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12 Pentagons, Part 3
====================

This is the third part in an investigation into answering the following question:

> A soccer ball is a (roughly spherical) figure made of pentagons and hexagons, each meeting 3 at a point. [T]here are 12 pentagons...how many hexagons can there be?

The [first post]() in the series proved the 12 pentagons portion and investigated (dodecahedral) Goldberg polyhedra. The [second post]() investigated (tetrahedral) Goldberg polyhedra as well as other tetrahedral solutions. Yet more unconventional solutions are presented below.


Introduction
------------


### Combinatorics of Goldberg-Coxeter Operators

I mentioned previously that these solutions cannot be created easily from seed polyhedra in a polyhedron viewer. This prompts the question of how to calculate the vertex, edge, and face counts without acquiring them from the output of such a program.

Each of the simple GC operators (*dk*, *c*, and *w*) has a combinatoric matrix form when applied to a vector of the vertex, edge, and face counts of a figure S.

$$
S = \begin{pmatrix} v \\ e \\ f \end{pmatrix} = \begin{pmatrix} 2F_6 + 20 \\ 3F_6 + 30 \\ F_6 + 12 \end{pmatrix} \\
dk = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 3 & 0 \\ 1 & 0 & 1 \end{pmatrix}, ~
c = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 4 & 0 \\ 0 & 1 & 1 \end{pmatrix}, ~
w = \begin{pmatrix} 1 & 4 & 0 \\ 0 & 7 & 0 \\ 0 & 2 & 1 \end{pmatrix}
$$

As a reminder, *tk* is a power of the *dk* operator. Powers of a matrix are more readily expressed when the matrix is diagonalized.

$$
\begin{align*}
dk &= \begin{pmatrix} 0 & 2 & 0 \\ 0 & 3 & 0 \\ 1 & 0 & 1 \end{pmatrix} =
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}
\begin{pmatrix} 1 & -2/3 & 0 \\ 1 & -1 & 1 \\ 0 & 1/3 & 0 \end{pmatrix}
\\
c &= \begin{pmatrix} 1 & 2 & 0 \\ 0 & 4 & 0 \\ 0 & 1 & 1 \end{pmatrix} =
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}
\begin{pmatrix} 1 & -2/3 & 0 \\ 1 & -1 & 1 \\ 0 & 1/3 & 0 \end{pmatrix}
\\
w &= \begin{pmatrix} 1 & 4 & 0 \\ 0 & 7 & 0 \\ 0 & 2 & 1 \end{pmatrix} =
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 7 \end{pmatrix}
\begin{pmatrix} 1 & -2/3 & 0 \\ 1 & -1 & 1 \\ 0 & 1/3 & 0 \end{pmatrix}
\end{align*}
$$

All of these operators share the same eigenvectors. This means that composition of these operators only modifies the diagonal matrix, specifically the upper-left and lower-right eigenvalues. Note also that:

- The (right) eigenvector (2, 3, 1), is shared by the general system governing the vertex, edge, and hexagon counts (simply look at the vector *S* to see this). In this system (not the one above), it has eigenvalue 0.
- The left eigenvector (1, -1, 1) has eigenvalue 1, which corresponds to the Euler characteristic being maintained.
- The left eigenvector, (3, -2, 0), when applied to general polyhedra corresponds to the edges and vertices added by the operation: *dk* forces all vertices to be degree-3 (hence the eigenvalue 0) and *c* and *w* maintain previous vertices while adding degree-3 ones (hence the eigenvalue 1).

All solutions share the same Euler characteristic and the condition 3V = 2E, so the rightmost multiplication simplifies things considerably.

$$
\begin{align*}
g_T S &= \begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & T \end{pmatrix}
\begin{pmatrix} 1 & -{2 \over 3} & 0 \\ 1 & -1 & 1 \\ 0 & {1 \over 3} & 0 \end{pmatrix}
\begin{pmatrix} v \\ e \\ f \end{pmatrix}
\\[8pt] &=
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & T \end{pmatrix}
\begin{pmatrix} v - {2 \over 3} e = 0 \\ v - e + f = 2 \\ {1 \over 3} e \end{pmatrix}
\\[8pt] &=
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 0 \\ 2 \\ {1 \over 3} T e \end{pmatrix}
\end{align*}
$$

Recall that each of the operators applied to the dodecahedron was a Goldberg polyhedron: *dkD* = *tI* = GC(1, 1), *cD* = GC(2, 0), and *wD* = GC(2, 1). The norm of each of these is 3, 4, and 7 respectively, which is also the eigenvalue T. In fact, all Goldberg-Coxeter operators, not just compositions of these three, simply set $T = \|a + bu\|$ (the eigenvalue x is T mod 3, which as previously mentioned, is never congruent to 2).

This eigenvalue problem also implies something about the possible values of *T*: it is closed under multiplication. For any *T*, all its powers are *also* possible *T*, as are products between *T*. In fact, the existence of chiral pairs ($a \neq b$) and the ability to compose opposites is what can make *T* nonunique for a given pair (e.g.: $ww = (2, 1) \odot (2, 1) = (5, 3)$, but $ww' = (2, 1) \odot (1, 2) = (7, 0)$).

Returning to the equation,

$$
\begin{align*}
\begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 3 \\ 1 & 1 & 1 \end{pmatrix}
\begin{pmatrix} 0 \\ 2 \\ {1 \over 3} T e \end{pmatrix} &=
\begin{pmatrix} {2 \over 3} Te \\ Te \\ {1 \over 3}Te + 2 \end{pmatrix}
\\[8pt] &=
\begin{pmatrix} {2 \over 3} T(3F_6 + 30) \\ T(3F_6 + 30) \\ {1 \over 3}T(3F_6 + 30) + 2 \end{pmatrix}
\\[8pt] &=
\begin{pmatrix} 2T(F_6 + 10) \\ 3T(F_6 + 10) \\ T(F_6 + 10) + 2 \end{pmatrix}
\\[12pt]
F'_6 &= T(F_6 + 10) - 10
\end{align*}
$$

This demonstrates the importance of both the constructed base solution figure *S* with $F_6$ hexagons and the possible values of *T*.


### A brief word about Symmetry

The dodecahedron, and by extension dodecahedral solutions, have icosahedral symmetry. A sub-symmetry of this, being tetrahedral symmetry, was exploited in the previous post. However, there are further sub-symmetries which have not been encountered. For example, some "smaller" symmetries of the dodecahedron are the vertices under rotation and inversion (dihedral, degree 3) and the faces under rotation and inversion (dihedral, degree 5). One of these two symmetries must be shared by a solution.

Rather than the typical construction based on paths between pentagons, this section will focus on certain base cases followed by a rudimentary application of the Conway operators above. The particular symmetry group of each figure will be mentioned in each section.


Deltahedra
----------

*Deltahedra* are polyhedra formed by equilateral triangles, a [table of which](https://en.wikipedia.org/wiki/Deltahedron) is available on Wikipedia. Equilateral triangles are important since three to five of them can be joined at a convex vertex, and if coplanar arrangements are allowed, up to six of them. Degree-5 and degree-6 vertices can be dualized or truncated to pentagons and hexagons respectively; additionally, if the former operation is used, then the triangles become degree-3 vertices.

All convex deltahedra besides the tetrahedron and the icosahedron contain degree-4 vertices. These pose a problem since they become quadrilaterals. Though selective (pyramidal) augmentation can correct these vertices to degree-5 (and those of degree-5 to degree-6), this section will focus on the coplanar entries on the table.


### Triangulated Rhombohedron (Dih3)

The figure made by cutting the short diagonals of a *trigonal trapezohedron* (a rhombohedron like the cube) is one such degree-4 vertex-free deltahedron. It can also be seen as either a *biaugmentation* (by tetrahedra on opposite faces) of an octahedron or as a [*gyroelongated triangular bipyramid*](https://en.wikipedia.org/wiki/Gyroelongated_bipyramid). Truncating the order-5 vertices produces an eggy-looking figure, and one for which the polyhedron viewer I typically use refuses to comply. Try using this viewer (which unfortunately does not support hotlinking recipes) with the recipe t5z applied to the cube. This figure inherits the dihedral symmetry of degree 3 that the vertices of the dodecahedron have.

