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Take a positive integer $k$. Let $S \subseteq \mathbb{N}$ be the set of squarefree, and define $\displaystyle \rho_k = \mu*\omega^k$, where $ \displaystyle f* g = \sum_{d \mid n} f(d)g(n/d)$ . Show that $\rho_k(n)$ is 0 for almost all $n \in S$.

Note: $\mu(n) $ is the moebius function and $\omega(n) = |\{p \mid n, p \text{ prime} \}|$

Michael Hardy
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frame95
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Let $n = p_1\cdots p_r$ be the prime factorization of $n\in S$. Then, its divisors are of the form $p_{i_1}\cdots p_{i_j}$ with $\{i_1,\ldots , i_j\}\subseteq \{1,\ldots , r\}$. Notice that $\mu(p_{i_1}\cdots p_{i_j}) = (-1)^j$ and $\omega^k(p_{i_1}\cdots p_{i_k}) = j^k$. Notice that there are $\binom{r}{j}$ divisors of $n$ which have $j$ prime factors.

Thus, your $\rho_k(n)$ is $$\rho_k(n) = \displaystyle\sum_{j=0}^r \binom{r}{j}(-1)^{r-j} j^k$$

Since this is the $r$-th forward difference of $x^k$ at $x=0$, for sufficiently large $r$ it must be $0$.

However, I don't know what notion of "almost all" you're using, because there is an infinite number of values for which $\rho_k$ isn't $0$ (eg at prime numbers), and I think that "almost all" usually refers to "all but finitely many".

Nacho Darago
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  • "Almost all" usually means "the complement of a set of measure zero". For the counting measure, this means all but finitely many, but in this case, it probably refers to a measure based on arithmetic progressions. – Andrew Dudzik Feb 03 '15 at 16:27
  • Yes, you are right, I forgot to write it. I meant as measure $\delta$ of a set $S \subseteq \mathbb{N}$ the one given by $\displaystyle \delta(S) = \lim_{n \to \infty} | S_n| /n $, where $S_n:= S \cap [1, n] $. And for "almost all" I mean $\delta(S) = 1$. – frame95 Feb 07 '15 at 02:11