Visualising a Difference of Two Cubes

For some time I've thought about a visual representation of a positive integer that is the difference of two cubes. For example, the number 973 can be represented thus:$$ \begin{align} 973 &= 1000 - 27 \\ &=10^3-3^3 \end{align} $$Given the limitations of Nano Bananas, I thought that a rought sketch of how I wanted to represent the number would help. This is shown in Figure 1.

Figure 1

I uploaded this sketch to Gemini along with the following prompt:

Use the uploaded image to create a realistic depiction of the cube shown following the heavy lines that drift the edges. The large cube has a smaller cube removed from it. The cube sits on a flat surface and is golden coloured with soft lighting highlighting the missing section of the cube. Maintain two point perspective and incorporate the numerals shown.

Figure 2 shows the result:

Figure 2

I asked Gemini:

Given today's gold price what would a cube of 100% pure gold with side of 10 millimetres be worth in US dollars. Also price a cube of gold with side of 3 millimetres.

The response:

To calculate the value of both cubes, we need today's gold price and the physical density of pure gold.

As of February 24, 2026, the spot price for 24k (100% pure) gold is approximately 166.69 USD per gram. The density of pure gold is 19.32 grams per cubic centimeter (g/cm³).

1. The 10 mm Cube

Volume: 10 mm × 10 mm × 10 mm = 1,000 mm³, which is exactly 1 cm³.

Mass: 1 cm³ × 19.32 g/cm³ = 19.32 grams.

Value: 19.32 grams × 166.69 USD/gram = 3,220.45 USD.

2. The 3 mm Cube

Volume: 3 mm × 3 mm × 3 mm = 27 mm³, which is 0.027 cm³.

Mass: 0.027 cm³ × 19.32 g/cm³ = 0.52164 grams.

Value: 0.52164 grams × 166.69 USD/gram = 86.95 USD.

Connecting this back to the golden object depicted in your image—a 10 mm cube with a 3 mm cubed corner removed—its total gold value today would be 3,220.45 USD minus 86.95 USD, which equals 3,133.50 USD.

There are 824 numbers up to 40000 that are the difference of two cubes (some in more ways than one) but only those from 28000 to 40000 are listed below (permalink will generate the full list):

28063, 28415, 28460, 28519, 28568, 28656, 28672, 28701, 28737, 28791, 28828, 28854, 29051, 29062, 29078, 29080, 29107, 29279, 29393, 29402, 29448, 29528, 29575, 29617, 29666, 29701, 29727, 29735, 29763, 29764, 29783, 29790, 30016, 30024, 30043, 30105, 30248, 30301, 30312, 30483, 30571, 30708, 30807, 30907, 30970, 31024, 31031, 31040, 31085, 31106, 31213, 31228, 31232, 31304, 31437, 31519, 31768, 31806, 31841, 31869, 31976, 32039, 32137, 32227, 32256, 32319, 32425, 32445, 32464, 32465, 32552, 32562, 32643, 32704, 32741, 32760, 32761, 32767, 32832, 32851, 32858, 32920, 32949, 32984, 33077, 33193, 33336, 33391, 33472, 33614, 33724, 33740, 33752, 33875, 34027, 34047, 34209, 34391, 34489, 34531, 34606, 34658, 34669, 34784, 34875, 34902, 34930, 34937, 35008, 35028, 35163, 35189, 35208, 35315, 35317, 35425, 35576, 35594, 35721, 35812, 35873, 35910, 35929, 35936, 35971, 36008, 36016, 36153, 36253, 36297, 36316, 36504, 36506, 36560, 36631, 36632, 36785, 36829, 37000, 37043, 37107, 37296, 37297, 37367, 37395, 37448, 37449, 37576, 37648, 37962, 37969, 37973, 38017, 38142, 38151, 38285, 38304, 38402, 38486, 38528, 38575, 38619, 38647, 38656, 38779, 38792, 38961, 39004, 39088, 39130, 39179, 39240, 39247, 39277, 39296, 39303, 39331, 39368, 39500, 39611, 39636, 39797, 39807, 39815, 39816, 39823

