Geometric Interpretation of Tetraprimes

Following on from my previous post titled Tetraprimes, I asked Gemini the following:

I’d like you to create a program in SageMath that will accept any positive integer with up to four not necessarily distinct prime factors as INPUT and if the integer has FOUR prime factors it should return 4D Hypervolume, 3D Surface Volume, 2D Total Face Area and 1D Total Edge Length. If it has THREE factors then 3D Volume, 2D Surface Area and 1D Edge Length are returned. If it has TWO factors then 2D Surface Area and 1D Edge Length are returned. If the input is prime or has more than four factors then this is stated and no output is returned. The factorisation of the numbers shown in the output should also be displayed.

Here is a permalink to the program that it created using 28302 as input. The output looks of the program is shown below:

Input: 28302
4D Hypervolume: 28302 = 2 * 3 * 53 * 89
3D Surface Volume: 48874 = 2 * 7 * 3491
2D Total Face Area: 21732 = 2^2 * 3 * 1811
1D Total Edge Length: 1176 = 2^3 * 3 * 7^2

The program also deals with triprimes and biprimes. Here is the output using 28303 and 28299 as inputs.

Input: 28303
3D Volume: 28303 = 11 * 31 * 83
2D Surface Area: 7654 = 2 * 43 * 89
1D Edge Length: 500 = 2^2 * 5^3

Input: 28299
2D Surface Area: 28299 = 3 * 9433
1D Edge Length: 18872 = 2^3 * 7 * 337

I've incorporated this algorithm into my daily number analysis. 

In the example shown above, we can see easily that 28303 can be interpreted as the volume of a rectangular prism with a surface area that is also a sphenic number and thus interpretable as the volume of another rectangular prism. This additional information for my daily number analysis will prove useful in other ways I'm sure.

Tetraprimes

While I have looked at numbers with four prime factors counted with multiplicity whose reversals also have this property, I've not actually looked at reversible tetraprimes. Numbers with four distinct prime factors are called tetraprimes. The first example of such a number is 1518 where we have:

• \(1518 = 2 \times 3 \times 11 \times 23 \)
• \( 8151 = 3 \times 11 \times 13 \times 19 \)

There are 273 such numbers in the range up to 40000 (permalink):

1518, 2046, 2226, 2262, 2418, 2478, 2618, 2622, 2814, 2838, 2886, 3135, 3927, 4170, 4182, 4386, 4389, 4746, 4785, 4935, 5313, 5394, 5406, 5478, 5565, 5655, 5838, 5874, 6018, 6045, 6222, 6402, 6438, 6474, 6486, 6690, 6699, 6834, 6846, 6882, 7293, 7458, 8106, 8142, 8151, 8162, 8346, 8382, 8385, 8547, 8742, 8745, 9834, 9966, 10434, 10506, 11022, 11346, 11814, 11946, 12243, 12441, 12738, 12765, 13026, 13299, 13542, 13629, 13695, 14105, 14118, 14421, 14469, 14574, 15114, 15873, 16005, 16107, 16359, 16665, 16786, 16962, 16995, 17017, 17358, 17589, 17655, 17754, 17922, 18183, 18258, 18447, 18462, 18546, 18615, 18879, 19434, 19437, 19446, 19578, 19662, 20022, 20055, 20085, 20118, 20145, 20163, 20190, 20262, 20310, 20355, 20382, 20405, 20526, 20553, 20559, 20562, 20658, 20746, 20769, 20774, 20878, 20922, 20958, 22002, 22011, 22074, 22098, 22154, 22242, 22290, 22458, 22515, 22533, 22578, 22695, 22710, 22737, 22755, 22854, 22902, 22946, 22962, 22971, 24115, 24123, 24178, 24186, 24198, 24222, 24270, 24297, 24339, 24465, 24530, 24531, 24582, 24585, 24618, 24645, 24662, 24717, 24766, 24846, 24890, 24906, 24915, 24969, 26013, 26103, 26106, 26187, 26202, 26286, 26358, 26394, 26445, 26481, 26490, 26502, 26535, 26554, 26634, 26642, 26691, 26733, 26895, 26922, 26961, 26978, 26985, 26994, 28083, 28086, 28194, 28245, 28302, 28329, 28410, 28446, 28455, 28497, 28509, 28542, 28545, 28578, 28623, 28626, 28635, 28785, 28866, 28886, 28905, 28974, 28985, 28990, 30162, 30426, 31038, 31062, 31146, 31494, 31515, 31614, 31955, 32142, 32574, 32637, 32682, 32745, 32835, 32865, 33198, 33285, 33522, 33594, 33654, 33666, 33762, 34017, 34221, 34386, 34419, 34518, 34554, 34626, 34854, 35502, 35673, 35706, 35826, 36102, 36195, 36218, 36645, 36966, 36993, 37037, 37074, 37158, 37365, 37605, 37653, 37785, 37851, 38055, 38082, 38181, 38445, 39045, 39666, 39738, 39963

