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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.

%I A379638 #17 Jan 16 2025 21:25:07
%S A379638 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,
%T A379638 0,9,548,835,324,49,8,0,0,3,892,3345,1842,539,56,9,0,0,0,1148,10445,
%U A379638 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
%N 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.
%C A379638 The length here is the longer of the two dimensions.
%H A379638 <a href="/index/Pol#polyominoes">Index entries for sequences related to polyominoes</a>.
%F A379638 T(n,k) = k*A379624(n,k).
%e A379638 Triangle begins:
%e A379638   1;
%e A379638   0,  2;
%e A379638   0,  2,   3;
%e A379638   0,  2,   9,    4;
%e A379638   0,  0,  24,   12,     5;
%e A379638   0,  0,  24,   84,    25,      6;
%e A379638   0,  0,  21,  236,   180,     30,      7;
%e A379638   0,  0,   9,  548,   835,    324,     49,     8;
%e A379638   0,  0,   3,  892,  3345,   1842,    539,    56,     9;
%e A379638   0,  0,   0, 1148, 10445,   9762,   3773,   824,    81,   10;
%e A379638   0,  0,   0, 1020, 27360,  42756,  22659,  6712,  1206,   90,   11;
%e A379638   0,  0,   0,  676, 59595, 165024, 116942, 46808, 11439, 1680,  121,  12;
%e A379638   ...
%e A379638 Illustration for n = 5:
%e A379638 The free polyominoes with five cells are also called free pentominoes.
%e A379638 For k = 1 there are no free pentominoes of length 1, so T(5,1) = 0.
%e A379638 For k = 2 there are no free pentominoes of length 2, so T(5,2) = 0.
%e A379638 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.
%e A379638    _ _     _ _       _ _     _ _ _     _         _           _       _ _
%e A379638   |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
%e A379638   |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
%e A379638   |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
%e A379638 .
%e A379638 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.
%e A379638    _         _       _
%e A379638   |_|      _|_|    _|_|
%e A379638   |_|     |_|_|   |_|_|
%e A379638   |_|_    |_|       |_|
%e A379638   |_|_|   |_|       |_|
%e A379638 .
%e A379638 For k = 5 there is only one free pentomino of length 5 as shown below, so T(5,5) = 5.
%e A379638    _
%e A379638   |_|
%e A379638   |_|
%e A379638   |_|
%e A379638   |_|
%e A379638   |_|
%e A379638 .
%e A379638 Therefore the 5th row of the triangle is [0, 0, 24, 12, 5].
%Y A379638 Row sums give A379629.
%Y A379638 Cf. A000105, A379624, A379625, A379627, A379637.
%K A379638 nonn,tabl
%O A379638 1,3
%A A379638 _Omar E. Pol_, Jan 16 2025