Difficult puzzle

A set of duples in an ordered list is generated by a certain symmetrical function in steps from 1 to 14.
{1 1} is the zeroth member.

The list starts:

{{1 1}.{2 3},{9 18},{40 107},{178 617},{752 3329},{2989 16712},{11019 75845},{38049 290024},{120169 629249},...}

Calculate the next 5 duples in the list; describe the step function and what it does.



  • skipper
    clue #1: the list is a set of permutations;
    clue #2: there are prime factors in one or both members of each ordered pair;
    clue #3: the second value in each duple is zero, for steps 12,13,14, only the first member has a value after step 11.
  • skipper
    ok, clue #4: the numbers in each pair in the list are ordered by n, the number of quarter or full (through pi/2 or pi) turns of a 2-slice group, which is found on the Pocket 2x2x2 cube. This has 6 colors but the list is for a 3-color map on the "2-cube" and Rubik's rotation group of cube products.

    So that I just copied a table of numbers from the wikipedia page for the Pocket Cube and divided them by 3, except the first, trivial pair of 1s, which should really be {1/3 1/3} to be consistent I suppose. Then, my 1/3 map or list is for the 3x3x3 and up, since higher numbers of slices all have the 2x2x2 cube in each corner, and each subgroup has 3 colors. Alternately each subgroup has 3 more colors 'opposite' each corner so that a 3x3x3 is 6x2x2x2 pockets, and so on.
    Trying to derive an ideal formula for the permutations in the 2-slice cube and map this to the shallow-slices on the icosahedron of the 4-color 'toy' I bought a while ago.

    These Rubik's puzzles and the various Platonic solids, sliced deeply or otherwise, represent another look at ancient problems and stuff like why there are a finite number of regular solids. The 4-color sphere/icosahedrally sectioned number puzzle is a dual space for the 3/6 colored cube, since the "missing red" means the hole left behind has up to 3 colors around it, this maps directly to the 3 colors on corner pieces of the sliced cube, colored up to 6.

    It's a toughie, but there are a finite number of permutations per n slices, starting with 3 or 6, or some combination as in bandaged and void cube puzzles. Since the cube is less symmetrical than the sphere, a shallow sliced cube with movable pieces would also require a removable piece like the Magic Number sphere (The Brain Racker) has. On the cube, pieces would have to be able to go "around a corner", so that slicing the whole cube is actually easier, unless maybe you used deformable sections or could roll and unroll them. An electronic cube-type puzzle would work too: make one section blank and allow one of the surrounding sections to fill it.

    The sectioned cube and sphere - shallow cut - are like 3d versions of the square moving-number puzzles; deep slices allow color permutations which are partitioned by n the number of rotations; each group depends on the previous group having been generated by n, but contains independent primes.
    For some reason the first prime appears after 3 full turns and is 107 (i.e. invariant) for 3 or 6 colors. What if there are 4 colors, or 2 (a 2-polytope)? What happens when the number of colors changes, IOW?
    Since, the 3x3x3 has 9 separate pieces per face, you can color each face up to 9?

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