It is a well-known fact that the generalized von Mangoldt function, defined by
$$\displaystyle \Lambda_k(n) = \sum_{d | n} \mu(d) \left(\log \frac{n}{d}\right)^k$$
vanishes whenever $n$ has more than $k$ distinct prime factors. I was able to prove this fact through a relatively lengthy and cumbersome combinatorial argument by proving first for the squarefree case, say when $n = p_1 \cdots p_s$ with $s > k$, and showed that the vanishing of $\Lambda_k(n)$ can be deduced from the following polynomial identity:
$$\displaystyle (x_1 + \cdots + x_s)^k – \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-1}} \left(\sum_{i \in S} x_i\right)^k + \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-2}} \left(\sum_{i \in S} x_i \right)^k – \cdots $$
$$ + (-1)^{s-1}\sum_{i=1}^s x_i^k = 0.$$
The general case is then done through a similar argument (which depends on the above identity) and strong induction.
However, I find the above argument to be somewhat insipid and not very 'number theoretic', as it is deduced from a general polynomial identity rather than using any properties of numbers. Is there any conceptually simpler, more number theoretic proof? It would be an added bonus if the proof is shorter than the above.
Best Answer
Write the Riemann zeta function as a product of its Euler factors $\zeta (s)=\prod_i E_{i}(s)$. Repeated application of the Leibniz rule shows $$\frac{\zeta^{(k)}(s)}{\zeta (s)}=\sum_{i_1+\cdots+i_k=k}\sum_{t_1,\dots,t_k}\frac{E_{t_1}^{(i_1)}(s)}{E_{t_1}(s)}\cdots \frac{E_{t_k}^{(i_k)}(s)}{E_{t_k}(s)}.$$ The left hand side is clearly the Dirichlet series for the generalized Von Mangoldt function, and the right hand side is a Dirichlet series that is supported on integers with at most $k$ prime factors.