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.