[math-fun] back to kindergarten
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
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 https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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?
Neil _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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?
Neil _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
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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
I've just had a go with some coloured sheets of A4. Four images, going from 2 folds to 5: http://checkmyworking.com/misc/folded-holes/folded-holes-2.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-3.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-4.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-5.jpg It went very wrong at 5 folds as the accumulating inaccuracies in the folds and the very small edge lengths made it very hard to cut the corners off accurately. Someone less dyspraxic than I am could probably do a much better job than this. On Tue, 5 Jul 2016 at 19:39 James Propp <jamespropp@gmail.com> wrote:
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
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I also just tried starting with a square piece of paper and alternating between the fold-all-four-corners-into-the-middle operation and the snip-all-four-corners operation; the number of holes appears to go 0,4,8,24,48,... Can anyone continue the sequence, guess the pattern, and/or prove it? Jim Propp On Tuesday, July 5, 2016, Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
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
I've just had a go with some coloured sheets of A4. Four images, going from 2 folds to 5:
http://checkmyworking.com/misc/folded-holes/folded-holes-2.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-3.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-4.jpg http://checkmyworking.com/misc/folded-holes/folded-holes-5.jpg
It went very wrong at 5 folds as the accumulating inaccuracies in the folds and the very small edge lengths made it very hard to cut the corners off accurately. Someone less dyspraxic than I am could probably do a much better job than this.
On Tue, 5 Jul 2016 at 19:39 James Propp <jamespropp@gmail.com <javascript:;>> wrote:
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 <javascript:;>> 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:;> <javascript:;>
On Tue, Jul 5, 2016 at 12:21 PM, Tom Karzes <karzes@sonic.net <javascript:;> <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:;> <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
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We can distinguish between two different orientations of holes (when playing fold-cut-fold-cut-... with an isosceles triangle); if we tally the two orientations separately, we get 0+0, 0+1, 2+1, 2+7, 14+7, 14+35, ... Jim Propp On Tuesday, July 5, 2016, Tom Karzes <karzes@sonic.net> wrote:
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 <javascript:;>> 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:;> <javascript:;>
On Tue, Jul 5, 2016 at 12:21 PM, Tom Karzes <karzes@sonic.net <javascript:;> <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:;> <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Problem for John' s grandson: For which primes p,q,r,s binomial(p,q) = binomial(r,s)? https://oeis.org/A058077
Another way to generalize this is to allow fan-folds instead of a single fold per step. So for instance, a given fold could divide the paper into 2, or 3, or 4, etc. partitions. If p is the number of partitions created per fold, you just change the 2 to p: prod(p^floor((n+i)/d)-1) for i = 0 to d-1 Alternatively: ((p^(n/d)-1)^(d-mod(n,d))) * ((p^(n/d+1)-1)^(mod(n,d))) where p is the number of partions created by a fold step (so you are making p-1 folds per step), n is the number of fold steps, and d is the number of dimensions. Tom Neil Sloane writes:
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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Neil Sloane -
Tom Karzes -
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