:::: {}
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Biaugmented octahedron
:::

::: {}
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Base case: order-5 truncation of left figure
:::

::: {}
![]()
Graph of the pole of the base case. On the opposite pole, blue and red hexagons are exchanged.
:::
::::

Conway  |   $F_6$  |  V  |  E
--------|----------|-----|-----
t5C = B |        6 |  32 |  48
dkB     |       38 |  96 | 144
cB      |       54 | 128 | 192
wB      |      102 | 224 | 336
tkB     |      134 | 288 | 432
$g_T B$ | 16T - 10 | 32T | 48T


### Triangular Frustum (Dih3)

This figure is formed by a group of four triangles as a base, three half-hexagons on the edges, and closed by a triangle atop. It can also be seen as an augmented octahedron, but with the tetrahedra placed around a single face (a *triaugmentation*). Unfortunately, this figure is not easily constructible from normal seed polyhedra, even in the alternative viewer. However, it is still possible to operate on its projection as a planar graph. Truncating all vertices except those of degree 3 ($t_5 t_6$) produces a figure with 8 hexagons and indeed, 12 pentagons. In this case, the pentagons gather in groups of 4 and are separated by three bands of hexagons joined by two hexagons at either end. The symmetry inherited by this figure is also dihedral of degree 3, but centered about a hexagon, rather than a complex of pentagons.

:::: {}
::: {}
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:::

::: {}
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::: {}
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:::

::: {}
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:::

Construction based on planar graph (1). First, the degree-6 vertices are truncated (2). Then, the graph is inverted (3) and the degree-5 vertices are truncated (4).
::::

::: {}
![]()
One-third of the figure, showing a group of four pentagons. Pentagons in green, band hexagons in red, cap hexagons in blue.
:::

Conway  |   $F_6$  |  V  |  E
--------|----------|-----|-----
F       |        8 |  36 |  54
dkF     |       44 | 108 | 162
cF      |       62 | 144 | 216
wF      |      116 | 252 | 378
tkF     |      152 | 324 | 486
$g_T F$ | 18T - 10 | 36T | 54T

***

The remaining entries of the Wikipedia table with no degree-4 vertices (save the hexagonal antiprism, which I will discuss in the next section) are easily accessible from the tetrahedron. For example, the "augmented tetrahedron" is simply a subdivided tetrahedron (*uT*) and the "(subdivided) truncated tetrahedron" is the hexakis truncated tetrahedron ($k_6 tT$).

Larger subdivisions (in the $u_n$ sense) of the above two cases are also coplanar deltahedra, so this can be performed before the truncation. Triangular subdivision can add degree-6 vertices, so for the triangulated rhombohedron, an additional order-6 truncation needs to be done. Doing so means $t_5 t_6$ is applied to either case to form the solution figure.

An example subidivision is shown below. Note how the 3 pentagons at the pole have been separated from another 3 by a triangle of hexagons. I am fairly sure it is possible for mismatched pole configurations to be joined to one another for a more selective subdivision. Other Goldberg-Coxeter operations other than *u* can also be used; *dwd* for example connects non-polar pentagons with spirals of hexagons.

:::: {}
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:::

::: {}
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::: {}
![]()
$F_6$ = 60, V = 140, E = 210
:::

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![]()
$F_6$ = 114, V = 248, E = 372
:::

Left: $t_5 t_6 uzC$, Right: $t_5 t_6 dwdzC$
::::


Other Solutions
---------------

This section is for polyhedra which are either more easily constructible or are not easily derived from deltahedra.


### Truncated Trapezohedron (Dih6)

A *trapezohedron* is the polyhedral dual of an antiprism; an example is a d10, which is the pentagonal case. Despite the name "*n*-gonal trapezohedron," the figure is made entirely from 2*n* kites with the long edges meeting at points and the short edges meeting with each other. Truncating the order-*n* vertices where the long edges of the kites meet produces a truncated trapezohedron, which possesses two regular *n*-gonal "caps" separated by a ring of pentagons. In a way, they can be considered as the pentagonal analogue to prisms (quadrilaterals) and antiprisms (triangles).

The dodecahedron itself can be realized as a (order-5) truncated pentagonal trapezohedron ($t_5 d A_5$), which emphasizes its degree-5 dihedral symmetry. The next largest truncated trapezohedron is the hexagonal case ($t_6 d A_6$), which contains and 2 hexagons and has dihedral symmetry of degree 6 (which itself contains that of degree 3).

