Erdős Problem #143

Erdős Problem #143

1) ROUND-2 OBJECTIVE

Path C (obstruction / correction). Round-1 already settled the weaker harmonic-sum statement \[ \sum_{\substack{x\in \mathcal A\\ x\le X}}\frac1x=o(\log X) \] under the integer-dilation separation hypothesis, using the 2025 theorem of Koukoulopoulos--Lamzouri--Lichtman. The unresolved part is the stronger convergence claim \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \] In this round I isolate the exact remaining obstruction by converting the open statement into an Abel-summation criterion and an equivalent exponential-block criterion, prove a broad strategy-barrier showing that the Round-1 information alone can never force the stronger convergence, and prove the stronger convergence for a genuine new subclass (rational sets with only finitely many reduced denominators).

2) Round-1 FOUNDATION USED

I use the following Round-1 facts as black boxes.

1. If $\mathcal A\subset (1,\infty)$ satisfies \[ |kx-y|\ge 1 \qquad (x\neq y\in\mathcal A,\ k\in\mathbb Z_{\ge 1}), \] then \[ H_{\mathcal A}(X):=\sum_{\substack{x\in\mathcal A\\ x\le X}}\frac1x=o(\log X). \]
2. The same hypothesis implies $1$-separation (take $k=1$), hence at most one element of $\mathcal A$ lies in each interval $[m,m+1)$.

I do not re-prove these facts here.

3) NEW INSIGHT / TOOL (ROUND-2)

There are four genuinely new ingredients in this round.

1. An exact Abel-summation criterion showing that, once Round-1 gives $H_{\mathcal A}(X)=o(\log X)$, the open convergence question is equivalent to the integral test \[ \int_2^\infty \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt<\infty. \]
2. An equivalent exponential-block formulation: if \[ a_m:=\left|\mathcal A\cap [e^m,e^{m+1})\right|,\qquad b_m:=e^{-m}a_m, \] then \[ H_{\mathcal A}(e^N)\asymp \sum_{m\le N} b_m, \qquad \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty \iff \sum_{m\ge 1}\frac{b_m}{m}<\infty. \] So the remaining gap after Round-1 is exactly a logarithmic weighted summability problem for exponential-block masses.
3. An explicit barrier family: for any nondecreasing $\omega(m)\to\infty$ with $\sum_{m\ge 2}\frac{1}{m\omega(m)}=\infty$, there exists a $1$-separated set $B\subset (1,\infty)$ such that \[ \sum_{\substack{b\in B\\ b\le e^N}}\frac1b=o(N) \qquad\text{but}\qquad \sum_{b\in B}\frac{1}{b\log b}=\infty. \] This rules out any proof strategy that uses only the Round-1 harmonic estimate and $1$-separation.
4. A new positive class: if $\mathcal A\subset \mathbb Q_{>1}$ satisfies the integer-dilation separation condition and only finitely many reduced denominators occur, then \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \]

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap is the step \[ H_{\mathcal A}(X)=o(\log X)\quad \Longrightarrow\quad \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty, \] which is not logically valid in general. So the plan is:

1. Convert the open series into an integral involving the already-known function $H_{\mathcal A}(X)$.
2. Discretise that integral on exponential blocks, so the open question becomes a completely explicit summability problem for the block masses $b_m$.
3. Prove an explicit counterexample family at the level of $1$-separated sets, showing that no argument depending only on Round-1 data can finish the problem.
4. Salvage a genuinely new positive result by exploiting arithmetic structure in one clean rational subclass.

5) WORK (ROUND-2)

For convenience write \[ P_{\mathcal A}(X):=\sum_{\substack{x\in\mathcal A\\ 2<x\le X}}\frac{1}{x\log x}. \]

Lemma 143.1 (Abel criterion).

For every countable $\mathcal A\subset (1,\infty)$ and every $X\ge 3$, \[ P_{\mathcal A}(X) = \frac{H_{\mathcal A}(X)}{\log X} + \int_2^X \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt + O_{\mathcal A}(1). \] Consequently, under the Round-1 conclusion $H_{\mathcal A}(X)=o(\log X)$, \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty \iff \int_2^\infty \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt<\infty. \]

Proof.