Notice how there is clustering about cubic numbers. For example, \(31^3=32768\) and so the following are members of the sequence:

• \(31^3-1^3=32767\)
• \(31^3-2^3=32760\)
• \(31^3-3^3=32741\)
• \(31^3-4^3=32704\)

However, looking back at the range of numbers above, notice how 32761 slips in (to form a consecutive pair with 32760) and this is because:$$32761=105^3-104^3$$This number is also mentioned below. These numbers that are differences of two cubes form OEIS A181123. The OEIS comments are interesting:

Because \(x^3-y^3 = (x-y)(x^2+xy+y^2)\), the difference of two cubes is a prime number only if \(x=y+1\), in which case all the primes are cuban, see A002407.

The difference can be a square (see A038597), but Fermat's Last Theorem prevents the difference from ever being a cube.

The numbers that are square and the difference of two cubes are 169, 784, 2401, 10816, 21609, 32761 and 35721 with squares as shown:$$ \begin{align} 169 &=13^2 =8^3-7^3\\ 784 &= 28^2 =10^3-6^3\\ 2401 &=49^2=14^3-7^3 \\ 10816 &=104^2=32^3-28^3 \\ 21609 &= 147^2=28^3-7^3 \\32761 &=181^2 =105^3-104^3 \\35761 &= 189^2=33^3-6^3 \end{align}$$There are 44 cuban primes in the range and they are:

7, 19, 37, 61, 127, 271, 331, 397, 547, 631, 919, 1657, 1801, 1951, 2269, 2437, 2791, 3169, 3571, 4219, 4447, 5167, 5419, 6211, 7057, 7351, 8269, 9241, 10267, 11719, 12097, 13267, 13669, 16651, 19441, 19927, 22447, 23497, 24571, 25117, 26227, 27361, 33391, 35317

With all these cuban primes, the difference between the bases of the primes must be 1 e.g. \(7=2^3-1^3\).

In conclusion, the main point of this post was to show one way in which a difference of two cubes can be represented physically. While Mathematics is highly abstract, it's still possible on occasion to make it more concrete.

Compressing a Hollow Cube

Rewatching the movie "Transformers", my attention was caught by the giant (presumably hollow) cube made up out of many tiny "unit" cubes, that reassembled itself into a much smaller solid cube. This got me thinking about what volume a hollow cube could be compressed into to form a solid cube. Let's consider a  hollow cube with a side of \(n\) units. It has a volume of \(n^3\) cubic units with a hollow centre of \( (n-2)^3 \) cubic units . Thus the number of unit cubes are:$$\begin{align} n^3 - (n-2)^3 &= 2 (n^2 + n(n-2) + (n-2)^2) \\ &= 2(n^2+n^2-2n+n^2-4n+4)\\ &=2(3n^2-6n+4) \end{align}$$Using this formula we can determine the size of the solid cube that can be formed from the unit cubes and how many blocks are left over from this process. See Table 1.

Table 1: permalink

From Table 1 we can see that 216 of the 218 unit cubes that comprise a hollow cube of side 7 units can be used to form a solid cube of side 6 units with only two cubes left over. See Table 2.

Table 2

Also we can see that 13824 of the 13826 unit cubes that comprise a hollow cube of side 49 units can be used to form a solid cube of side 24 units with only two cubes left over. See Table 3.

Table 3

Looking beyond the figures in Table 1, we find that hollow cubes with sides of 163, 385, 751 and 1297 units also compress to solid cubes with sides of 54, 96, 150 and 216 units with two cubes left over (permalink).

