I'm considering an extended problem of kissing number in $\mathbb{R}^2$.

Suppose I have a given disc $\mathcal{D}$ of radius 1/2 and infinitely many discs all of radius 1/2 and all these discs and my given disc $\mathcal{D}$ do not overlap. I assume these discs are all closely packed together and I place the center of my given disc $\mathcal{D}$ at the origin.

I'm interested in the distance between the center of other discs and the origin.

Since it is well-known that the kissing number in $\mathbb{R}^2$ is 6, which means that there are only 6 discs whose centers have distance 1 to the origin. I call these 6 discs the "layer 1". Then, besides these 6 discs, all other discs must have distance more than 1. In fact, the shortest distance for these discs is $\sqrt{3}$ and I call all these discs with distance $\sqrt{3}$ the "layer 2". Hopefully, the second layer also has 6 discs. Then, besides the first and the second layer, we can also find some discs that have the shortest distance. For "layer 3", the distance is 2 and there are also 6 discs in "layer 3". We continue define "layer 4", "layer 5",...

Now my question is that, is there a formula $f(n)$ to calculate the number of discs in "layer $n$"? Or at least a good bound on $f(n)$? I'm also interested in the distance of "layer $n$", which I call $\tau(n)$.

I have calculated the first few values of $f(n),\tau(n)$: $$f(1)=6,f(2)=6,f(3)=6,f(4)=12,f(5)=6,...$$ $$\tau(1)=1,\tau(2)=\sqrt{3},\tau(3)=2,\tau(4)=\sqrt{7},\tau(5)=3,...$$

For the calculation of $\tau$, I think it is equivalent to the following problem: define $$G=\{\sqrt{a^2+b^2+ab}:a=0,1,2,...;b=0,1,2,...\}$$ Then, if we rank all elements in $G$ in increasing order, it should give $\tau$ if we set the smallest element in $G$ (which is 0) to be $\tau(0)$. Here $\sqrt{a^2+b^2+ab}$ is just the length of the diagonal of a parallelogram with $a,b$ as the side length and $\pi/3$ as the angle.

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    $\begingroup$ The centers form a lattice of equilateral triangles, right? I believe that questions about the locations and number of lattices points have been well-studied, though I regret I can't cite a reference offhand. But see oeis.org/A035019 $\endgroup$ – Gerry Myerson Apr 10 '18 at 23:17
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    $\begingroup$ And oeis.org/A003136 $\endgroup$ – Gerry Myerson Apr 11 '18 at 0:33
  • $\begingroup$ @GerryMyerson I think your comments would make a fine answer to this question. $\endgroup$ – j.c. Apr 11 '18 at 8:25

$a^2+ab+b^2=(a+\omega b)(a+\omega'b)=N(a+\omega b)$, where $\omega,\omega'$ are $(-1\pm\sqrt{-3})/2$ and $N$ is the norm in the field ${\bf Q}(\omega)$, takes on all and only the values of the form $xy^2$, where $x$ has no prime factor $p\equiv2\bmod3$. These values are $0,1,3,4,7,9,12,13,16,19,21,25,27,28,\dots$, tabulated at http://oeis.org/A003136. The associates of $\alpha=a+b\omega$ are $\alpha,\omega\alpha,\omega'\alpha,-\alpha,-\omega\alpha,-\omega'\alpha$, so the number of times any nonzero value $m$ is taken on will be a multiple of 6 (of course, this is obvious from the geometry). Which multiple depends on the prime factorization of $m$, actually, on the number of prime factors of $m$ (in ${\bf Q}(\omega)$). The multiplicity of $m$ is tabulated at http://oeis.org/A035019. Both OEIS pages have links to the literature.

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