gpt pro 5.4 attempt on Math-689

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.