We can summarise the results for the "2 left over hollow to solid cubes" as follows:

(7, 6) gives 62.97 % ratio of volumes (solid to hollow)
(49, 24) gives 11.75 % ratio of volumes (solid to hollow)
(163, 54) gives 3.636 % ratio of volumes (solid to hollow)
(385, 96) gives 1.550 % ratio of volumes (solid to hollow)
(751, 150) gives 0.7968 % ratio of volumes (solid to hollow)
(1297, 216) gives 0.4619 % ratio of volumes (solid to hollow)

Naturally as the side length of the hollow cube increases, the ratio of the volume of the solid cube to its hollow counterpart decreases rapidly as more and more empty space is formed inside. For example, in the case of the 751 hollow cube, the ratio of the solid to hollow volumes in less than 1%. It can be noted that none of the sides of the "2 left over hollow to solid cubes" have any factors in common.

Let's see if there's a pattern in the sides of the hollow and solid cubes:

• 7 = 7 and 6 = 2 * 3
• 49 = 7^2 and 24 = 2^3 * 3
• 163 = 163 and 54 = 2 * 3^3
• 385 = 5 * 7 * 11 and 96 = 2^5 * 3
• 751 = 751 and 150 = 2 * 3 * 5^2
• 1297 = 1297 and 216 = 2^3 * 3^3

It can be seen that the solid cubes all contain "6" as a factor. Let's investigate further. Is it possible that a hollow cube with integer sides can be collapsed into a solid cube of integer side \(x\) with no unit cubes left over? If it were possible then we would have:$$ \begin{align} 2(3n^2-6n+4) &= x^3\\6n^2-12n+8 &= x^3\\x &= (6n^2-12n+8)^{1/3}  \end{align}$$Now testing up to \(n=10000\), the expression on the LHS of the equation above never produces a whole number. I'll test for larger values of \(n\) later.

What about when there are two unit cubes left over. In that case we have:$$ \begin{align} 2(3n^2-6n+4) &= x^3+2\\6n^2-12n+6 &= x^3 \\ x &= (6n^2-12n+6)^{1/3} \end{align} $$Now in this case, up to \(n=10000\), we get all the solutions shown above plus some more. These are shown in Table 4.

Table 4: permalink

So the key equation to work with is \(6n^2-12n+(8-a) \) where \(a\) is the number of unit cubes left over. Up to 10000, there are no solutions for \(a=0\) and \(a=1\) but with \(a=2\) we get the solutions shown in Table 4. Going back to Table 1 and using the values of \(a\) shown there in the column "blocks wasted" will produce other tables of solutions as shown in Table 4. There's more to be discovered here but this post is at least a start.

The Horn Torus

The Horn Torus is a solid that if formed by the revolution of a circle around a point on its circumference. Suppose the starting circle has a diameter of 1 unit and thus a circumference of \(\pi\) units. Consider a small rotation of length \( \delta x\) along the outer circumference that produces a wedge-shaped solid that is equivalent to a cylinder of curved surface area \(\pi  \delta x/2\). The total surface area of the resultant torus is given by:$$\begin{align} \text{Surface Area }&=\lim_{\delta x \rightarrow 0} \sum_0^{2\pi} \pi \, \delta x/2 \\ &=\int_0^{2\pi} \! \! \pi/2 \, dx\\&=\bigg [ \pi \, x/2 \bigg ]_0^{ 2\pi}\\&=\pi^2 \end{align}$$This is beautifully simple. If we envisage \(\pi\) as the area of a circle with unit radius then \( \pi^2 \) can be envisaged as the surface area of a horn torus with unit tube diameter. 

While we're here, we may as well calculate the volume of a horn torus with a unit tube diameter. The calculation of volume is very similar to that of the surface area. Consider a small rotation of length \( \delta x\) along the outer circumference that produces a wedge-shaped solid that is equivalent to a cylinder of cross-sectional area \( \pi/4\) and thickness \( \delta x/2\). The volume of this wedge shape is \(\pi \, \delta x/8\). The total volume of the resultant torus is given by:$$\begin{align} \text{Volume }&=\lim_{\delta x \rightarrow 0} \sum_0^{2\pi} \pi \, \delta x/8 \\ &=\int_0^{2\pi} \!  \! \pi/8 \, dx\\&=\bigg [ \pi \, x /8 \bigg ]_0^{ 2\pi}\\&=\pi^2/4\end{align}$$

