That's interesting. This case seems a little more complicated than the rectangular case, since with the triangle you alternate between horizontal/vertical folds vs. diagonal folds, and the resulting fold pattern is more complex, but I think the basic argument is essentially the same. Note that with the triangle, the first hole doesn't appear until after you've made 3 folds, so if you start counting at zero folds, you have three leading zeros in the sequence: 0,0,0,1,3,9,21,... Tom James Propp writes:
The same sequence (0,0,1,3,9,21,49,...) turns up when you start with an isosceles right triangular piece of paper and repeatedly fold it in half, snipping corners as you go.
Is there an easy way to see why the two questions have the same answer?
Jim Propp
On Tuesday, July 5, 2016, Neil Sloane <njasloane@gmail.com> wrote:
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 <javascript:;>
On Tue, Jul 5, 2016 at 12:21 PM, Tom Karzes <karzes@sonic.net <javascript:;>> 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?
Neil _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun