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COMMENTS
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The folds are always made so the longer side becomes the shorter side.
We could have counted not only the holes but also all the notches: 4,6,9,15,25,45,81,153,289 ... which has the formula a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) and appears to match the sequence A183978. - Philippe Gibone, Jul 06 2016
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? - James Propp, Jul 05 2016
Reply from Tom Karzes, Jul 05 2016: (Start)
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,... (End)
Also the number of subsets of {1,2,...,n} that contain both even and odd numbers. For example, a(3)=3 and the 3 subsets are: {1,2}, {2,3}, {1,2,3}; a(4)=9 and the 9 subsets are {1,2}, {1,4}, {2,3}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}. (See comments in A052551 for the number of subsets of {1,2,...,n} that contain only odd and even numbers). - Enrique Navarrete, Mar 26 2018
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FORMULA
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u(0) = 0; v(0) = 0; u(n+1) = v(n); v(n+1) = 2u(n) + 1; a(n) = u(n)*v(n).
a(n) = (2^ceiling(n/2) - 1)*(2^floor(n/2) - 1).
Proof from Tom Karzes, Jul 05 2016: (Start)
Let r be the number of times you fold along one axis and s be the number of times you fold along the other axis. So r is ceiling(n/2) and s 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^r) by (2^s) rectangles. The interior grid lines will have diamond-shaped holes where they intersect (assuming diagonal cuts).
There are (2^r-1) internal grid lines along one axis and (2^s-1) along the other. The total number of internal grid line intersections is therefore (2^r-1)*(2^s-1), or (2^ceiling(n/2)-1)*(2^floor(n/2)-1) as claimed. (End)
From Colin Barker, Jun 22 2016, revised by N. J. A. Sloane, Jul 05 2016: (Start)
It follows that:
a(n) = (2^(n/2)-1)^2 for n even, a(n) = 2^n+1-3*2^((n-1)/2) for n odd.
a(n) = 3*a(n-1)-6*a(n-3)+4*a(n-4) for n>3.
G.f.: x^2 / ((1-x)*(1-2*x)*(1-2*x^2)).
a(n) = (1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))). (End)
a(n) = A000225(n) - 2*A052955(n-2) for n > 1. - Yuchun Ji, Nov 19 2018
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