A128101 Triangle read by rows: T(n,k) is the number of ways to tile a 4 X n rectangle with k pieces of 2 X 2 tiles and 4(n-k) pieces of 1 X 1 tiles (0<=k<=2*floor(n/2)).

1, 1, 1, 3, 1, 1, 6, 4, 1, 9, 16, 8, 1, 1, 12, 37, 34, 9, 1, 15, 67, 105, 65, 15, 1, 1, 18, 106, 248, 250, 108, 16, 1, 21, 154, 490, 726, 522, 176, 24, 1, 1, 24, 211, 858, 1736, 1824, 994, 260, 25, 1, 27, 277, 1379, 3604, 5148, 4090, 1770, 385, 35, 1, 1, 30, 352, 2080

Offset: 0

Author

Emeric Deutsch, Feb 19 2007

Keywords

• nonn
• tabf

Comments

Row 2n has 2n+1 terms; row 2n+1 has 2n+1 terms.

Examples

			Triangle starts:
1;
1;
1,3,1;
1,6,4;
1,9,16,8,1;
1,12,37,34,9;
1,15,67,105,65,15,1;
		

References

• S. Heubach, Tiling an m X n area with squares of size up to k X k (m <= 5), Congressus Numerantium 140 (1999), pp. 43-64.

Links

• R. J. Mathar, Tiling n x m rectangles with 1 x 1 and s x s squares arXiv:1609.03964 (2016)

Crossrefs

Cf. A054854 (row sums), A008585, A080855, A128102.

Programs

Maple

G:=(1-t*z)/(1-z-t*z-2*t*z^2-t^2*z^2+t^2*z^3+t^3*z^3): Gser:=simplify(series(G,z=0,15)): for n from 0 to 12 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 12 do seq(coeff(P[n],t,j),j=0..2*floor(n/2)) od; # yields sequence in triangular form

Mathematica

CoefficientList[#, t]& /@ CoefficientList[(1 - t z)/(1 - z - t z - 2 t z^2 - t^2 z^2 + t^2 z^3 + t^3 z^3) + O[z]^12, z]  // Flatten (* Jean-François Alcover, Aug 07 2018 *)

Formula

G.f.=(1-tz)/(1-z-tz-2tz^2-t^2*z^2+t^2*z^3+t^3*z^3).

Sum (T(n,k), k=0..2*floor(n/2) ) = A054854(n).

T(n,1)=3(n-1)=A008585(n-1).

T(n,2)=A080855(n-2).

Sum(k*T(n,k), k=0..2*floor(n/2)) = A128102(n).

T(n,3) = (n-3)*(9*n^2-63*n+124)/2, n>=3. - R. J. Mathar, Aug 23 2016

T(n,4) = (3*n-13)*(9*n^3-123*n^2+602*n-1024)/8, n>=4. - R. J. Mathar, Aug 23 2016