I've also considered biprimes as rectangles and triprimes as rectangular prisms but what about tetraprimes. I asked Gemini the following:

A triprime with prime factors of \(a\), \(b\) and \(c\) can be conceived of as a rectangular prism with sides \(a\), \(b\), \(c\), volume \(a \, b \, c\) and surface area equal to \(2 \times (ab + bc + ac) \). Are there equivalent formulae for volume and surface area of a four dimensional object derived from a tetraprime with four prime factors.

Here was Gemini's response:

The direct geometrical equivalent of a tetraprime with four prime factors \(a\), \(b\), \(c\), and \(d\) is a four-dimensional rectangular orthotope (commonly called a hyperrectangle).

Because a 4D object possesses boundaries in multiple lower dimensions (3D, 2D, and 1D), the equivalent to your 3D formulas expands into a hierarchy of measurements. Just as a 3D prism is bounded by 2D faces, a 4D orthotope is bounded by 3D cells, 2D faces, and 1D edges.

Here are the equivalent geometric formulas derived from a tetraprime:

• 4D Hypervolume (The equivalent of 3D Volume): The total 4D spatial content of the object is simply the product of all four prime factors, which is exactly equal to the tetraprime itself.$$V_4 = a \cdot b \cdot c \cdot d$$

• 3D Surface Volume (The structural equivalent of 3D Surface Area): In 3D geometry, "surface area" measures the (n-1)-dimensional boundary of the object. For a 4D object, its outer boundary is three-dimensional. A hyperrectangle is enclosed by 8 solid 3D rectangular prisms (called "cells"). There are two cells for each possible triplet of the dimensions.$$S_3 = 2(abc + abd + acd + bcd)$$

• 2D Total Face Area (The literal equivalent of 2D Area): If you are looking specifically for flat, two-dimensional surface area, a hyperrectangle has 24 flat 2D rectangular faces. For every pair of dimensions, there are 4 identical faces throughout the 4D structure.$$A_2 = 4(ab + ac + ad + bc + bd + cd)$$

• 1D Total Edge Length: For complete geometric breakdown, a hyperrectangle has 32 one-dimensional edges. There are 8 parallel edges for each of the four prime factors.$$L_1 = 8(a + b + c + d)$$

By substituting your four prime factors into \(V_4\) and \(S_3\), you achieve the exact mathematical analogies for volume and surface area as you step up from three dimensions into four.

I'll investigate this further in a subsequent blog.

Dali and the Unfolded Hypercube

Figure 1

In April of 1984, as a birthday present from my father, I acquired a TR-80 computer from Tandy with 32Kb of RAM and a keyboard but no disk drive. Data input and output were via a cassette tape. Display was via a connected TV. Using this I began my BASIC programming and one of my first creations was a simulation of a projectile being launched to hit a target. The essential formulae for the projectile's motion are:$$x(t)=v_0 \cos(\theta) \cdot t \text{ and } y(t)=v_0 \sin(\theta) \cdot t - 0.5 g t^2$$where we have:

• \(x(t)\) is horizontal position of projectile after time \(t\) with \(x=0\) when \(t=0\)
• \(y(t)\) is vertical position of projectile after time \(t\) with \(y=0\) when \(t=0\)
• \(v_0\) is initial velocity at time \(t=0\)
• \(\theta\) is angle of projection
• \(g\) is deceleration due to gravity \(\approx \) -9.8 \(ms^{-2}\)

Figure 1 shows the situation with an angle of 60° and an initial velocity fo 6 \(ms^{-1}\). What has all this to do with unfolding hypercubes? Well, April of 1984 marked my return to Mathematics after a long absence. In August of that year, I began my training as a Mathematics teacher and one of the assigned projects was an assignment on some aspect of Mathematics. I chose Mathematics and Art, a choice which led to me to the unfolded hypercube. In the University library, I came across Salvator Dali's 1954 Crucifixion (Corpus Hypercubus). See Figure 2.

