Abstract
It is known that \(\sum \nolimits _{i =1}^\infty {1/ (2i+1)^2}={\pi ^2/8}-1\). We can ask what is the smallest \(\epsilon \ge 0\) such that all squares of sides 1 / 3, 1 / 5, 1 / 7,...can be packed into a rectangle of area \({\pi ^2/8}-1+\epsilon \). We show that the proof of Paulhus\('\) key Lemma for the best known result is false and we give new upper estimate \(\epsilon <4.43\times 10^{-10}\).
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Joós, A., Bálint, V. Packing of odd squares revisited. J. Geom. 110, 10 (2019). https://doi.org/10.1007/s00022-018-0464-9
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DOI: https://doi.org/10.1007/s00022-018-0464-9