# Was this arithmetic Möbius/Mangoldt function ever used for something?

Definition: Let $$n=\prod_k p_k^{c_k}$$, with $$p_k \in \mathbb P$$ and $$A(n)=\sum_{d|n} \mu(d)\Lambda(d)=\sum_\limits{c_k\neq 0} \log p_k ,$$

with the Möbius function $$\mu(n)$$, which is:

• $$μ(n) = \pm 1$$ if $$n$$ is a square-free positive integer with an even/odd number of prime factors.
• $$μ(n) = 0$$ if $$n$$ has a squared prime factor.

And the von Mangoldt function $$Λ(n)$$, defined as $$\Lambda(n) = \begin{cases} \log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

Properties of $$A(n)$$:

• $$A(\cdot)$$ basically takes every number $$n$$ to a sum of logarithms of prime factors, e.g. $$A(12)=A(2^23)=\log 2 + \log 3=A(6);\\ A(4)=A(2)=\log 2.$$

• The arithmetic function is additive, since $$A(nm)=A(n)+A(m)$$ only for co-prime $$n$$ and $$m$$, e.g. $$\log 2 + \log 3=A(6)=A(2\cdot 3) = A(2)+A(3)= \log 2 + \log 3$$

• You can apply the Möbius Inversion to it. You'll get: $$\mu(n)\Lambda(n)=\sum_{d|n} \mu(d) A(n/d),$$ e.g. (where I write only terms with $$\mu(d)\neq 0$$) $$0=\mu(12)\Lambda(12)=\mu(6)A(2)+\mu(3)A(4)+\mu(2)A(6)+\mu(1)A(12)\\ =\log 2-\log 2-(\log 2+\log 3)+(\log 2+\log 3)=0$$

Was the additive arithmetic function $$A(n)$$ ever used in any context?

## 2 Answers

Given the unexplanatory tone of many number theory expositions, it is, in fact, reasonable to ask this question, I think... insofar as there is a good explanation, apart from the artifactual one.

That is, the Riemann Explicit Formula, as reformulated a bit by von Mangoldt, is an exact (not merely asymptotic) equality of a sum over primes and a sum over zeros of the zeta function. The sum over primes, at best, is not quite just $\sum_p 1$, but, perhaps most simply and most naturally, a sum of $\log p$ over prime powers $p^m$. That is, the "left-hand side" is $\sum_{m,p:p^m<T}\log p$. That is, one side is a sum over prime powers under a given bound, and the weight of the counting is $\log p$. This is an immediate and transparent artifact of the complex analysis the extracts the equality. No imagination required ... beyond the profound insight to do the thing at all... :)

The understandable confusion arises from both unexplanatory sequels that presume one knows the context, as well as from knock-off derivative work that takes whatever is found in books and papers and unthinkingly fools around with it.

Thus, the short answer is that there are compelling reasons for the appearance of this weighting scheme. Sure, it is not what the most-naive/ideal program would ask for, but it is somehow the perfectly correct answer. The more naive question of "counting primes" is somehow incorrectly posed... in the sense that one must go from the clear (with RH or not...) statement of the explicit formula to messier assertions about the naive-but-awkward formulation.

Sure, we could say that the more-sophisticated answer dodges the original question. Or, we could say that the original question was in some sense doomed, because it could not admit as simple an answer as we hoped, while the more-complicated question allowed a good answer.

• hmm, I don't get what you trying to say... – draks ... Sep 10 '15 at 10:44
• In brief, the von Mangoldt function arises naturally, fairly inevitably, rather than just being made up. The fact that it's not what we might wish it to be is somewhat irrelevant. That is, the most obvious "count primes" function is not as natural as counting weighted by the von Mangoldt function. That kind of thing. This becomes clear in the Riemann Explicit formula, in Guinand-Weil extension of it, and so on. – paul garrett Sep 10 '15 at 18:01

I cite from here:

Let $\mathrm{rad}(n)$ denote the radical of an integer $n$, which is the product of the distinct prime numbers dividing n. Or equivalently, $$\mathrm{rad}(n)=\prod_{\scriptstyle p\mid n\atop p\text{ prime}}p.$$ Assume $\mathrm{rad}(1)=1$, so that $\mathrm{rad}(n)$ is multiplicative.

So $$A(n)=\log\mathrm{rad}(n) \;\text{ or }\; \mathrm{rad}(n)=\exp{\left(\sum_{d|n} \mu(d)\Lambda(d)\right)}.$$ Then Wiki:Radical of an integer states that:

One of the most striking applications of the notion of radical occurs in the abc conjecture, which states that, for any $ε > 0$, there exists a finite $K_ε$ such that, for all triples of coprime positive integers $a, b,$ and $c$ satisfying $a + b = c$, $$c < K_\varepsilon\, \operatorname{rad}(abc)^{1 + \varepsilon}$$ Furthermore, it can be shown that the nilpotent elements of $\mathbb{Z}/n\mathbb{Z}$ are all of the multiples of $\operatorname{rad}(n)$.

More references at OEIS:A007947...