Ignore finitely many elements of $\mathcal A$ in $(1,2]$, since they contribute only $O_{\mathcal A}(1)$. Let \[ U(t):=\sum_{\substack{x\in\mathcal A\\ 2<x\le t}}\frac1x. \] Take $f(t):=1/\log t$ on $[2,\infty)$, so \[ f'(t)=-\frac{1}{t(\log t)^2}. \] Abel summation gives \[ \sum_{\substack{x\in\mathcal A\\ 2<x\le X}}\frac1x\,f(x) = U(X)f(X)-\int_2^X U(t)f'(t)\,dt. \] Since $U(t)=H_{\mathcal A}(t)+O_{\mathcal A}(1)$ for $t\ge 2$, this becomes \[ P_{\mathcal A}(X) = \frac{H_{\mathcal A}(X)}{\log X} + \int_2^X \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt + O_{\mathcal A}(1), \] as claimed.

If $H_{\mathcal A}(X)=o(\log X)$, then $H_{\mathcal A}(X)/\log X\to 0$, so the convergence of $P_{\mathcal A}(X)$ is equivalent to the convergence of the integral. $\square$

This immediately yields a quantitative sufficient condition.

Corollary 143.2 (integrable saving suffices).

Assume the Round-1 hypothesis and suppose there is a positive function $L(X)$ such that \[ H_{\mathcal A}(X)\ll \frac{\log X}{L(X)} \qquad (X\ge X_0), \] and \[ \int_{X_0}^\infty \frac{dt}{t\log t\,L(t)}<\infty. \] Then \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \] In particular, it would be enough to prove any bound of the form \[ H_{\mathcal A}(X)\ll \frac{\log X}{(\log\log X)^{1+\delta}} \qquad (\delta>0). \]

Proof.

Insert the assumed upper bound into Lemma 143.1: \[ \int_2^\infty \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt \ll \int_{X_0}^\infty \frac{dt}{t\log t\,L(t)}<\infty. \] Now apply Lemma 143.1 again. $\square$

Lemma 143.3 (exponential-block reformulation).

For $m\ge 1$, let \[ I_m:=[e^m,e^{m+1}),\qquad a_m:=|\mathcal A\cap I_m|,\qquad b_m:=e^{-m}a_m. \] Then for every $N\ge 1$, \[ e^{-1}\sum_{m\le N} b_m \le H_{\mathcal A}(e^{N+1})-H_{\mathcal A}(e) \le \sum_{m\le N} b_m, \] and \[ e^{-1}\sum_{m\le N}\frac{b_m}{m+1} \le P_{\mathcal A}(e^{N+1}) \le \sum_{m\le N}\frac{b_m}{m}. \] Hence \[ H_{\mathcal A}(e^N)\asymp \sum_{m\le N} b_m, \] and \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty \iff \sum_{m\ge 1}\frac{b_m}{m}<\infty. \]

Proof.

If $x\in I_m$, then \[ e^{-m-1}\le \frac1x \le e^{-m} \] and, for $m\ge 1$, \[ \frac{e^{-m-1}}{m+1}\le \frac{1}{x\log x}\le \frac{e^{-m}}{m}. \] Summing over $x\in \mathcal A\cap I_m$ gives the blockwise bounds \[ e^{-1}b_m \le \sum_{x\in \mathcal A\cap I_m}\frac1x \le b_m, \qquad \frac{e^{-1}}{m+1}b_m \le \sum_{x\in \mathcal A\cap I_m}\frac{1}{x\log x}\le \frac1m b_m. \] Summing over $m\le N$ proves the displayed inequalities and the equivalence. $\square$

By Round-1, $H_{\mathcal A}(e^N)=o(N)$, so Lemma 143.3 implies \[ \sum_{m\le N} b_m=o(N). \] Thus the entire unresolved part of the problem is now compressed into the single question \[ \sum_{m\le N}b_m=o(N) \quad\stackrel{?}{\Longrightarrow}\quad \sum_{m\ge 1}\frac{b_m}{m}<\infty \] for the special block masses coming from integer-dilation-separated sets.

Theorem 143.4 (explicit barrier family).

Let $\omega:\mathbb N\to [1,\infty)$ be nondecreasing, with $\omega(m)\to\infty$, and assume \[ \sum_{m\ge 2}\frac{1}{m\omega(m)}=\infty. \] Then there exists a $1$-separated set $B\subset (1,\infty)$ such that \[ \sum_{\substack{b\in B\\ b\le e^N}}\frac1b \ll \sum_{m\le N}\frac{1}{\omega(m)} = o(N), \] but \[ \sum_{b\in B}\frac{1}{b\log b}=\infty. \]

Proof.

