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I need some help to find a mistake in my proof.

I have to prove that $\sum_{n\leq x}2^{\Omega(n)}\sim cx\log^2x$ for $x\rightarrow+\infty$, where $\Omega(p_1^{k_1}\cdot\ldots\cdot p_j^{k_j})=k_1+\ldots+ k_j$.

I defined $F(s)=\sum_{n\geq 1}\frac{2^{\Omega(n)}}{n^s}$ and calculated its Euler's product as follows (here, I use the fact that $\Omega(p^k)=k$ for all primes $p$) \begin{equation} \prod_p\left(1+\sum_{k\geq1}\frac{2^k}{p^{ks}}\right)=\prod_p\left(1+\frac{2}{p^s}\frac{1}{1-\frac{2}{p^s}}\right)=\prod_p\left(\frac{p^s}{p^s-2}\right). \end{equation}

Let $H(s)$ s.t. $F(s)=H(s)\zeta^{2}(s)$, so that \begin{equation} H(s)=\prod_p\left(\frac{p^s}{p^s-2}\right)\left(1-\frac{1}{p^s}\right)^{2}=\prod_p\left(1+\frac{1}{p^s(p^s-2)}\right) \end{equation} and this product converges for $\Re(s)>\frac{1}{2}$.

Then, I defined $h(n)$ such that $H(s)=\sum_{n\geq 1}\frac{h(n)}{n^s}$ (converging for $\Re(s)>1/2$). In this way, by $F=H\zeta^2$, you get the expression $2^{\Omega(n)}=h\ast d(n)$, where $\zeta^2(s)=\sum_{n\geq1}\frac{d(n)}{n^s}$.

So, \begin{equation} \sum_{n\leq x}2^{\Omega(n)}=\sum_{n\leq x}\sum_{m|n}h(m)d\left(\frac{n}{m}\right)=\sum_{d\leq x}h(d)\sum_{m\leq \frac{x}{d}}d(m). \end{equation} Then, I used $\sum_{n\leq x}d(n)=x\log x+O(x)$, to get \begin{equation} \sum_{n\leq x}2^{\Omega(n)}=x\log x\sum_{d\leq x}\frac{h(d)}{d}-x\sum_{d\leq x}\frac{h(d)\log d}{d}+O(x). \end{equation} Now, $H(1)=\sum_{n\geq 1}\frac{h(n)}{n}$ and $H'(1)=-\sum_{n\geq 1}\frac{h(n)\log n}{n}$, so that I get

\begin{equation} \sum_{n\leq x}2^{\Omega(n)}=H(1)x\log x+... \end{equation}

I can't get $Cx\log^2x$ out in any way. Please, help me!

Gary
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UnusualMathem
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  • When you compute the Euler product for $H$ you compute $F(s)\zeta ^2 (s)$. However, $H(s)=F(s)/\zeta ^2 (s)$. – Gary Mar 31 '20 at 21:26
  • Sorry, I typed $\zeta^{-2}$, but actually, in the calculation I use $\zeta^{2}$. I corrected the mistake in the text here :) – UnusualMathem Mar 31 '20 at 21:38
  • Hi, in fact the asymptotic should not be $\sim c x (\log x)^2$ but $\sim c x \log x$. Philosophically: if $n$ is square-free then $2^{\Omega(n)}$ equals $d(n)$, the divisor function. The divisor function has average asymptotic $x \log x $, thus, the same should happen with $2^\Omega$. – Mr. No Apr 01 '20 at 16:26
  • Well, actually, I was wrong xD, I posted the correct solution :) – UnusualMathem Apr 01 '20 at 16:36
  • Does this answer your question? Asymptotic estimate for the sum $\sum_{n\leq x} 2^{\omega(n)}$ – Kimball Dec 09 '22 at 14:28
  • @Kimball: The proposed duplicate is related but not the same. Here $\Omega(n)$ counts prime factors up to multiplicity while your target involves $\omega(n)$ counting distinct prime factors. Obviously $\omega(n) \le \Omega(n)$ but more is needed to make a conclusion of the form sought here. – hardmath Dec 10 '22 at 14:50
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    @hardmath Ah, right. It was late and both $\Omega$ and $\omega$ hashed to the same thing in my head. – Kimball Dec 11 '22 at 09:41

1 Answers1

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I think I solved the problem: the function $F(s)$ defined above has infinite poles relatively to every $s=\log_p2+\frac{2k\pi I}{\log p}$ and an essential singularity in $0$, so I can't express $F(s)=H(s)\zeta^2(s)$ where $H$ is a holomorphic function in $\Re(s)>1/2$.

I suppose I have to work on the summation $\sum_{n\leq x}2^{\Omega(n)}$ in order to isolate the terms relative to those $n=2^{k}m$ for $k\geq 1$. This is an easy computation that follows from the splitting of the sum $\sum_{2^k m\leq x}2^{\Omega(2^km)}=\sum_{k\leq \log_px}\sum_{m\leq x/2^k}...$, etc.

Thank you, anyway :)

UnusualMathem
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