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Let $\mu_{k}(n)$ be the MÖBIUS function of order $k$, defined by $$ \mu_{k}(n)= \begin{cases} 1 &\text{if }\: n=1,\\ 0 &\text{if }\: p^{k+1}\mid n, \\ (-1)^r & \text{if }\: n = p^k_1· · · p^k_r\prod_{i>r} p_i^{α_i}, \quad0<\alpha_i<k, \\ 1 & \text{otherwise}. \end{cases} $$

Under this definition, what could be this $$\sum_{n\leq x}\mu_k(n)^2\:?$$ Any small hint is welcome!

Bernard
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Math123
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1 Answers1

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This $\mu_k(n)^2$ is simply the indicator function of the $(k+1)$-free numbers—it equals $1$ if $n$ is not divisible by the $(k+1)$st power of a prime and $0$ otherwise. It is well known that $$ \sum_{n\le x} \mu_k(n)^2 \sim \frac x{\zeta(k+1)}, $$ generalizing the case $k=1$.

Indeed, one can even show using standard multiplicative-function techniques that for $k\ge2$, $$ \sum_{n\le x} \mu_k(n) \sim x \prod_p \bigg( 1 - \frac2{p^k} + \frac1{p^{k+1}} \bigg), $$ where the product is over all primes $p$. (The case $k=1$ is harder, essentially equivalent to the prime number theorem.) From these two results one can work out the density of those numbers $n$ for which $\mu_k(n)=1$ or $\mu_k(n)=-1$ or $\mu_k(n)=0$.

Greg Martin
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    Well-known ?? What is the abscissa of convergence of $(\sum \mu_m m^{-s})(\sum_n \mu_k(n)^2 n^{-s})$ and its value at $1$ ? The OP probably wants to know that the result follows from there. – reuns Aug 13 '20 at 10:42