%I #15 Aug 07 2018 09:37:38
%S 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,
%T 248,250,108,16,1,21,154,490,726,522,176,24,1,1,24,211,858,1736,1824,
%U 994,260,25,1,27,277,1379,3604,5148,4090,1770,385,35,1,1,30,352,2080
%N 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)).
%C Row 2n has 2n+1 terms; row 2n+1 has 2n+1 terms.
%D 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.
%H R. J. Mathar, <a href="http://arxiv.org/abs/1609.03964">Tiling n x m rectangles with 1 x 1 and s x s squares</a> arXiv:1609.03964 (2016)
%F G.f.=(1-tz)/(1-z-tz-2tz^2-t^2*z^2+t^2*z^3+t^3*z^3).
%F Sum (T(n,k), k=0..2*floor(n/2) ) = A054854(n).
%F T(n,1)=3(n-1)=A008585(n-1).
%F T(n,2)=A080855(n-2).
%F Sum(k*T(n,k), k=0..2*floor(n/2)) = A128102(n).
%F T(n,3) = (n-3)*(9*n^2-63*n+124)/2, n>=3. - _R. J. Mathar_, Aug 23 2016
%F T(n,4) = (3*n-13)*(9*n^3-123*n^2+602*n-1024)/8, n>=4. - _R. J. Mathar_, Aug 23 2016
%e Triangle starts:
%e 1;
%e 1;
%e 1,3,1;
%e 1,6,4;
%e 1,9,16,8,1;
%e 1,12,37,34,9;
%e 1,15,67,105,65,15,1;
%p 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
%t 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 *)
%Y Cf. A054854 (row sums), A008585, A080855, A128102.
%K nonn,tabf
%O 0,4
%A _Emeric Deutsch_, Feb 19 2007