Reversible Sphenic Numbers

There is the reversible prime, known as an emirp. There is the reversible semiprime, known as an emirpimes, and then there is the reversible sphenic number, known as an cinehps. This is a rather ugly term so I'll just use the term reversible sphenic number. An example of an emirp is 17 whose reversal, 71, is also prime. An example of an emirpimes is 26 = 2 x 13 whose reversal, 62 = 2 x 31, is also a semiprime. The first example of a cinehps is 165 = 3 x 5 x 11 whose reversal, 561 = 3 x 11 x 17, is also a sphenic number.

These numbers form OEIS A270175:

A270175

Cinehps numbers: sphenic numbers whose reversal is a different sphenic number.

Note that palindromic sphenic numbers are excluded. The initial members of the sequence are:

165, 246, 285, 286, 366, 418, 435, 438, 498, 534, 561, 582, 609, 642, 663, 682, 759, 814, 834, 894, 906, 957, 1002, 1023, 1034, 1066, 1095, 1113, 1131, 1185, 1209, 1239, 1245, 1265, 1311, 1342, 1353, 1374, 1398, 1419, 1443, 1446, 1479, 1515, 1526, 1542, 1545, ...

Up to the one million mark, these numbers total 5.28% of the range. One of the concepts associated with a sphenic number is that of the sphenic brick. Let's consider a sphenic number \(n\) whose factors are \(a,b,c\). The sphenic brick is the three dimensional cuboid with volume of \(n\) cubic units and linear dimensions of \(a,b\) and \(c\) units.

Such a brick has an associated surface area and a thought occurred to me. Are there any reversible sphenic number pairs that each have the same surface area? A little investigation revealed that there are. They are rare birds indeed however, and there are only eight ( in four pairs) in the range up to three million (366 and 663, 3245 and 5423, 3685 and 5863, 921239 and 932129). Here are the details and here is the permalink to the calculation. Surface areas are shown in bold red.

366  = 2 x 3 x 61 --> 622 and 663 = 3 x 13 x 17 --> 622

3245 = 5 x 11 x 59 --> 1998 and 5423 = 11 x 17 x 29 --> 1998

3685 = 5 x 11 x 67 --> 2254 and 5863 = 11 x 13 x 41 --> 2254

921239 = 11 x 89 x 941 --> 190158 and 932129 = 11 x 101 x 839 --> 190158

None of the sphenic bricks associated with these numbers look like bricks because they are all very elongated but that's the term that is used for these shapes. Figure 1 shows that the 11 x 89 x 941 looks more like a plank than a brick.

Figure 1

There may well be more beyond the three million mark but SageMathCell timed out above that. Anyway, fascinating that such numbers exist with the first of them being 366, the number of days in a leap year. It can be noted that 366 is the only even number. Placed in sequence on the number line we have:

366, 663, 3245, 3685, 5423, 5863, 921239, 932129

This sequence of terms could be described thus:

Non-palindromic sphenic numbers which, when reversed, are also sphenic numbers with the members of both pairs having identical sphenic brick surface areas. 

Not surprisingly this sequence does not appear in the OEIS, nor will it, as I have ceased to contribute.

ADDENDUM:

The above idea can be applied to semiprimes as well. In this case we will be working with two dimensional rectangles. The two factors of the semiprime form the length and width of the associated rectangle. In the range up to one million, reversible semiprimes comprise 6.06% of the total range but there are only 26 reversible semiprimes with the property that their associated areas are equal. See permalink for calculation.

