How to derive A001187 formula combinatorically

Question

I’ve learned recently that the formula for A001187 is defined recursively. The formula is: $$n2^{\binom n2}=\sum_{k=0}^{n} \binom{n}{k}kd_k2^{\binom{n-k}{2}}$$ Where $d_k$ is the number of connected labeled graphs with $k$ nodes. That way is hard to understand for me, is there a simpler way to derive this using combinatorics from the graph?

I was thinking something like the following but I don’t think it’s correct: For $k$ from $1$ to $n-1$, we choose $k$ vertices. We then find the connected graph for this $k$ vertices $d_k$. Then the exponent accounts for the rest of the possible edges.

Is there a more formal and correct way to derive this?

This is a formula not for $d_n$ but for $n2^{n(n-1)/2}$, right? And the sum should have $n$ as the upper bound, I guess? — colt_browning, Nov 09 '23 at 01:21
Yes, this is quite confusing. Had it not been for @colt_browning's comment, I would have assumed that that expression was for $d_n$. — Brian Tung, Nov 09 '23 at 04:50
Assuming $d_k$ is the $k$-th entry in the sequence ($a(k)$ in the parlance of that site), the formula given in the OEIS entry is $$n2^{n(n-1)/2} = \sum_{k=1}^n \binom nk k d_k 2^{(n-k)(n-k-1)/2}$$which can be converted to $$d_n = 2^{n(n-1)/2} - \frac 1n\sum_{k=1}^{n-1} \binom nk k d_k 2^{(n-k)(n-k-1)/2}$$ — Paul Sinclair, Nov 09 '23 at 20:11

score 1 · Accepted Answer · answered Nov 12 '23 at 05:56

It can be solved by double-counting. Denote $V(G)$ as a graph's vertex set, and $G(S)$ be the induced subgraph on vertex set $S$.

For each graph $G\subseteq K_n$, if it has connected components $G_1,G_2,\dots,G_k$, we assign $V(G_1),\dots,V(G_n)$ the number $|V(G_1)|,\dots,|V(G_n)|$ respectively.

On the one hand, for each graph, since we have assigned a total of $n$, the total assigned number is $n2^{\binom n2}$.

On the other hand, for each $S\subseteq V(G)$, we count how many amount is assigned in total. For $S$ to be a legit connected component, it cannot connect to other nodes. So for all other $n-|S|$ nodes, there is $2^{\binom{n-|S|}2}$ connection methods. Also, $G(S)$ has to be connected, so there are $d_{|S|}$ possible connected graph. For each $|S|$, there are $\binom n{|S|}$ such $S$. Finally, it receives $|S|$ from assignment. So, in sum, if we sum the assigned number by the set, we have the total assigned number is $\binom n{|S|}|S|d_{|S|}2^{\binom{n-|S|}2}$. Therefore, summing through $|S|$, we can prove that desired identity

$$n2^{\binom n2}=\sum_{|S|}\binom n{|S|}|S|d_{|S|}2^{\binom{n-|S|}2}$$

How to derive A001187 formula combinatorically

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