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If we have $n$ rigid circles of the same radius we can form 'chains' on the plane by placing them in such a way that they intersect (here two circles intersect if and only if they have $2$ common points).

The diagrams below illustrate the cases $n=2,3,4$. The number of circles intersecting a given circle is shown for each chain in an ascending sequence. The circles are unlabeled.

enter image description here

I counted $15$ different chains for $n=5$, but I'm not sure it's all of them.

So, I have two questions:

Is it possible to compute the number of different chains for $n$ unlabeled circles? Is there a closed form expression?

It's obvious that this number can't exceed the number of possible combinations $k_1k_2\dots k_n$ where $k_1 \leq k_2 \leq \dots \leq k_n$ and $1 \leq k_j \leq n-1$. Which is equal to $\left( \begin{array}( 2n-2 \\ ~~n-2 \end{array} \right)$. But the number of chains is likely much smaller for large $n$.

Is it possible to explicitly label all the chains for a given $n$ by the number of cirlces intersecting each circle as I've done above for $n=2,3,4$?

If there is some study for this problem, I would like a reference. And what about chains formed with rigid rings in 3D?

I'm not even sure what area of mathematics studies such things. It's probably not knot theory, because the links here are rigid, right?

As a response to comments, I agree that the rigidity of the rings makes the problem too complicated. For example, for $n=7$ we still can have the chain labeled $1111116$, because $2\pi>6$, however for $n=8$ there is no way to obtain $11111117$, because $7$ rings won't be able to intersect the $8$th without at least two of them intersecting each other.

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    Possibly related: http://math.stackexchange.com/questions/1307018 – 2017-01-19
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    This link http://oeis.org/search?q=number%20of%20connected%20graphs&sort=&language=english&go=Search may be of some use, although the restriction that they are of a rigid radius compounds the problem. – 2017-01-19
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    It would seem to me this is equivalent to the number of nonisomorphic graphs on $n$ vertices. You let each vertex represent a circle and then draw an edge between two vertices if the circles they represent intersect. This is a difficult problem, but https://oeis.org/A000088 lists some references. – 2017-01-19
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    @LaarsHelenius, I was originally interested in this problem for rigid circles in $3D$, but that case seemed even more complicated. All the same, there should be definite rules on which configuration is possible – 2017-01-19
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    I love combinatorics, but I'm filing this one under... too hard! It can clearly be simplified by asking the ways to arrange $n$ points in which no point is more than $r$ from at least one other point. If we fix the first point and $r_{ab}$ is the distance from point $a$ to point $b$ then all positions of points in which one $r_{ax}$ is in the same set in relation to all $x\in B$ forms an equivalence class. Then continue for all $n$ points. Finally, reduce all permutations of the points to combinations. – 2017-01-19
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    @Yuriy S: Since the chains you are considering always include all the circles, the parallel graph theory problem would actually be the number of $connected$ non-isomorphic graphs. Then it might be helpful to realize a connected graph always has a spanning tree, so you could start by counting the number of nonisomorphic unlabeled spanning trees. To get started on that, consider looking at Harary and Palmer's $Graphical~Enumeration.$ Also https://oeis.org/A000055 has additional references. – 2017-01-19
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    @LaarsHelenius, thank you very much for your recommendations, I will look it up. – 2017-01-19
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    Can two circles overlap each other while overlapping the same circle in a line? Hard to describe without a diagram, but take 112 - can the green and red circles overlap *within* the yellow circle? If so, i'm not sure we can map to connected graphs... (*edit* I did mean 112, yes). – 2017-01-19
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    @gilleain, they can, but it will be the same as 222. Wait, there is no 122, did you mean 112? – 2017-01-19

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