The pairs (up to one million) are:

14269 = 19 x 751 --> 1540 and 96241 = 157 x 613 --> 1540

15167 = 29 x 523 --> 1104 and 76151 = 271 x 281 --> 1104

16237 = 13 x 1249 --> 2524 and 73261 = 61 x 1201 --> 2524

18449 = 19 x 971 --> 1980 and 94481 = 107 x 883 --> 1980

18977 = 7 x 2711 --> 5436 and 77981 = 29 x 2689 --> 5436

36679 = 43 x 853 --> 1792 and 97663 = 127 x 769 --> 1792

140941 = 97 x 1453 --> 3100 and 149041 = 103 x 1447 --> 3100

150251 = 347 x* 433 --> 1560 and 152051 = 383 x 397 --> 1560

196891 = 401 x 491 --> 1784 and 198691 = 431 x 461 --> 1784

302363 = 211 x 1433 --> 3288 and 363203 = 263 x 1381 --> 3288

308459 = 173 x 1783 --> 3912 and 954803 = 937 x 1019 --> 3912

319853 = 317 x 1009 --> 2652 and 358913 = 379 x 947 --> 2652

958099 = 761 x 1259 --> 4040 and 990859 = 839 x 1181 --> 4040

It can be noted that all these reversible semiprimes are odd.

If we are even more adventurous we can consider numbers with four distinct factors as being associated with 4-dimensional hypercuboids. When unfolded into 3-dimensions, the eight cuboids are the equivalent of the surface area of our sphenic bricks. In the range up to three million, about 1.48% of numbers are reversible four factor numbers with the first one being 1518 = 2 x 3 x 11 x 23 and 8151 = 3 x 11 x 13 x 19. However, no number pairs emerge with identical volumes of unfolded cuboids. Figure 1 shows an unfolded hypercube and an unfolded hypercuboid will look similar but will have four pairs of cuboids each with different volumes, instead of consisting of eight identical cubes. Permalink (which may need double checking).

Figure 1

Sphenic Brick Trajectories

I've made mention of sphenic numbers in three earlier posts. Specifically:

• Sphenic Numbers on Monday, 25th June 2018
• Sphenic Numbers Revisited on Tuesday, 1st January 2019
• An Unhappy Family on Monday, 10th June 2019

I've long championed the association between the surface area of a sphenic brick and its volume. Consider a sphenic number such as 170 that factors to 2 * 5 * 17. It can be considered to represent a rectangular prism with volume 170 cubic units and dimensions of 2, 5 and 7 units. The surface area of such a prism is 258 square units. In previous posts, I've examined the ratio of surface area to volume but today a thought struck me. What if the surface area itself in a sphenic number? This would mean that the surface area could be linked to another rectangular prism.

This is indeed the case for 170 because its surface area of 258 = 2 * 3 * 43 and can thus be linked to a prism with volume of 258 cubic units and dimensions of 2, 3 and 43 units. This prism has a surface area of 442 square units. The obvious question is: can this process be continued? Well 442 = 2 * 13 * 17 and so the answer is yes. The resulting prism has a surface area of 562 square units but 562 = 2 * 281 and so this is where things stopped.

I then got to thinking about the maximum number of iterations possible up to a certain limit. To investigate this, I needed to develop a robust algorithm and I spent most of the day tinkering with one. In the end, using SageMathCell, I succeeded. Here's a permalink to the coding window and below are the runs of eight iterations up to 40,000:

[7386, 12322, 12970, 18178, 19018, 20194, 22042, 22882, 25642]
[8078, 10414, 11086, 12142, 14062, 14734, 15502, 16942, 17902]
[9514, 10066, 12970, 18178, 19018, 20194, 22042, 22882, 25642]
[9515, 5646, 9422, 12142, 14062, 14734, 15502, 16942, 17902]
[9562, 12322, 12970, 18178, 19018, 20194, 22042, 22882, 25642]
[10634, 12322, 12970, 18178, 19018, 20194, 22042, 22882, 25642]
[15085, 10414, 11086, 12142, 14062, 14734, 15502, 16942, 17902]
[15110, 21174, 35302, 39094, 46246, 51190, 71686, 73942, 87430]
[15654, 26102, 27910, 39094, 46246, 51190, 71686, 73942, 87430]
[23313, 18110, 25374, 42302, 48862, 57790, 80926, 84862, 86590]
[27363, 26102, 27910, 39094, 46246, 51190, 71686, 73942, 87430]
[28217, 10414, 11086, 12142, 14062, 14734, 15502, 16942, 17902]
[30173, 10414, 11086, 12142, 14062, 14734, 15502, 16942, 17902]
[30441, 21566, 22782, 37982, 48862, 57790, 80926, 84862, 86590]
[32331, 26606, 27822, 46382, 59662, 64942, 71854, 75886, 83950]
[35121, 26606, 27822, 46382, 59662, 64942, 71854, 75886, 83950]

