A234275 Expansion of (1+2*x+9*x^2-4*x^3)/(1-x)^2.
1, 4, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392, 400, 408, 416, 424, 432
Offset: 0
References
- J. H. Conway, H. Burgiel and Chaim Goodman-Strauss, The Symmetries of Things, A K Peters, Ltd., 2008, ISBN 978-1-56881-220-5. See p. 191.
- S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- J. Choi, N. Pippenger, Counting the Angels and Devils in Escher's Circle Limit IV, arXiv preprint arXiv:1310.1357, 2013.
- Tom Karzes, Tiling Coordination Sequences
- N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
- N. J. A. Sloane, Overview of coordination sequences of Laves tilings [Fig. 2.7.1 of Grünbaum-Shephard 1987 with A-numbers added and in some cases the name in the RCSR database]
- Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
- S. Wolfram, A New Kind of Science
- Index entries for sequences related to cellular automata
- Index to 2D 5-Neighbor Cellular Automata
- Index to Elementary Cellular Automata
- Index entries for linear recurrences with constant coefficients, signature (2,-1).
Crossrefs
For partial sums see A265056.
List of coordination sequences for Laves tilings (or duals of uniform planar nets): [3,3,3,3,3.3] = A008486; [3.3.3.3.6] = A298014, A298015, A298016; [3.3.3.4.4] = A298022, A298024; [3.3.4.3.4] = A008574, A296368; [3.6.3.6] = A298026, A298028; [3.4.6.4] = A298029, A298031, A298033; [3.12.12] = A019557, A298035; [4.4.4.4] = A008574; [4.6.12] = A298036, A298038, A298040; [4.8.8] = A022144, A234275; [6.6.6] = A008458.
Programs
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Mathematica
Join[{1, 4}, LinearRecurrence[{2, -1}, {16, 24}, 60]] (* Jean-François Alcover, Jan 08 2019 *)
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PARI
Vec(-(4*x^3-9*x^2-2*x-1)/(x-1)^2 + O(x^100)) \\ Colin Barker, Jul 10 2015
Formula
a(n) = A022144(n), n>1. - R. J. Mathar, Jan 11 2014
From Colin Barker, Jul 10 2015: (Start)
a(n) = 8*n, n>1.
a(n) = 2*a(n-1) - a(n-2) for n>3.
G.f.: -(4*x^3-9*x^2-2*x-1) / (x-1)^2.
(End)
Comments