1. Introduction C4(n) is a formula for an upper bound of the number of distinguishable configurations of an n×n×n×n Rubik's Cube, which will be derived in this paper. It will be assumed that the reader is familiar with a 4-dimensional Rubik's Cube. Online, one can find the free computer program Magic Cube 4D, developed by Melinda Green, Don Hatch, and Jay Berkenbilt, which is a completely interactive representation of a 4-dimensional Rubik's Cube, and which was the inspiration for this paper and much of my other work.1 An FAQ page has been provided to help familiarize new users with the necessary concepts of higher dimensions and how Rubik's Cubes would function in these spaces. Additionally, a solution guide has been provided by Roice Nelson, who is another pioneer in the research of higherdimensional puzzles. His creations include the free programs MagicCube5D, which was written along with Charlie Nevill, and Magic120Cell, which are representations of a 5-dimensional Rubik's Cube and a puzzle based on the 120-cell, respectively.2,3 I would like to thank Roice in particular for his continual support and encouragement, which includes both hosting this paper and my other work on his website, and proofreading this paper while it was being developed. Roice found many oversights and errors, all of which have been corrected, and provided simplifications and new ideas. His creations MagicCube5D and Magic120Cell have also inspired me, and my work is focused on these programs as well. It should also be mentioned that my discoveries would not have been possible without the previous investigations of H. J. Kamack and T. R. Keane in their paper, "The Rubik Tesseract"; it was used extensively in developing sections 3 and 4 of this paper.4 Eric Balandraud's article, "Calculating the Permutations of 4D Magic Cubes", was also helpful, and greatly assisted me in examining the properties of 4-dimensional Rubik's Cubes.5

2. The Plan Here is C4(n):

15 n ,2 2 n ,2 $ n mod 2 2 n ,3 2

C4 n =

15! 12 6

n ,4 2

24! 32! 226 633

n mod 2

64! 63 3 2

96! 95 2 2424 48! 6! 8

n ,2 2 2

192! 2448

64! 8! 8

n ,2 2 2

n ,2 2

n mod 2

96! 12!

n ,3 2

n mod 2

n ,3 2

8

n ,4 2

$

n mod 2 n , 5 n , 3 n , 1 $

192! 8 24!

n mod 2 , 1 n , 4 n , 3 n , 2 24

We will deduce this formula in two stages. First, we will calculate the specific values of C4(n) for 2 n 8. Then we will generalize our findings and build the formula up term by term.

3. The 24 Cube A 24 Rubik's Cube consists of 16 corner pieces, each with four stickers (which are also called facelets). There is an important note to be made for cubes regarding the number of pieces per edge. If it is odd (an odd cube), we can fix the central 1-colored pieces in place and observe how the other pieces permute and orient around them. This is valid because the central pieces never move relative to each other, and we can therefore fix them in space. Note also that we never have to consider making a slice move that repositions the central pieces because this is equivalent to rotating all of the other parallel layers in the opposite direction and reorienting the entire cube. For cubes with an even number of pieces per edge (an even cube), such as the 24, we will need a way to fix the cube in space so that we do not inadvertently count extra configurations due to the fact that the entire cube can rotate in 4-space. This can be accomplished by fixing a corner piece in place, and using it as a point of reference in the same manner as the central pieces for an odd cube. Now, back to the 24 cube. We count 24 = 16 pieces, all of which are corners with four facelets each. To determine the number of permutations the corner pieces can attain, we examine what takes place when we make a 90 degree face rotation. Observe that we never need to consider other...