Figure 2: source

It is a large oil painting, with dimensions of 194.3 cm × 123.8 cm (76.5 in × 48.7 in). Figure 3 shows the net of a hypercube (or tesseract).

Figure 3: source

What reminded me of my youthful investigation into the realm of the hypercube was a Mathologer video that I watched today titled The Iron Man hyperspace formula really works.

Figure 4 shows a screenshot that demonstrates information about the number of vertices, edges and faces in an \(n\)-dimensional cube where \(n\) varies from 0 to 4. Looking at it, we see that a zero dimensional "cube" consists of only a single point or vertex. The number of vertices is shown by the coefficients of the \(x^0\) terms, coloured red. A one dimensional "cube", corresponding to a line, has two vertices and one edge. The number of edges is shown by the coefficients of the \(x^1\) terms, coloured green. A two dimensional "cube", corresponding to a square, has four vertices, 4 edges and one face. The number of faces is shown by the coefficients of the \(x^2\) terms. The familiar three dimensional cube has 8 vertices, 12 edges and 6 faces, plus one cube. The number of cubes is shown by the coefficients of the \(x^3\) terms. The four dimensional hypercube has 16 vertices, 32 edges, 24 faces, 8 cubes and one hypercube. The vertices, edges, faces and cubes can be counted using the net shown in Figure 2.

Figure 4: source

Figure 5 shows another screenshot where a more general version of Euler's polyhedron formula is displayed. It begins with one dimension where a line of finite length does indeed have two end points or vertices. Thus V=2. For two dimensions, the number of vertices equals the number of edges e.g. consider a square. Thus V=E. For three dimensions, we have the familiar V-E+F=2 and for the hypercube we have V-E+F-C=0 because as can be seen in Figure 4, V=16, E=32, F=24 and C=8.

Figure 5: source

However, getting back to Dali's hypercube, I just finished watching an interesting talk about Dali from an American mathematician who met the artist on a number of occasions. The video is titled Math Encounters -- Encountering Salvador Dali in the Fourth Dimension. 

The video was uploaded in March of 2014 and the mathematician, Tom Banchoff, is still alive and now aged 83. He born on April 7th 1938. He has written a book, titled Beyond the Third Dimension Geometry, Computer Graphics, and Higher Dimensions, a digital copy of which I've managed to acquire. It was first published in 1990. The cover is shown in Figure 6.

Figure 6

The author makes mention of his book in this talk and mentions also Flatland which is a classic that I remember reading many years ago but exactly when I can't recall. Here is a little more about this 1884 book:

This masterpiece of science (and mathematical) fiction is a delightfully unique and highly entertaining satire that has charmed readers for more than 100 years. The work of English clergyman, educator and Shakespearean scholar Edwin A. Abbott (1838-1926), it describes the journeys of A. Square, a mathematician and resident of the two-dimensional Flatland, where women-thin, straight lines-are the lowliest of shapes, and where men may have any number of sides, depending on their social status. 

Through strange occurrences that bring him into contact with a host of geometric forms, Square has adventures in Spaceland (three dimensions), Lineland (one dimension) and Pointland (no dimensions) and ultimately entertains thoughts of visiting a land of four dimensions—a revolutionary idea for which he is returned to his two-dimensional world. Charmingly illustrated by the author, Flatland is not only fascinating reading, it is still a first-rate fictional introduction to the concept of the multiple dimensions of space. "Instructive, entertaining, and stimulating to the imagination." — Mathematics Teacher.

Figure 7 shows the frontispiece to the original publication. Dali was probably familiar with this book as he was very interested in such dimensional adventures. 

Figure 7

Speaking of Flatland's two dimensions, what would a hypercube look like in Flatland? One way it could appear is shown in Figure 8. I got this image from an interesting web page about hypercubes.

Figure 8: source

Figure 9 gives a clearer idea of why this is so. The shadow of the tesseract, itself the 3D shadow of the hypercube, is shown falling on a two dimensional surface.

Figure 9: source

To end this post, I'll return to Mathologer and his YouTube channel. In July of 2017, he posted an interesting video about hypercube shadows that's well worth a look.

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.