Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots,x\}.$

Question

I apologise for the long-windedness of this question.

Let $a$ be a positive real constant and let $d(n)$ denote the number of divisors of $n.$ Define $$ S_a(x)=\sum_{n\leq x} d(n)^a. $$ For $a=1,$ the following is well known $$ S_1(x)=\sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal O}(\sqrt{x}) $$ while for more general $a$, one has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ More accurate estimates are available via the Selberg-Delange method, though the details get quite technical.

My question is the following. Let the subset $A$ be defined by $A\subset \{1,2,\ldots,x\},$ (assume $x$ is integer or use the floor function) with $$\#A\geq \frac{x}{2}+ c \frac{\log x}{\log \log x}:=Z(x).$$ Now define $$ S^{A}_{a}(x)=\sum_{n\leq x:n \in A} d(n)^a. $$ I want to lower bound this sum, as $A$ varies subject to the size condition, i.e., to derive a lower bound to $$ M:=\min \left\{ S_a^A(x): A \subset \{1,\ldots,x\}, \#A =Z(x) \right\}. $$ For $a=1,$ I hope a lower bound of the form $M\gg x \log x$ may be possible if the Poisson approximation is good enough. Essentially this would say that half the points achieve a constant factor of the full sum, even when restricted to those points with the fewest number of divisors.

From a Poisson approximation point of view, it would seem that approximately half the points $n\in \{1,\ldots,x\}$ have $\omega(n)\leq \log\log n,$ with the relevant Poisson distribution having mean $\log\log n.$

I am unsure if the known techniques are strong enough to address such a delicate size specification, compared to say $\lceil x/2 \rceil,$ but some comments related to Selberg-Delange have a discussion of how a multiplicative $O(1+(\log x)^{-c'})$ factor can be shown for various sums, which may imply that it is indeed possible.

Answer 1

Even if you take $Z(x)=(1-\varepsilon)x$ for some fixed $0<\varepsilon<1$, you are going to get

$$ M=x(\ln x)^{a\ln 2+o(1)}. $$

To prove this, observe that both $\omega(n)$ and $\Omega(n)$ have normal order $\ln\ln n$ (here $\Omega$ and $\omega$ are numbers of prime factors with and without multiplicity respectively). Also note that

$$ 2^{\omega(n)}\leq d(n)\leq 2^{\Omega(n)} $$

for every $n$, which implies that for any $\delta>0$ for all but $o(x)$ numbers $n\leq x$ we have

$$ (\ln x)^{\ln 2-\delta}\leq d(n) \leq (\ln x)^{\ln 2+\delta}. $$

Therefore, for any $A\subset [1,x]\cap \mathbb N$ with $|A|\geq cx$ for some $c>0$ we have

$$ S_a^A\geq (|A|-o(x))(\ln x)^{a(\ln 2-\delta)}, $$

so that $M\gg x(\ln x)^{a(\ln 2-\delta)}$. On the other hand, if you throw all the $n$ with $d(n)>(\ln x)^{\ln 2+\delta}$ out of your interval, you will still have $x-o(x)$ numbers left. This implies that for any $\varepsilon>0$ and $x$ large enough there is $A$ with $|A|\geq (1-\varepsilon)x$ such that

$$ S_a^A\leq |A|(\ln x)^{a(\ln 2+\delta)}. $$

Answer 2

There is a cheap way via Cauchy's inequality in case you do no want to use Erdos--Kac. I will only do it for $a=1$ just to outline the idea. $$\frac{x}{2} (1+o(1) ) \leq \sum_{n \leq x } 1_A(x)= \sum_{n \leq x } 1_A(x)\sqrt{\tau(n)} \tau(n)^{-1/2}\leq (S_1^A(x))^{1/2} (\sum_{n\leq x } \tau(n)^{-1} )^{1/2} .$$ Now use $$ \sum_{n\leq x } \tau(n)^{-1} \ll \frac{x}{\sqrt{\log x } }$$ to get $$ M \gg x (\log x)^{1/2}.$$ This is a long way from $M\gg x \log x$ but it is better than the trivial bound $M \geq x/2 (1+o(1) ) $. You might try to use H"older's inequality instead to find a better logarithmic exponent.