When packing disks into a square, is it best to be greedy?

The problem is to pack $$n$$ non-overlapping disks (not necessarily of the same size) of greatest total area into a unit square. The case $$n=1$$ is obvious: just place a disk of radius $$\frac12$$ centrally in the square, to occupy an area $$\frac14\pi\approx78.54\%$$ of the square. For $$n=2$$, we can do no better than follow up the previous case and add a disk (of radius $$\frac32-\surd2$$) in a corner to touch the existing disk and two sides of the square, thus altogether covering an area $$(\frac92-3\surd2)\pi\approx80.85\%$$ of the square. Similarly, for $$n=3,4,5$$, we can do no better than place further disks of that size in the remaining corners of the square.

For $$n=6,...,13$$, it seems optimal to build on the case $$n=5$$ and place further disks in the eight regions bounded by a side of the square, the disk of radius $$\frac12$$, and a disk of radius $$\frac32-\surd2$$. Clearly, we can keep going like this, packing the next disk into whatever remaining space will accommodate the largest disk—the greedy method. But the question arises:

Is there an $$n$$ for which the packing of maximal total area is not the greedy packing described above?

• For $n=1, 4, 16, 64. . . 2^{2(n-1)}$ total area is constant. The neasure of radii of circles between each four circle reduces with increasing n. So I think maximal total area is for n=1 plus some more circles in four corners. finally if you want circles with same diameter n must be a member of above sequence. Commented Sep 5, 2020 at 7:49
• @Sil : If your comment were posted as an answer, it would probably be the accepted answer. Commented Sep 18, 2020 at 20:08
• @JohnBentin Okay turned it into one, although probably not satisfying :)
– Sil
Commented Sep 18, 2020 at 20:24

Conjecture 3: For all $$n\in \mathbb{N}$$ and $$k \geq 3$$, the greedy arrangement solves the problem of finding an arrangement of $$n$$ circles in a tangential $$k$$-gon with the maximal total area.