Tom, Thanks! That clears up the whole problem. I updated the 2-D version, A274230, with your proof, and I created A274626 for the 3-D version where I gave your general formula for d dimensions. All that remains is to find someone with a pair of scissors and a scanner to create a couple of nice illustrations for the 2-D sequence Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Tue, Jul 5, 2016 at 12:21 PM, Tom Karzes <karzes@sonic.net> wrote:
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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