a(1) = 0 as a 2 x 2 square, with area 4, cannot be tiled with distinct rectangles with perimeters that sum to 4.
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.
a(3) = 8. The possible tilings for the 6 x 6 square, with area 36, excluding those equivalent by symmetry, are:
.
+---+---+---+---+---+---+ +---+---+---+---+---+---+
| | | |
+---+---+---+---+---+---+ + +
| | | |
+ + +---+---+---+---+---+---+
| | | |
+ + + +
| | | |
+ + + +
| | | |
+ + + +
| | | |
+---+---+---+---+---+---+ +---+---+---+---+---+---+
.
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.
a(4) = 1024. And example tiling of the 8 x 8 square, with area 64, is:
.
+---+---+---+---+---+---+---+---+
| | | |
+ + +---+---+
| | | |
+ + + +
| | | |
+---+---+---+---+---+---+---+---+
| |
+ +
| |
+ +
| |
+ +
| |
+ +
| |
+---+---+---+---+---+---+---+---+
.
where (2*1 + 2*3) + (2*5 + 2*3) + (2*2 + 2*1) + (2*2 + 2*2) + (2*8 + 2*5) = 64.
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