Tough algebraic sequence of integers

Replies

• 

  skipper

  First obvious point, the formula for N can be reduced to (K-1)!.D[sup]6[/sup], eliminating the denominator.
  This can be rewritten: (K-1)!.D[sup](K-2)[/sup].
  Second detail: I should include that each k is 3-dimensional, it has a x,y and z. There are 8 elements k and subsets of 4(k) share a x,y or z.
  
  Call this sharing "S"; each set of k[sub]i[/sub] equal to n(k) = 4 is a selection of exactly 1/2 of the set of 8 elements. There are so many combinations or permutations that select 4 elements so they each share a dimension x,y or z. You might envisage this as each of 4 elements k, having "the same orientation or color", whatever works out.
  
  I might, since there are 8 elements, need to look into octonions or quaternions...? Those numbers or q-values, need to be re-encoded somehow so the formula for N gets "distributed" by each of the q(n) "functions", as shown.
  
  I could (or should) have q0 = m0 and set m0 to 1; then I can do:
  m0 = 1
  m1 = m0 + 4
  m2 = m1 + 16
  ...
  
  But this only gets me so far with nice numbers (factors or powers of 2)
• 

  skipper

  Here's some more clues:
  The list or series is bidirectional; the 'integers' are as noted, actually the number of states (eigenstates) in each subset q.
  K = k1+k2+...+k8 is an algebraic manifold. This is easy to visualize as the number of 'plies' or 'folds' needed to get from q(n) to q(n+1) or backwards. So that, q0 is "one-fold", q1 is "six fold" relative to q0, etc.
  
  The concepts of distance and area are equivalent. You can orient K/2 in say, the X dimension so that K/2 = k1+k2+k3+k4 \ k5+k6+k7+k8 (the "\"is a set-theoretic "exclude" operator, so K/2 is the set excluding 1/2 of the k elements). Then if k1(x) = k2(x) = k3(x) = k4(x), K/2 is "flat" or torsion-free.
  
  And q1 corresponds to q0 + 5, so q1 - 5 = [q0]. This says q1 has 5 more states than q0, and q0 + q1 = 7 (another number that pops up in the formula for N).
  Call the torsion state of q1 "T[sup]1[/sup]", and q0 "the origin"; derive logarithmic rules for the series (i.e. find an integrable formulation for N/n in terms of q).
  
  q2 is then T[sup]2[/sup]; q3 is T[sup]3[/sup].
  At q4 (T[sup]4[/sup]) there's a clunky "prime addend" to contend with, I can't factor it to less than a value of 17...
  
  You can also consider the q subsets as ordered (by n) and partially ordered by prime factors (some of the q cosets of N, are "prime-free" and factor out nicely with powers of 2 and 3; although these are both primes they're also conventional values for all kinds of other physical things - square areas when 2 is an exponent - cubes when 3 is the exponent etc).
  
  K[sub]8[/sub] is 8(3d) many-folded (a manifold). I'm trying to find a Riemann sum which is integrable. Note the "curve" in the series; if you plot the n along the x axis (abscissa) the q(n) are along the ordinal and it looks like a set of rectangles (a Riemann sum). But there are those primes "in the way" of a straightforward recursive descent or recurrence algorithm = an integral.
  
  😀
  
  There are more details, but the series, being the "external derivative" of K, divided up into q(n) partitions is simply a set of (subsets of) integers.
  Say I also reveal that the manifold has a general swap operation in it - any two k's can swap places or orientations. The manifold can be considered a "component, with slots in it" so the k are "sub-components that go in the slots". There's another more well-known name for this.
  
  p.s. told you I'd be looking at this torsion thing again...
• 

  skipper

  It's a problem of "least and most" in terms of magnitude. This is what Riemann was using as an abstraction of distance or area (so that one can be "folded" into the other, a "distance of lines" is an area, a "distance of areas" is a volume, a distance of volumes is 4 dimensional).
  
  The least value is q[sub]0[/sub], or q[sup]0[/sup], which means it's idempotent (q[sup]2[/sup] = q).
  logb(1) is a constant, b is any base so numbers have b dimensions logarithmically.
  The next least value is q[sub]1[/sub] to which the above applies in base 1. If I use two dimensions i and j to represent the six states in q[sub]1[/sub], it has a square product in base b, which is ij, so I map i and j to the dimensions of a hexagon (this happens to divide a certain 20-sided polyhedron in two). I map each vertex to i or j, and alternate this - I swap i and j, and write the formula that swaps a vertex for an edge, or two edges, or two vertices - a general swap operator for the hexagonal state space in q[sub]1[/sub].
  