![]()

Base case: truncated hexagonal trapezohedron

Conway                                                                         |   $F_6$  |  V  |  E
-------------------------------------------------------------------------------|----------|-----|-----
[$t_6dA_6 = T_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300t6dA6) |        2 |  24 |  36
[$dkT_6$](https://levskaya.github.io/polyhedronisme/?recipe=dkK300t6dA6)       |       26 |  72 | 108
[$cT_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300cK30t6dA6)      |       38 |  96 | 144
[$wT_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300wK30t6dA6)      |       86 | 168 | 252
[$tkT_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300tkK30t6dA6)    |       98 | 216 | 324
$g_T T_6$                                                                      | 12T - 10 | 24T | 36T


Medially-Separated Dodecahedron (Dih5)
--------------------------------------

As the "truncated pentagonal trapezohedron", the dodecahedron can be separated into halves along an equator (specifically, along the [Petrie polygon](https://en.wikipedia.org/wiki/Petrie_polygon)). This produces two halves with the pole configuration of the dodecahedron, joined together with hexagons. Organizing the pentagons into two pairs of 6 is similar to the earlier case with the triangulated cube, but with a different arrangement of pentagons. The resultant figure has dihedral symmetry of degree 5. Interestingly, it is also the first solution polyhedron with an odd number of faces.

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Dodecahedral graph
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Pole configuration of solution
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Complete base solution. Note how the outer face has been rotated when compared with the dodecahedral graph
:::
::::

Conway          |   $F_6$  |  V  |  E
----------------|----------|-----|-----
$D_p$           |        5 |  30 |  45
$dkD_p$         |       35 |  90 | 135
$cD_p$          |       50 | 120 | 180
$wD_p$          |       95 | 210 | 315
$tkD_p$         |      125 | 270 | 405
$g_T D_p$       | 15T - 10 | 30T | 45T


Truncated Gyro-Pyramids (Dih5, Dih6)
------------------------------------

In the previous post, I used the Conway operator *g*, which is called "gyro". It is the dual operator to snubbing, intuition for which can be seen by looking at the [snub cube](https://en.wikipedia.org/wiki/Snub_cube). A *gyro-pyramid* (i.e., the gyro operator applied to pyramids) can roughly be described as a ring of 2*n* pentagons which alternate in orientation, the points of which have an additional edge connecting to one of two antipodal endpoints. Truncating these these adds a face with the same number of edges as the original pyramid, surrounded by hexagons. Solution polyhedra can be found by examining the pentagonal ($t_5 g Y_5$) and hexagonal ($t_6 g Y_6$) cases. The former has dihedral symmetry of degree 5 and the latter has that of degree 6 (which contains degree 3).

:::: {}
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Base case 1: truncated pentagonal gyro-pyramid
:::

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Base case 2: truncated hexagonal gyro-pyramid
:::
::::

Conway                                                                            |   $F_6$  |  V  |  E
----------------------------------------------------------------------------------|----------|-----|-----
[$t_5gY_5 = G_5$](https://levskaya.github.io/polyhedronisme/?recipe=K300t5K30gY5) |       10 |  40 |  60
[$dkG_5$](https://levskaya.github.io/polyhedronisme/?recipe=dkK300t5K30gY5)       |       50 | 120 | 180
[$cG_5$](https://levskaya.github.io/polyhedronisme/?recipe=K300dudt5K30gY5)       |       70 | 160 | 240
[$wG_5$](https://levskaya.github.io/polyhedronisme/?recipe=K300wK30t5K30gY5)      |      130 | 280 | 420
[$tkG_5$](https://levskaya.github.io/polyhedronisme/?recipe=tkK300t5K30gY5)       |      170 | 360 | 540
$g_T G_5$                                                                         | 20T - 10 | 40T | 60T
[$t_6gY_6 = G_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300t6K30gY6) |       14 |  48 |  72
[$dkG_6$](https://levskaya.github.io/polyhedronisme/?recipe=dkK300t6K30gY6)       |       62 | 144 | 216
[$cG_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300dudK40t6gY6)       |       86 | 192 | 288
[$wG_6$](https://levskaya.github.io/polyhedronisme/?recipe=K300wK30t6K30gY6)      |      158 | 336 | 504
[$tkG_6$](https://levskaya.github.io/polyhedronisme/?recipe=tkK300t6K30gY6)       |      206 | 432 | 648
$g_T G_6$                                                                         | 24T - 10 | 48T | 72T

