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A123521
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Triangle read by rows: T(n,k)=number of tilings of a 2 X n grid with k pieces of 1 X 2 tiles (in horizontal position) and 2n-2k pieces of 1 X 1 tiles (0<=k<=n).
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3
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1, 1, 1, 2, 1, 1, 4, 4, 1, 6, 11, 6, 1, 1, 8, 22, 24, 9, 1, 10, 37, 62, 46, 12, 1, 1, 12, 56, 128, 148, 80, 16, 1, 14, 79, 230, 367, 314, 130, 20, 1, 1, 16, 106, 376, 771, 920, 610, 200, 25, 1, 18, 137, 574, 1444, 2232, 2083, 1106, 295, 30, 1, 1, 20, 172, 832, 2486, 4744, 5776
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OFFSET
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0,4
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COMMENTS
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Also the triangle of the coefficients of the squares of the Fibonacci polynomials. Row n has 1+2*floor(n/2) terms. Sum of terms in row n = [fibonacci(n+1)]^2 (A007598).
From Michael A. Allen, Jun 24 2020: (Start)
T(n,k) is the number of tilings of an n-board (a board with dimensions n X 1) using k (1/2,1/2)-fence tiles and 2(n-k) half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal). A (1/2,1/2)-fence is a tile composed of two 1/2 X 1 pieces separated by a gap of width 1/2.
T(n,k) is the (n,(n-k))-th entry of the (1/(1-x^2),x/(1-x)^2) Riordan array.
(-1)^(n+k)*T(n,k) is the (n,(n-k))-th entry of the (1/(1-x^2),x/(1+x)^2) Riordan array (A158454). (End)
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REFERENCES
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Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
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LINKS
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Table of n, a(n) for n=0..67.
Kenneth Edwards and Michael A. Allen, New Combinatorial Interpretations of the Fibonacci Numbers Squared, Golden Rectangle Numbers, and Jacobsthal Numbers Using Two Types of Tile, arXiv:2009.04649 [math.CO], 2020.
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FORMULA
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G.f.: G=(1-t*z)/[(1+t*z)*(1-z-2*t*z+t^2*z^2)]. G=1/(1-g), where g=z+t^2*z^2+2*t*z^2/(1-t*z) is the g.f. of the indecomposable tilings, i.e., of those that cannot be split vertically into smaller tilings. The row generating polynomials are P[n]=(F[n])^2, where F[n] are the Fibonacci polynomials defined by F[0]=F[1]=1, F[n]=F[n-1]+tF[n-2] for n>=2. They satisfy the recurrence relation P[n]=(1+t)(P[n-1]+t*P[n-2])-t^3*P[n-3].
T(n,k) = T(n-2,k-2)+binomial(2*n-k-1,2*n-2*k-1). - Michael A. Allen, Jun 24 2020
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EXAMPLE
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T(3,1)=4 because the 1 X 2 tile can be placed in any of the four corners of the 2 X 3 grid.
Triangle starts:
1;
1;
1,2,1;
1,4,4;
1,6,11,6,1;
1,8,22,24,9;
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MAPLE
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G:=(1-t*z)/(1+t*z)/(1-z-2*t*z+t^2*z^2): Gser:=simplify(series(G, z=0, 14)): for n from 0 to 11 do P[n]:=sort(coeff(Gser, z, n)) od: for n from 0 to 11 do seq(coeff(P[n], t, k), k=0..2*floor(n/2)) od; # yields sequence in triangular form
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MATHEMATICA
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Block[{T}, T[0, 0] = T[1, 0] = 1; T[n_, k_] := Which[k == 0, 1, k == 1, 2 (n - 1), True, T[n - 2, k - 2] + Binomial[2 n - k - 1, 2 n - 2 k - 1]]; Table[T[n, k], {n, 0, 11}, {k, 0, 2 Floor[n/2]}]] // Flatten (* Michael De Vlieger, Jun 24 2020 *)
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CROSSREFS
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Cf. A007598, A158454.
Sequence in context: A306614 A264336 A322038 * A322115 A294217 A123246
Adjacent sequences: A123518 A123519 A123520 * A123522 A123523 A123524
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch, Oct 16 2006
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STATUS
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approved
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