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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.
5
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
1,3
COMMENTS
The length here is the longer of the two dimensions.
FORMULA
T(n,k) = k*A379624(n,k).
EXAMPLE
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].
CROSSREFS
Row sums give A379629.
Sequence in context: A238156 A281260 A390159 * A182406 A160706 A087509
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Jan 16 2025
STATUS
approved