OFFSET
0,4
COMMENTS
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
Also the number of integer compositions of n + 1 with an odd part other than the first or last. The complementary compositions are counted by A052955(n>0) = A027383(n) + 1. - Gus Wiseman, Feb 05 2022
Also the number of unit squares in the (n+1)-st iteration in the version of the dragon curve where the rotation directions alternate, so that any clockwise rotation is followed by a counterclockwise rotation, and vice versa (see image link below). - Talmon Silver, May 09 2023
LINKS
Paolo P. Lava, Table of n, a(n) for n = 0..1000
Philippe Gibone, Illustration of a(0)-a(4) (idealized).
Talmon Silver, Illustration of alternating dragon.
N. J. A. Sloane, Illustration for a(4) = 9 (scan of an actual cut-up piece of paper)
N. J. A. Sloane, Illustration for a(5) = 21 (scan of an actual cut-up piece of paper)
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).
FORMULA
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) = A079667(2^(n-1)) for n >= 1. - J. M. Bergot, Jan 18 2021
E.g.f.: cosh(x) + cosh(2*x) - 2*cosh(sqrt(2)*x) + sinh(x) + sinh(2*x) - 3*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, Apr 06 2022
MAPLE
A274230:=n->(1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))): seq(A274230(n), n=0..50); # Wesley Ivan Hurt, Jul 07 2016
MATHEMATICA
Table[(2^Ceiling@ # - 1) (2^Floor@ # - 1) &[n/2], {n, 0, 31}] (* Michael De Vlieger, Jun 30 2016 *)
PROG
(Magma) [(2^Ceiling(n/2)-1)*(2^Floor(n/2)-1 ): n in [0..35]]; // Vincenzo Librandi, Jul 02 2016
(PARI) a(n)=2^n+1-(n%2+2)<<(n\2) \\ Charles R Greathouse IV, Jul 05 2016
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
Philippe Gibone, Jun 15 2016
STATUS
approved