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j.c.
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One problem How many non-homeomorphic collections of $N$ circles in combinatorial topology$\mathbb{R}^3$ are there?

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YCor
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Let's have a finite collection of N$N$ circles S^1$\mathbb{S}^1$ in \mathbb{R}^3$\mathbb{R}^3$. ( TheseThese circles could not intersect.) Every circle could be "hooked on" other circle and it could be "hooked" for simplicity only once. My question, which I need to solve, is how many combinations of non-homeomorphic structures will I obtain? For example N=2$N=2$: I have 2$2$ combinations, two unhooked circles and two hooked circles; I know already that this question could be translated to the language of graph theory. Do You know, if anyone has solved such a problem? Thank You!

Let's have a finite collection of N circles S^1 in \mathbb{R}^3. ( These circles could not intersect.) Every circle could be "hooked on" other circle and it could be "hooked" for simplicity only once. My question, which I need to solve, is how many combinations of non-homeomorphic structures will I obtain? For example N=2: I have 2 combinations, two unhooked circles and two hooked circles; I know already that this question could be translated to the language of graph theory. Do You know, if anyone has solved such a problem? Thank You!

Let's have a finite collection of $N$ circles $\mathbb{S}^1$ in $\mathbb{R}^3$. (These circles could not intersect.) Every circle could be "hooked on" other circle and it could be "hooked" for simplicity only once. My question, which I need to solve, is how many combinations of non-homeomorphic structures will I obtain? For example $N=2$: I have $2$ combinations, two unhooked circles and two hooked circles; I know already that this question could be translated to the language of graph theory. Do You know, if anyone has solved such a problem? Thank You!

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