  With this formula I descend recursively into the "K-space". Since there are large-ish prime factors like 8969 up ahead, what I will need is a way to count prime numbers less than this, or any larger primes that factor out, as addends (remainders) or factors in ratios, there are "missing" prime numbers for some reason, so gaps exist in the function generating them which I want to find - to integrate the series iteratively or recursively, by folding products into sums or sums into products.
  
  In topological terms, it's a kind of switching fabric, a "torsion bundle", which is tension-free; stresses are all along glide-surfaces = glide transforms that translate. The Hamiltonian is algebraic and "stressed" - the k "bundle" might be considered a set of keys - 8 keys of thirds in an "octave of k's"; each k, a third, can be diminished (flattened) or natural (normal), or augmented (sharpened). Like a six-string guitar you can tune by playing.
• 

  skipper

  OK; I'm actually looking at the Rubik's cube (group), for the 2x2x2 "Pocket Cube" puzzle.
  The number of symmetry groups in the cube puzzles (up to 7x7x7) is "too large to list".
  
  The reason I'm chasing prime factors or sums, is because of the Riemann zeta function (and Fermat primes, and other kinds). There is a known "prime generator" that gives all the primes less than a given integer (which may not be prime). This is a kind of selector that discards integers that have divisors other than 1 (and themselves), a prime "delta function".
  
  Using the "remainder" approach, note that the initial state q[sub]0[/sub] means there are (3[sup]6[/sup])7! - 1 remaining "integer" states; this factors into a "small" and a "large" prime number - call these a and b (for no good reason other than: "you can"). See if this yields a recursion which is integral (has integer sums/products).
  
  a = 17, b = 216127; I'll see if I can find all the primes between a and b and if this will be useful (is there a "prime number symmetry" in the Rubik Group?); below 17 I can generate the list of primes "in my head" - these are "trivial" to calculate.
  
  BION (believe it or not) the Rubik's Cube group explains certain cosmological principles, the model has been exploited to explore certain invariants and relations (including general relativity, the Higgs mechanism that generates "mass energy" - a condensation - and various other gee-whiz stuff).
• 

  Kaustubh Katdare

  No one here? Hello? Looks like the topic is too tough for most of the brains here 😛
• 

  Saandeep Sreerambatla

  I got struck in the first post.
  
  "The total number N = 3,674,160 up to n = 14; "
  
  Skipper,
  
  What does this mean actually , can you explain it a bit please.
• 

  skipper

  N is the total space of states, in the "manifold". It's a thing you can fold together (by rotating corners in SO(3) in a Cartesian space). Note: you can rotate the K as K[sup]0[/sup] (the whole thing), or as K[sup]0[/sup]/2. The 3x3x3 has K[sup]0[/sup]/3, the 4x4x4 has K[sup]0[/sup]/4 possible rotations additionally.
  
  I suppose I could characterize K as an "addition matrix" or something. Factor N, and some strange numbers appear, which are large and irreducible in any base including 2. These are primes or pseudoprimes (coprimes, Gauss primes, Fermat primes, all kinds of primes that belong in sequences). I believe I need to use the Riemann zeta to count these and see if any are missing; for instance, q[sub]6[/sub] has 8969 states (all identical in some aspect, or "variable" of K).
  
  The quotients are isolates, all the q in q(n) are like equivalent distances or areas of "numbers"; of course, these numbers can have an arbitrary number of dimensions (7 say, or something simpler).
  The idea is to fit "dimensions" to the set of integers so the primes line up in a nice, defined sequence.
  
  This is, in fact, exactly what you do when you scramble or unscramble a cube - the numbers are "colored areas"; equivalently the numbers are "distances from a color", or any abstraction that fits the model, or that the model fits, like say QM and the quark domain (quarks have 3 + 3 colors).
  
  Oh yeah, another clue is, if I do make a = 17, Gauss (ma main man) found a way to construct (with ruler and compass, i.e. a way to 'figure' a number) a 17-sided polygon = division of the unit circle. This problem is about dividing "space and time" with numbers = "colored areas that associate, via rotations in SO(3)".
  
