What is the sidelength of the smallest square in which one can fit $n$ non-overlapping squares of sidelengths $1,2,3,4,...,n$ ?

And what is the sidelength of the smallest cube in which one can fit $n$ non-intersecting cubes of sidelengths $1,2,3,4,...,n$ ?

All squares/cubes have all sides paralell, no rotation

  • 3
    $\begingroup$ For squares, there is information at maa.org/editorial/mathgames/mathgames_12_01_03.html The series is oeis.org/A005842. I didn't find a series for cubes and my Google search turned up travel tips on packing. $\endgroup$ – Ross Millikan Nov 2 '11 at 20:01
  • $\begingroup$ I'll accept asymptotics $\endgroup$ – TROLLHUNTER Nov 2 '11 at 20:56
  • $\begingroup$ The sum of the squares up to $n^2$ is $n(n+1)(2n+1)/6$, so when $n$ gets very large I would expect them to pack into about the square root of that, $\sqrt{n^3/6}$. Similarly for cubes, the volume is $n^2(n+1)^2/4$ and I would expect the cube root of that. This is just because you have lots of little pieces, so shouldn't have to waste that much space, no better than that. $\endgroup$ – Ross Millikan Nov 2 '11 at 21:18

For 180 dollars, you can get a 37 squares solution and there is some discussion of cubes and here's a discussion of 70 squares.


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