Evaluate weighted stirling number sum - $\sum_{i=0}^n k*c(n,k)$ How to evaluate
$$\sum_{k=0}^n k*c(n,k)$$
Where $c(n,k)=|s(n,k)|$ is the stirling number of the first kind without a sign.
 A: Differentiation of the falling factorial is definitely the way to go, as @Phicar noted. Given that we know
$$x^{\underline{n}}=\sum _{k=0}^n S_n^k\,x^k$$
All we have to do is let $x \to -x$,  multiply by $(-1)^n$, and differentiate both sides and to get
$$(-1)^n(-x)^{\underline{n}}(H_{-n-x}-H_{-x})=\sum _{k=0}^n k\,x^{k-1}S_n^k(-1)^{n-k}=\sum _{k=0}^n k\,x^{k-1}c(n,k)$$
Now just let $x=1$. The LHS will simplify significantly when you do.
A: It turns out that
$$
\sum_{k=1}^n kc(n,k)=c(n+1,2),
$$
which I shall prove bijectively. Both sides have a natual combinatorial interpretation:

*

*$\sum_{k=1}^nkc(n,k)$ is equal to the number of ways to choose a permutation on $\{1,\dots,n\}$, and then to single out one of the cycles as "special".


*$c(n+1,2)$ enumerates permutations of $\{0,1,\dots,n\}$ with exactly two cycles.
I will give a bijection between these two sets, which completes the proof.
Given a permutation of $\{0,1,\dots,n\}$ with exactly two cycles, we need to define a permutation $\pi$ of $\{1,\dots,n\}$, such that one cycle of $\pi$ is singled out as special. Call the two cycles $\sigma$ and $\tau$, where $\sigma$ is the cycle containing $0$. We will let $\tau$ be the "special" cycle. All that remains is to define the action of $\pi$ on the nonzero elements of $\sigma$. We can write $\sigma$ with the $0$ first, so in the form
$$
\sigma=(0,\;\sigma_1,\;\sigma_2,\;\dots\;\sigma_k),
$$
For each $i\in \{1,\dots,k\}$, let $\sigma_i'$ be the $i^\text{th}$ smallest element of $\{\sigma_1,\dots,\sigma_k\}$. We then define the action of $\pi$ on the nonzero elements of $\sigma$ by defining
$$
\pi(\sigma_i')=\sigma_i,\qquad \text{for each }i\in \{1,\dots,k\}
$$
A: We seek to verify that
$$\sum_{k=0}^n k {n\brack k} = {n+1\brack 2}.$$
This is
$$n! [z^n] \sum_{k=1}^n k
\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
Here we can extend to infinity due to the coefficient extractor and the
fact that $\log\frac{1}{1-z} = z +\cdots$:
$$n! [z^n] \log\frac{1}{1-z} \sum_{k\ge 1}
\frac{1}{(k-1)!} \left(\log\frac{1}{1-z}\right)^{k-1}
\\ = n! [z^n] \log\frac{1}{1-z}
\exp\log\frac{1}{1-z}
= n! [z^n] \frac{1}{1-z} \log\frac{1}{1-z}
= n! \times H_n.$$
On the other hand we have
$${n+1\brack 2} = (n+1)! [z^{n+1}] \frac{1}{2}
\left(\log\frac{1}{1-z}\right)^2
\\ = (n+1)! \frac{1}{2}
\sum_{q=1}^n \frac{1}{q} \frac{1}{n+1-q}
\\ = n! \frac{1}{2}
\sum_{q=1}^n \left(\frac{1}{q} + \frac{1}{n+1-q}\right)
= n! \times H_n$$
and we have the claim. We can prove that $\exp\log\frac{1}{1-z} =
\frac{1}{1-z}$ using the combinatorial class $\mathcal{P}$ of
permutations which are sets of cycles:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \textsc{SET}(\textsc{CYC}(\mathcal{Z}))$$
as presented in Analytic Combinatorics by Flajolet amd Sedgewick.