  One important thing to bear in mind here is, the list of states as integer values (with a curvature, like a curve-fitting interpolation) is a presentation of K, in quotient terms - the quotients all divide N[sub]K[sub]8[/sub][/sub].
  
  In fact the states in each q "enfold" the previous q; the transition from q[sub]0[/sub] to q[sub]1[/sub] is equivalent to extending a unit (square or circle - perhaps it's both?) to a six-sided figure (six squares or circles). I can easily construct a hexagon as mentioned, with q[sub]0[/sub] as the origin , or I can fold two circles (a doubled circle = the Mobius loop) and copy 3 of them - arranging them geometrically; this transform is equivalent in the circle group, to making a 2-torus that crosses itself at 3 points in SO(3), which is the trefoil knot. viz:
  
  
  
  
  Trefoil Knot
  
• 

  skipper

  Now, the Indian hero Ramunajan turns up, with his factorization of pi - the ratio of a circle to its radius.
  Interestingly this is where a certain Greek engineer comes into the picture "my mate", Stefanides from Athens, with his map of Z, the integer group (we can rotate integers, expand them and contract them, do anything at all including counting "universes"). This is where Hamilton and his free cycles also turn up - three "crossings in space and time". Mr Panagiotis also has a good map of the heliotropy of this - his home-made helioskopos record, a recent one which he presented at a conference of astronomers earlier this decade. Gravity shows up at the heliocenter - p(0). And some physics.
  
  And also of course, graph theory, coloring problems, SAT, P not NP, and polynomials in general - I want a transfer Hamiltonian, H, that gets me from q(n) to q(n-1) or q(n+1) - this is what I want "in a triangle, or convex hull".
  Also called an alternating sum, sort of wedge-shaped, an alternating wedge (I need a delta, I have "n and (-)1"), then there are Jones polynomials and that trefoil.
• 

  skipper

  Ah cha, suppose we place q[sub]15[/sub] at infinity since 0 states are generated beyond q[sub]15[/sub]; which I will label the edge or extremal of G, the graph.
  Then Q = q[sub]n[/sub]: n :={0,1,2,...,14} is the set of states in each subgroup q[sub]n[/sub];
  So we have the set S of states, in G, and Q or G(Q,S). Q are V the set of vertices, S are distances along or, areas on the open disk: D \ {0}.
  
  Subsets of S are "in space and time" = the set of all signals in Q, the rational numbers isomorphic to Z/nZ where n is 15; that is infinity, is 15 moves away, in the K-space.
• 

  skipper

  Ok, so this extreme is beyond q[sub]14[/sub] in the quotient space. This is isomorphic to a path with 14 connections in it - the 15th connection goes to infinity; this is equivalent to saying "after 14 moves on the Pocket Rubik's cube, there are no states left to generate, you can only go back to the previous quotient, q[sub]13[/sub]".
  
  A graph with 14 connections is equivalent to a graph with 15, where one "doesn't complete" since one end is joined to infinity. K[sub]6[/sub] has 15 connections; leaving these in place, elect 1 of 6 vertices that will "lose" it's 5th degree (of connectedness). Now there is a vertex on K[sub]6[/sub] with in-degree 4, and out-degree 5, where the 5th "goes to infinity". and 4 vertices with degree 5, one of which is disconnected from the graph; it needs to be re-connected - perhaps I can connect it to a triangle which is K[sub]3[/sub], with 3 nodes and give one vertex an extra edge, connected to the K*[sub]6[/sub] or modified version? Then I do the same with the vertex at infinity - I connect a triangle on the perimeter of the cone at infinity, to the triangle connected to the vertex (1 of 4) that requires completion.
  
  Then I can permute the vertices in the following way, on the triangles: Tijk, Tjki, Tkij. The Q[sub]8[/sub] group of quaternions/octonions.
  
