A317888 - OEIS

We can prove that the corner cells are filled with values from 2 to 2^n+1, because in order to prove that none of the corner cells have to skip a value so as not to repeat a value within any of the other cells, we must prove that the non-corner cells all have values greater than 2^n+1 in at least one arrangement of corner cell values. 2(2^n-n+2) can describe the lowest value of cells in between two corner cells in every dimension from one to infinity, excluding the third dimension, because in the ideal arrangement of the three-dimensional grid, 12 must appear between 3 and 4, while the formula returns the value 14. Regardless, 12 is still greater than 2^n+1, which is 9 in three dimensions. Every value for 2(2^n-n+2) is greater than 2^n+1 as can clearly be observed from a graph. The formula 2(2^n-n+2) was derived from the fact that the smallest non-corner values appear next to 2 in every dimension except the third, and next to 3 in the first three dimensions. So with the exclusion of the third dimension due to its limited number of corner cells, these values must be equal to the product of 2 and the lowest number that is paired with 2. The lowest number that is paired with 2 in an ideal arrangement of values is always equal to one less than the number of dimensions, n-1, subtracted from the highest corner value, 2^n+1, thus resulting in 2^n+1-(n-1), or in its simplified form, 2^n-n+2. When multiplying this by two, we receive the lowest non-corner value possible in every dimension except the third.