Is $\sum_{n\leq x}{z^{\Omega(n)}} = O(x^{\frac12 + \varepsilon})$ equivalent to the Riemann hypothesis for all roots of unity $z\neq1$?

$$\Omega(n)$$ is the number of prime divisors of $$n$$, counted with multiplicity.

For $$z=-1$$, $$z^{\Omega(n)} = \lambda(n)$$ is the Liouville function, and it's known that $$\sum_{n\leq x}\lambda(n) = O(n^{\frac12 + \varepsilon})$$ is equivalent to the Riemann hypothesis.

Is this the case for all roots of unity other than 1? Is this a known result?

If we look at the Dirichlet series $$F_z(s) = \sum_{n=1}^{\infty} \frac{z^{\Omega(n)}}{n^s} = \prod_p {\frac{1}{1 - z p^{-s}}}$$, we have for the product of roots of unity of order $$n$$: $$\prod_{k=1}^n F_{e^{\frac kn2\pi i}}(s) = \prod_p {\frac{1}{1 - p^{-n s}}} = \zeta(n s)$$, and using Mobius inversion we can look only at the primitive roots of unity of order $$n$$, $$\prod_{k\leq n\\ (k,n)=1} F_{e^{\frac kn2\pi i}}(s) = \prod_{d | n} \zeta(d s)^{\mu(\frac nd)}$$, but I'm not sure how to continue from here and treat each of the roots separately.

Heuristically, $$\Omega(n)$$ is normal with increasing mean and variance, so for any order of a root of unity $$k$$ I'd expect $$\Omega(n) \mod k$$ to be uniform as $$n\to \infty$$, so $$\sum z^{\Omega(n)}$$ should be approximately a random walk in 2D and thus $$O(n^{\frac12 + \varepsilon})$$.

• The generating function has an essential singularity at $s=1$ except in the case where $z=1$ (where you get a pole) and $z=-1$ (where you get a zero -- and hence are holomorphic). This means that you won't get the kind of cancellation you are suggesting. The usual thing to do in this situation is to pull out the essential singularity as the zeta factor $\zeta(s)^z$ and then apply the Selberg-Delange method. Commented Oct 21, 2023 at 6:36
• @AnuragSahay From what I understand, the Selberg-Delange method gives bounds of $O(x (\log x)^k)$ (although much more precisely). How can I use it to prove a bound of $O(x^{\frac12 + \varepsilon})$? Commented Oct 21, 2023 at 8:41
• Such a bound is definitely false - it would imply that $F_z$ is holomorphic around $s=1$, which can't be true since as Anurag notes it is equal to $\zeta(s)^z$ times a holomorphic Euler product. I suspect the random walk model runs into biases due to the distinction between the Poisson model and the Poisson-Dirichlet model for the prime factors of a random number, but haven't worked out the details. Commented Oct 21, 2023 at 15:10
• @TerryTao I see, thanks, I'll definitely try the Selberg-Delange method. Commented Oct 21, 2023 at 15:24
• The 'true' main term of $\sum_{n \le x} z^{\Omega(n)}$ is not elegant but can be written as an Hankel integral of $F_z(s)x^s/s$ around the essential singularity at $s=1$; let us call this integral $M(x)$. RH is in fact equivalent to $\sum_{n \le x} z^{\Omega(n)}$=M(x)+O_{\varepsilon}(x^{1/2+\varepsilon})$. This is quite similar to Lucia's answer to this MO question: mathoverflow.net/q/35927/31469 Commented Oct 23, 2023 at 21:38 1 Answer Summarizing the discussion in the comments ($$+\epsilon$$ more): For roots of unity other than $$-1$$, one does not get square-root cancellation. In fact, an application of the Selberg-Delange method (see, for example, Chapter 5 of Tenenbaum's book) gives that if $$z \in S^1 \setminus \{-1\}$$, then $$\sum_{n\leqslant X} z^{\Omega(n)} \sim \frac{H_z(1)}{\Gamma(z)}\cdot\frac{X}{(\log X)^{1-\Re z}} , (\star)$$ where $$H_z(s) = \zeta(s)^{-z} F_z(s),$$ is an Euler product that is convergent (and hence holomorphic) for $$\Re(s) > 1/2$$. Note that this does not actually contradict the equidistribution of $$\Omega(n)$$ modulo $$k$$ for any $$k$$ which you alluded to in your question, since Weyl's criterion tells you that such equidistribution is equivalent to the assertion that $$\sum_{n\leqslant X} e\big(\frac{a}{k}\Omega(n)\big) = o(X),$$ for $$k \nmid a$$ and $$e(t) = e^{2\pi it}$$, which follows from $$(\star)$$ with $$z = e(a/k)$$ since the real part of a nontrivial root of unity is strictly less than $$1$$. Finally, see this question and its responses, which is basically about the same question. In particular, see this paper linked in so-called friend Don's answer, which seems to be the first in the literature to address your question. • Minor comment: One can establish cancellation in$\sum_{n\le X}z^{\Omega(n)}$($|z|=1\$) by appealing to Halász's Theorem (which avoids Hankel contour and in a sense more elementary than Selberg's proof). Commented Oct 23, 2023 at 16:11
• Aha, thanks for the comment. Commented Oct 23, 2023 at 20:30