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I have constructed a counting argument which is fallacious, but I don't know why it is incorrect. I would appreciate it if one of the users can point the flaw in my argument.

The question: How many connected graphs can be formed by $n$ labelled vertices?

My argument: Let $G_{n}$ represent the number of connected graphs that can be formed by $n$ labelled vertices. Notice then that $G_{1} = 1$. Now consider $n$ labelled vertices. I claim that $\forall n \geq 2, G_{n} = (2^{n - 1} - 1)G_{n - 1}.$ The base case for $n = 2$ indeed yields that $G_{n} = 1.$ Assume that $\forall k$ s.t. $2 \leq k \leq n$ that $G_{k} = (2^{k - 1} - 1)G_{k - 1}$ holds. Consider the case of $G_{n + 1}.$ We know that $(n + 1)^{st}$ vertex can be connected to atmost $n$ other vertices. This yields $2^{n}$ different possibilities of edge connections from the $(n + 1)^{st}$ vertex (either including an edge between the $(n + 1)^{st}$ vertex and the $i^{th}$ vertex or not including it where $1 \leq i \leq n$). However, we must account for the case where the vertex in consideration is not connected to any of the other $n$ vertices (which is $1$ case). Hence, $G_{n + 1} = (2^{n} - 1)G_{n}$.

With this in mind, where is the error in my thought process?

gnagar1996
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I think your error is that you are only counting the connected graphs on $n$ vertices, with a new $(n+1)th$ vertex added and some edges added. But there are also disconnected graphs on $n$ vertices, which become connected through the addition of the $(n+1)th$ vertex and two (or more) edges.

E.g. for $n=3$, you don't add in the graph with edges $\{(1,3), (2,3)\}$, because you haven't considered the disconnected graph on $2$ vertices in your recurrence; thus your total is $G_3 = (2^2 - 1)G_2 = 3$ instead of $4$.

Chas Brown
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