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How many ways are there to tile an $n\times n$ square with exactly $n$ rectangles, each of which has integer sides and area $n$?

The sequence $C(n)$ begins 1, 2, 2, 9, 2, 46, 2, 250, 37. Clearly $C(p) = 2$ for prime $p$. The value $C(8) = 250$ was provided to me by Sjoerd Visscher, but I cannot vouch for it personally, not having seen the details of his enumeration.

OEIS was no help.

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$C(8)=250$ is correct, but $C(9)=2\left(\binom90+\binom71+\binom52\right)+1=37$. Here's code that computes $C(n)$ up to $n=23$. (The computation for $n=24$ didn't complete after a couple of minutes.) The first terms are $1,2,2,9,2,46,2,250,37,254,2,31052,2,1480,896,306174,2,2097506,2,6025516,6638,59‌​930,2$. (P.S.: I get a display bug in that line; the penultimate number is $59930$, without a space.) –  joriki Apr 12 '12 at 7:21
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I've submitted this sequence as OEIS sequence A182106 (it's pending review). –  joriki Apr 12 '12 at 8:01
    
For small $n$ one can break this down to a calculation of the form $\pm k + 2\sum{n-2i\choose i}$ as in @joriki's $n=9$ example, but as $n$ increases this will stop working in many cases. –  MJD Apr 13 '12 at 14:26
    
The OEIS sequence has been approved and published. –  joriki Apr 16 '12 at 15:30

1 Answer 1

Sorry to poke a dead post but it was near the top of the "unanswered questions" queue for me and it's a decent problem.

Working locally should provide a good avenue of attack on this problem.

For instance, a relatively straightforward analysis (which I'll post if there is interest) yields:

$$ C(p^2) = 2\left(\sum_{k=0}^{p} {p^2-k(p-1) \choose k}\right)-1$$

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