So starting with 7386, there is then a run of eight sphenic numbers generated by the volume-area iteration. The run ends at 25642 which is not a sphenic number. Similarly for the other chains shown and it should be noted that several chains merge into others. For example, 7386, 9514 and 9562, all end with 25642. So, how many iterations are possible? Well, up to \( \textbf{ten million}\), there are three chains of 15 iterations, all ending in \( \textbf{10186102} \) (permalink):

8710117, 1384374, 2307302, 2394582, 3990982, 4009942, 4044454, 4470262, 4758790, 6662326, 7873702, 8021302, 8362822, 9649462, 9733222, 10186102

The factorisations are:

  number     factor

  8710117    13 * 613 * 1093
  1384374    2 * 3 * 230729
  2307302    2 * 53 * 21767
  2394582    2 * 3 * 399097
  3990982    2 * 467 * 4273
  4009942    2 * 239 * 8389
  4044454    2 * 19 * 106433
  4470262    2 * 31 * 72101
  4758790    2 * 5 * 475879
  6662326    2 * 11 * 302833
  7873702    2 * 107 * 36793
  8021302    2 * 47 * 85333
  8362822    2 * 13 * 321647
  9649462    2 * 233 * 20707
  9733222    2 * 43 * 113177
  10186102   2 * 23 * 79 * 2803

9469213, 1768854, 2948102, 2958742, 3118822, 4009942, 4044454, 4470262, 4758790, 6662326, 7873702, 8021302, 8362822, 9649462, 9733222, 10186102

The factorisations are:

number     factor

  9469213    13 * 61 * 11941
  1768854    2 * 3 * 294809
  2948102    2 * 787 * 1873
  2958742    2 * 37 * 39983
  3118822    2 * 7 * 222773
  4009942    2 * 239 * 8389
  4044454    2 * 19 * 106433
  4470262    2 * 31 * 72101
  4758790    2 * 5 * 475879
  6662326    2 * 11 * 302833
  7873702    2 * 107 * 36793
  8021302    2 * 47 * 85333
  8362822    2 * 13 * 321647
  9649462    2 * 233 * 20707
  9733222    2 * 43 * 113177
  10186102   2 * 23 * 79 * 2803

9749077, 1768854, 2948102, 2958742, 3118822, 4009942, 4044454, 4470262, 4758790, 6662326, 7873702, 8021302, 8362822, 9649462, 9733222, 10186102

The factorisations are:

  number     factor

  9749077    13 * 73 * 10273
  1768854    2 * 3 * 294809
  2948102    2 * 787 * 1873
  2958742    2 * 37 * 39983
  3118822    2 * 7 * 222773
  4009942    2 * 239 * 8389
  4044454    2 * 19 * 106433
  4470262    2 * 31 * 72101
  4758790    2 * 5 * 475879
  6662326    2 * 11 * 302833
  7873702    2 * 107 * 36793
  8021302    2 * 47 * 85333
  8362822    2 * 13 * 321647
  9649462    2 * 233 * 20707
  9733222    2 * 43 * 113177
  10186102   2 * 23 * 79 * 2803

Sphenic Numbers

Today I turned 25285 days old and, as I discovered in Numbers Aplenty, 25285 is a sphenic number. The definition given on that site is:

A number \(n\)  is called sphenic if it is the product of 3 distinct primes. 

