A379627
Total area of the bounding boxes of the free polyominoes with n cells.
Original entry on oeis.org
1, 2, 7, 26, 95, 353, 1387, 5699, 23592, 99020, 415796, 1748722, 7336978, 30741661, 128510698, 536268877, 2233642396, 9288889882
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:
_
| | _ _ _ _ _ _
| | | | | | | | _ _
| | | | | | | | | |
|_| |_ _| |_ _| |_ _| |_ _|
.
4 X 1 3 X 2 3 X 2 3 X 2 2 X 2
.
The total area of the bounding boxes is 4 + 6 + 6 + 6 + 4 = 26, so a(4) = 26.
.
A379623
Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).
Original entry on oeis.org
1, 1, 1, 1, 1, 4, 1, 5, 6, 1, 12, 22, 1, 18, 71, 18, 1, 37, 193, 138, 1, 60, 490, 661, 73, 1, 117, 1221, 2547, 769, 1, 200, 3011, 8417, 5189, 255, 1, 379, 7393, 26164, 25920, 3743, 1, 669, 18025, 78074, 108834, 32038, 950, 1, 1250, 43847, 229881, 408217, 201956, 16819
Offset: 1
Triangle begins:
1;
1;
1, 1;
1, 4;
1, 5, 6;
1, 12, 22;
1, 18, 71, 18;
1, 37, 193, 138;
1, 60, 490, 661, 73;
1, 117, 1221, 2547, 769;
1, 200, 3011, 8417, 5189, 255;
1, 379, 7393, 26164, 25920, 3743;
1, 669, 18025, 78074, 108834, 32038, 950;
1, 1250, 43847, 229881, 408217, 201956, 16819;
...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there is only one free pentomino of width 1 as shown below, so T(5,1) = 1.
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.
For k = 2 there are five free pentominoes of width 2 as shown below, so T(5,2) = 5.
_ _ _
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.
For k = 3 there are six free pentominoes of width 3 as shown below, so T(5,3) = 6.
_ _ _ _ _ _ _ _ _ _
_|_|_| |_|_|_| |_| |_|_ _|_|_ |_|_|
|_|_| |_| |_|_ _ |_|_|_ |_|_|_| |_|_
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.
Therefore the 5th row of the triangle is [1, 5, 6] and the row sum is A000105(5) = 12.
.
For free polyominoes of length k see
A379624.
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.
A379628
Total area between the free polyominoes with n cells and their bounding boxes.
Original entry on oeis.org
0, 0, 1, 6, 35, 143, 631, 2747, 12027, 52470, 227993, 985522, 4235295, 18114067, 77112058, 327000797, 1381807943, 5821692946
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:
_
|_| _ _ _
|_| |_| |_|_ |_|_ _ _
|_| |_|_ |_|_| |_|_| |_|_|
|_| |_|_| |_| |_| |_|_|
.
The bounding boxes are respectively as shown below:
_
| | _ _ _ _ _ _
| | | | | | | | _ _
| | | | | | | | | |
|_| |_ _| |_ _| |_ _| |_ _|
.
4 x 1 3 X 2 3 X 2 3 X 2 2 X 2
.
The total area of the free tetrominoes is 5*4 = 20.
The total area of the bounding boxes is 4 + 6 + 6 + 6 + 4 = 26.
The total area between the bounding boxes and the free tetrominoes is 26 - 20 = 6, so a(4) = 6.
.
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].
.
Column 1 and leading diagonal give
A000004.
A379629
Sum of the lengths of the free polyominoes with n cells.
Original entry on oeis.org
1, 2, 5, 15, 41, 139, 474, 1773, 6686, 26043, 101814, 402297, 1593124, 6332329, 25200575, 100462874, 400908688, 1601541747
Offset: 1
For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
_
_ _ _ |_|
_ _ |_| |_|_ |_|_ |_|
|_|_| |_|_ |_|_| |_|_| |_|
|_|_| |_|_| |_| |_| |_|
.
There are no free tetrominoes of length 1, there is only one free tetromino of length 2, there are three free tetrominoes of length 3 and there is only one free tetromino of length 4, hence the sum of the lengths is 0 + 2 + 3 + 3 + 3 + 4 = 15, so a(4) = 15.
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.
_ _ _ _ _ _ _ _ _ _ _ _ _ _
|_|_| |_|_| _|_|_| |_|_|_| |_| |_|_ _|_|_ |_|_|
|_|_| |_|_ |_|_| |_| |_|_ _ |_|_|_ |_|_|_| |_|_
|_| |_|_| |_| |_| |_|_|_| |_|_| |_| |_|_|
.
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:
_
|_| _ _ _
|_| |_| |_|_ |_|_ _ _
|_| |_|_ |_|_| |_|_| |_|_|
|_| |_|_| |_| |_| |_|_|
.
The bounding boxes are respectively as shown below:
_
| | _ _ _ _ _ _
| | | | | | | | _ _
| | | | | | | | | |
|_| |_ _| |_ _| |_ _| |_ _|
.
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.
.
Cf.
A379628 (total area of the regions).
Cf.
A000105,
A057766,
A379623,
A379624,
A379625,
A379626,
A379627,
A379629,
A379637,
A379638,
A380282.
A381703
Irregular triangle read by rows in which every row of length A071764(n) lists A(n,w,h) = the number of free polyominoes of size n, width w and height h (for w <= h, and all possible w,h pairs).
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 6, 1, 1, 6, 5, 7, 15, 1, 2, 11, 5, 7, 39, 25, 18, 1, 1, 10, 19, 7, 3, 59, 96, 35, 77, 61, 1, 3, 22, 28, 7, 1, 42, 210, 188, 49, 181, 383, 97, 73, 1, 1, 15, 52, 40, 9, 21, 255, 550, 332, 63, 266, 1304, 822, 155, 529, 240, 1, 3, 45, 90, 53, 9, 4, 212, 954, 1231, 529, 81, 251, 2847, 3548, 1551, 220, 2413, 2366, 410, 255
Offset: 1
Triangle begins:
n
1: 1
2: 1
3: 1 1
4: 1 1 3
5: 1 2 3 6
6: 1 1 6 5 7 15
7: 1 2 11 5 7 39 25 18
8: 1 1 10 19 7 3 59 96 35 77 61
9: 1 3 22 28 7 1 42 210 188 49 181 383 97 73
10: 1 1 15 52 40 9 21 255 550 332 63 266 1304 822 155 529 240
...
Any row contains an irregular array that shows the number of polyominoes having width w and height h. E.g., row 6 contains the array:
h/w 1 2 3
1
2
3 1 7
4 6 15
5 5
6 1
.
There are 5 polyominoes of size 6 with width 2 and height 5, so A(6,2,5)=5:
.
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