For large $m$, let \[ L_m:=\left\lfloor \frac{e^m}{\omega(m)}\right\rfloor, \qquad n_m:=\lceil e^m\rceil, \qquad B_m:=\{n_m,n_m+1,\dots,n_m+L_m-1\}. \] Because $L_m<e^m$ and \[ n_{m+1}-n_m \sim (e-1)e^m, \] the sets $B_m$ are pairwise disjoint for all large $m$. Let \[ B:=\bigcup_{m\ge m_0} B_m \] for $m_0$ large enough that the disjointness already holds. Since $B$ is a union of disjoint intervals of integers, it is $1$-separated.

For $b\in B_m$ we have $e^m\le b\le 2e^m$ for large $m$, hence \[ \frac{1}{2e^m}\le \frac1b\le \frac{1}{e^m}, \qquad \frac{1}{2e^m(m+\log 2)}\le \frac{1}{b\log b}\le \frac{1}{e^m m}. \] Therefore the contribution of the $m$th block satisfies \[ \sum_{b\in B_m}\frac1b \asymp \frac{L_m}{e^m}\asymp \frac{1}{\omega(m)}, \] and \[ \sum_{b\in B_m}\frac{1}{b\log b}\asymp \frac{L_m}{e^m m}\asymp \frac{1}{m\omega(m)}. \] Summing over $m\le N$ gives \[ \sum_{\substack{b\in B\\ b\le e^N}}\frac1b \ll \sum_{m\le N}\frac{1}{\omega(m)}. \] Since $1/\omega(m)\to 0$, Cesàro's theorem gives \[ \sum_{m\le N}\frac{1}{\omega(m)}=o(N). \] On the other hand, \[ \sum_{b\in B}\frac{1}{b\log b} \asymp \sum_{m\ge m_0}\frac{1}{m\omega(m)} = \infty. \] This proves the theorem. $\square$

Taking $\omega(m)=\log(m+2)$ gives an explicit $1$-separated set with \[ \sum_{b\le e^N,\ b\in B}\frac1b \asymp \frac{N}{\log N}=o(N) \] but \[ \sum_{b\in B}\frac{1}{b\log b}=\infty. \] So the Round-1 estimate $H_{\mathcal A}(X)=o(\log X)$, even together with $1$-separation, is far from enough to imply the desired convergence.

Theorem 143.5 (bounded reduced denominators).

Assume $\mathcal A\subset \mathbb Q_{>1}$ satisfies the integer-dilation separation condition, and assume only finitely many reduced denominators occur among the elements of $\mathcal A$. Then \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \]

Proof.

Let $D$ be the finite set of reduced denominators occurring in $\mathcal A$. For each $q\in D$, define \[ B_q:=\{a\in\mathbb N:\ \gcd(a,q)=1,\ a/q\in \mathcal A\}. \] Fix $q\in D$. I claim that $B_q$ is a primitive set of integers.

Indeed, if $a,b\in B_q$ are distinct and $b=ka$ for some integer $k\ge 2$, then \[ \frac{b}{q}=k\frac{a}{q}, \] so taking $x=a/q$ and $y=b/q$ violates the defining condition $|kx-y|\ge 1$ (the left-hand side would be $0$). Thus no element of $B_q$ divides another, so $B_q$ is primitive.

By Erdős's classical theorem on primitive sets, \[ \sum_{a\in B_q}\frac{1}{a\log a}<\infty. \] Split the sum over $a/q\in\mathcal A$ into the finitely many terms with $a<2q$ and the rest. For $a\ge 2q$ we have $\log(a/q)\ge \frac12\log a$, hence \[ \frac{1}{(a/q)\log(a/q)} = \frac{q}{a\log(a/q)} \le \frac{2q}{a\log a}. \] Therefore \[ \sum_{\substack{a/q\in\mathcal A\\ a\ge 2q}} \frac{1}{(a/q)\log(a/q)} \le 2q\sum_{a\in B_q}\frac{1}{a\log a} < \infty. \] Since there are only finitely many $a<2q$, the whole $q$-slice converges. Summing over the finite denominator set $D$ gives \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \] $\square$

