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A380282 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells having k regions between the polyominoes and their bounding boxes, n >= 1, k >= 0.

%I A380282 #47 Feb 14 2025 23:16:45
%S A380282 1,1,1,1,2,1,2,1,4,5,1,1,2,6,18,7,2,1,13,50,34,10,2,25,144,146,50,2,2,
%T A380282 48,402,574,240,18,1,2,97,1168,2142,1120,122,4,1,201,3368,7813,4920,
%U A380282 738,32,3,420,9977,28010,20946,4015,225,4,1,904,29856,99610,86400,20221,1561,37,1
%N A380282 Irregular triangle read by rows: T(n,k) is the number of free polyominoes with n cells having k regions between the polyominoes and their bounding boxes, n >= 1, k >= 0.
%C A380282 The regions include any holes in the polyominoes.
%H A380282 John Mason, <a href="/A380282/b380282.txt">Table of n, a(n) for n = 1..116</a> (first 18 rows (16 rows from _Pontus von Brömssen_))
%H A380282 <a href="/index/Pol#polyominoes">Index entries for sequences related to polyominoes</a>.
%F A380282 T(n,0) = A038548(n). - _Pontus von Brömssen_, Jan 24 2025
%e A380282 Triangle begins:
%e A380282    1;
%e A380282    1;
%e A380282    1,   1;
%e A380282    2,   1,     2;
%e A380282    1,   4,     5,     1,     1;
%e A380282    2,   6,    18,     7,     2;
%e A380282    1,  13,    50,    34,    10;
%e A380282    2,  25,   144,   146,    50,     2;
%e A380282    2,  48,   402,   574,   240,    18,    1;
%e A380282    2,  97,  1168,  2142,  1120,   122,    4;
%e A380282    1, 201,  3368,  7813,  4920,   738,   32;
%e A380282    3, 420,  9977, 28010, 20946,  4015,  225,  4;
%e A380282    1, 904, 29856, 99610, 86400, 20221, 1561, 37,  1;
%e A380282    ...
%e A380282 Illustration for n = 5:
%e A380282 The free polyominoes with five cells are also called free pentominoes.
%e A380282 For k = 0 there is only one free pentomino having no regions into its bounding box as shown below, so T(5,0) = 1.
%e A380282    _
%e A380282   |_|
%e A380282   |_|
%e A380282   |_|
%e A380282   |_|
%e A380282   |_|
%e A380282 .
%e A380282 For k = 1 there are four free pentominoes having only one region into their bounding boxes as shown below, so T(5,1) = 4.
%e A380282    _
%e A380282   |_|      _ _     _ _      _
%e A380282   |_|     |_|_|   |_|_|    |_|
%e A380282   |_|_    |_|_|   |_|_     |_|_ _
%e A380282   |_|_|   |_|     |_|_|    |_|_|_|
%e A380282 .
%e A380282 For k = 2 there are five free pentominoes having two regions into their bounding boxes as shown below, so T(5,2) = 5.
%e A380282      _       _
%e A380282    _|_|    _|_|    _ _ _    _        _ _
%e A380282   |_|_|   |_|_|   |_|_|_|  |_|_     |_|_|
%e A380282   |_|       |_|     |_|    |_|_|_     |_|_
%e A380282   |_|       |_|     |_|      |_|_|    |_|_|
%e A380282 .
%e A380282 For k = 3 there is only one free pentomino having three regions into its bounding box as shown below, so T(5,3) = 1.
%e A380282      _ _
%e A380282    _|_|_|
%e A380282   |_|_|
%e A380282     |_|
%e A380282 .
%e A380282 For k = 4 there is only one free pentomino having four regions into its bounding box as shown below, so T(5,4) = 1.
%e A380282      _
%e A380282    _|_|_
%e A380282   |_|_|_|
%e A380282     |_|
%e A380282 .
%e A380282 Therefore the 5th row of the triangle is [1, 4, 5, 1, 1] and the row sums is A000105(5) = 12.
%e A380282 .
%Y A380282 Row sums give A000105.
%Y A380282 Row lengths give A380286.
%Y A380282 Cf. A379623, A379627, A379628, A379637, A380283, A380284, A380285.
%Y A380282 Cf. A038548.
%K A380282 nonn,tabf
%O A380282 1,5
%A A380282 _Omar E. Pol_, Jan 18 2025
%E A380282 Terms a(23) and beyond from _Pontus von Brömssen_, Jan 24 2025