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Source. Thomas F. Bloom, On a density conjecture about unit fractions, arXiv:2112.03726v2 (12 October 2023). Printed and PDF page numbers agree.

Use $R(A)$, $A_q$, $Q_A$ and $R(A;q)$ as defined in Lemma 6; $Q_A$ consists of exact prime powers, and $\omega(n)$ counts distinct prime divisors. Unqualified sums over $q$ are sums over prime powers.

Statement (Proposition 3, pp. 15-17). Let $N$ be sufficiently large, let $N\geq M\geq N^{1/2}$, and suppose $A\subseteq[M,N]$ satisfies

$$\frac{99}{100}\log\log N\leq\omega(n)\leq2\log\log N \qquad(n\in A),$$ $$R(A)\geq(\log N)^{-1/101},$$

and, for every $q\in Q_A$,

$$R(A;q)\geq(\log N)^{-1/100}.$$

Then at least one of the following alternatives holds.

1. There is $B\subseteq A$ such that

$$R(B)\geq\frac13R(A) \qquad\text{and}\qquad \sum_{q\in Q_B}\frac1q\leq\frac23\log\log N.$$

2. For every interval $I$ of length at most

$$MN^{-2/\log\log N},$$

either

$$\#\{n\in A:\text{ no element of }I\text{ is divisible by }n\} \geq\frac M{\log N}, \tag{3.2a}$$

or the following holds. Define $D_I$ to be the set of $q\in Q_A$ such that

$$\#\{n\in A_q:\text{ no element of }I\text{ is divisible by }n\} <\frac{M}{2q(\log N)^{1/100}}. \tag{3.2b-threshold}$$

Then some $x\in I$ is divisible by every $q\in D_I$.

If

$$\sum_{q\in Q_A}\frac1q\leq\frac23\log\log N,$$

then alternative 2 is guaranteed.

Rewritten proof. It is enough to fix an arbitrary interval $I$ of the stated maximum length and show that either alternative 1 already holds or the assertion in alternative 2 holds for this $I$. Let

$$A_I=\{n\in A:n\text{ divides some element of }I\}.$$

If $|A\setminus A_I|\geq M/\log N$, then (3.2a) holds. Assume from now on that

$$|A\setminus A_I|<M/\log N. \tag{3.5}$$

Define

$$E_I=\left\{q\in Q_A: R(A_I;q)>\frac1{2(\log N)^{1/100}}\right\}.$$

If $q\in D_I$, the elements of $A_q\setminus(A_I)_q$ are fewer than the quantity in (3.2b-threshold). Since every such element is at least $M$,

$$\begin{aligned} R(A_I;q) &> R(A;q)- \left(\frac{M}{2q(\log N)^{1/100}}\right)\frac qM\\ &\geq\frac1{2(\log N)^{1/100}}. \end{aligned}$$

Therefore

$$D_I\subseteq E_I. \tag{3.6}$$

For each $q\in E_I$, apply Lemma 5 to $A_I$, choosing $k$ by

$$(\log N)^{1/k}=2\log\log N. \tag{3.7}$$

For large $N$, this $k$ lies in the range required by Lemma 5, and the lower bound defining $E_I$ is stronger than $(\log N)^{-1/2}$. Lemma 5 supplies an integer $d_q$ for which

$$qd_q>|I|, \qquad \omega(d_q)<\frac1{500}\log\log N, \tag{3.8}$$

and

$$\sum_{\substack{n\in A_I\\qd_q\mid n\\(qd_q,n/qd_q)=1}} \frac{qd_q}{n} \gg \frac1{(\log N)^{1/100}(\log\log N)^2}. \tag{3.9}$$

Indeed, (3.7) turns the lower bound on $qd_q$ from Lemma 5 into

$$qd_q>M N^{-1/(2\log\log N)} >MN^{-2/\log\log N}\geq|I|,$$

and it turns $(\log N)^{2/k}$ into $4(\log\log N)^2$. Also $5/\log k<1/500$ for sufficiently large $N$.

Every $n$ counted in (3.9) divides some element of $I$. All these elements of $I$ divisible by $qd_q$ must be the same, because two distinct multiples of $qd_q>|I|$ cannot lie in $I$. Denote the unique such element by $x_q$, and define

$$A_I^{(q)}= \left\{\frac{n}{qd_q}:n\in A_I,\ qd_q\mid n, (qd_q,n/qd_q)=1\right\}.$$

Equation (3.9) says, for sufficiently large $N$, that

$$R(A_I^{(q)})\geq(\log N)^{-1/99}.$$

If $m=n/(qd_q)\in A_I^{(q)}$, then the coprimality in its definition, the lower bound on $\omega(n)$, and (3.8) give

$$\omega(m)\geq\frac{97}{99}\log\log N,$$

while $\omega(m)\leq2\log\log N$. Lemma 4 with $\epsilon=2/99$ therefore gives

$$\sum_{r\in Q_{A_I^{(q)}}}\frac1r \geq\frac{95}{99}e^{-1}\log\log N. \tag{3.10}$$

Every exact prime-power component $r$ of a member $m$ of $A_I^{(q)}$ is also an exact prime-power component of the corresponding $n$, so $Q_{A_I^{(q)}}\subseteq Q_A$. Also $m\mid n\mid x_q$. Hence (3.10) implies

$$\sum_{\substack{r\mid x_q\\r\in Q_A}}\frac1r \geq\frac{95}{99}e^{-1}\log\log N \geq0.35\log\log N. \tag{3.11}$$

When $u\neq v$ are elements of $I$, Lemma 3 and the bound $|u-v|\leq N$ give, for large $N$,

$$\sum_{q\mid(u,v)}\frac1q \leq C\log\log\log N =o(\log\log N) \leq0.01\log\log N. \tag{3.12}$$

If $\sum_{q\in Q_A}1/q\leq(2/3)\log\log N$, equations (3.11)-(3.12) show that two distinct values of $x_q$ are impossible: the union of their prime-power supports would have reciprocal mass at least $(0.35+0.35-0.01)\log\log N>(2/3)\log\log N$. Thus all $x_q$ with $q\in E_I$ coincide. If $D_I$ is empty, the requested divisibility condition is vacuous. Otherwise $E_I$ is nonempty by (3.6), and since $q\mid x_q$ for each $q\in E_I$, their common value is divisible by every member of $D_I$. This proves both the last sentence of the proposition and the assertion of alternative 2 in the small-total-mass case.

