A379625
Triangle read by rows: T(n,k) is the number of free polyominoes with n cells whose difference between length and width is k, n >= 1, k >= 0.
Original entry on oeis.org
1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 6, 2, 3, 0, 1, 7, 16, 6, 5, 0, 1, 25, 39, 27, 11, 5, 0, 1, 80, 120, 97, 45, 19, 7, 0, 1, 255, 425, 307, 191, 71, 28, 7, 0, 1, 795, 1565, 1077, 706, 347, 115, 40, 9, 0, 1, 2919, 5217, 4170, 2505, 1454, 574, 171, 53, 9, 0, 1, 10378, 18511, 15164, 10069, 5481, 2740, 919, 257, 69, 11, 0, 1
Offset: 1
Triangle begins:
1;
0, 1;
1, 0, 1;
1, 3, 0, 1;
6, 2, 3, 0, 1;
7, 16, 6, 5, 0, 1;
25, 39, 27, 11, 5, 0, 1;
80, 120, 97, 45, 19, 7, 0, 1;
255, 425, 307, 191, 71, 28, 7, 0, 1;
795, 1565, 1077, 706, 347, 115, 40, 9, 0, 1;
2919, 5217, 4170, 2505, 1454, 574, 171, 53, 9, 0, 1;
10378, 18511, 15164, 10069, 5481, 2740, 919, 257, 69, 11, 0, 1;
...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 0 there are six free pentominoes with length 3 and width 3 as shown below, thus the difference between length and width is 3 - 3 = 0, so T(5,0) = 6.
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For k = 1 there are two free pentominoes with length 3 and width 2 as shown below, thus the difference between length and width is 3 - 2 = 1, so T(5,1) = 2.
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For k = 2 there are three free pentominoes with length 4 and width 2 as shown below, thus the difference between length and width is 4 - 2 = 2, so T(5,2) = 3.
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For k = 3 there are no free pentominoes whose difference between length and width is 3, so T(5,3) = 0.
For k = 4 there is only one free pentomino with length 5 and width 1 as shown below, thus the difference between length and width is 5 - 1 = 4, so T(5,4) = 1.
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Therefore the 5th row of the triangle is [6, 2, 3, 0, 1] and the row sum is A000105(5) = 12.
Note that for n = 6 and k = 1 there are 15 free polyominoes with length 4 and width 3 thus the difference between length and width is 4 - 3 = 1. Also there is a free polyomino with length 3 and width 2 thus the difference between length and width is 3 - 2 = 1, so T(6,1) = 15 + 1 = 16.
.
Row sums except the column 1 give
A259087.
Cf.
A109613,
A379623,
A379624,
A379626,
A379627,
A379628,
A379629,
A379637,
A379638,
A380283,
A380284.
A380284
Triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and length k, and their bounding boxes, n >= 1, k >= 1.
Original entry on oeis.org
0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 16, 5, 0, 0, 0, 14, 48, 9, 0, 0, 0, 12, 145, 89, 9, 0, 0, 0, 3, 354, 453, 138, 13, 0, 0, 0, 0, 608, 1930, 876, 203, 13, 0, 0, 0, 0, 804, 6348, 4930, 1598, 276, 17, 0, 0, 0, 0, 721, 17509, 22575, 10197, 2554, 365, 17, 0, 0, 0, 0, 454, 40067, 91007, 54691, 18984, 3955, 462, 21, 0
Offset: 1
Triangle begins:
0;
0, 0;
0, 1, 0;
0, 0, 5, 0;
0, 0, 16, 5, 0;
0, 0, 14, 48, 9, 0;
0, 0, 12, 145, 89, 9, 0;
...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there are no free pentominoes of length 1, hence there are no regions, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, hence there are no regions, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 1, 3, 2, 1, 2, 4, 2, hence the total number of regions is 1 + 1 + 3 + 2 + 1 + 2 + 4 + 2 = 16, so T(5,3) = 16.
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For k = 4 there are three free pentominoes of length 4 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 2, 2, hence the total number of regions is 1 + 2 + 2 = 5, so T(5,4) = 5.
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For k = 5 there is only one free pentomino of length 5 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,5) = 0.
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Therefore the 5th row of the triangle is [0, 0, 16, 5, 0].
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Column 1 and leading diagonal give
A000004.
A379626
Sum of the widths of the free polyominoes with n cells.
Original entry on oeis.org
1, 1, 3, 9, 29, 91, 322, 1206, 4600, 17931, 70577, 279652, 1110758, 4424120, 17647314, 70484576, 281750598, 1127181327
Offset: 1
For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
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There is only one free tetromino of width 1 and there are four free tetrominoes of width 2, hence the sum of the widths is 1 + 2 + 2 + 2 + 2 = 9, so a(4) = 9.
A379638
Triangle read by rows: T(n,k) is the sum of the lengths of the free polyominoes with n cells and length k, n >= 1, k >= 1.
Original entry on oeis.org
1, 0, 2, 0, 2, 3, 0, 2, 9, 4, 0, 0, 24, 12, 5, 0, 0, 24, 84, 25, 6, 0, 0, 21, 236, 180, 30, 7, 0, 0, 9, 548, 835, 324, 49, 8, 0, 0, 3, 892, 3345, 1842, 539, 56, 9, 0, 0, 0, 1148, 10445, 9762, 3773, 824, 81, 10, 0, 0, 0, 1020, 27360, 42756, 22659, 6712, 1206, 90, 11, 0, 0, 0, 676, 59595, 165024, 116942, 46808, 11439, 1680, 121, 12
Offset: 1
Triangle begins:
1;
0, 2;
0, 2, 3;
0, 2, 9, 4;
0, 0, 24, 12, 5;
0, 0, 24, 84, 25, 6;
0, 0, 21, 236, 180, 30, 7;
0, 0, 9, 548, 835, 324, 49, 8;
0, 0, 3, 892, 3345, 1842, 539, 56, 9;
0, 0, 0, 1148, 10445, 9762, 3773, 824, 81, 10;
0, 0, 0, 1020, 27360, 42756, 22659, 6712, 1206, 90, 11;
0, 0, 0, 676, 59595, 165024, 116942, 46808, 11439, 1680, 121, 12;
...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there are no free pentominoes of length 1, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, hence the sum of the lengths is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 8*3 = 24, so (5,3) = 24.
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For k = 4 there are three free pentominoes of length 4 as shown below, hence the sum of the lengths is 4 + 4 + 4 = 3*4 = 12, so T(5,4) = 12.
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For k = 5 there is only one free pentomino of length 5 as shown below, so T(5,5) = 5.
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Therefore the 5th row of the triangle is [0, 0, 24, 12, 5].
A380285
Total number of regions between the free polyominoes with n cells and their bounding boxes.
Original entry on oeis.org
0, 0, 1, 5, 21, 71, 255, 961, 3630, 13973, 53938, 209641, 815784, 3183642, 12439291, 48686549, 190787588, 748645732
Offset: 1
Illustration for n = 4:
The free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
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The bounding boxes are respectively as shown below:
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From left to right the number of regions between the free tetrominoes and their bounding boxes are respectively 0, 1, 2, 2, 0. Hence the total number of regions is 0 + 1 + 2 + 2 + 0 = 5, so a(4) = 5.
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Cf.
A379628 (total area of the regions).
Cf.
A000105,
A057766,
A379623,
A379624,
A379625,
A379626,
A379627,
A379629,
A379637,
A379638,
A380282.
Showing 1-5 of 5 results.
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