  Where is this going? Large primes, and Godel numbers that descend into Z, also logarithms and e, the function of numbers. here's an appetizer, the "god number" for n-dimensional spaces
  
  

  Magic 120-cell
  
  File:120 Cell Zoomed In.png File:120 Cell Zoomed In.png
  120-cell virtual puzzle, close in view in solved state
  
  
  File:120 Cell.png File:120 Cell.png
  120-cell virtual puzzle, solved
  
  
  Geometric shape: 120 Cell or hecatonicosachoron The 120-cell is a 4-D geometric figure (4 Polytope) composed of 120 Dodecahedron, which in turn is a 3-D figure composed of 12 Pentagon. The 120-cell is the 4-D analogue of the dodecahedron in the same way that the tesseract (4-cube) is the 4-D analogue of the cube. The 4-D 120-cell software sequential move puzzle from Gravitation3d is therefore the 4-D analogue of the Megaminx dodecahedral 3-D puzzle.
  The puzzle is rendered in only one size, that is three cubies on a side, but in six colouring schemes of varying difficulty. The full puzzle requires a different colour for each cell, that is 120 colours. This large number of colours adds to the difficulty of the puzzle in that some shades are quite difficult to tell apart. The easiest form is two interlocking tori, each torus forming a ring of cubies in different dimensions. The full list of colouring schemes is as follows;
  
  

  • 2-colour tori.
  • 9-colour 4-cube cells. That is, the same colouring scheme as the 4-cube.
  • 9-colour layers.
  • 12-colour rings.
  • 60-colour antipodal. Each pair of diametrically opposed dodecahedron cells is the same colour.
  • 120-colour full puzzle.

  The controls are very similar to the 4-D Magic Cube with controls for 4-D perspective, cell size, sticker size and distance and the usual zoom and rotation. Additionally, there is the ability to completely turn off groups of cells based on selection of tori, 4-cube cells, layers or rings.
  Gravitation3d has created a "Hall of Fame" for solvers, who must provide a log file for their solution. As of January 2009, Noel Chalmers remains the only person to have solved the puzzle.N Dimensional Sequential Move Puzzles Cite Note Solved120C 3
  Piece CountN Dimensional Sequential Move Puzzles Cite Note 120Cell 4 Number of vertices 600 Number of 4-colour pieces 600 Number of edges 1,200 Number of 3-colour pieces 1,200 Number of faces 720 Number of 2-colour pieces 720 Number of cells 120 Number of 1-colour pieces 120 Number of 4-cells 1 Number of 0-colour pieces 1 Number of coloured pieces 2,640 Number of stickers 7,560 Achievable combinations:N Dimensional Sequential Move Puzzles Cite Note 120Cell 4
  This calculation of achievable combinations has not been mathematically proven and can only be considered an upper bound. Its derivation assumes the existence of the set of algorithms needed to make all the "minimal change" combinations. There is no reason to suppose that these algorithms will not be found since puzzle solvers have succeeded in finding them on all similar puzzles that have so far been solved.

• 

  skipper

  Right, I have a better picture of this whole puzzle thing now.
  We can do these puzzles - i.e. solve a scrambled one, because our brains work like the mathematical formulas do (in fact, our brains inveneted the formulas).
  A formula is a Turing number. Enumerating (E) a number or the number of digits (N) in a number are an automorphism - a number is ultimately a list of "ones".
  
  It's all related to the problem of least and most descent (see Bernoulli, Euler &c.).
  In the puzzles this is how many moves are required to get to the "start" state, or the identity in Z, the "integral object".
  The list I posted is for quarter turns; there is a second list of full turns which is shorter. It has 11 quotients in it. 11 is coincidentally the number of dimensions in some string theories.
  
  f and q are isomorphic to N -1 = g[sub]K[/sub] the generator of the "God number" of moves, which is 17(216127). The 2x2x2 universe, or Turing # of ones in the symbol.
  11 and 14 divide this number as full and quarter turn maxima to get from "infinity" or the edge of G, to the center or "home". You need to build a few other numbers to do this.
  That is, full and quarter turns generate the same number, but in different subspaces, so they 'double' the 2-sliced cube. The cube doubling problem, P, is in #P. Cube doubling goes back to Euclid, Descartes etc. You double the algebraic states (with quotients). There is an isomorphism of the 54 states in the 2nd q-space for the shorter-path function f; the 3-slice cube has this many facelets 6(9) - note how I "bundle" the two numbers, to imply multiplication.
  
