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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.

%I A380284 #37 Feb 14 2025 18:06:58
%S A380284 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,
%T A380284 354,453,138,13,0,0,0,0,608,1930,876,203,13,0,0,0,0,804,6348,4930,
%U A380284 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
%N 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.
%C A380284 The regions include any holes in the polyominoes.
%C A380284 The first 28 terms were calculated by hand.
%H A380284 John Mason, <a href="/A380284/b380284.txt">Table of n, a(n) for n = 1..171</a> (first 18 rows)
%H A380284 <a href="/index/Pol#polyominoes">Index entries for sequences related to polyominoes</a>.
%e A380284 Triangle begins:
%e A380284   0;
%e A380284   0,  0;
%e A380284   0,  1,  0;
%e A380284   0,  0,  5,   0;
%e A380284   0,  0, 16,   5,  0;
%e A380284   0,  0, 14,  48,  9,  0;
%e A380284   0,  0, 12, 145, 89,  9,  0;
%e A380284   ...
%e A380284 Illustration for n = 5:
%e A380284 The free polyominoes with five cells are also called free pentominoes.
%e A380284 For k = 1 there are no free pentominoes of length 1, hence there are no regions, so T(5,1) = 0.
%e A380284 For k = 2 there are no free pentominoes of length 2, hence there are no regions, so T(5,2) = 0.
%e A380284 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.
%e A380284    _ _     _ _       _ _     _ _ _     _         _           _       _ _
%e A380284   |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
%e A380284   |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
%e A380284   |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
%e A380284 .
%e A380284 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.
%e A380284    _         _       _
%e A380284   |_|      _|_|    _|_|
%e A380284   |_|     |_|_|   |_|_|
%e A380284   |_|_    |_|       |_|
%e A380284   |_|_|   |_|       |_|
%e A380284 .
%e A380284 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.
%e A380284    _
%e A380284   |_|
%e A380284   |_|
%e A380284   |_|
%e A380284   |_|
%e A380284   |_|
%e A380284 .
%e A380284 Therefore the 5th row of the triangle is [0, 0, 16, 5, 0].
%e A380284 .
%Y A380284 Column 1 and leading diagonal give A000004.
%Y A380284 Column 2 gives A063524.
%Y A380284 Row sums give A380285.
%Y A380284 Cf. A000105, A379624, A379625, A379627, A379628, A379629, A379638, A380282, A380283.
%K A380284 nonn,tabl
%O A380284 1,9
%A A380284 _Omar E. Pol_, Jan 18 2025
%E A380284 More terms from _John Mason_, Feb 14 2025