I was trying to simplify the expression $\displaystyle \sum_{j=k}^n j(j-1) {n \brack j} {j \brace k}$ for $0 \leq k \leq n$, where ${n \brack j}$ and ${j \brace k}$ denote the Stirling number of the first and second kind respectively. I ended up with the following equation, where $H_n$ denotes the n-th harmonic number:
$$2 \cdot \Biggl[ \frac{n!}{(k-2)!} \displaystyle \sum_{j=1}^{n-k+1} H_j \frac{(-1)^{j+1}}{j+1} \binom{n+1}{k+j} + \frac{n!}{(k-1)!} \displaystyle \sum_{j=1}^{n-k} H_j \frac{(-1)^{j+1}}{j+1} \binom{n}{k+j} \Biggr] $$
Next up, I wanted to simplify the sums. After shifting the index by $k$,
$$ \displaystyle \sum_{j=k+1}^{n} (-1)^{j-k+1} \binom{n}{j} \frac{H_{j-k}}{j-k+1} $$
it looks a lot like the following equation, where $H_n^{(2)} = \sum_{j=1}^n \frac{1}{j^2}$:
$$\sum_{j=k+1}^n(-1)^{j-1}{n\choose j} \frac{H_j}{j}=H_n^{(2)} - H_k^{(2)}.$$
which can be proven according to the second answer in A finite sum involving the binomial coefficients and the harmonic numbers. But of course, some terms are shifted, while the binomial coefficient stays the same. I tried solving it similarly to the equation above using the binomila formula, but wasn't able to find a solution due to the shifted terms. At this point I'm not sure whether there is an easier form, but I thought someone might recognize it from somewehere or know a way to prove it. I will keep on trying to simplify it and update the question if I find a way.
The reason the expression is interesting is that it can be used to calculate the Variance of the Lah-distribution, which is defined in Definition 1.1 in https://arxiv.org/abs/2105.11365. Although the authors do calculate a formula for the second moment in Corollary 3.6, I have used a different method to arrive at the above mentioned formula and am therefore interested whether you can find a nicer form for the second moment than Corollary 3.6 gives. I have tested my formula numerically and it is correct, so simplifying it might be worthwhile.