6) ADVERSARIAL VERIFICATION

1. Potential issue: the Abel formula near $x=1$. I explicitly removed the finitely many elements in $(1,2]$ and absorbed them into $O_{\mathcal A}(1)$, so the logarithmic singularity at $1$ causes no problem.
2. Potential issue: The barrier set does not satisfy the original integer-dilation condition. Correct, and this is intentional. The theorem is not offered as a counterexample to the problem. Its role is to rule out an entire proof strategy: any argument that only sees $1$-separation plus the Round-1 harmonic estimate cannot possibly finish the problem.
3. Potential issue: could the bounded-denominator proof miss cross-denominator interactions? No. The series splits over denominator classes, and positivity means each class can be handled separately. Cross-denominator interactions are irrelevant for convergence once each slice converges.
4. Potential issue: primitive-set input for $B_q$. The only property needed is that no element of $B_q$ divides another; this was checked directly.
5. Potential issue: block criterion uses natural logarithms. Yes, deliberately. The blocks $[e^m,e^{m+1})$ are exactly the right ones for the weight $1/\log x$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 does not prove or disprove the full real-variable convergence statement \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty. \] However, it makes three substantive advances beyond Round-1:

1. It isolates the exact remaining obstruction: \[ \sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty \iff \int_2^\infty \frac{H_{\mathcal A}(t)}{t(\log t)^2}\,dt<\infty, \] and equivalently \[ \sum_{m\ge 1}\frac{b_m}{m}<\infty \quad\text{for}\quad b_m=e^{-m}\left|\mathcal A\cap[e^m,e^{m+1})\right|. \]
2. It proves a broad strategy barrier: the Round-1 information $H_{\mathcal A}(X)=o(\log X)$ and $1$-separation is insufficient on its own, even in very strong quantitative forms.
3. It proves the full desired convergence for a new class: rational sets with finitely many reduced denominators.

So the open core of Problem #143 is now sharply localised: one must exploit genuinely arithmetic features of the integer-dilation condition that are invisible to purely block-counting arguments.

8) COMPLETION ESTIMATE

COMPLETION: 68%

9) REFERENCES

1. D. Koukoulopoulos, Y. Lamzouri, J. D. Lichtman, Erdős's integer dilation approximation problem and GCD graphs, arXiv:2502.09539, 2025.
2. T. F. Bloom, Erdős Problem #143, erdosproblems.com/143, accessed 2026-03-07.
3. J. D. Lichtman, A proof of the Erdős primitive set conjecture, arXiv:2202.02384, especially the historical introduction recalling Erdős's 1935 primitive-set theorem.

Erdős Problem #143: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

If $\mathcal A\subset(1,\infty)$ satisfies $|kx-y|\ge1$ for distinct $x,y\in\mathcal A$ and every positive integer $k$, must $\sum_{x\in\mathcal A}1/(x\log x)$ converge?

All supplied versions considered: 143.tex. The source reduces convergence to weighted block masses and treats rational sets with finitely many denominators. There are repairable errors: its logarithmic comparison needs $a\ge q^2$, not $a\ge2q$; its Abel lemma needs local finiteness rather than mere countability. The hypothesis supplies local finiteness, since distinct points have distance at least one.

2) WORK

A simpler corrected bounded-denominator proof.

Suppose every denominator divides an integer $Q$. The set $B=\{Qx:x\in\mathcal A\}\subset\mathbb N$ is primitive: if $b'=kb$ for distinct members, then $y=kx$, violating the hypothesis. By the classical primitive-set convergence theorem, \[ \sum_{b\in B}{1\over b\log b}<\infty. \] For $b\ge Q^2$ (and $b>1$), $\log(b/Q)\ge\tfrac12\log b$, so \[ {1\over (b/Q)\log(b/Q)}\le {2Q\over b\log b}. \] The finitely many smaller $b$ contribute a finite amount. Thus the desired series converges in this subclass. Taking the least common multiple of the finitely many reduced denominators obtains the source's conclusion using one primitive set rather than separate slices.

Exact summation statement.

Define $U(t)=\sum_{2<x\le t,\ x\in\mathcal A}1/x$. Local finiteness makes this finite, and integration by parts gives exactly \[ \sum_{2<x\le X}{1\over x\log x}={U(X)\over\log X}+\int_2^X{U(t)\over t(\log t)^2}\,dt. \] There is no uncontrolled boundary term, since $U(2)=0$. The one-separation assumption allows at most two points in $(1,2]$, so restoring these terms is legitimate. This repairs the stated generality without extending convergence to arbitrary real sets.

