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On the subject of an ultimate cube...

There can not possibly be an ultimate cube just like (because) there is

no "ultimate", ie., largest, integer. But if you can solve the

(N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You

would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in

the NxNxN group and continue from there like MBW did when looking for

God's algorithm in the 3x3x3 group.

The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1)

cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the

edge moves and the center cubie orientation moves - that is, as

Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding

the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the

edges and the centers, and along with them came the edge moves (to form

the group of the 3x3x3 cube) and the center orientations (to form the

3x3x3 super-group). The edge moves alone are a proper subgroup of the

cube group and the cube group is a proper subgroup of the super-group.

A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4

cube. If you constrain the cube so that the central 2 slices can not be

moved independently of one another then the 2 central edge pieces act

exactly like the edges of a 3x3x3 cube and the 4 face center pieces act

exactly like the face centers of the 3x3x3 cube. When the central slices

are allowed to move independently of one another permutations are added

to the 3x3x3 group and super-group to make up the 4x4x4 group and

super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4

groups.

The pattern continues as the value of N increases with the N+1 group

being larger than the N group and properly containing the N group. So

the answer is no, there is no ultimate cube.