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A092440 a(n) = 2^(2n+1) - 2^(n+1) + 1.

%I A092440 #44 Feb 16 2025 08:32:52
%S A092440 1,5,25,113,481,1985,8065,32513,130561,523265,2095105,8384513,
%T A092440 33546241,134201345,536838145,2147418113,8589803521,34359476225,
%U A092440 137438429185,549754765313,2199021158401,8796088827905,35184363700225
%N A092440 a(n) = 2^(2n+1) - 2^(n+1) + 1.
%C A092440 Arises from enumeration of domino tilings of Aztec Pillow-like regions.
%C A092440 Each beginning with 1, the subsequences of A046899 are 1; 1,2; 1,3,6; 1,4,10,20 and so forth.  Create triangles with the sides being equal to each of these subsequences; the interior members T(i,j)=T(i-1,j-1) + T(i-1,j).  The sum of all members for each triangle will reproduce the terms of this sequence.  Example using the fourth subsequence 1,4,10,20 will give row(1)=1; row(2)=4,4; row(3)=10,8,10; row(4)=20,18,18,20 giving a sum for all members of 113, the fourth term in the sequence. - _J. M. Bergot_, Oct 17 2012
%C A092440 Also, the number of active (ON, black) cells at stage 2^n-1 of the two-dimensional cellular automaton defined by "Rule 510", based on the 5-celled von Neumann neighborhood. - _Robert Price_, May 04 2016
%C A092440 Let M be some square matrix of rank 2^n, containing the positive real value X everywhere except on the diagonal; let Y be some complex value with phase 3*Pi/4 everywhere else (thus all coefficients on the diagonal). Then, for M to be a unitary matrix, X must be 1/sqrt(a(n)). - _Thomas Baruchel_, Aug 10 2020
%D A092440 J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).
%D A092440 S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.
%H A092440 Robert Price, <a href="/A092440/b092440.txt">Table of n, a(n) for n = 0..500</a>
%H A092440 J. Propp, <a href="http://faculty.uml.edu/jpropp/articles.html">Publications and Preprints</a>
%H A092440 J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), <a href="http://www.msri.org/publications/books/Book38/contents.html">New Perspectives in Algebraic Combinatorics</a>
%H A092440 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>
%H A092440 S. Wolfram, <a href="http://wolframscience.com/">A New Kind of Science</a>
%H A092440 <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%H A092440 <a href="https://oeis.org/wiki/Index_to_2D_5-Neighbor_Cellular_Automata">Index to 2D 5-Neighbor Cellular Automata</a>
%H A092440 <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>
%H A092440 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-14,8).
%F A092440 a(n) = 2^(2n+1) - 2^(n+1) + 1.
%F A092440 From _Colin Barker_, Nov 22 2012: (Start)
%F A092440 a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).
%F A092440 G.f.: -(4*x^2-2*x+1)/((x-1)*(2*x-1)*(4*x-1)). (End)
%F A092440 a(n) = A000225(n)^2 + (A000225(n) + 1)^2. - _César Aguilera_, May 28 2023
%t A092440 Table[2^(2n + 1) - 2^(n + 1) + 1, {n, 0, 200}] (* _Robert Price_, May 04 2016 *)
%o A092440 (PARI) a(n)=2^(2*n+1)-2^(n+1)+1 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y A092440 Cf. A092437, A092438, A092439, A092441, A092442, A092443.
%K A092440 easy,nonn
%O A092440 0,2
%A A092440 Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004