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%I #5 Feb 20 2023 08:21:22
%S 0,1,8,1024,620448
%N Number of ways to tile a 2n X 2n square using rectangles with distinct dimensions such that the sum of the rectangles perimeters equals the area of the square.
%C All possible tilings are counted, including those identical by symmetry. Note that distinct dimensions means that, for example, a 1 x 3 rectangle can only be used once, regardless of if it lies horizontally or vertically.
%C Only squares with even edges lengths are possible since the area of a square with odd edge lengths is odd, while the perimeter of any rectangle is even.
%e a(1) = 0 as a 2 x 2 square, with area 4, cannot be tiled with distinct rectangles with perimeters that sum to 4.
%e a(2) = 1 as a 4 x 4 rectangle, with area 16, can be tiled with a 4 x 4 square with perimeter 4 + 4 + 4 + 4 = 16.
%e a(3) = 8. The possible tilings for the 6 x 6 square, with area 36, excluding those equivalent by symmetry, are:
%e .
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%e | | | |
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%e | | | |
%e + + + +
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%e +---+---+---+---+---+---+ +---+---+---+---+---+---+
%e .
%e where for the first tiling (2*6 + 2*1) + (2*6 + 2*5) = 36 while for the second tiling (2*6 + 2*2) + (2*6 + 2*4) = 36. Both of these tilings can occur in 4 ways, giving 8 ways in total.
%e a(4) = 1024. And example tiling of the 8 x 8 square, with area 64, is:
%e .
%e +---+---+---+---+---+---+---+---+
%e | | | |
%e + + +---+---+
%e | | | |
%e + + + +
%e | | | |
%e +---+---+---+---+---+---+---+---+
%e | |
%e + +
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%e +---+---+---+---+---+---+---+---+
%e .
%e where (2*1 + 2*3) + (2*5 + 2*3) + (2*2 + 2*1) + (2*2 + 2*2) + (2*8 + 2*5) = 64.
%Y Cf. A360499, A360498, A360725, A360256, A182275, A004003, A099390, A065072.
%K nonn,more
%O 1,3
%A _Scott R. Shannon_ and _N. J. A. Sloane_, Feb 20 2023