Joerg Arndt <jj@suse.de> [Feb 25, 2006 12:05]:
* David W. Cantrell <DWCantrell@sigmaxi.org> [Feb 22. 2006 22:46]:
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You might be interested in <http://www.pisquaredoversix.force9.co.uk/Tiling.htm>. BTW, before Clive found that, I had found another conjectured packing, using an algorithm slightly different from his. If interested, see <http://groups.google.com/group/sci.math/msg/d92d4a0ea46b4762> and other messages in that thread and its parent.
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Going along the top side: Pi^2/6-(1/1+1/2+1/7) == 0.00207692399108358 How does the series continue?
You're asking me about Clive's packing. Offhand, I don't know how that series continues. Ask Clive, or implement his algorithm and see for yourself how it continues.
Similarly, half way down: Pi^2/6-(1/1+1/3+1/4+1/17) == 0.00277720410312840
But, if you wish, I can fairly easily answer similar questions about the packing produced by my simple as-far-to-the-left-and-down-as-possible algorithm. Here's a picture of squares of side lengths 1, 1/2, 1/3,..., 1/99 packed in a rectangle 1 by pi^2/6. The bounding rectangle is shown as a dashed line. <http://img518.imageshack.us/img518/1297/990ng.gif> And below is a detail of the lower right corner after squares down to a side length 1/999 have been packed. (BTW, to the immediate left of the square of side length 1/67 is that of side length 1/8, visible in the previous figure.) <http://img119.imageshack.us/img119/2027/999lowerright9up.gif> So, at the bottom of the packing produced by my algorithm, we have, thus far, a gap of width Pi^2/6 - (1/1 + 1/2 + 1/8 + 1/67 + 1/293 + 1/993) = 0.0005886+. David