OFFSET
0,1
COMMENTS
For general n, fold a rectangular sheet of paper (A4, say) in half (fold lower half up), and again into half (left half to the right), and again (lower half up), and again (left half to the right)... altogether n folds. Cut along the diagonal top left - bottom right of the resulting small rectangle. Count the pieces.
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).
FORMULA
a(n) = (2^n+2^(n/2)*(1+(-1)^n+3*sqrt(2)*(1-(-1)^n)/4)+2)/2 for n>1. (Johan Nilsson)
a(0) = 2, a(1) = 3, a(n+1) = 2*a(n)-2^(floor((n-1)/2))-1.
G.f.: -(2*x^5-x^4+5*x^3-4*x^2-3*x+2)/((x-1)*(2*x-1)*(2*x^2-1)). - Alois P. Heinz, Apr 23 2015
a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>5. - Colin Barker, Feb 05 2020
E.g.f.: (1/4)*(-2 - 2*x + 2*cosh(2*x) + 4*cosh(sqrt(2)*x) + 4*sinh(x) + 4*cosh(x)*(1 + sinh(x)) + 3*sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, Feb 05 2020
EXAMPLE
n=1: Take a rectangular sheet of paper and fold it in half. Cutting along the diagonal of the resulting rectangle yields 3 smaller pieces of paper.
n=0: Cutting the sheet of paper (without any folding) along the diagonal yields two pieces.
MAPLE
2, seq(floor((2^n+2^(n/2)*(1+(-1)^n+3*sqrt(2)*(1-(-1)^n)/4)+2)/2), n=1..25);
MATHEMATICA
Table[Floor[(2^n + 2^(n/2)*(1 + (-1)^n + 3 Sqrt[2]*(1 - (-1)^n)/4) + 2)/2], {n, 0, 25}] (* Michael De Vlieger, Apr 24 2015 *)
PROG
(PARI) concat(2, vector(30, n, round((2^n+2^(n/2)*(1+(-1)^n+3*sqrt(2)*(1-(-1)^n)/4)+2)/2))) \\ Derek Orr, Apr 27 2015
(PARI) Vec((2 - 3*x - 4*x^2 + 5*x^3 - x^4 + 2*x^5) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)) + O(x^35)) \\ Colin Barker, Feb 05 2020
(MAGMA) [2, 3, 5, 8] cat [Floor((2^n+2^(n/2)*(1+(-1)^n+3*Sqrt(2)*(1-(-1)^n)/4)+2)/2):n in [4..40]]; // Vincenzo Librandi, May 05 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Dirk Frettlöh, Apr 22 2015
STATUS
approved