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Partitions into ALMOST equal rectangles?
In how many ways can N blocks be arranged into two rectangles so that the first rectangle is one unit wider than the second, which is one unit taller than the first?
The first value of N for which this is non-zero is 4 = 2×1+1×2. <count is 1>
The next N with a non-zero count is 7 = 3×1+2×2 = 2×2+1×3 <count is 2>
The first N with a count of 3 is 40 = 14×1+13×2 = 4×5+5×4 = 2×13+1×14
This question was originally asked by Izjed:
http://ca.answers.yahoo.com/activity?show=s4xnCYKV...
but went unanswered.
The original distinguished "height" from "width", (and allowed neither to be zero in either rectangle), so the counts I gave are the ones I want.
1 Answer
- gianlinoLv 79 years agoFavorite Answer
Looks to me like the count of 7 is 1 and count of of 40 is 2. The unions should be non isometric right?
Anyway, set A = mn + (m + 1)(n - 1).
Then 2A = 4mn +2n - 2m - 2 = (2m + 1 )(2n - 1) - 1.
So all you need is to count the distinct factorizations of 2A + 1
For A = 40 , 2 A + 1 = 81 and you indeed have only 2 factorizations 9*9 and 3*27.
To have 3 non isometric decompositions you can use A = 52 2A+1 = 3*5*7
(m,n) = (1,18), (2,11),(3,8)
52 = 1*18 + 2*17 = 2*11 + 3*10 = 3*8 + 4*7