3) ADVERSARIAL VERIFICATION

The comparison fails at $a=2q$ when $q$ is large: $\log(a/q)=\log2$ whereas $\tfrac12\log a$ grows. Replacing the cutoff by $q^2$ fixes the proof. A countable set can accumulate near one, so countability alone cannot justify discarding finitely many small elements. Bounded denominators are still a restrictive hypothesis.

4) FINAL

UNRESOLVED. No control of the integral is established for a general dilation-separated real set. The repaired subclass theorem does not bridge that remaining gap.

5) SOURCES CHECKED

Read all of 143.tex; checked Lichtman, A proof of the Erdős primitive set conjecture, https://arxiv.org/abs/2202.02384, on 2026-09-23 for the primitive-set input. The argument imports that known convergence theorem and does not label it novel.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 12%

Erdős Problem #143 (Integer dilation approximation)

1) FORMAL RESTATEMENT

Let $\mathcal A\subset (1,\infty)$ be a countably infinite set. Assume the integer-dilation separation condition \begin{equation} \label{wsref-0-65-71-3a-31-34-33-2d-73-65-70} \forall x\neq y\in \mathcal A\ \forall k\in\mathbb Z_{\ge 1}:\quad |kx-y|\ge 1. \end{equation} Define the truncated harmonic sum \[ H_{\mathcal A}(X):=\sum_{\substack{x\in\mathcal A\\ x\le X}}\frac1x\qquad (X\ge 1), \] and the "primitive-set analogue" sum \[ P_{\mathcal A}:=\sum_{x\in\mathcal A}\frac{1}{x\log x}, \] where $\log$ is the natural logarithm (so $x>1$ ensures $\log x>0$).

The question asks whether (eq:143-sep) forces $\mathcal A$ to be sparse, in particular whether it implies either \begin{align*} \text{(Q1)}\quad & P_{\mathcal A}<\infty,\\ \text{(Q2)}\quad & H_{\mathcal A}(X)=o(\log X)\quad\text{as }X\to\infty. \end{align*}

2) QUICK LITERATURE / CONTEXT CHECK

• If $\mathcal A\subset\mathbb N$ consists of integers, then (eq:143-sep) is equivalent to $\mathcal A$ being primitive (no element divides another): for integers, $|kx-y|\ge 1$ always holds unless $kx=y$. For primitive integer sets, Erdős proved $\sum_{n\in\mathcal A}\frac{1}{n\log n}<\infty$, and later works of Behrend and Erdős--Sárkőzy--Szemerédi gave quantitative bounds for $\sum_{n\le X,\,n\in\mathcal A}\frac1n$.
• For general real sets $\mathcal A\subset(1,\infty)$, the second implication $Q2$ is now known to hold: Koukoulopoulos--Lamzouri--Lichtman $2025$ proved that if $\limsup_{X\to\infty}\frac{1}{\log X}\sum_{x\in\mathcal A\cap[1,X]}\frac1x>0$, then for every $\varepsilon>0$ there exist infinitely many distinct $\alpha,\beta\in\mathcal A$ and a positive integer $n$ with $|n\alpha-\beta|<\varepsilon$. The contrapositive yields $H_{\mathcal A}(X)=o(\log X)$ under (eq:143-sep).
• The convergence question $Q1$ for general real sets satisfying (eq:143-sep) appears to remain open.

3) ATTACK PLAN

Proof track.

1. Extract immediate structural consequences of (eq:143-sep) (e.g. $1$-separation from the case $k=1$) to obtain baseline bounds.
2. Use the strongest known theorem (Koukoulopoulos--Lamzouri--Lichtman) as a black box to deduce $Q2$ rigorously.
3. For $Q1$, try to adapt "primitive set" arguments: one would like to partition $\mathcal A$ into classes where "multiplicative structure" forces a convergent majorant, but in the real setting there is no prime factorization. A plausible route is to convert (eq:143-sep) into a graph/Diophantine approximation statement and seek a "density increment" or "energy" inequality strong enough to upgrade $Q2$ to $Q1$.

Disproof track.

