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This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

EDIT : So, as Tim explained, this is not difficult as you are bound by the 8 biggest cube to be at most 21. And it's easy to see that there is a trivial solution for 21 (as there are a lot of empty spaces between the largest cube placed in the corners) : as you fit 21,20,19,18,17,16,14,15 in 8 consecutive corners, then between 14 and 15 you can place 13, between 14 and 16, you place 12 and 11 (it's ok as 15 and 17 don't use all their half), between 17 and 15 place 10 and 9 and all remaining cubes are easy to place.

So We have to buy a square large enough to fit 21 squares : this is a 58x58 square (as explained in A005842).

Here a solution, that I found, I hope it's ok :

This is a very nice problem !

This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

EDIT : So, as Tim explained, this is not difficult as you are bound by the 8 biggest cube to be at most 21. And it's easy to see that there is a trivial solution for 21 (as there are a lot of empty spaces between the largest cube placed in the corners) : as you fit 21,20,19,18,17,16,14,15 in 8 consecutive corners, then between 14 and 15 you can place 13, between 14 and 16, you place 12 and 11 (it's ok as 15 and 17 don't use all their half), between 17 and 15 place 10 and 9 and all remaining cubes are easy to place.

So We have to buy a square large enough to fit 21 squares : this is a 58x58 square (as explained in A005842).

Here a solution, that I found, I hope it's ok :

This is a very nice problem !

Comment added by Penguino. An example of a cube-packing exposure schedule is shown in [a] below (the second number on each cube is the height of its top surface). To more clearly demonstrate that there is no solution for N=22, the 14-cube cannot fit in an orthogonal orientation with the 8 larger cubes (as noted by Tim). It can fit if rotated (about the time axis), but then it intersects with the 13-cube and 12-cube (see [b] below). There are other options (for example you can put 14-cube square on the side and rotate 15-cube), but a little exploration shows they all have a similar problem.

This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

EDIT : So, as Tim explained, this is not difficult as you are bound by the 8 biggest cube to be at most 21. And it's easy to see that there is a trivial solution for 21 (as there are a lot of empty spaces between the largest cube placed in the corners) : as you fit 21,20,19,18,17,16,14,15 in 8 consecutive corners, then between 14 and 15 you can place 13, between 14 and 16, you place 12 and 11 (it's ok as 15 and 17 don't use all their half), between 17 and 15 place 10 and 9 and all remaining cubes are easy to place.

So We have to buy a square large enough to fit 21 squares : this is a 58x58 square (as explained in A005842).

Here a solution, that I found, I hope it's ok :

This is a very nice problem !

This problem is only about parking space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.

This problem is only about packing space-time cube of size $1,2,...,n$ in a cube of size $2n$. $$\sum_1^n k^3=\left(\frac{n(n+1)}{2}\right)^2$$

As the combined volumes of the $n$ cubes is a polynomial of degree 4 compared to the size of the large cube (of side $2n$) that is $8n^3$ (polynomial of degree 3 ) then there are only a finite number of possible solutions where the volume of the main large cube can contains all the small cubes. So we need $$(2n)^3-\left(\frac{n(n+1)}{2}\right)^2\ge0$$ Hence $n<30$. Is there a valid packing for 29 or even 28 ? I don't know, it seems to be a difficult problem.