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A379637 Irregular triangle read by rows: T(n,k) is the sum of the widths of the free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).

%I A379637 #15 Jan 16 2025 21:24:59
%S A379637 1,1,1,2,1,8,1,10,18,1,24,66,1,36,213,72,1,74,579,552,1,120,1470,2644,
%T A379637 365,1,234,3663,10188,3845,1,400,9033,33668,25945,1530,1,758,22179,
%U A379637 104656,129600,22458,1,1338,54075,312296,544170,192228,6650,1,2500,131541,919524,2041085,1211736,117733
%N A379637 Irregular triangle read by rows: T(n,k) is the sum of the widths of the free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).
%C A379637 The width here is the shorter of the two dimensions.
%H A379637 <a href="/index/Pol#polyominoes">Index entries for sequences related to polyominoes</a>.
%F A379637 T(n,k) = k*A379623(n,k).
%e A379637 Triangle begins:
%e A379637   1;
%e A379637   1;
%e A379637   1,    2;
%e A379637   1,    8;
%e A379637   1,   10,     18;
%e A379637   1,   24,     66;
%e A379637   1,   36,    213,     72;
%e A379637   1,   74,    579,    552;
%e A379637   1,  120,   1470,   2644,     365;
%e A379637   1,  234,   3663,  10188,    3845;
%e A379637   1,  400,   9033,  33668,   25945,    1530;
%e A379637   1,  758,  22179, 104656,  129600,   22458;
%e A379637   1, 1338,  54075, 312296,  544170,  192228,   6650;
%e A379637   1, 2500, 131541, 919524, 2041085, 1211736, 117733;
%e A379637   ...
%e A379637 Illustration for n = 5:
%e A379637 The free polyominoes with five cells are also called free pentominoes.
%e A379637 For k = 1 there is only one free pentomino of width 1 as shown below, so T(5,1) = 1.
%e A379637    _
%e A379637   |_|
%e A379637   |_|
%e A379637   |_|
%e A379637   |_|
%e A379637   |_|
%e A379637 .
%e A379637 For k = 2 there are five free pentominoes of width 2 as shown below, hence the sum of the widths is 2 + 2 + 2 + 2 + 2 = 5*2 = 10, so T(5,2) = 10.
%e A379637    _           _         _
%e A379637   |_|        _|_|      _|_|      _ _       _ _
%e A379637   |_|       |_|_|     |_|_|     |_|_|     |_|_|
%e A379637   |_|_      |_|         |_|     |_|_|     |_|_
%e A379637   |_|_|     |_|         |_|     |_|       |_|_|
%e A379637 .
%e A379637 For k = 3 there are six free pentominoes of width 3 as shown below, hence the sum of the widths is 3 + 3 + 3 + 3 + 3 + 3 = 6*3 = 18, so T(5,3) = 18.
%e A379637      _ _     _ _ _     _         _           _       _ _
%e A379637    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
%e A379637   |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
%e A379637     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
%e A379637 .
%e A379637 Therefore the 5th row of the triangle is [1, 10, 18].
%e A379637 .
%Y A379637 Row lengths give A110654.
%Y A379637 Row sums give A379626.
%Y A379637 Cf. A000105, A379623, A379625, A379627, A379638.
%K A379637 nonn,tabf
%O A379637 1,4
%A A379637 _Omar E. Pol_, Jan 16 2025