There are 6 ways to form a rectangle from 3 rectangles with same area:
+-----+ +-+-+-+ +-----+ +--+--+ +-+---+ +---+-+
| | | | | | | | | | | | | | | | |
+-----+ | | | | +--+--+ | | | | | | | | |
| | | | | | | | | | | | | +---+ +---+ |
+-----+ | | | | | | | +--+--+ | | | | | |
| | | | | | | | | | | | | | | | |
+-----+ +-+-+-+ +--+--+ +-----+ +-+---+ +---+-+
So a(3)=6.
b(n) in the given formula is the sum of the appropriate tilings from certain 'frames'. A number that appears in a subrectangle in a frame is the number of rectangles into which the subrectangle is to be divided. Tilings are also counted that are from a reflection and/or half-turn of the frame.
For n = 6 there are 3(X2) frames:
+---+-+-+ +-+-----+ +-+-----+
| | | | | | | | | |
| | | | | +---+-+ | | 2 |
+-+-+ | | | | | | | | |
| | | | | | +---+ | | +---+-+
| | +-+-+ | | | | | | | |
| | | | +-+---+ | +-+---+ |
| | | | | | | | | |
+-+-+---+ +-----+-+ +-----+-+
2 ways 2 ways 8 ways
The only other frames which yield desired tilings are obtained by rotating each frame above by 90 degrees and scaling it to fit a rectangle with the inverse aspect ratio.
So b(6) = 2(2+2+8) = 24, and a(6) = b(6)+4*a(5)+2*a(4)-4*a(3)-2*a(2) = 24+4*88+2*21-4*6-2*2 = 390.
For n = 7 we can use 7(X2) frames:
+---+--+
| | |
| | |
| 4 |3 |
| | |
| | |
| | |
+---+--+
63 ways [of creating tilings counted by b(7)]
+---+--+ +-+----+ +--+---+ +-----++ +--+---+ +----+-+
| | | | | | | | | ++----+| | | | ++-+-+ |
| +-++ | +---++ |2 | 2 | || || | +-+-+ || | | |
| 3 | || |2| || | +--++ || || |2 | | | || | | |
| | || | | 2 || | | || || 3 || | | | | || +-+-+
| | || | | || +--+--+| || || +--+-+2| || | |
+---+-+| +-+---+| | || |+----++ | | | |+-+---+
+-----++ +-----++ +-----++ ++-----+ +----+-+ ++-----+
24 ways 16 ways 12 ways 10 ways 8 ways 4 ways
As for n = 6, these are only half the frames and tilings.
So b(7) = 2(63+24+16+12+10+8+4) = 274, and a(7) = b(7)+4*a(6)+2*a(5)-4*a(4)-2*a(3) = 274+4*390+2*88-4*21-2*6 = 1914.
(End)
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