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Search: id:A123521
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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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+0
1
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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
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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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).
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FORMULA
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G.f.=G=(1-tz)/[(1+tz)(1-z-2tz+t^2*z^2)]. G=1/(1-g), where g=z+t^2*z^2+2tz^2/(1-tz) 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]+tP[n-2])-t^3*P[n-3].
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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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CROSSREFS
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Cf. A007598.
Sequence in context: A034372 A078121 A119732 this_sequence A123246 A122518 A129704
Adjacent sequences: A123518 A123519 A123520 this_sequence 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 (deutsch(AT)duke.poly.edu), Oct 16 2006
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