  Correction; I should be more exact about the role of the generator g (and its subgroups) in G which has N states; then g "subtracts one from itself" to become g*, which has N - 1 states; or algebraically g: g -> g*; N: N -> N -1
  
  The generator in the subgraph (beyond q[sub]0[/sub] = f[sub]0[/sub] = 1) or g*, has the 17(216127) number T = Tp,q; Here the Tp are prime factors (in decimal) and q are the quotients (f composes q so although f has a distinct subspace in G*(V,E), the reduced graph with a connection missing :- the " -1" , q is still the subspace of quotients. The list is for the "inner" states which f, the "outer" list is a composition of, i.e. the set {f o g} for g the generators in Q.
• 

  skipper

  Note (ii): this enumeration of ALL states in the Pocket (2-slice) cube with 6 colors, would be possible if you purchased about 3 million of them and set them to different enumerations (then count all the different ones, discard these and repeat; when no more can be added to the heap of different configurations, you have exhausted N).
  
  Thus, the method of exhaustion to find the integral NUMBER (God's #) of possible steps in the Turing enumeration. A Turing test of a number is whether it can be written in a finite number of symbols. A practical test of a number is to express it as a decimal (which can't be completed for certain ratios - although they have an exact integer representation, they can't be written as decimal fractions or binary ones, or any other base including base[sub]1[/sub]).
  
  This notion of not being able to write a fraction (like say the fractional part of pi, or e) is the same notion of not being able to reach N - it is at infinity or an infinite # of steps = # of decimal places away. Thus the incomplete or incompleted (static) graph; make a connection that has a single point "missing" or at infinite calculable distance - whatever distance means to the calculation. In the case of algebraic numbers, these are those with a large height like the above n-dimensional number in the 120-piece puzzle.
• 

  skipper

  Just to keep everyone up to date: I've realized that "numbers" when it comes to numbered (or colored, or "shaped") puzzles, are "precisely abstract".
  
  I'll explain that; abstraction is, as you know the process of "taking" something and "fitting" it in another paradigm - we abstract the idea of a fluid to explain certain properties of electricity "flowing", or of a gas to consider "expansion", we can expand numbers too, and make them "flow".
  
  The least/most descent problem in the Rubik's universe is recursion. The notion of transporting a "coherent state" around the cube or sphere, is analogous to the idea of transporting an equation (formula) from one setting to another (from say, liquid flow to gas flow, in terms of expansion/compression, i.e. a contraction that makes something vanish. In the case of condensing a gas, pressure vanishes).
  Look more closely at this compressive process - it generates a liquid from a gas. The generator is "applied pressure", the thing generated "stores" the generator g.P (the pressure generator) in the liquid, as a density. Density (of states) in R, the universe of rotations, on a cube with 12 outer edges, and extra "inner edges" means it's completed as a graph G, in a K-space. There are 8 elements k, in K, each is 3-colored (a 3-color map). It has fluid and solid "character" - the cube is solid, the numbers "flow", if numbers = colors adjacent to each other.
  
  Transport and parallel transport are phase changes in a "phase space", this phase is apparent when you compare the two algebraic "curve-fittings" of the outer and inner quotients.
  
  These have an Euler totient; you the operator (in terms of momentum, etc) accumulate this (in T, as steps t1, t2, ...), and with luck you unlock the particular path p, around the quotient space. Starting at the outermost ring is 14 moves minimum; unless you "cheat" and drop into (i.e. find a solution for q) the f quotient/totient space which has Min(11) moves, and there are Max(infinite) moves. This is possible because at n = 14, in q[sub]14[/sub] you can only get to q[sub]13[/sub], there, you can go two ways, to the 12th or back to the 14th, there is no 15th q, it lives in the missing connection in the K[sub]6[/sub] you have to build. This manifold (the solution) retains (hopefully) the particular holonomy you want to keep "straight", i.e. parallel to the quotient space "surface" S. I think I can say "f is a minor (key, theme, construction in G) of Q, in T the space of quotients" 😀
  
  S[sub]xy[/sub] is a single 2x2 "matrix" of X's. so k here is the "color" of each X, etc. Each Y is = 4X, etc.
  
  And here is "the map". The phase I'm talking about is the max of q, the min of f, at n = 10,11
  Note those last 2 numbers have binary values too.
  