For example, 370 is a sphenic number because it is the product of the 3 primes 2, 5 and 37. 

Sphenic numbers are quite common: up to  \(10^8\) there are 20,710,806 sphenic numbers. 

The sum of the reciprocals of the sphenic numbers diverges, while the sum of the reciprocal of their squares converges to 0.003696244... , which can be expressed as  \((P(2)^3-3\,P(2)\,P(4)+2\,P(6))/6\), where $$P(s)=\sum_{p\mathrm{\ prime}}\frac{1}{p^s}$$is the so-called prime Zeta function 

The first sphenic numbers are 30, 42, 66, 70, 78, 102, 105, 110, 114, 130, 138, 154, 165, 170, 174, 182, 186, 190, 195, 222, 230, 231, 238, 246, 255, 258, 266, 273, 282, 285, 286, 290, 310

The sum of the reciprocals of the squares of the sphenic numbers does indeed approach 0.003696244 as can be seen by taking the numbers from 30 to 310 and applying the following SAGE code:

INPUT: 

sphenic=[30, 42, 66, 70, 78, 102, 105, 110, 114, 130, 138, 154, 165, 170, 174, 182, 186, 190, 195, 222, 230, 231, 238, 246, 255, 258, 266, 273, 282, 285, 286, 290, 310]
sum=0
for n in sphenic:
    sum+=1/n^2
print(sum.n())

OUTPUT: 

0.00320320889263633 

The following is a modification of the Wikipedia entry for sphenic numbers:

In number theory, a sphenic number is a natural number|positive integer that is the product of three distinct prime numbers. Thus a sphenic number is a product ''pqr'' where ''p'', ''q'', and ''r'' are three distinct prime numbers. This definition is more stringent than simply requiring the integer to have exactly three prime factors. For instance, \(60 = 2^2 × 3 × 5 \) has exactly 3 prime factors, but is not sphenic. 

The smallest sphenic number is 30 = 2 × 3 × 5, the product of the smallest three primes. The largest known sphenic number is
$$(2^{77232917}− 1) \cdot (2^{74,207,281} − 1) \cdot (2^{57,885,161} − 1)$$It is the product of the three largest known primes. 

All sphenic numbers have exactly eight divisors.  If we express the sphenic number as \(n = p \cdot q \cdot r \), where ''p'', ''q'', and ''r'' are distinct primes, then the set of divisors of ''n'' will be { 1, p, q, r, pq, pr, qr, n }. 

The converse does not hold. For example, 24 is not a sphenic number, but it has exactly eight divisors. 

All sphenic numbers are by definition squarefree, because the prime factors must be distinct. 

The Möbius function of any sphenic number is -1. 

The first case of two consecutive sphenic integers is 230 = 2×5×23 and 231 = 3×7×11. The first case of three is 1309 = 7×11×17, 1310 = 2×5×131, and 1311 = 3×19×23. There is no case of more than three, because every fourth consecutive positive integer is divisible by 4 = 2×2 and therefore not square-free. 

The numbers 2013 (3×11×61), 2014 (2×19×53), and 2015 (5×13×31) are all sphenic. The next three consecutive sphenic years will be 2665 (5×13×41), 2666 (2×31×43) and 2667 (3×7×127) (see OEIS A165936).

The OEIS sequence A007304 (sphenic numbers, products of 3 distinct primes) mentions that a sphenic brick is a rectangular parallelopiped whose sides are components of a sphenic number, namely whose sides are three distinct primes. For example, the distinct prime triple (3,5,7) produces a 3 x 5 x 7 unit brick which has volume 105 cubic units. 

From my early days of investigating numbers, I had considered triprimes (number that factor into three, not necessarily distinct, primes) as having unique representations as rectangular prisms. I was interested in the relationship between a prisms volume and its surface area, coining the term "cubicity". Here is a snapshot of a February 2014 Instagram post: