Perfect packing of cubes
From the known formula \(\sum_{i=1}^{\infty} 1/i^2 = {\pi}^2/6\) the question has arisen what is the smallest \(\varepsilon\) such that the squares of sides of length \(n^{-1}\) can be packed into a rectangle of area \({\pi}^2/6+\varepsilon\). If such a packing exists for \(\varepsilon = 0\), it is called perfect. \textit{A. Chalcraft} [J. Comb. Theory, Ser. A 92, No. 2, 158--172 (2000; Zbl 0977.52023)] generalized the question and packed the squares of sides of length \(n^{-t}\) for \(n\in\{1, 2, \dots\}\) into a square of appropriate area. In the paper under review, the problem is generalized to the 3-dimensional space. From the formula \(\sum_{i=1}^{\infty} i^{-3t} = \zeta(3t)\), the author asks whether the cubes of edge length \(n^{-t}\) where \(n\in\{1, 2, \dots\}\) can be packed into a right rectangular prism of volume \(\zeta(3t)\). The author generalizes the algorithm of Chalcraft [loc. cit.] and proves that for any \(t\) in the range \(0.36273 \leq t \leq 4/11\) the cubes of edge length \(n^{-t}\) can be packed perfectly into the right rectangular prism of size \(1 \times 1 \times \zeta(3t)\).
scientific article; zbMATH DE number 7011034