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A059684
Triangle T(n,k) giving number of 4 X k polyominoes with n cells (n >= 4, 1<=k<=n-3).
2
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, 1304, 822, 155, 0, 0, 4, 488, 2847, 3548, 1551, 220, 0, 0, 1, 321, 4441, 10323, 8239, 2680, 313, 0, 0, 0, 122, 5008, 21995, 29442, 16821, 4327, 415, 0, 0, 0, 35, 4168, 36035, 79155, 71742, 31576
OFFSET
4,3
COMMENTS
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
LINKS
R. C. Read, Contributions to the cell growth problem, Canad. J. Math., 14 (1962), 1-20.
EXAMPLE
Triangle starts:
1;
0,3;
0,6,15;
0,2,39, 30;
0,1,59,148, 61;
0,0,42,349, 383, 97;
0,0,21,519,1304, 822, 155;
0,0, 4,488,2847, 3548, 1551, 220;
0,0, 1,321,4441,10323, 8239, 2680, 313;
0,0, 0,122,5008,21995,29442,16821, 4327,415;
0,0, 0, 35,4168,36035,79155,71742,31576,...
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
CROSSREFS
Cf. A059680 (flipped or rotated considered distinct).
Sequence in context: A057381 A144091 A019145 * A270509 A083350 A002043
KEYWORD
nonn,easy,nice,tabl
AUTHOR
N. J. A. Sloane, Feb 05 2001
EXTENSIONS
Changed 518 to 519 (correcting Read...) and added values for n>=11 cells. R. J. Mathar, May 12 2019
STATUS
approved