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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.

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I think the fundamental underlying fact is that if you do $(k+1)$-fold repeated differences on a polynomial of order $k$, then you get 0. That is $\sum_{\epsilon_1,\ldots,\epsilon_{k+1}}f(x+\epsilon_1d_1+\ldots+\epsilon_{k+1}d‌​_{k+1})=0$, where then $\epsilon_i$ run over $\{\pm 1\}$. In your case, the polynomial is $f(t)=t^k$. – Anthony Quas Nov 25 '13 at 23:20
up vote 13 down vote accepted

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.

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Everything I know about these generalized von Mangoldt functions goes through the identity $$ \Lambda_{k+1}(n) = (\Lambda_k \ast \Lambda)(n) + \Lambda_k(n) \log n $$ (where $\ast$ denotes Dirichlet convolution); this identity isn't obvious, but can be proved by identifying the Dirichlet series corresponding to both sides. But in particular, the desired statement about the support of $\Lambda_k$ follows easily by induction from this recursive identity.

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I just wanted to add that the specific Dirichlet series identity follows from differentiating $\zeta^{(k)}(s)/\zeta(s)$. We have that $$\frac{d}{ds}\left(\frac{\zeta^{(k)}(s)}{\zeta(s)}\right)=\frac{\zeta^{(k+1)}(s‌​)}{\zeta(s)}-\frac{\zeta'(s)}{\zeta(s)}\frac{\zeta^{(k)}(s)}{\zeta(s)}, $$ and by looking at the coefficents we obtain $$\Lambda_k(n)\log n= \Lambda_{k+1}(n)-(\Lambda_k*\Lambda)(n).$$ – Eric Naslund Nov 26 '13 at 17:29

This is a supplement to the responses so far. Here is a proof avoiding Dirichlet series that $$ \Lambda_{k+1}=\Lambda_k\log+\Lambda\ast\Lambda_k. $$ By definition, $$ \Lambda_{k+1}=\mu\ast(\log^k\log)=(\mu\ast\log^k)\log+(-\mu\log)\ast\log^k $$ $$ =(\mu\ast\log^k)\log+(-\mu\log)\ast 1\ast\Lambda_k =\Lambda_k\log+\Lambda\ast\Lambda_k. $$ In the first line, we used that $\log(n/d)=\log(n)-\log(d)$. In the second line, we used that $\Lambda=(-\mu\log)\ast 1$, which is the special case of the first line for $k=0$.

P.S. From this identity one can prove by induction on $k$ that $\Lambda_k(n)\geq 0$, with equality iff $n$ has more than $k$ distinct prime factors.

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Whether you use Dirichlet series or not it seems like the point is that pointwise multiplication by $\log n$ is a derivation with respect to Dirichlet convolution. If we denote this derivation by $D$ then the first line above is $\mu \ast (D \log^k) + (D \mu) \ast (\log^k) = D (\mu \ast \log^k)$ which is of course just the Leibniz rule. – Qiaochu Yuan Nov 26 '13 at 21:49
@Qiaochu: Good point, thanks! – GH from MO Nov 26 '13 at 21:50

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