+5x+15x^2+35x^3+69x^4+121x^5+195x^6+295x^7 + ...$$

Pentatope Numbers

Today I turned 26176 days old and one of the properties of this number, as listed in the Online Encyclopaedia of Integer Sequences (OEIS), is that it's a member of OEIS A118411:

A118411

Numerator of sum of reciprocals of first n pentatope numbers A000332.

 The sequence runs:

1, 6, 19, 136, 83, 119, 656, 73, 190, 121, 1816, 559, 679, 815, 3872, 1139, 886, 513, 2360, 2023, 2299, 2599, 11696, 3275, 7306, 1353, 5992, 1653, 5455, 5983, 26176, ...

I didn't know what pentatope numbers were so I needed to find out. Firstly, this type of number is a particular example of a more general type of number, the polytope number. It can exist in any number of dimensions and so we can speak of an \(n\)-polytope where \(n\) represents the number of dimensions. However, we get into deep water quickly if we wax too general so I'll bring it back to specifics for the moment by looking at Pascal's triangle (Figure 1).

Figure 1: Derivation of pentatope numbers from
    a left-justified Pascal's triangle. Source.

So a pentatope number could be defined as any number in the fifth cell of any row of Pascal's triangle starting with the 5-term row 1 4 6 4 1. The terms are given by OEIS A000332:

1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, ...

Pascal's triangle lists the binomial coefficients and the pentatope numbers (let's refer to them as \(P_n\)), can be expressed in terms of these coefficients, specifically:$$P_n=\binom{n+3}{4} \text{ where }n \geq 1$$This reads as the number of ways to choose \(n+3\) things four at a time. An alternative representation involves the use of the rising factorial, so that we can write:$$P_n=\frac {n^{ \overline 4}}{4!}=\frac{n(n+1)(n+2)(n-3)}{24} \text { where }n \geq 1$$The infinite sum of the reciprocals of all pentatope numbers is \(\frac{4}{3} \) and this takes us back to OEIS A118411, the sequence that started this post. In other words:$$ \sum^ {\infty}_{n=1} \frac{4!}{n(n+1)(n+2)(n+3)}=\frac{4}{3}$$This sequence only shows the numerators, so let's write some SageMath code to generate the actual fractions (permalink). See Figure 2.

Figure 2

The result is:

(1, 1), (6, 5), (19, 15), (136, 105), (55, 42), (83, 63), (119, 90), (656, 495), (73, 55), (190, 143), (121, 91), (1816, 1365), (559, 420), (679, 510), (815, 612), (3872, 2907), (1139, 855), (886, 665), (513, 385), (2360, 1771), (2023, 1518), (2299, 1725), (2599, 1950), (11696, 8775), (3275, 2457), (7306, 5481), (1353, 1015), (5992, 4495), (1653, 1240), (5455, 4092), (5983, 4488), (26176, 19635)

Now \(\frac{4}{3}=1. \dot{3} \) and even \( \frac{6}{5}=1.2 \) is getting close to this. If we look at the final fraction in the above list, we see that: $$\frac{26176}{19635}=1.33312961548256$$ Clearly the fractions are getting closer to \(1. \dot{3}\). The generating function for the pentatope numbers is:$$ \frac{x}{(1-x)^5}=x+5x^2+15x^3+35x^4+...$$Referring back to Figure 1, it can be seen that the pentatope numbers are the cumulative sums of the tetrahedral number, just as the tetrahedral numbers are the cumulative sums of the triangular numbers and the triangular numbers are the cumulative sums of the natural numbers. 

To quote from this source:

"Pentatope" is a recent term. Regarding the fifth row, Pascal wrote that ... since there are no fixed names for them, they might be called triangulo-triangular numbers. Pentatope numbers exists in the 4D space and describe the number of vertices in a configuration of 3D tetrahedrons joined at the faces.

This is a big topic and a lot more could be said of about pentatopes and polytopes but that will do for now as we are focusing on the pentatope numbers.