1. Attempt to construct $\mathcal A$ with relatively large local density (e.g. about one point per interval of length $\log\log x$) while arranging fractional parts so that no integer dilation approximates another element within $1$.
2. Heuristic: to make $\sum \frac{1}{x\log x}$ diverge, one needs $\mathcal A$ to have counting function $A(X)=|\mathcal A\cap[1,X]|$ decaying slowly enough, e.g. $A(X)\asymp X/\log\log X$ would already force divergence of $\int \frac{A'(t)}{t\log t}\,dt$. So a counterexample would need to be much denser than the known examples (like primes) but still satisfy (eq:143-sep).

4) WORK

Phase 1 sanity checks (tiny examples).

• Integer example: If $\mathcal A$ is the set of prime numbers, then (eq:143-sep) holds (no prime is a multiple of another prime), and one has $H_{\mathcal A}(X)=\sum_{p\le X}\frac1p\sim\log\log X$ (hence $o(\log X)$) and $\sum_{p}\frac{1}{p\log p}<\infty$.
• Trivial upper bound: Any $\mathcal A$ satisfying (eq:143-sep) is $1$-separated, hence has at most one element per unit interval.

Known theorem (used as a black box).

We record the 2025 theorem in a form tailored to this problem.

Status of $Q1$.

The argument above settles $Q2$ but does not appear to imply $Q1$. Indeed, there exist abstract sequences $(a_n)$ with $\sum_{a_n\le X} \frac1{a_n}=o(\log X)$ but $\sum_n \frac{1}{a_n\log a_n}=\infty$ (for instance, choosing $a_n$ so that $|\{a_n\le X\}|\asymp X/\log\log X$). The open difficulty is to show that (eq:143-sep) forbids such "near $X/\log\log X$" densities.

5) VERIFICATION

• Quantifiers: Corollary cor:143-Q2 matches $Q2$ exactly: $\lim_{X\to\infty} H_{\mathcal A}(X)/\log X=0$.
• Edge cases: $x>1$ is used only to ensure $\log x>0$ in $Q1$. The $Q2$ deduction works already for $\mathcal A\subset[1,\infty)$.
• Self-pairs: Theorem thm:KLL-contrapos requires $\alpha\ne\beta$, matching (eq:143-sep).

6) FINAL

UNRESOLVED. The stronger convergence claim $Q1$ remains open (to the best of my current knowledge), but $Q2$ is resolved affirmatively.

1. (i) Strongest fully proved partial result obtained:
   • $Q2$ holds: $\sum_{x\in\mathcal A,\,x\le X}\frac1x=o(\log X)$ (Corollary cor:143-Q2).
   • Trivial baseline: $H_{\mathcal A}(X)\le 1+\log X$ for $X\ge 2$ (Corollary Corollary (Baseline harmonic bound)).
2. (ii) Exact first gap: I do not have a proof upgrading $Q2$ to $\sum_{x\in\mathcal A}\frac{1}{x\log x}<\infty$, nor an explicit construction of $\mathcal A$ satisfying (eq:143-sep) for which that sum diverges.
3. (iii) Top 3 next moves:
   1. Try to refine the KLL "GCD graph" machinery to obtain a quantitative bound on the counting function $A(X)=|\mathcal A\cap[1,X]|$, strong enough to force $\int \frac{dA(t)}{t\log t}<\infty$.
   2. Attempt a transference principle: discretize $\mathcal A$ by integer parts and fractional parts, and show that (eq:143-sep) enforces an "approximate primitivity" property that controls $\sum 1/(x\log x)$.
   3. Explore constructive disproof: build $\mathcal A$ by choosing one point in many unit intervals via a Lovász local lemma / entropy-compression argument, arranging to avoid all forbidden neighborhoods $|kx-y|<1$.
4. (iv) What a minimal counterexample would likely look like: a set with counting function roughly $A(X)\approx X/(\log\log X)$ (or slightly smaller), spread across unit intervals, with carefully chosen fractional parts so that for each $x\in\mathcal A$ the nearest multiple $kx$ stays at distance $\ge 1$ from every other point. Such a set would make $\sum \frac{1}{x\log x}$ diverge while remaining consistent with $H_{\mathcal A}(X)=o(\log X)$.

7) COMPLETION ESTIMATE

9/10. The remaining gap $Q1$ appears to require genuinely new ideas beyond the current $Q2$ technology.