%I #22 May 12 2019 10:17:07
%S 1,0,3,0,6,15,0,2,39,30,0,1,59,148,61,0,0,42,349,383,97,0,0,21,519,
%T 1304,822,155,0,0,4,488,2847,3548,1551,220,0,0,1,321,4441,10323,8239,
%U 2680,313,0,0,0,122,5008,21995,29442,16821,4327,415,0,0,0,35,4168,36035,79155,71742,31576
%N Triangle T(n,k) giving number of 4 X k polyominoes with n cells (n >= 4, 1<=k<=n-3).
%C Note that for k=4 (polyominoes with square bounding rectangle) these are not the free polyominoes, because Read does not apply the full symmetry group of order 8 to reduce the fixed polyominoes for d_q(n), but only the symmetry group of order 4 (excluding the 90 deg rotations). The free polyominoes with square bounding rectangles are his z_4(n) instead. - _R. J. Mathar_, May 12 2019
%H R. C. Read, <a href="https://doi.org/10.4153/CJM-1962-001-2">Contributions to the cell growth problem</a>, Canad. J. Math., 14 (1962), 1-20.
%e Triangle starts:
%e 1;
%e 0,3;
%e 0,6,15;
%e 0,2,39, 30;
%e 0,1,59,148, 61;
%e 0,0,42,349, 383, 97;
%e 0,0,21,519,1304, 822, 155;
%e 0,0, 4,488,2847, 3548, 1551, 220;
%e 0,0, 1,321,4441,10323, 8239, 2680, 313;
%e 0,0, 0,122,5008,21995,29442,16821, 4327,415;
%e 0,0, 0, 35,4168,36035,79155,71742,31576,...
%e There are T(5,2)=3 out of 12 pentominoes that fill the 4X2 shape: the L, N and Y. The F, T, V, W, X, and Z require both dimensions >= 3; the P and U would fit but not touch all sides; the I requires one dimension of 5. - _R. J. Mathar_, May 08 2019
%Y Cf. A059680 (flipped or rotated considered distinct).
%K nonn,easy,nice,tabl
%O 4,3
%A _N. J. A. Sloane_, Feb 05 2001
%E Changed 518 to 519 (correcting Read...) and added values for n>=11 cells. _R. J. Mathar_, May 12 2019