OFFSET
0,1
COMMENTS
The actual Jerusalem cube fractal cannot be built using a simple integer grid. However, one can create an approximate one by choosing the cube side length to be a Pell number (see link).
In practice, the first two terms represent the level 0 because they both consist of cubes (1 X 1 X 1 and 2 X 2 X 2, respectively). The "cross" shape appears at index 2, which is usually considered as the first iteration (for example, the "hole" shape in the Menger Sponge is visible at level 1).
The limit of a(n+1)/a(n) is equal to 2*(2+sqrt(7)) as n approaches infinity.
REFERENCES
Eric Baird, L'art fractal, Tangente 150 (2013), 45.
Thomas Hull, Project Origami: Activities for Exploring Mathematics, A K Peters/CRC Press, 2006.
LINKS
Eric Baird, The Jerusalem Cube
Malachi B-J. Brown, Business Card Origami
Robert Dickau, Cross Menger (Jerusalem) Cube Fractal
Origami Resource Center, Jerusalem Cube Fractal (Level 1)
Franck Ramaharo, An approximate Jerusalem square whose side equals a Pell number, arXiv:1801.00466 [math.CO], 2018.
Wikipedia, Cube de Jérusalem [In French]
Index entries for linear recurrences with constant coefficients, signature (12, -16, -80, -48)
FORMULA
a(n) = (3/14)*(7*(2 - 2*sqrt(2))^n + 7*(2 + 2*sqrt(2))^n + (21 - 5*sqrt(7))*(4 - 2*sqrt(7))^n + (21 + 5*sqrt(7))*(4 + 2*sqrt(7))^n).
a(n) = 12*a(n-1) - 16*a(n-2) - 80*a(n-3) - 48*a(n-4), n > 4.
G.f.: 12*(1 - 6*x + 8*x^3)/((1-4*x-4*x^2)*(1-8*x-12*x^2)) .
E.g.f.: (3/14)*(7*exp((2 - 2*sqrt(2))*x) + 7*exp((2 + 2*sqrt(2))*x) + (21 - 5*sqrt(7))*exp((4 - 2*sqrt(7))*x) + (21 + 5*sqrt(7))*exp((4 + 2*sqrt(7))*x)).
EXAMPLE
a(2) = 672 because 456 business cards are needed for the squeleton and 216 more for the panels.
MATHEMATICA
LinearRecurrence[{12, -16, -80, -48}, {12, 72, 672, 6048}, 20]
PROG
(Maxima) makelist((3/14)*(7*(2 - 2*sqrt(2))^n + 7*(2 + 2*sqrt(2))^n + (21 - 5*sqrt(7))*(4 - 2*sqrt(7))^n + (21 + 5*sqrt(7))*(4 + 2*sqrt(7))^n), n, 0, 20), ratsimp;
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Franck Maminirina Ramaharo, Oct 18 2018
STATUS
approved