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%I #19 Dec 24 2022 22:25:05
%S 1,0,0,1,0,4,0,0,1,2,0,0,8,0,0,1,0,0,20,0,0,12,0,0,1,0,16,0,0,54,0,0,
%T 16,0,0,1,4,0,0,112,0,0,104,0,0,20,0,0,1,0,0,108,0,0,352,0,0,170,0,0,
%U 24,0,0,1,0,48,0,0,664,0,0,800,0,0,252,0,0,28,0,0,1,8,0,0,704,0,0,2280,0,0,1520,0,0,350,0,0,32,0,0,1,0,0,416,0,0,4064,0,0,5820,0,0,2576,0,0,464,0,0,36,0,0,1
%N Triangle read by rows: T(n,k) is the number of tilings of a 2 X n board with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares) that have k 1 X 1 tiles.
%C Row n has 2n+1 entries.
%C Sum of entries in row n = A127864(n).
%C Sum_{k>=0} k*T(n,k) = A127865(n).
%H P. Chinn, R. Grimaldi and S. Heubach, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Heubach/heubach40.html">Tiling with L's and Squares</a>, Journal of Integer Sequences, Vol. 10 (2007), Article 07.2.8
%F G.f.: 1/(1-t^2*z - 4*t*z^2 - 2*z^3).
%F The trivariate g.f. with z marking length, t marking 1 X 1 tiles, and s marking L-shaped tiles is 1/(1-t^2*z-4*t*s*z^2-2*s^2*z^3).
%F From _Robert Israel_, Aug 15 2014: (Start)
%F T(n+3,k+2) = T(n+2,k) + 4*T(n+1,k+1) + 2*T(n,k+2).
%F T(n,0) = 2^(n/3) if n == 0 (mod 3), T(n,0) = 0 otherwise.
%F T(n,1) = (n+1)*2^((n+4)/3)/3 if n == 2 (mod 3), T(n,1) = 0 otherwise.
%F (End)
%e T(2,1)=4 because we can place the 1 X 1 tile in any corner of the 2 X 2 board.
%e Triangle starts:
%e 1;
%e 0, 0, 1;
%e 0, 4, 0, 0, 1;
%e 2, 0, 0, 8, 0, 0, 1;
%e 0, 0, 20, 0, 0, 12, 0, 0, 1;
%p G := 1/(1-t^2*z-4*t*z^2-2*z^3): Gser := simplify(series(G, z = 0, 15)): for j from 0 to 13 do P[j] := sort(coeff(Gser, z, j)) end do: for j from 0 to 13 do seq(coeff(P[j], t, i), i = 0 .. 2*j) end do; # yields sequence in triangular form
%Y Cf. A127864, A127865, A245966.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Aug 15 2014