0
$\begingroup$

$f(n)$ - Number of (non-oriented) simple graph with $n$ vertices numbered $1,2,\dots n$ in which for every vertex $v$, deg$(v)=2$.

I've been trying to connect the $n^{th}$ vertex choosing $2$ out of $n-1$ vertices for $n>3$ where for $n=3$ $f=1$, and $0$ for $n<3$. However this solution seems fail for $n=6$.

I'm not sure how to approach this problem differently

I Appreciate the help

Thanks

  • 0
    If every vertex has degree 2, it means that connected components of such graphs are cycles of at least 3 vertices. – 2017-01-09

1 Answers 1

0

Idea: count the number of different $n$-th integer partition that don't contain the number $1$ or $2$, up to rotation and reflection (?).

I recommend you to read about the combinatorics definition of bracelet.

Here also an interesting visual tool to display different necklaces.