Hardy-Ramanujan theorem for $\Omega(n)$

Question

From the Theorem we have that $$\frac{1}{x}\sum_{n\leq x}(\omega(n)-\log\log n)^2=(1+o(1))\log\log x,$$ and for all $\epsilon>0$ $$|\{n\leq x:|\omega(n)-\log\log n|\geq\epsilon\log\log x\}|<<\epsilon x.$$ In every reference I've read It's written that the same results holds for $\Omega(n)$, but for me It's not clear! Do the same estimates of $\omega(n)$ hold for $\Omega(n)$?

Answer 1

The answer for the first estimate is yes. I base my explanation on Problem 26 in Section 8.3 of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991. In the following, $n$, $k$, $j$, and $i$ are positive integers; $x$ is a real number; and $p$ and $q$ are primes.

\begin{align} \sum_{n \le x} (\Omega(n) - \omega(n))^2 & = \sum_{n \le x} \left( \sum_{\substack{p^k|n \\ k > 0}} 1 - \sum_{p|n} 1 \right)^2 & \text{(by definition of } \Omega \text{ and } \omega) (1)\\ & = \sum_{n \le x} \left( \sum_{\substack{p^k|n \\ k > 1}} 1 \right)^2\\ & = \sum_{n \le x} \left( \sum_{\substack{p^k|n \\ k > 1}} 1 \right) \left( \sum_{\substack{q^j|n \\ j > 1}} 1 \right) & (2)\\ & = \sum_{n \le x} \sum_{\substack{p^k|n \\ k > 1}} \sum_{\substack{q^j|n \\ j > 1}} 1 & \text{(by the distributive law)}\\ & = \sum_{\substack{p \ne q \\ p^kq^j \le x \\ k > 1, \, j > 1}} \left\lfloor \frac{x}{p^kq^j} \right\rfloor + \sum_{\substack{p^k \le x \\ k > 1}} (2k - 3) \left\lfloor \frac{x}{p^k} \right\rfloor. \end{align} The last step might require some explanation: We grouped the products of two $1$s according to whether $p = q$. If not, we see from Equation ($2$) that we have $\lfloor x/(p^kq^j) \rfloor 1$s. Note that I lumped the two indices of summation $p^k \le x, q^j \le x$ into $p^kq^j \le x$ to remove terms for which the floor function returns zero. If $p = q$, consider Equation ($1$): If $p^2|n$ and $p^3 \nmid n$, then $(\Omega(n) - \omega(n))^2$ contributes $(2 - 1)^2 = 1^2$ to the sum, and there are $\lfloor x/p^2 \rfloor - \lfloor x/p^3 \rfloor$ such $n$. Continuing for greater powers of $p$, we have \begin{align} \sum_{\substack{p^k|n \\ k > 1}} (\Omega(n) - \omega(n))^2 & = 1^2 \left( \left\lfloor \frac{x}{p^2} \right\rfloor - \left\lfloor \frac{x}{p^3} \right\rfloor\right) + 2^2 \left( \left\lfloor \frac{x}{p^3} \right\rfloor - \left\lfloor \frac{x}{p^4} \right\rfloor\right) + \cdots\\ & = 1^2 \left\lfloor \frac{x}{p^2} \right\rfloor + (2^2 - 1^2) \left\lfloor \frac{x}{p^3} \right\rfloor + (3^2 - 2^2) \left\lfloor \frac{x}{p^4} \right\rfloor + \cdots\\ & \quad + ((i - 1)^2 - (i - 2)^2) \left\lfloor \frac{x}{p^i} \right\rfloor + \cdots\\ & =\sum_{\substack{p^k \le x \\ k > 1}} (2k - 3) \left\lfloor \frac{x}{p^k} \right\rfloor \qquad \qquad \qquad \text{(because $(i - 1)^2 - (i - 2)^2 = 2i - 3$).} \end{align}

Continuing with the original chain of equations, we have \begin{align} \sum_{n \le x} (\Omega(n) - \omega(n))^2 & = \sum_{\substack{p \ne q \\ p^kq^j \le x \\ k > 1, \, j > 1}} \left\lfloor \frac{x}{p^kq^j} \right\rfloor + \sum_{\substack{p^k \le x \\ k > 1}} (2k - 3) \left\lfloor \frac{x}{p^k} \right\rfloor\\ & = \sum_{\substack{p \ne q \\ p^kq^j \le x \\ k > 1, \, j > 1}} \left( \frac{x}{p^kq^j} + O(1) \right) + \sum_{\substack{p^k \le x \\ k > 1}} (2k - 3) \left( \frac{x}{p^k} + O(1) \right)\\ & = x \left( \sum_{\substack{p \ne q \\ p^kq^j \le x \\ k > 1, \, j > 1}} \frac{1}{p^kq^j} + \sum_{\substack{p^k \le x \\ k > 1}} \frac{2k - 3}{p^k} \right)\\ & \quad + O\left( \sum_{\substack{p \ne q \\ p^kq^j \le x \\ k > 1, \, j > 1}} \frac{1}{p^kq^j} + \sum_{\substack{p^k \le x \\ k > 1}} \frac{2k - 3}{p^k} \right)\\ & = O(x) + O(1) \qquad \qquad \text{(because the sums converge and so are constants)}\\ & = O(x). \end{align} To check the reasonableness of that intermediate result, I fit a line to $f(x) = \sum_{n \le x} (\Omega(n) - \omega(n))^2$ versus $x$ for $x = 1, 2, 3, \, \ldots, 32000$ and got $f(x) \approx 2.2751x - 299.97$ with $R^2 \approx 1$.

Now \begin{align} \sum_{n \le x} ((\Omega(n) - \omega(n)) + (\omega(n) - \log \log n)) & = O(\sqrt x) + \sqrt{(1+o(1)) x \log \log x}\\ \sum_{n \le x} (\Omega(n) - \log \log n) & = \sqrt{(1+o(1)) x \log \log x}\\ \sum_{n \le x} (\Omega(n) - \log \log n)^2 & = (1+o(1)) x \log \log x\\ \frac{1}{x} \sum_{n \le x} (\Omega(n) - \log \log n)^2 & = (1+o(1)) \log \log x. \end{align} That answers the first part of your question.

After I wrote that answer, I found another proof in Martin Klazar's lecture notes "Introduction to Number Theory," Section 4.3, which also includes the answer to the second part of your question.