The pentagonal case demonstrates something interesting: an appeal to the equation $12 = 10 + 2 = 5 \cdot 2 + 2$. The similar equation $12 = 7 + 5$ is unlikely to bear fruit, as an odd number of pentagons cannot alternate up and down.


Japanese Floor Tiling
---------------------

*Tatami* are a Japanese traditional style of floor mat, used even today in Japan as an intuitive measure for the surface area of living spaces. A single mat has an aspect ratio of 2:1, but the complexity comes in how they are arranged. Layouts are termed "inauspicious" (不祝儀敷き, *fushūgi-jiki*) when there are points where four mats meet, while  "auspicious" (祝儀敷き, *shūgi-jiki*) layouts have all mats meet in threes. For a sample of the fascination mathematicians have with these arrangements, you need only [search for "tatami" in the OEIS](http://oeis.org/search?q=tatami&language=english&go=Search) to find dozens of combinatorial sequences.

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Inauspicious 4x4 layout
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Auspicious 4x4 layout
:::

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Possible topologies for a single mat
:::
::::

Topologically, each *tatami* mat in an arrangement can be thought of as either a quadrilateral, a pentagon or a hexagon. Obviously when considered alone, one mat is a rectangle, as is each mat in a stack of them like in the inauspicious layout above. When two mats are affixed to one side of the mat, it becomes a pentagon, as in the auspicious layout above; all 8 mats are pentagons. Finally, when done to both sides, it becomes a hexagon.

<p class="has-text-align-center">
<a title="Tomruen, CC BY-SA 4.0 <https://creativecommons.org/licenses/by-sa/4.0>, via Wikimedia Commons" href="https://commons.wikimedia.org/wiki/File:Conway_wC.png"><img width="192" alt="Conway wC" src="https://upload.wikimedia.org/wikipedia/commons/f/f4/Conway_wC.png"></a>
<a title="See page for author, CC BY-SA 4.0 <https://creativecommons.org/licenses/by-sa/4.0>, via Wikimedia Commons" href="https://commons.wikimedia.org/wiki/File:Tearoom_layout.svg"><img style="background-color: white !important; padding: 5px" width="192" alt="Tearoom layout" src="https://upload.wikimedia.org/wikipedia/commons/thumb/c/c0/Tearoom_layout.svg/256px-Tearoom_layout.svg.png"></a>
</p>
::: {}
Images were retrieved from Wikimedia and belong to their respective owners.
:::

Naturally, the interplay between the latter two elements, as well as the condition that all mats meet in threes has a direct application to the problem at hand. In fact, the graph of the whirl operation on the surface of a cube (left) clearly makes the shape of one of these arrangements (with a half-mat included, right). Two things pose a small issue for auspicious floor layouts:

- when interpreted as a graph, the external face very quickly accumulates a large number of edges and
- each of the mats at the corners of the room have a single degree-2 vertex

Both of these can be (partially) amended by connecting pairs of corner vertices, which creates additional faces while decreasing the number of edges on the perimeter.

The standard 4x6 auspicious layout is shown below, with corners connected and pentagons/hexagons identified. Next to it is an equivalent graph with a regular hexagon as an outer face. As a 3D figure, it has only three hexagons and degree-3 dihedral symmetry, another odd number.

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::: {}
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:::

Conway        |   $F_6$  |  V  |  E
--------------|----------|-----|-----
$M$           |        3 |  26 |  39
$dkM$         |       29 |  78 | 117
$cM$          |       42 | 104 | 156
$wM$          |       81 | 182 | 273
$tkM$         |      107 | 234 | 351
$g_T M$       | 13T - 10 | 26T | 39T

While the chamfer and whirl are both multiples of 3, the dual-kis somewhat strangely bears 29 and 107, which are both prime. In fact, up to $(dk)^{11}$, the values are either prime or semiprime, and up to $(dk)^{28}$, the values are either 1-, 2-, or 3-, almost primes.


