In *Multiplicative Number Theory - Vol. I* by Montgomery and Vaughan the following result is proved.


**Theorem 7.20** Let $A(x,r)$ denote the number of $n\leq x$ such that $\Omega(n)\leq r \log \log x,$ and let $B(x,r)$ denote the number of $n\leq x$ for which $\Omega(n)\geq r \log \log x.$ If $0<r\leq 1$ and $x\geq 2$ then
$$
A(x,r)\ll x(\log x)^{r-1-r \log r}.
$$
If $1\leq r \leq R<2$ and $x\geq 2$ then
$$
B(x,r)\ll\_{R} x(\log x)^{r-1-r \log r}.
$$
How does the $R$-dependent proportionality constant vary as $R$ ranges over $[1,2)$?


Also, the indicated proof techniques are the same for both $A(x,r)$ and $B(x,r),$ but $r<R$ is required for $B(x,r)$. What happens if $r\geq 2$? Surely then even fewer $n$ satisfy the inequality $\Omega(n)\geq r \log \log n$ so the bound on $B(x,r)$ still holds. So why the restriction $r<R$?



Proportionality constant in Montgomery-Vaughan Theorem 7.20
**Edit:** *My only goal is to mark this as answered.*


As explained by Greg Martin in a comment:


Certainly $(,)$
is a decreasing function of $$,
but that doesn't imply that the bound on $(,)$
continues to hold for $\geq 2$.


The reason that $<2$
is required in the given proof is that it proceeds via upper bounds for
$$
\sum\_{n \leq x} \Omega(n),
$$
and this sum genuinely changes character when $>2$, since the integers $$
that are powers of 2 (or nearly so) are then increased by the map $n \mapsto \Omega(n)$
; for example, the largest power of $2$ less than $$ already gives a contribution of


$$x^{(\log r)/\log 2}$$
to the sum.