In general, Mertens' estimate gives

$$\sum_{q\in Q_A}\frac1q \leq(1+o(1))\log\log N \leq1.01\log\log N. \tag{3.13}$$

Equations (3.11)-(3.12) imply that there are at most two distinct values among the $x_q$: three distinct values would have union mass at least $(3\cdot0.35-3\cdot0.01)\log\log N=1.02\log\log N$, contradicting (3.13). If some $x\in I$ is divisible by all $q\in D_I$, alternative 2 holds for this interval. Otherwise the $x_q$ assume exactly two values, say $w_1,w_2$.

For $i=1,2$, set

$$A^{(i)}=\{n\in A:n\mid w_i\}, \qquad A^{(0)}=A\setminus(A^{(1)}\cup A^{(2)}).$$

Every member of $Q_{A^{(1)}}$ divides $w_1$. Subtracting the reciprocal mass of the prime powers of $Q_A$ that divide $w_2$, and restoring those that also divide $w_1$, gives

$$\begin{aligned} \sum_{q\in Q_{A^{(1)}}}\frac1q &\leq\sum_{q\leq N}\frac1q -\sum_{\substack{q\mid w_2\\q\in Q_A}}\frac1q +\sum_{q\mid(w_1,w_2)}\frac1q\\ &\leq \left(1-\frac{95}{99}e^{-1}+o(1)\right)\log\log N\\ &\leq\frac23\log\log N \end{aligned} \tag{3.14}$$

for large $N$, by (3.10), (3.12), and Mertens' estimate. The same bound holds for $Q_{A^{(2)}}$.

Because

$$R(A^{(0)})+R(A^{(1)})+R(A^{(2)})\geq R(A),$$

alternative 1 follows with $B=A^{(1)}$ or $B=A^{(2)}$ unless

$$R(A^{(0)})\geq\frac13R(A). \tag{3.15}$$

Assume (3.15). Let $A'$ be the set of all $n\in A_I\cap A^{(0)}$ such that every $q\in Q_A$ with $n\in A_q$ belongs to $E_I$. By (3.5), the definition of $E_I$, and Mertens' estimate,

$$\begin{aligned} R(A^{(0)}\setminus A') &\leq\frac{|A\setminus A_I|}{M} +\sum_{q\in Q_A\setminus E_I}\frac1qR(A_I;q)\\ &\ll\frac{\log\log N}{(\log N)^{1/100}}. \end{aligned} \tag{3.16}$$

This is $o((\log N)^{-1/101})$. Equations (3.15)-(3.16) and the hypothesis on $R(A)$ therefore imply

$$R(A')\gg(\log N)^{-1/101}. \tag{3.17}$$

Since $n\geq M$ for all $n\in A'$, (3.17) gives the cardinality estimate

$$|A'|\geq M R(A')\gg M(\log N)^{-1/101}. \tag{3.18}$$

Every $n\in A'$ divides at least one integer in $I$. Pigeonholing over the integers of $I$, whose number is $O(MN^{-2/\log\log N})$, produces an $x\in I$ for which, with

$$A''=\{n\in A':n\mid x\},$$

one has

$$|A''|\gg N^{2/\log\log N}(\log N)^{-1/101} \geq N^{3/(2\log\log N)} \tag{3.19}$$

for sufficiently large $N$. Necessarily $x\neq w_1,w_2$, because $A'\subseteq A^{(0)}$.

For $n\in A''$, each exact prime-power component $q$ of $n$ lies in $E_I$, so it divides either $w_1$ or $w_2$. Hence $n\mid w_1w_2$ as well as $n\mid x$, and therefore

$$n\mid(x,w_1w_2) \leq(x,w_1)(x,w_2) \leq|x-w_1|\,|x-w_2| \leq N^2.$$

Thus all members of $A''$ are divisors of one fixed integer $m\leq N^2$. The cited divisor bound gives

$$|A''|\leq\tau(m) \leq N^{(1+o(1))\,2\log 2/\log\log N},$$

contradicting (3.19), since $2\log 2<3/2$. This contradiction rules out (3.15), so alternative 1 holds whenever the interval assertion fails. As $I$ was arbitrary, the proposition follows.

Dependencies and source detail

Lemma 3, Lemma 4, and Lemma 5. The external estimates are Mertens' prime-power sum (p. 3) and the maximal-order divisor bound cited on p. 17 to Montgomery and Vaughan, Theorem 2.11: uniformly for positive integers $m\le N^2$, $\tau(m)\le N^{(1+o(1))2\log2/\log\log N}$.

On p. 17 the source prints $|A'|\gg M/(\log N)^{-1/101}$; the preceding reciprocal-mass estimate gives the multiplicative negative power in (3.18) above. The source's following pigeonhole estimate agrees with that corrected expression.

Bears on

• Problem 298
• Problem 299