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.