Truncated Pyramid

I was surprised to discover that the number associated with my diurnal age today (27930) has a connection to the volume of a truncated square pyramids. Firstly, let's recall the formula for the volume of such figure:$$V=\frac{1}{3} (a^2+ab+b^2) \, h$$where \(a\) is the side length of the square base, \(b\) is the side length of the top square and \(h\) is the height of the truncated pyramid. The formula is easily confirmed by integration from first principles without relying on the formula for the volume of a pyramid. 

Now 27930 is a member of OEIS A027444: \( \text{a}(n) = n^3 + n^2 + n\) with the following interesting comment attached to the entry:

For \(n>1\), a(\(n\)) is the volume of a truncated square pyramid with height \(n\) and base  lengths \(n+2\) and \(n-1\). See Figure 1.

Figure 1

Now if we substitute \(a=n+2\), \(b=n-1\) and \(n=h\) into our earlier formula we find the volume \(V\) of the truncated pyramid becomes:$$V=n^3+n^2+n$$When \(n=30\), the volume is 27930 cubic units. Here are the volumes for various values of \(n>1\):

  n    n-1   n+2   V 

  2    1     4     14

  3    2     5     39

  4    3     6     84

  5    4     7     155

  6    5     8     258

  7    6     9     399

  8    7     10    584

  9    8     11    819

  10   9     12    1110

  11   10    13    1463

  12   11    14    1884

  13   12    15    2379

  14   13    16    2954

  15   14    17    3615

  16   15    18    4368

  17   16    19    5219

  18   17    20    6174

  19   18    21    7239

  20   19    22    8420

  21   20    23    9723

  22   21    24    11154

  23   22    25    12719

  24   23    26    14424

  25   24    27    16275

  26   25    28    18278

  27   26    29    20439

  28   27    30    22764

  29   28    31    25259

  30   29    32    27930

  31   30    33    30783

  32   31    34    33824

  33   32    35    37059

Johnson Solids J1 and J8

The number associated with my diurnal age today, 27126, introduced me to the notion of Johnson solids via its membership in OEIS A227221:

A227221

Volume of Johnson square pyramid placed upright on cube (rounded down) with edge lengths equal to \(n\).

The members of this sequence, up to 40000, are:

1, 9, 33, 79, 154, 266, 423, 632, 900, 1235, 1644, 2135, 2714, 3390, 4170, 5061, 6071, 7206, 8475, 9885, 11443, 13157, 15034, 17082, 19307, 21718, 24322, 27126, 30137, 33363, 36812

The formula for the area is given by \(  (1+\dfrac{ \sqrt{2}}{6}) \times n\) where \(n\) is the edge length. For 27216, \(n=28\). This shape is known as an elongated square pyramid and represent Johnson solid J8. A Johnson solid is a convex polyhedron with all edges equal and there 92 distinct types. The equilateral square pyramid sitting on one of the faces of the cube is Johnson solid J1. See Figures 1 and 2.

Figure 1: Johnson solid J1 (source)

Figure 2: Johnson solid J8 (source)

Figure 3 shows a octahedron, one of the five Platonic solids, that can be considered a square bipyramid, i.e. two Johnson square pyramids connected base-to-base.

Figure 3: octahedron (source)

Figure 4 shows a tetrakis hexahedron that can be constructed from a cube with Johnson square pyramids added to each face. It is a Catalan solid.

Figure 4: tetrakis hexahedron (source)

Pyramidal Numbers

Today I turned 26588 days old and one of the properties of this number is that it's a 15-gonal or pentadecagonal number and a member of OEIS A177890:

A177890

15-gonal (or pentadecagonal) pyramidal numbers:                 a(\(n\)) = \( \dfrac{n (n+1)  (13n-10)}{6}\)

The initial members of the sequence are:

0, 1, 16, 58, 140, 275, 476, 756, 1128, 1605, 2200, 2926, 3796, 4823, 6020, 7400, 8976, 10761, 12768, 15010, 17500, 20251, 23276, 26588, 30200, 34125, 38376, 42966, 47908, 53215, 58900, 64976, 71456, 78353, 85680, 93450, 101676, 110371, 119548, 129220

GENERAL FORMULA

There is a general formula for \(P(n)\), the \(n\)-th polygonal pyramidal number, that uses \(T(n)\), the \(n\)-th triangular number, and \(s\) representing the number of sides of the polygon. Here it is:$$P(n,s)=T(n) \times \frac{(s-2) \times n - (s-5)}{3}$$In the case of 15-gonal numbers, the formula becomes:$$ \begin{align} P(n)&=T(n) \times \frac{13n - 10}{3}\\&=\frac{n(n+1)(13n-10)}{6} \text{ where }T(n)=\frac{n(n+1)}{2}\end{align}$$GENERATING FUNCTION

The generating function for pyramidal numbers is given by:$$G(x,s)=x \, \frac{(s-3)x+1}{(1-x)^4}$$In the case of the pentadecagonal numbers, this results in:$$G(x)=x \, \frac{12x+1}{(1-x)^4}$$PARTICULAR EXAMPLE

Figure 1 shows how a square pyramidal number is constructed:

Figure 1: source

Note that the 3-dimensional pyramidal numbers are constructed from 2-dimensional polygons stacked one on top of the other. In Figure 1, these polygons are squares. The sequence of square pyramidal numbers is given by:$$P_n^{^ {\,4}}=\frac{n(n+1)(2n+1)}{6}$$Figure 2 shows an actual pile of cannonballs forming a square-based pyramid.

