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How to find an estimate for the sum $\sum_{n\leq x} 2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime factors of $n$.

Since $2^{\omega(n)}$ is multiplicative, computing its value at prime power, we see that $2^{\omega(n)}=\sum_{d\mid n}\mu^2(d)$. Then \begin{align} \sum_{n\leq x}2^{\omega(n)}&=\sum_{n\leq x}\sum_{d\mid n}\mu^2(d)=\sum_{d\leq x}\mu^2(d)\sum_{d\mid n} 1\\ &=\sum_{d\leq x}\mu^2(d)\left\lfloor \frac{x}{d}\right\rfloor=\sum_{d\leq x}\mu^2(d)(\frac{x}{d}+O(1))\\ &=x\sum_{d\leq x}\frac{\mu^2(d)}{d}+O(\sum_{d\leq x}\mu^2(d)) \end{align}

I get stuck here, the series $\sum_{n=1}^\infty\frac{\mu^2(n)}{n}$ is not convergent, I don't know how to estimate the first term.

qwr
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Xiang Yu
  • 5,193

2 Answers2

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It is not difficult to see that $$\sum_{d\leq x}\mu^{2}\left(d\right)=x\frac{6}{\pi^{2}}+O\left(\sqrt{x}\right)\tag{1}$$ (for a proof see here) so by Abel's summation we have $$\sum_{d\leq x}\frac{\mu^{2}\left(d\right)}{d}=\frac{\sum_{d\leq x}\mu^{2}\left(d\right)}{x}+\int_{1}^{x}\frac{\sum_{d\leq t}\mu^{2}\left(d\right)}{t^{2}}dt $$ hence using $(1)$ we have $$ \begin{align*} \sum_{d\leq x}\frac{\mu^{2}\left(d\right)}{d}= & \frac{6}{\pi^{2}}+O\left(\frac{1}{\sqrt{x}}\right)+\frac{6}{\pi^{2}}\int_{1}^{x}\frac{1}{t}dt+O\left(\int_{1}^{x}\frac{1}{t^{3/2}}dt\right)\\ = & \frac{6}{\pi^{2}}\log\left(x\right)+O\left(1\right) \end{align*} $$ hence $$\sum_{n\leq x}2^{\omega\left(n\right)}=\frac{6}{\pi^{2}}x\log\left(x\right)+O\left(x\right).$$

Marco Cantarini
  • 33,458
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I might be too late to the party, but here's an argument that gives the true main term and a better error term. The key ingredient comes from Dirichlet's error term for his divisor problem.

The Dirichlet series $\sum_n 2^{\omega(n)}/n^s$ has Euler product $$\prod_p\left(1+\frac2{p^s}+\frac2{p^{2s}}+\frac2{p^{3s}}+\cdots\right)=\prod_p\frac{1+p^{-s}}{1-p^{-s}}=\frac{\zeta(s)^2}{\zeta(2s)},$$ hence the identity $$2^{\omega(n)}=\sum_{de^2=n}\tau(d)\mu(e)$$ where $\tau$ stands for the number of divisors. Summing over $n\le x$, $$\begin{aligned} \sum_{n\le x}2^{\omega(n)}&=\sum_{de^2\le x}\tau(d)\mu(e)\\ &=\sum_{e\le x^{1/2}}\mu(e)\sum_{d\le x/e^2}\tau(d)\\ &=\sum_{e\le x^{1/2}}\mu(e)\left(\frac{x}{e^2}\ln\frac{x}{e^2}+(2\gamma-1)\frac{x}{e^2}+O(x^{1/2}/e)\right)\\ &=x\ln x\sum_{e\le x^{1/2}}\frac{\mu(e)}{e^2}+x\sum_{e\le x^{1/2}}\frac{\mu(e)}{e^2}(2\gamma-1-2\ln e)+O(x^{1/2}\ln x)\\ &=\frac{x\ln x}{\zeta(2)}+x\left(\frac{2\gamma-1}{\zeta(2)}-2\frac{\zeta'(2)}{\zeta(2)^2}\right)+O(x^{1/2}\ln x). \end{aligned}$$ (The $\zeta'(2)$ comes from differentiating both sides of $1/\zeta(s)=\sum_n \mu(n)/n^s$.)

pzq_alex
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