Put
$$
L=\log\log x,
\qquad
S=\sum_{p\leq x}\frac1p=L+O(1)
$$
by part a. Double-counting divisibility gives the first moment of the <prime omega function>:
$$
\sum_{n\leq x}\omega(n)
=\sum_{p\leq x}\left\lfloor\frac xp\right\rfloor
=xS+O(\pi(x)).
$$
Moreover,
$$
\omega(n)^2=\omega(n)+2\sum_{\substack{p<q\\pq\mid n}}1,
$$
so
$$
\sum_{n\leq x}\omega(n)^2
\leq xS+2x\sum_{p<q\leq x}\frac1{pq}
\leq xS+xS^2.
$$
Expanding the square and using the given bound $\pi(x)\ll x/\log x$ now gives the <Turán normal-order theorem for distinct prime divisors> estimate
$$
\sum_{n\leq x}(\omega(n)-L)^2
\ll x(S-L)^2+xS+L\pi(x)
\ll xL.
$$

By the <Chebyshev inequality>, the number of $n\leq x$ for which
$$
|\omega(n)-L|>\tfrac12L^{3/4}
$$
is $O(x/L^{1/2})=o(x)$. Discard the $O(\sqrt x)$ integers below $\sqrt x$. For $\sqrt x<n\leq x$,
$$
|\log\log n-L|\leq\log2
$$
and $(\log\log n)^{3/4}\sim L^{3/4}$. Hence, for all sufficiently large $x$, every remaining integer counted in the question also satisfies the preceding inequality. Therefore
$$
\boxed{\#\{n\leq x:|\omega(n)-\log\log n|>(\log\log n)^{3/4}\}=o(x)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-150/1/b/solution