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.

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

Digitally Distinct (2D) and Doubly Digitally Distinct (3D) Numbers

A Digitally Distinct Number or 2D number is a term that I concocted to describe a number that has:

• no repeated digits  
• an additive digital root that is different to any of its digits

The number associated with my diurnal age today, 26870, is one such number since it clearly has no repeated digits and its additive digital root is 5.

The numbers 0 to 9 do not qualify because they are identical to their additive digital roots. However, 12 has an additive digital root of 3 and thus it is the first 2D number and begins a run of seven consecutive such numbers viz. 12, 13, 14, 15, 16, 17 and 18. The percentage of such numbers declines with their size. Here is a summary:

• 0 -10 0.00%
• 0 - 100 56.0%
• 0 - 1000 50.4%
• 0 - 10000 31.0%
• 0 - 100000 16.2%
• 0 - 1000000 6.99%

Once a number has more than nine digits, it cannot be a 2D number because at least one digit would then repeat. The upper limit must be below 987,654,320, a number that has an additive digital root of 8 and is thus not a 2D number. I excluded 987,654,321 because additive digital roots lie between 1 and 9 and all those digits are taken. The question that must be asked is what is the largest 2D number? It can contain no more than nine digits and one of those must be zero. Testing revealed that:

• no nine digit number containing the digit 9 can be a 2D number
• 876,543,210 qualifies as a 2D number since it has a digital root of 9

So it is that 876,543,210 is the largest 2D number although all of the 8 x 8! = 322,560 (leading zeros not allowed) possible permutations are of course 2D numbers.

In the range between 26500 and 27000, the percentage of 2D numbers is 17.2%. The numbers are (with my diurnal age shown in bold):

26503, 26504, 26508, 26509, 26513, 26514, 26517, 26518, 26530, 26531, 26539, 26540, 26541, 26548, 26549, 26571, 26578, 26580, 26581, 26584, 26587, 26589, 26590, 26593, 26594, 26598, 26703, 26704, 26708, 26715, 26730, 26740, 26748, 26749, 26751, 26758, 26780, 26784, 26785, 26789, 26794, 26798, 26803, 26805, 26807, 26809, 26814, 26815, 26830, 26834, 26839, 26841, 26843, 26845, 26847, 26850, 26851, 26854, 26857, 26859, 26870, 26874, 26875, 26879, 26890, 26893, 26895, 26897, 26904, 26905, 26908, 26935, 26938, 26940, 26945, 26947, 26950, 26953, 26954, 26958, 26974, 26978, 26980, 26983, 26985, 26987

Here is a permalink to the algorithm that I used to generate these numbers. 

An interesting extension is to consider the multiplicative digital root which is the single digit reached when multiplying the digits of the number together (the results can range from 0 to 9). I've concocted the term Doubly Digitally Distinct or 3D for numbers that satisfy the following criteria:

• no repeated digits
• an arithmetic digital root that is different to any of its digits
• a multiplicative digital root that is different to any of its digits and also to the arithmetic digital root

Applying these criteria to the same range of numbers as earlier (26500 to 27000), we find 11.6% of numbers satisfy. These are:

26513, 26514, 26517, 26518, 26531, 26539, 26541, 26548, 26549, 26571, 26578, 26581, 26584, 26587, 26589, 26593, 26594, 26598, 26715, 26748, 26749, 26751, 26758, 26784, 26785, 26789, 26794, 26798, 26814, 26815, 26834, 26839, 26841, 26843, 26845, 26847, 26851, 26854, 26857, 26859, 26874, 26875, 26879, 26893, 26895, 26897, 26935, 26938, 26945, 26947, 26953, 26954, 26958, 26974, 26978, 26983, 26985, 26987

The number 26870 does not qualify as a 3D number because its multiplicative digital root is 0 and that is one of the digits of the number. In fact, any number containing a zero cannot be a 3D number. However, the nearby 26874 and 26875 both qualify as they have additive digital roots of 9 and 1 respectively and multiplicative digital roots of 0. A similar table to that shown above but this time for 3D numbers looks like this.

• 0 -10  0.00%
• 0 - 100 33.0%
• 0 - 1000  26.7%
• 0 - 10000 14.9%
• 0 - 100000 7.61%
• 0 - 1000000 2.78%

Here is a permalink that can be used to generate the statistics in this table. The first 3D number is 23 and it begins a run of three consecutive such numbers: 23, 24 and 25. We see that:

• 23 has an additive digital root of 5 and a multiplicative digital root of 6
• 24 has an additive digital root of 6 and a multiplicative digital root of 8
• 25 has an additive digital root of 7 and a multiplicative digital root of 0

However, the next number 26 has an additive digital root of 8 and a multiplicative digital root of 2 which is one of the digits of the original number. Thus it does not meet the criteria. The question remains as to what is the largest 3D number. It cannot contain more than eight distinct digits. Testing revealed that:

• no eight digit number containing the digit 9 is a 3D number
• 87,654,321 qualifies as a 3D number
• It has a arithmetic digital root of 9
• it has a multiplicative digital root of 0

So 87,654,321 is the largest 3D number although any of the 8! = 40,320 permutations of those digits will also be a 3D number.

ADDENDUM 

October 30th 2020

It occurred to me that it would also be interesting to look at the "complement" of 2D and 3D numbers. The complement of 2D numbers I will define as numbers that have at least one repeated digit and whose arithmetic digital root is one of the digits of the number. The complement of 3D numbers I will define as numbers that have at least one repeated digit and whose arithmetical digital root and multiplicative digital roots are digits of the number.

Here is a permalink to an algorithm that will identify complementary 3D numbers in the range up to 40,000. I have also made an entry in my Bespoken For Sequences. Such numbers comprise 8.01% of the range. Here are the initial members: 0, 100, 118, 181, 188, 200, 299, 300, 400, 500, 600, 700, 800, 811, 818, 881, 899, 900, 909, 929, 989, 990, 992, 998, 1000.

Numbers like 1000 clearly qualify for membership so let's take the less obvious 998. The number has one repeated digit (9) and its arithmetic digital root is 8 while its multiplicative digital root is also 8. Thus it qualifies too. Clearly such complementary 2D and 3D numbers have no upper bound unlike the 2D and 3D numbers themselves. 

A Spider-Fly Problem with a Surprising Solution

I’ve long been familiar with the spider and fly problem and its solution. Although, after reading this article, I realised that I had been focused on an incorrect solution. This was indeed surprising to realise that I’d been deluded all these years. The problem can be stated as follows: a spider and a fly are on opposite walls of a 30 × 12 × 12 meter room. The spider is 1 meter above the floor, the fly is 1 meter below the ceiling. They are both 6 meters from adjacent walls, as shown in Figure 1. If the fly does not move, what is the shortest distance the spider can crawl to reach it?

Figure 1

A sensible first attempt would be to travel straight up (or down) and across. For example, straight up the spider’s wall (11 meters), along the roof (30 meters) and down to the fly (1 meter). See Figure 2. This gives a total distance of 42 meters.

Figure 2

What I believed to be the shortest path is shown in Figure 3 and it is clearly not the shortest path!

Figure 3

In fact the shortest path requires the spider to cross five of the six internal surfaces and this is shown in Figure 4. The shortest path can be seen to be 40 metres.

Figure 4

The site from which this information is taken has some nice animated gifs of the rectangular prism’s unfolding, so the reader is encouraged to visit. The author of the article is Russell Lim, a high school teacher in Melbourne.

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.