  

  The number f of positions that require n full twists and number q of positions that require n quarter turn twists are:
  (note the "broadening, of f next to q )
  
  n f q
  0 1 1
  1 9 6
  2 54 27
  3 321 120
  4 1847 534
  5 9992 2256
  6 50136 8969
  7 227536 33058
  8 870072 114149
  9 1887748 360508
  10 623800 930588
  11 2644 1350852
  ------------------------- at 11, f exhausts the quotient space - has reached
  the Euler limit for T the totient (this has an inner sum, you are at one of
  the "terms" in the state-space of the totient function f)...
  
  remaining q
  12 782536
  13 90280
  14 276

• 

  skipper

  Dang, I doubled the cube.
  
  
  
  
  It's a V8, the shaft has three subshafts; it outputs a "power distribution", which has the greatest height (algebraically) at n = 9 (f is the outer radius of curvature for distribution, then "rolls off" at n = 10.11, at 12 it is at max "absorbance" and the generator can't make any more numbers. The inner q efficiency is max at n = 11 = [n(f) + 2)] and the phase represents a tuning of the engine to max efficiency, at f[sub]9[/sub] "borrowing" from the "look-ahead" at n = 11 in the Q for the q-generators - these contribute to f (f and q are parallel quotients, they both run out of steam or headroom at a max in n, the ordering).
  
  The inner q "engine" is composed into two by f, or "f doubles q"; this is over a cubic polynomial = V8 number generator. Hence, two pictures...😁
• 

  skipper

  So, yadayada; I have, and so does anyone who can buy these puzzles, an algebraic phase-space, of which the above map is for the "simplest" form you can make/buy. Why is 8 the min # of elements?
  
  
  
  When you do this you reveal something about the "innards" and where the totient lives. There are as many corners in this configuration as in a closed config. - a "semi-quarter" turn, what the hey I'll just call it a semiquaver, q a quaver and f a "composition of quavers"; there are 8 keys in the "scale", an octave.
  
  The K graph can turn its edges into corners, the next form or the 3-slice has an inner "layer" or ring - if you skew the original Rubik's cube it has 12 corners you can see - the middle projects 4 extra "semi-corners" into space. (see if you can do this or not). Of course I mean when you have the puzzle rotated so the centre is aligned to the centre of the "top" face.
  
  That phase between maxima is 2m; an interesting integer, 2. Max(n[ 00 f ]) - the "double O bound" or most efficient (in terms of expanding Q) is at n = 9, Max( n [ 00 q]) is at n = 11. The 10th lives between the 9th and 11th "intervals" in Q. Thirds are major or minor, there are 3 or 4 respectively in an octave (see if you can play an octave of 3rds, on a piano, then try it on a Pocket Cube).
  
  Bwa haa! I think I've done my Topology 101 assignment; but there are so many papers that refer to the "causal lattice" aspects of these things - note the above is justone geometry that has the same space or subspace. The cube has two spheres (of influence); geometrically these are the circumscribed one that each extreme corner touches, and the inscribed which each edge touches (the I or identity). The centroid of the inner sphere is Z[sub]0[/sub]; of the outer sphere (C), its Z[sub]1[/sub]; Z[sub]0,1[/sub] has 8 corners touching C and 6 edges touching I, there is a c[sub]i[/sub] with an index x[sub]i[/sub] and x[sub]j[/sub]. If y = x[sub]ij[/sub] then Z[sub]1[/sub] = xy[sub]k[/sub], where k is an element in K.
• 

  Kaustubh Katdare

  Skipper - my honest confession - I'm going to take a day or two (or maybe more) to absorb this thread!
  
  Keep up the great work!
• 

  skipper

  Big_K, have you played with Rubik's puzzles, or those Chinese cube puzzles?
  
  I first solved a 3x3x3 a long time ago and I can still recall that sense of satisfaction you get; I knew it went back together, and I just had to uncover the moves. Back then I didn't think about Euler or Cayley graphs or polynomials but that's what my brain did anyways - it found the solution in K (whatever K actually is - I mean, it looks like a cube, right?)
• 

  Kaustubh Katdare

  I've solved the 3x3x3 cube using the known techniques. Yet to try the 5x5x5.
• 

  skipper

  This is really about symmetry groups - i saw I need to do a spot of group theory; I wouldn't pass a test with the notation, which there is an excuse here, for not using the stuff.
  When you write a program though, your'e using an algebra to construct another more abstract one; you hope it finds a path through the "problem space".
  
