I've generated some sample hole counts for 3 and 4 dimensions. Here are the counts (up to 20 folds) for 3 dimensions: holes(3, 0) = 0 holes(3, 1) = 0 holes(3, 2) = 0 holes(3, 3) = 1 holes(3, 4) = 3 holes(3, 5) = 9 holes(3, 6) = 27 holes(3, 7) = 63 holes(3, 8) = 147 holes(3, 9) = 343 holes(3, 10) = 735 holes(3, 11) = 1575 holes(3, 12) = 3375 holes(3, 13) = 6975 holes(3, 14) = 14415 holes(3, 15) = 29791 holes(3, 16) = 60543 holes(3, 17) = 123039 holes(3, 18) = 250047 holes(3, 19) = 504063 holes(3, 20) = 1016127 Here are the counts (up to 20 folds) for 4 dimensions: holes(4, 0) = 0 holes(4, 1) = 0 holes(4, 2) = 0 holes(4, 3) = 0 holes(4, 4) = 1 holes(4, 5) = 3 holes(4, 6) = 9 holes(4, 7) = 27 holes(4, 8) = 81 holes(4, 9) = 189 holes(4, 10) = 441 holes(4, 11) = 1029 holes(4, 12) = 2401 holes(4, 13) = 5145 holes(4, 14) = 11025 holes(4, 15) = 23625 holes(4, 16) = 50625 holes(4, 17) = 104625 holes(4, 18) = 216225 holes(4, 19) = 446865 holes(4, 20) = 923521 Tom Tom Karzes writes:
First consider the two-dimensional case.
Let a be the number of times you fold along one axis and b be the number of times you fold along the other axis. So a is ceil(n/2) and b is floor(n/2), where n is the total number of folds.
When unfolded, the resulting paper has been divided into a grid of (2^a) by (2^b) rectangles. The interior grid lines will have diamond-shaped holes where they intersect (assuming diagonal cuts). There are (2^a-1) internal grid lines along one axis and (2^b-1) along the other. The total number of internal grid line intersections is therefore (2^a-1)*(2^b-1), or (2^ceil(n/2)-1)*(2^floor(n/2)-1), which is the formula given on the OEIS page.
In d dimensions, assuming the axes for folding are selected in a round-robin fashion, the number of times a given dimension is folded is:
floor((n+i)/d)
where i runs from 0 (for the last dimension to be folded) through d-1 (for the first dimension to be folded).
The corresponding number of internal dividing lines/planes/etc. is (2^floor((n+i)/d)-1). The number of internal d-way intersections, which corresponds to the number of holes, is:
prod(2^floor((n+i)/d)-1) for i = 0 to d-1
where d is the number of dimensions and n is the total number of folds.
Tom
Neil Sloane writes:
Take a sheet of paper, fold it n times (in the naive way), then cut off the 4 corners. How many holes?
Answer: https://oeis.org/A274230, surprisingly a new sequence
There are some conjectures for a formula which should not be hard to prove?
My question is, is there a 3-D analog? Start with a very large soft brick. Dan?
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