

A123521


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 2n2k pieces of 1 X 1 tiles (0<=k<=n).


0



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

0,4


COMMENTS

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).


REFERENCES

Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169177.


LINKS

Table of n, a(n) for n=0..67.


FORMULA

G.f.=G=(1tz)/[(1+tz)(1z2tz+t^2*z^2)]. G=1/(1g), where g=z+t^2*z^2+2tz^2/(1tz) 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[n1]+tF[n2] for n>=2. They satisfy the recurrence relation P[n]=(1+t)(P[n1]+tP[n2])t^3*P[n3].


EXAMPLE

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;


MAPLE

G:=(1t*z)/(1+t*z)/(1z2*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


CROSSREFS

Cf. A007598.
Sequence in context: A306614 A264336 A322038 * A322115 A294217 A123246
Adjacent sequences: A123518 A123519 A123520 * A123522 A123523 A123524


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Oct 16 2006


STATUS

approved



