Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that:

$$s(n,2) = (n-1)!(1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n-1})$$

-I haven't dealt with Taylor series expansion in a long time and not quite sure how (brand new to)Stirling numbers would play into this proof.Any help is appreciated.


HINT: If you split $[n]$ into two cycles, one of length $k$ and the other of length $n-k$, there are $\binom{n}k$ ways to choose the elements of the $k$-cycle, $(k-1)!$ ways to arrange them in a cycle, and $(n-k-1)!$ ways to arrange the remaining elements in a cycle. That gives you a total of


permutations. Now sum over the possible values of $k$. Be careful, though: you’ll be counting every permutation twice.

You will probably also find it useful to notice that



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.