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Statement

Write $s(n)=\sigma(n)-n$ for the sum of the proper divisors of $n$.

Theorem 1.2 (manuscript p. 2), quoted: "Let $\epsilon=\epsilon(x)$ be a fixed function tending to 0 as $x\to\infty$. Suppose that $\mathcal{A}$ is a set of at most $x^{1/2+\epsilon(x)}$ positive integers. Then, as $x\to\infty$,

$$\#\{n\leq x:s(n)\in\mathcal{A}\}=o_\epsilon(x),$$

uniformly in the choice of $\mathcal{A}$."

In other words: for each such function $\epsilon$ there is a function $\delta_\epsilon(x)\to0$, depending on $\epsilon$ alone, such that every set $\mathcal{A}$ of positive integers with total cardinality $|\mathcal{A}|\leq x^{1/2+\epsilon(x)}$ has $\#\{n\leq x:s(n)\in\mathcal{A}\}\leq\delta_\epsilon(x)\,x$. The hypothesis bounds the whole of $\mathcal{A}$, not $|\mathcal{A}\cap[1,x]|$.

Infinite targets (derived here). The abstract (p. 1) draws the consequence that the conjecture of Erdős, Granville, Pomerance and Spiro holds for infinite sets with counting function $O(x^{1/2+\epsilon(x)})$, and the example after the theorem (p. 2) uses $s(n)<2n\log\log n$ for all large $n$. The truncation behind it is recorded here, not stated as a theorem in the paper. Let $B$ be a fixed set with $|B\cap[1,y]|\leq y^{1/2+o(1)}$. For large $x$ every $n\leq x$ has $s(n)<2x\log\log x$, so only the finite set $A_x=B\cap[1,2x\log\log x]$ matters, and $|A_x|\leq x^{1/2+o(1)}$ since the factor $2\log\log x$ is absorbed into the $o(1)$ exponent. Theorem 1.2 applied to $A_x$ gives $\#\{n\leq x:s(n)\in B\}=o(x)$, so $s^{-1}(B)$ has density zero.

Source. Paul Pollack, Carl Pomerance, and Lola Thompson, Divisor-Sum Fibers, Mathematika 64(2) (2018), 330--342, DOI 10.1112/S0025579317000535. Theorem 1.2 is on p. 2 of the 11-page author manuscript that the source card identifies; the published pagination is not asserted as a locator for that copy.

Read depth. Claims checked: the statement was read clause by clause against the manuscript, and the proof in Section 2 (pp. 3--4) was read for its structure only, not verified.

Proof pointer

Section 2, pp. 3--4. After replacing $\epsilon(x)$ by $\max\{\epsilon(x),1/\log\log x\}$, the proof sets aside an exceptional set of $o(x)$ inputs $n\leq x$: those with no prime factor at most $\log x$, those with a divisor in $(x^{1/2-10\epsilon(x)},x^{1/2+10\epsilon(x)})$, those with squarefull part above $x^{2\epsilon(x)}$, and those with $n\leq\sqrt x$. For the remaining $n$, it writes $n=de$ with $d$ the largest divisor of $n$ not exceeding $\sqrt x$, finds $\gcd(d,e)=1$ and $s(e)$ small, and uses $s(de)=\sigma(d)s(e)+s(d)e$ to put $s(e)$ in a determined residue class. Summing over $d$ shows that each target value has $\ll x^{1/2-9\epsilon(x)}$ non-exceptional preimages, and summing over the at most $x^{1/2+\epsilon(x)}$ targets gives the theorem. This is a map of the proof, not a reconstruction of it.

Dependencies

Ford's theorem on the distribution of divisors (Ann. of Math. 168 (2008), the paper's [8, Theorem 1]) for the divisor class; the maximal order of the divisor function; ideas the authors credit to Booker (arXiv:1610.07471, the paper's [3]), whose arguments by themselves, the authors say, almost immediately give the weaker bound with $x^{1/2-\epsilon}$ in place of $x^{1/2+\epsilon(x)}$, for fixed $\epsilon>0$ (p. 2).

Bears on

• Problem 955: the problem asks whether $s^{-1}(A)$ has density zero for every density-zero $A$. Through the truncation above, the theorem gives this for every fixed $A$ with counting function at most $y^{1/2+o(1)}$; it says nothing about denser density-zero sets.