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 A092440 a(n) = 2^(2n+1) - 2^(n+1) + 1. 28
 1, 5, 25, 113, 481, 1985, 8065, 32513, 130561, 523265, 2095105, 8384513, 33546241, 134201345, 536838145, 2147418113, 8589803521, 34359476225, 137438429185, 549754765313, 2199021158401, 8796088827905, 35184363700225 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Arises from enumeration of domino tilings of Aztec Pillow-like regions. 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(10=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 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 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 REFERENCES 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). S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170. LINKS Robert Price, Table of n, a(n) for n = 0..500 J. Propp, Publications and Preprints J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics Eric Weisstein's World of Mathematics, Elementary Cellular Automaton S. Wolfram, A New Kind of Science Index entries for linear recurrences with constant coefficients, signature (7,-14,8). FORMULA a(n) = 2^(2n+1)-2^(n+1)+1. From Colin Barker, Nov 22 2012: (Start) a(n) = 7*a(n-1)-14*a(n-2)+8*a(n-3). G.f.: -(4*x^2-2*x+1)/((x-1)*(2*x-1)*(4*x-1)). (End) MATHEMATICA Table[2^(2n + 1) - 2^(n + 1) + 1, {n, 0, 200}] (* Robert Price, May 04 2016 *) PROG (PARI) a(n)=2^(2*n+1)-2^(n+1)+1 \\ Charles R Greathouse IV, Sep 24 2015 CROSSREFS Cf. A092437, A092438, A092439, A092441, A092442, A092443. Sequence in context: A290920 A267228 A183926 * A196985 A124534 A261383 Adjacent sequences:  A092437 A092438 A092439 * A092441 A092442 A092443 KEYWORD easy,nonn AUTHOR Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004 STATUS approved

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Last modified August 6 00:46 EDT 2021. Contains 346493 sequences. (Running on oeis4.)