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A380283 Irregular triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and width k, and their bounding boxes, n >= 1, 1 <= k <= ceiling(n/2).

%I A380283 #36 Mar 11 2025 16:05:59
%S A380283 0,0,0,1,0,5,0,7,14,0,19,52,0,34,173,48,0,74,503,384,0,134,1368,1918,
%T A380283 210,0,282,3642,7742,2307,0,524,9552,26843,16267,752,0,1064,24889,
%U A380283 87343,84789,11556,0,2017,64200,272599,370799,103336,2833,0,4009,164826,838160,1445347,678863,52437
%N A380283 Irregular triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and width k, and their bounding boxes, n >= 1, 1 <= k <= ceiling(n/2).
%C A380283 The regions include any holes in the polyominoes.
%H A380283 John Mason, <a href="/A380283/b380283.txt">Table of n, a(n) for n = 1..90</a>
%e A380283 Triangle begins:
%e A380283   0;
%e A380283   0;
%e A380283   0,     1;
%e A380283   0,     5;
%e A380283   0,     7,      14;
%e A380283   0,    19,      52;
%e A380283   0,    34,     173,      48;
%e A380283   0,    74,     503,     384;
%e A380283   0,   134,    1368,    1918,      210;
%e A380283   0,   282,    3642,    7742,     2307;
%e A380283   0,   524,    9552,   26843,    16267,      752;
%e A380283   0,  1064,   24889,   87343,    84789,    11556;
%e A380283   0,  2017,   64200,  272599,   370799,   103336,    2833;
%e A380283   0,  4009,  164826,  838160,  1445347,   678863,   52437;
%e A380283   0,  7663,  420373, 2539843,  5240853,  3659815,  560348,  10396;
%e A380283   0, 15031, 1068181, 7631249, 18171771, 17199831, 4373770, 226716;
%e A380283   ...
%e A380283 Illustration for n = 5:
%e A380283 The free polyominoes with five cells are also called free pentominoes.
%e A380283 For k = 1 there is only one free pentomino of width 1 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,1) = 0.
%e A380283    _
%e A380283   |_|
%e A380283   |_|
%e A380283   |_|
%e A380283   |_|
%e A380283   |_|
%e A380283 .
%e A380283 For k = 2 there are five free pentominoes of width 2 as shown below, and from left to right there are respectively 1, 2, 2, 1, 1 regions between the pentominoes and their bounding boxes, hence the total number of regions is 1 + 2 + 2 + 1 + 1 = 7, so T(5,2) = 7.
%e A380283    _           _         _
%e A380283   |_|        _|_|      _|_|      _ _       _ _
%e A380283   |_|       |_|_|     |_|_|     |_|_|     |_|_|
%e A380283   |_|_      |_|         |_|     |_|_|     |_|_
%e A380283   |_|_|     |_|         |_|     |_|       |_|_|
%e A380283 .
%e A380283 For k = 3 there are six free pentominoes of width 3 as shown below, and from left to right there are respectively 3, 2, 1, 2, 4, 2 regions between the pentominoes and their bounding boxes, hence the total number of regions is 3 + 2 + 1 + 2 + 4 + 2 = 14, so T(5,3) = 14.
%e A380283      _ _     _ _ _     _         _           _       _ _
%e A380283    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
%e A380283   |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
%e A380283     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
%e A380283 .
%e A380283 Therefore the 5th row of the triangle is [0, 7, 14].
%e A380283 .
%Y A380283 Column 1 gives A000004.
%Y A380283 Row lengths give A110654.
%Y A380283 Row sums give A380285.
%Y A380283 Cf. A000105, A379623, A379625, A379626, A379627, A379628, A379637, A380284.
%K A380283 nonn,tabf
%O A380283 1,6
%A A380283 _Omar E. Pol_, Jan 18 2025
%E A380283 More terms from _John Mason_, Feb 14 2025