### Tatamified Projections

While other rectangular *tatami* arrangements contain either too few or too many edges, further graphs of solutions can be generated by using mats with non-rectangular arrangements (connecting corners as necessary), or more easily, nonstandard aspect ratios. Some of these are shown below:

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Dodecahedral graph, as a *tatami* arrangement with a single 4x1 mat
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Medially-separated dodecahedron, as seen above
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Truncated icosahedron graph. Note the path of 2x1 mats
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Triacontahedron (not a base case shown above), initially similar to tI
:::
::::


Final Tabulation and Closing
----------------------------

The solutions examined have been collected in the table below. Roughly, the possible values for $F_6$ are 2, 3, 4, 5, 6, 8, 10, 12, 14, 18, 20, 24, 26, 28, 29..., though there is nothing exhaustive about the examples considered. In particular, it should be possible to generate further values from selective subdivisions of deltahedra, or by using selective GC operators which still preserve certain symmetries.


| Symmetry    | Classification                       |    $F_6$   | Example values
|-------------|--------------------------------------|------------|-----------------------------------------
| Icosahedral | Dodecahedral Goldberg                | $10T - 10$ | 20, 30, 60, 80, 110, 120, 150, 180, 200
| Tetrahedral | Tetrahedral Goldberg Antitruncations | $2T - 14$  | 4, 10, 12, 18, 24, 28, 36, 40, 42, 48, 58, 60, 64
|             | Edge-preserving                      | $4((n+1)^2 - 3$        | 24, 52, 88, 132, 184, 244, 312, 388, 472, 564
|             | Other                                | $2TT' - 4T' - 10$      | 32, 46, 50, 56, 70, 74, 78, 88, 92, 102
|             |                                      | $(4(n+1)^2 - 2)T - 10$ | 172, 176, 184, 238, 424, 548
| $Dih_3$     | Rhombohedron                         | $16T - 10$ | 6, 38, 54, 102, 134, 182, 198
|             | Triangular Frustum                   | $18T - 10$ | 8, 44, 62, 116, 152, 206, 224, 278, 332
|             | [Tatamihedron](https://en.wikipedia.org/wiki/26-fullerene_graph) | $13T - 10$ | 3, 29, 42, 81, 107, 146, 159, 198, 237, 263, 315
| $Dih_5$     | Medially-separated Dodecahedron      | $15T - 10$ | 5, 35, 50, 95, 125, 170, 185, 230, 275, 305
|             | Truncated Pentagonal Gyropyramid     | $20T - 10$ | 10, 50, 70, 130, 170, 230, 250
| $Dih_6$     | Truncated Hexaagonal Trapezohedron   | $12T - 10$ | 2, 26, 38, 74, 98, 134, 146, 182, 218, 242, 290, 314
|             | Truncated Hexaagonal Gyropyramid     | $24T - 10$ | 14, 62, 86, 158, 206, 278, 302

T, T' are members of the sequence 1, 3, 4, 7, 9, 12, 13, 16, 19, 21... ([OEIS A003136](http://oeis.org/A003136))

\*: This is the result of a GC operator applied to $2T - 14$, which requires *T* > 7

Small naturals which do not appear on the list are 1, 7, 9, 11, and 13. Without constructing them, I am unsure whether they can exist. Some possible symmetries are $7 = 5 + 2$ (five equatorial and two polar hexagons, Dih5) and $11 = 9 + 2$ (three hexagons along three lines of longitude or three triangles of hexagons pointing toward a pole, Dih3).

***

Despite the Goldberg-Coxeter construction for dodecahedra being well-known, the 12 pentagons rule applies to a much broader class of polyhedra. In fact, due to the GC construction, any polyhedron satisfying the condition can be expanded and twisted into larger and larger solutions, which have little to do with soccer balls.

Polyhedron images were generated using [polyHédronisme](https://levskaya.github.io/polyhedronisme/) and Dr. Andrew J. Marsh's [polyhedron generator](https://drajmarsh.bitbucket.io/poly3d.html). Nets and graphs were created with GeoGebra. Other images were retrieved from Wikimedia and belong to their respective owners.
