I am also interested in knowing if there are any good reasons to pick one ordering of polygons over some other ordering.
I'll post what little I know, but I'm really hoping someone else will post something much better.
I have heard of 3 "standards" for ordering these polygons, but as far as I can tell they were arbitrarily picked:
Your icosahedron-based method,
the quadrilateralized spherical cube, and
Tegmark's icosahedron-based method for pixelizing the celestial sphere.
All three are examples of geodesic grids.
Ordering polygons on a polyhedron
icosahedron-based method for pixelizing the celestial sphere
"What is the best way to pixelize a sphere?" by Max Tegmark
gives a format that appears extremely similar to your proposal:
Tegmark starts with an icosahedron, just as you do,
and recursively subdivides it, just as you do.
So I find it hard to believe that his method is "computationally and geometrically far more complex than what [you] are proposing" or that it has significantly "greater distortion in either area or geometry".
Given some specified resolution (the same as your depth of recursion)
and some point on a sphere, Tegmark's system
assigns that point to a "pixel number" (the number indicating what major icosahedral face it is on, and which particular minor sub-triangle on that face that point is in).
quadrilateralized spherical cube
Wikipedia: "quadrilateralized spherical cube"
The all-sky, skyward looking, unfolded cube is reassembled by
arranging the faces in a sideways T with the bar on the right side:
0
4 3 2 1
5
where square face 0 is centered on the North pole,
square face 5 is centered on the South pole,
the vernal equinox -- at latitude=0, longitude=0 --
lies at the center of square face 1.
Ecliptic longitude increases from face 1 to face 4.
other polyhedra
Tegmark mentions that "it is clearly desirable to use as small faces as possible".
Subdividing each face of a (30-faced) rhombic triacontahedron into two nearly-equilateral isosceles triangles gives the (60-faced) pentakis dodecahedron, as you mentioned,
which one might expect to give less error than starting with a (20-faced) icosahedron.
Further subdividing each of those faces into two irregular triangles gives the (120-faced) disdyakis triacontahedron (dual to the truncated icosidodecahedron).
I've been told that the strictly convex polyhedron with the maximum number of identical faces (exactly equal-shaped and exactly equal-area) is that disdyakis triacontahedron.
However, apparently none of these shapes gives any improvement over your icosahedron proposal.
Tegmark's code immediately breaks up each equilateral triangle of the 20-sided icosahedron into 6 identical irregular triangles.
The resulting 20*6 = 120 identical triangles are the same triangles as the faces of the disdyakis triacontahedron, right?
Ordering polygons on a recursively subdivided polygon
The quadrilateralized spherical cube recursively subdivides each square -- each division of 4 by area adding 2 more bits of localization information -- using a Morton code (Z-order curve).
The Maidenhead Locator System uses a system similar to the Morton code.
You might also want to look at the military grid reference system.
Other ways of ordering such subdivided square (i.e., drawing a path that hits the center of each square exactly once) include the
Sierpiński curve, the Hilbert curve,
the Peano curve,
the Moore curve, etc.
There exist other space-filling curves based on recursively subdividing a triangle -- each division of 4 by area adding 2 more bits of localization area.
(Do any of them have popular names?)
There are 2 popular ways of subdividing squares and triangles:
( a, b )
"Class I"/"alternate" method:
each of the original triangles (or squares) is replaced with n^2 triangles (or squares) that exactly fit inside the original triangle (or square).
(n=1 gives the original base shape, n=2 gives a shape with 4 times as many faces).
The small triangles have edges (very close to) parallel to the original big triangles.
"Class II"/"triacon" method, employed on almost every major geodesic dome project in the 1950s ( a p. 30 ),
each of the original triangles is replaced with 3*n^2 triangles.
The small triangles have edges (very close to) right angles to the original big triangles.
(Unreliable sources tell me the triacon method can be applied to the cube or rhombic solids,
each of the original squares or rhombuses is replaced with 2*n^2 rhombuses).
(n=1 converts the icosahedron to the pentakis dodecahedron I mentioned earlier; n=8 converts the icosahedron to an approximation of the Epcot "Spaceship Earth"; n=1 converts the cube into the rhombic dodecahedron, etc.).
EDIT:
The most common application of numbering the sides of a platonic solid is while making dice.
One particular way of putting numbers of the sides of an icosahedron is shown on DiceCollector's page, but I don't know if that is particularly "standard".
Have you considered asking
"What is the proper way to number the sides of a 1d20 die?" or
"What is the proper way to number the sides of a 1d120 die?",
or both,
on the https://rpg.stackexchange.com/ ?
Best Answer
You don't need a formula. Remember that there are precisely $5$ regular convex polyhedrons:
So the most faces that a regular convex polyhedron can have is $20$, which is the icosahedron. If you want to find an estimate of the number of faces you can divide a sphere in to, given the number of points (vertices), you can make your own imaginary polyhedron using Euler's polyhedra formula. Which states that the number of vertices minus the number of edges plus the number of faces equals $2$. $$ V - E + F = 2 $$
Notice that the number of edges equals one half of the number of sides of the face multiplied by the number of faces. This is true because when the faces are connected to form a polyhedron, each side is shared with another face, so each side must only be counted once as an edge. So if we let s equal the number of sides of whatever face you decide to use (remember you can only choose from a triangle, square, or pentagon) then we can write: $$ \frac12 sF = E $$
Plug this in to Euler's formula to get: $$ V - \frac12 sF + F = 2\\ V + F \left(-\frac12 s + 1\right) = 2\\ F = \frac{2 - V}{-\frac12 s+1} $$ Plug in your number of vertices (points) and the number of sides of the face you decide to use to obtain an estimated number of faces you can divide your surface area in to. Note that the larger the number of points on the sphere, the more accurate your answer will be. This is true because this solution was obtained from assuming that whatever face you chose to use is planar. So the more points you have, the more faces you have, so the faces take up a smaller area of the sphere, in turn making each face less hyperbolic and more planar.