  So when you do this on a sliced-up, colored puzzle or put Chinese blocks together you do just that; it looks all squiggly and hard to understand; terms like "fibered space. tangent space. bundle over the sphere", and "symplectic volume form", "0-forms and 1-forms", but it's all pretty standard trig and calc, with a bit of modernization. I don't want to pass exams just now I'm only "playing around".
  
  Check this out: some math-dude into music embedded two circles of 5ths from the diatonic scale, into a torus; a dash of geometry and algebra later and he had a Mobius strip; the puzzles have one too because you can rotate edges, in 4x4 cycles (grouped by those generators that generate, you know, cycles),
• 

  skipper

  What the chart from the wiki page shows is the distribution of the two functions in Q, the "total space", which is the rationals; it counts up N-1 of these. By "it" I mean the Turing machine or "arithmetic" that produces the chart.
  The chart itself is of a brute-force, breadth-first traversal of G, where |G| the "absolute value" is the N-1 term - the number between 1 and the nth digit of N.
  
  So each column is a copy of |G|, but enumerated by a g, and a f with different bounds - f is an integral of g, or a conjugation (it gets to N first because it uses Q in larger chunks of arithmetic). The inner functor is also a conjugate, related to a symmetry of "the squares group", which is that the derivative of x[sup]2[/sup] is 2x; this is captured in:
  
  h(x): x -> x,x; g(x): x,x -> x[sup]2[/sup] (x*x)
  h[sup]-1[/sup]: x,x -> x; g[sup]-1[/sup]: x[sup]2[/sup] -> 2x -> (x,x).
  
  so f o g, the composition, gives:
  
  g(h(x)), h[sup]-1[/sup](g(x)); h[sup]-1[/sup](h(x)), ...
  
  At the transition point, n: n -> n+1, when n = 2, the next doubling is either the double |G|, i.e. the |G|[sub]f[/sub] + |G|[sub]g[/sub] at the limits for f,g or inside the numbers in some other way. Squares of 17 turn up when you factor the 9th quotient of |G|[sub]f[/sub].
  
  A square of primes - note 17 is the 7th prime. It's like the primes and coprimes are tones or fundamentals; poles you can find in a transfer function, you need the zeros for the polynomial so you can cancel the primes "tonality" and find a solution in sub-prime time...
  😀
• 

  skipper

  Oh man, this is getting complex; the two series (f and g) are generator groups. Group theory (or symmetry groups) is, like, a really big topic.
  
  It turns out the cubes and related puzzles are like transcendental numbers (or trans-dimensional). The two lists are for full and quarter moves of up to 3 slices, in groups. You can restrict this to moves for only two slices, say the "top" one, and the "right" one. This is the 2-slice, or two-generator group (squares group).
  
  I found a list for what I think are quarter moves (U,R in Singmaster notation with their corresponding inverse moves U[sup]-1[/sup], R[sup]-1[/sup]), for the 2-slice group. Note "group" therefore means a group of the elements (there are 8), which [the 2-generator group] excludes exactly 2 of, in a row or column - so you act only on 6/8 of the elements. The new list is 17 spaces long (it's another "spectral chart'), which number is an important prime since, you get 17 by adding 1 to 16, or by subtracting 1 from 18; that is from 2[sup]4[/sup] or 2*3[sup]2[/sup]; 4 is a square, 2 is its root.
  
  

    r       |   vertices     |        
            |  at distance r |  log B(r) / log(r)
  
    0              1                   −
    1              4                   −
    2             10              2.32193
    3             24              2.46497
    4             58               2.6427
    5            130              2.84243
    6            271              3.0277                      7            526              3.19162
    8            980              3.33333
    9            1750             3.46023
    10           2731             3.57449
    11           3905             3.6604
    12           5229             3.72191
    13           5848             3.76469
    14           4792             3.77948
    15           2375             3.7576
    16           508              3.70136
    17            18              3.62837
  
  Table 1: The number of vertices at distance r from the origin

  What happens if you use full moves (U[sup]2[/sup],R[sup]2[/sup])? With r = 1 at least, you should still have 4 combinations of 2 sides of the cube...?
  whoops, no because half of the full-turns would get you back to "no move" so there are 2 at r =1 for f (the generator of states in the semicircle on each face).