In the article "Primes in tuples I" by Goldston, Pintz and Yıldırım, it is claimed that the Dirichlet convolution

\begin{equation} \sum_{d\mid n}\mu(d)\left(\log \frac{n}{d}\right)^k \end{equation} vanishes if $n$ has more than $k$ distinct prime factors. Here $\log$ denotes the natural logarithm. I have tried to show this, but I haven't been successful. I attempted a "brute force" solution by simply trying to evaluate the sum, but that was not successful. Clearly, the function isn't multiplicative either.

The Dirichlet series of this function is the $k$:th derivative of the reciprocal of the zeta function (Edit: This is wrong, see the accepted answer), but I don't see how that can help me. I would be grateful for any help.

  • $\begingroup$ It is possibly to define arithmetic derivative for arithmetic function that obeys product rule. You may find synopsis in Apostol's book. $\endgroup$
    – TravorLZH
    Jul 1, 2021 at 7:07

1 Answer 1


Usually we define

\begin{equation} \Lambda_k= \sum_{d\mid n}\mu(d)\left(\log \frac{n}{d}\right)^k \end{equation}

where $\Lambda_0=\delta_0$ and $\Lambda_1$ is the usual Von Mangoldt function.

Now the actual generating Dirichlet series for $\Lambda_k$ is given by $(-1)^k\frac{\zeta^{(k)}}{\zeta}$ as one can easily see since if $L$ is the logarithm operator, $\Lambda_k=\mu*L^k$, while in Dirichlet terms $L$ is the negative differential operator and $L^k$ is then $(-1)^k$ times the $k$th derivative of the function generated by $1$ which is $\zeta$

From this, it immediately follows that:

(noting that $L\Lambda_k$ is generated by $-\frac{d}{ds}[(-1)^k\frac{\zeta^{(k)}}{\zeta}]$ and that $\Lambda*\Lambda_k$ is generated by $(-1)\frac{\zeta'}{\zeta} \times (-1)^k\frac{\zeta^{(k)}}{\zeta}$)

$\Lambda_{k+1}=L\Lambda_k+\Lambda*\Lambda_k$ and now the result follows by induction:

Cases $0,1$ are clear, and since $\Lambda$ is supported on prime powers and $\Lambda_k$ on numbers with at most $k$ distinct prime factors by the inductive hypothesis, their convolution is supported on numbers with at most $k+1$ distinct prime factors pretty much by the definition of the convolution, while $L\Lambda_k$ is already supported only on numbers with at most $k$ distinct prime factors.

  • $\begingroup$ Thank you, a very clear explanation. $\endgroup$ Jul 1, 2021 at 7:26
  • $\begingroup$ happy to be of help $\endgroup$
    – Conrad
    Jul 1, 2021 at 15:46

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