Figure 2: source

On the topic of cannonballs, the cannonball problem can be stated as follows:

The cannonball problem asks for the sizes of pyramids of cannonballs that can also be spread out to form a square array, or equivalently, which numbers are both square and square pyramidal. Besides 1, there is only one other number that has this property: 4900, which is both the 70th square number and the 24th square pyramidal number. Source.

SUMS OF RECIPROCALS

The sums of the reciprocals of the pyramidal polygonal numbers all converge. Here are some examples:

• triangular pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n(n+1)(n+2)}=\dfrac{3}{2} \)
• square pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n(n+1)(2n+1)}=6(3-4 \log(2)) \)
• pentagonal pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n^2(n+1)}=\dfrac{\pi^2}{3}-2 \)

More results are listed at this site. There is a rather formidable general formula for the sum of the reciprocals of the pyramidal numbers. I don't claim to understand it but here it is:$$ - \, \frac{6 [ s-5 +(s-2)(\psi ( \frac{3}{s-2} ) +\gamma)]}{(s-5)(2s-7)} $$where \( \psi(x) \) is the digamma function and \( \gamma \) is the  Euler-Mascheroni constant.

Alternating sums of reciprocals are also convergent. An example is the alternating sum of reciprocals of square pyramidal numbers where we have:$$6\sum_{n=1}^{\infty}  \dfrac{(-1)^{n-1}}{n(n+1)(2n+1)}=6(\pi-3)$$

House Numbers

When turning 25884 days old today, I discovered that 25884 is a house number. This was new to me and not surprising because the previous such number was 23101. I turned 23101 days old over 7.6 years ago. So I needed to discover what house numbers are. They are named not after some mathematician named House but because they can be represented as houses. A diagram is helpful at this point. See Figure 1.

Figure 1

Clearly such numbers are figurate but consist of two conjoined shapes: a cube on top of which sits a square-based pyramid. Here a definition from Numbers Aplenty:

The \(n\)-th house number \(h_n\) is a figurate number made by a cube of side \(n+1\), surmounted by a square pyramidal number with side \(n\), thus:$$h_n = (n+1)^3 + \sum_{k=1}^nk^2$$An alternative way of representing this is as: $$h_n=\frac{8n^3+21n^2+19n+6}{6}$$The generating function for this sequence is:$$\frac{1+5x+2x^2}{(x-1)^4}$$In Figure 2, the SageMath code is shown that makes use of this function to generate the coefficients of the polynomial that are the terms of the sequence (the output is truncated for display purposes):

Figure 2: permalink

Figure 3 shows the SageMath code to generate the coefficients by themselves, removed from the polynomial. The output has been truncated to fit the display. These are the house numbers. The sequence of house numbers forms  OEIS A051662. Below is the list of terms, up to and including 25884:

1, 9, 32, 78, 155, 271, 434, 652, 933, 1285, 1716, 2234, 2847, 3563, 4390, 5336, 6409, 7617, 8968, 10470, 12131, 13959, 15962, 18148, 20525, 23101, 25884, ...

Figure 3: permalink

The next house number is 28882, so it's quite some way off. There's nothing very deep about house numbers but they form an interesting variation on the figurate numbers that I'm familiar with. They are the sum of two different and independent figurate numbers that have been combined to form a composite shape, in this case a house. There's no reason that one couldn't combine other shapes together from these or different components.

For example, adding a square-based triangular pyramid to each of the six faces of the cube would produce a sort of stellated cube. The formula would be:$$h_n = (n+1)^3 + 6 \sum_{k=1}^nk^2$$ADDENDUM: May 7th 2020

Today I turned 25967 days old and this number turns out to be a 24-hedral number with formula:$$(2n + 1)(8n^2 + 14n + 7) \text{ where }n=11$$It can be considered as being made up out of a central cube and six square pyramids. The central cube number is given by OEIS A016755:$$CCub(n) = (2n + 1)^3$$The figurate number represented by the square pyramids is given by OEIS A000447:$$ P^{(4)}(n) = \frac{(1 + n)(1 + 2n)(3 + 2n)}{3}$$Therefore, for our 24-hedral number, we have:$$24H(n) = CCub(n) + 6P^{(4)}(n)$$ $$24H(n)=(2n+1)^3 + \frac{6(1+n)(1+2n)(3+2n)}{3}$$ $$24H(n)= (2n+1)(8n^2 +14n+7)$$Figure 4 shows the example for when \(n=11\):

Figure 4: link