A359025 Number of inequivalent tilings of a 9 X n rectangle using integer-sided square tiles.
1, 1, 30, 163, 2403, 32097, 459957, 6542578, 93604244
Offset: 0
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a(4) = 6, because there are 6 tilings of a 4 X 4 rectangle using right trominoes and 2 X 2 tiles: .___.___. .___.___. .___.___. .___.___. .___.___. .___.___. | . | . | | ._|_. | | ._| . | | ._|_. | | ._|_. | | . |_. | |___|___| |_| . |_| |_| |___| |_| ._|_| |_|_. |_| |___| |_| | . | . | | |___| | | |___| | | |_| . | | . |_| | | |___| | |___|___| |___|___| |___|___| |___|___| |___|___| |___|___|
a:= n-> (Matrix([[4, 1, 0, 1, 0$5]]). Matrix(9, (i,j)-> if i=j-1 then 1 elif j=1 then [1, 1, 9, 1, -3, -22, -16, 0, -4][i] else 0 fi)^n)[1,4]: seq(a(n), n=0..30);
Series[ (-6*x^3 - x + 1) / (4*x^9 + 16*x^7 + 22*x^6 + 3*x^5 - x^4 - 9*x^3 - x^2 - x + 1), {x, 0, 31}] // CoefficientList[#, x] & (* Jean-François Alcover, Jun 18 2013, after Alois P. Heinz *)
LinearRecurrence[{1,1,9,1,-3,-22,-16,0,-4},{1,0,1,4,6,16,37,92,245},40] (* Harvey P. Dale, Nov 09 2024 *)
a(2) = 4, because there are 4 tilings of a 5 X 2 rectangle using right trominoes and 2 X 2 tiles: .___. .___. .___. .___. | . | | . | | ._| |_. | |___| |___| |_| | | |_| | ._| |_. | |___| |___| |_| | | |_| | . | | . | |___| |___| |___| |___|
a:= n-> (Matrix(14, (i, j)-> `if`(i=j-1, 1, `if`(i=14, [-80, -160, 308, -88, -2, 396, -453, -10, 190, -12, -57, 2, 13, 0][j], 0)))^n. <<0, 1/4, 0, 1, 0, 4, 0, 16, 0, 136, 0, 1128, 384, 8120>>)[4,1]: seq(a(n), n=0..30);
a[n_] := (MatrixPower[ Table[ If[i == j-1, 1, If[i == 14, {-80, -160, 308, -88, -2, 396, -453, -10, 190, -12, -57, 2, 13, 0}[[j]], 0]], {i, 1, 14}, {j, 1, 14}], n] . {0, 1/4, 0, 1, 0, 4, 0, 16, 0, 136, 0, 1128, 384, 8120})[[4]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Dec 05 2013, translated from Alois P. Heinz's Maple program *)
concat(0, Vec(x*(4 + 9*x - 10*x^2 - 8*x^3 + 6*x^4 + x^5) / ((1 + x)^2*(1 - 3*x + x^4)^2) + O(x^30))) \\ Colin Barker, Jun 07 2020