cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A320660 Number of business cards required to build an origami level n Jerusalem cube.

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%I A320660 #17 Mar 06 2022 10:00:48
%S A320660 12,72,672,6048,55488,511872,4738560,43943424,407890944,3787941888,
%T A320660 35186122752,326885842944,3037038034944,28217571901440,
%U A320660 262178452930560,2436006721486848,22634041833160704,210303674768424960,1954034324430913536,18155901427591938048
%N A320660 Number of business cards required to build an origami level n Jerusalem cube.
%C A320660 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).
%C A320660 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).
%C A320660 The limit of a(n+1)/a(n) is equal to 2*(2+sqrt(7)) as n approaches infinity.
%D A320660 Eric Baird, L'art fractal, Tangente 150 (2013), 45.
%D A320660 Thomas Hull, Project Origami: Activities for Exploring Mathematics, A K Peters/CRC Press, 2006.
%H A320660 Eric Baird, <a href="http://alt-fractals.blogspot.com/2011/08/jerusalem-cube.html">The Jerusalem Cube</a>
%H A320660 Malachi B-J. Brown, <a href="http://www.spencerandbrown.com/mbb/origami/buscard/">Business Card Origami</a>
%H A320660 Robert Dickau, <a href="https://robertdickau.com/jerusalemcube.html">Cross Menger (Jerusalem) Cube Fractal</a>
%H A320660 Origami Resource Center, <a href="https://www.origami-resource-center.com/jerusalem-cube-fractal-level-1.html">Jerusalem Cube Fractal (Level 1)</a>
%H A320660 Franck Ramaharo, <a href="https://arxiv.org/abs/1801.00466">An approximate Jerusalem square whose side equals a Pell number</a>, arXiv:1801.00466 [math.CO], 2018.
%H A320660 Wikipedia, <a href="https://fr.wikipedia.org/wiki/Cube_de_J%C3%A9rusalem">Cube de Jérusalem</a> [In French]
%H A320660 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12, -16, -80, -48)
%F A320660 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).
%F A320660 a(n) = 12*a(n-1) - 16*a(n-2) - 80*a(n-3) - 48*a(n-4), n > 4.
%F A320660 G.f.: 12*(1 - 6*x + 8*x^3)/((1-4*x-4*x^2)*(1-8*x-12*x^2)) .
%F A320660 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)).
%F A320660 a(n) = 3*( A084128(n) -2*A239549(n) +3*A239549(n+1) ). - _R. J. Mathar_, Mar 06 2022
%e A320660 a(2) = 672 because 456 business cards are needed for the squeleton and 216 more for the panels.
%t A320660 LinearRecurrence[{12, -16, -80, -48}, {12, 72, 672, 6048}, 20]
%o A320660 (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;
%Y A320660 At the n-th level, the cube side length is A000129(n+1), the squeleton requires 6*A239549(n+1) business cards, and each face requires A057087(n) units for the panels.
%Y A320660 Cf. A212596 (Origami Menger sponge), A304960 (Origami Mosely snowflake sponge).
%K A320660 nonn,easy
%O A320660 0,1
%A A320660 _Franck Maminirina Ramaharo_, Oct 18 2018