# Bound for sums of bounded multiplicative functions that are zero at primes

Let $h:\mathbb{N}\rightarrow\mathbb{C}$ be a bounded multiplicative function with $h(p)=0$. The motivation for this question is just a general enquiry and, since I suppose it has already been considered, it seems worthwhile asking it here.

What is the best possible bound on the modulus of the sum $$S(x)=\sum_{n\leq x}h(n),$$that is, for all such $h$?

Using some basic facts about Euler products it follows that $S(x)\in O(x^{1/2+\epsilon})$, $\epsilon>0$, but can $\epsilon$ be replaced by something explicit?

• The number of square-full integers up to $x$ is $O(x^{\frac 12})$. Nov 13 '14 at 0:32

To expand on Lucia's comment, we note that we can assume without loss of generality that each coefficient of $h$ is bounded by $1$, in which case $S(x)$ is largest when $h(p^k) = 1$ for all $k \geq 2$ (and of course $h(p) = 0$). So $S(x) = \#\left\{n \leq x : p | n \implies p^2 | n\right\}.$ This is the number of powerful numbers less than $x$, which Golomb has shown to be asymptotic to $\sum_{n = 1}^{\infty}\frac{\mu^2(n)}{n^{3/2}} x^{1/2} = \frac{\zeta(3/2)}{\zeta(3)} x^{1/2},$ with $\frac{\zeta(3/2)}{\zeta(3)} \approx 2.173.$ One can see this directly by noting that the Dirichlet series with coefficients given by the multiplicative function $h(p^k) = 1$ for all $k \geq 2$, $h(p) = 0$, satisfies $\sum_{n = 1}^{\infty} \frac{h(n)}{n^s} = \prod_p \sum_{\substack{k = 0 \\ k \neq 1}}^{\infty}\frac{1}{p^{ks}} = \prod_p \left(\frac{1}{1 - p^{-s}} - \frac{1}{p^s}\right),$ and some manipulation shows that this is equal to $\prod_p \frac{1 - p^{-6s}}{(1 -p^{-2s})(1 - p^{-3s})} = \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)},$ which is holomorphic for $\Re(s) > 1/2$ and has simple poles at $s = 1/2$, $s = 1/3$, and then at the nontrivial zeroes of $\zeta(6s)$ in the strip $0 < \Re(s) < 1/6$, plus at the trivial zeroes of $\zeta(6s)$ to the left of $\Re(s) = 0$. So Cauchy's residue theorem implies that $\begin{split} S(x) & = \frac{1}{2\pi i} \int^{\sigma + i\infty}_{\sigma - i\infty} \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} \frac{x^s}{s} ds \\ & = \frac{\zeta(3/2)}{\zeta(3)} x^{1/2} + \frac{6 \zeta(2/3)}{\pi^2} x^{1/3} + o(x^{1/6}). \end{split}$