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Math-5362-A6 · revision r1 · on Math-5362

gpt pro 5.2 attempt on Math-5362, version 2

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. 1) ROUND-2 OBJECTIVE
  2. 2) ROUND-1 FOUNDATION USED
  3. 3) NEW INSIGHT / TOOL (ROUND-2)
  4. 4) ATTACK PLAN (ROUND-2)
  5. 5) WORK (ROUND-2)
  6. 5.1 A strengthened dyadic summability lemma
  7. 5.2 A concrete corollary matching $(\log N)^{1+\varepsilon}$
  8. 5.3 External input: Bloom–Sisask bound on $r_3(N)$
  9. 5.4 Full deduction of the $k=3$ Erdős conjecture
  10. 5.5 A clean criterion reducing the (k)-case to a summability condition on $r_k$
  11. 5.6 Why the Round‑1 strategy still does not reach $k\ge 4$
  12. 6) ADVERSARIAL VERIFICATION
  13. 7) FINAL (EXACTLY ONE)
  14. 8) COMPLETION ESTIMATE (MANDATORY)
  15. 9) REFERENCES
  16. Round 3: Log-weighted resolution and minimal correction
  17. 1) ROUND-3 OBJECTIVE
  18. 2) ROUND-1/2 FOUNDATION USED
  19. 3) NEW INSIGHT / TOOL (ROUND-3)
  20. 4) ATTACK PLAN (ROUND-3)
  21. 5) WORK (ROUND-3)
  22. 5.1 Weighted dyadic decomposition
  23. 5.2 From $k$--AP--freeness to dyadic upper bounds
  24. 5.3 Convergence for $\varphi(n)=\log n$ using current $r_k$ bounds
  25. 5.4 The corrected statement: divergence of $\sum_{n\in A}1/(n\log n)$ forces arbitrarily long progressions
  26. 5.5 A sharp barrier for this method
  27. 6) ADVERSARIAL VERIFICATION
  28. 7) FINAL (EXACTLY ONE)
  29. 8) COMPLETION ESTIMATE (MANDATORY)
  30. 9) REFERENCES
  31. References

1) ROUND-2 OBJECTIVE

(C) Obstruction/correction path. Given the Round‑1 reduction (“divergence ⇒ infinitely often $|A\cap[1,N]|$ beats any summably small density profile; APs follow if one can beat $r_k(N)$”), the most promising strict advance is to close the gap for the only length currently reachable unconditionally: $k=3$. Concretely, I will:

1. Strengthen Round‑1 Lemma 2 to a general “summability criterion” (still elementary, dyadic). 2. Combine it with the Bloom–Sisask bound $r_3(N)\ll N/(\log N)^{1+c}$ to give a complete, gap‑free proof of the $k=3$ case of the Erdős conjecture (indeed, arbitrarily large 3‑APs). 3. Precisely isolate why the same method does not currently extend to $k\ge 4$ with the best known $r_k(N)$ bounds.

This is a genuine advance beyond Round‑1: it turns the Round‑1 framework into a full theorem for $k=3$, and it refines the “gap” to an explicit summability condition on $r_k$.

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2) ROUND-1 FOUNDATION USED

I will rely on the following Round‑1 results as black boxes:

* Lemma 1 (dyadic decomposition upper bound): With (B_j:=A$\cap$$2^j,2^{j+1}]$, \[ \sum_{n\in A\cap[2,\infty)}\frac1n\ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j}. \] * Lemma 3 (definition link): If $|A\cap[1,N]|>r_k(N)$ then (A) contains a nontrivial (k)-term AP in $[1,N]$.

I will not restate their proofs.

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3) NEW INSIGHT / TOOL (ROUND-2)

New tool: a general dyadic “summability criterion” that strictly strengthens Round‑1 Lemma 2.

Instead of hard-coding $(\log N)(\log\log N)^2$, we prove:

* If $|A\cap[1,N]|$ is eventually bounded by (N) times any profile whose dyadic reciprocals form a convergent series, then $\sum_{n\in A}1/n$ converges.

This lets us match the known best upper bound on $r_3(N)$ (Bloom–Sisask) and deduce the $k=3$ Erdős conjecture as a corollary.

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4) ATTACK PLAN (ROUND-2)

Round‑1 gap: need an upper bound on $r_k(N)$ strong enough that the divergence‑forced density spikes (from Lemma 2) exceed $r_k(N)$ infinitely often.

What we prove now:

1. A strengthened lemma: if for some $\varepsilon>0$, \[ |A\cap[1,N]|\ \le\ C\frac{N}{(\log N)^{1+\varepsilon}} \quad\text{for all large }N, \] then $\sum_{n\in A}1/n<\infty$. (This is a direct dyadic summability argument.)

2. Use Bloom–Sisask: \[ r_3(N)\ \ll\ \frac{N}{(\log N)^{1+c}} \] for some absolute $c>0$. ([arXiv][1]) Hence any 3‑AP‑free (A) satisfies the above cardinality bound, forcing $\sum_{n\in A}1/n<\infty$. Contraposition yields the $k=3$ Erdős statement.

3. Finally, show arbitrarily large 3‑APs exist by applying the same argument to dyadic blocks.

Why this overcomes Round‑1 obstacles: for $k=3$ we now do have a decay exponent $>1$ in $\log N$, which makes the relevant dyadic series summable. For $k\ge 4$, current bounds do not yield such summability, and we will make that explicit.

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5) WORK (ROUND-2)

5.1 A strengthened dyadic summability lemma

Let $A\subseteq\mathbb N$. Define dyadic blocks (B_j:=A$\cap$$2^j,2^{j+1}]$ as in Round‑1.

Lemma 4 (general dyadic summability criterion). Let $(u_j)*{j\ge 0}$ be a sequence of nonnegative reals with $\sum*{j\ge 0}u_j<\infty$. If there exists $J_0$ such that for all $j\ge J_0$, \[ \frac{|B_j|}{2^j}\ \le\ u_j, \] then $\sum_{n\in A}\frac1n<\infty$.

Proof. By Round‑1 Lemma 1, \[ \sum_{n\in A\cap[2,\infty)}\frac1n\ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j} = \sum_{j< J_0}\frac{|B_j|}{2^j} + \sum_{j\ge J_0}\frac{|B_j|}{2^j}. \] The first sum is finite (finitely many terms). The second is $\le \sum_{j\ge J_0}u_j<\infty$. Add the possible $n=1$ term. ∎

This lemma is strictly more flexible than Round‑1 Lemma 2; Lemma 2 is the special case $u_j\asymp 1/(j(\log j)^2)$.

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5.2 A concrete corollary matching $(\log N)^{1+\varepsilon}$

Corollary 5. Fix $\varepsilon>0$. Suppose there exist constants $C>0$ and $N_0\ge 3$ such that for all $N\ge N_0$, \[ |A\cap[1,N]|\ \le\ C\frac{N}{(\log N)^{1+\varepsilon}}. \] Then $\sum_{n\in A}\frac1n<\infty$.

Equivalently (contrapositive): if $\sum_{n\in A}1/n=\infty$, then for every $C>0$ and $N_0$ there exists some $N\ge N_0$ with \[ |A\cap[1,N]|\ >\ C\frac{N}{(\log N)^{1+\varepsilon}}. \]

Proof. Choose $J_0$ so that $2^{J_0}\ge N_0$. For $j\ge J_0$, \[ |B_j|\ \le\ |A\cap[1,2^{j+1}]| \ \le\ C\frac{2^{j+1}}{(\log 2^{j+1})^{1+\varepsilon}} = C\frac{2^{j+1}}{((j+1)\log 2)^{1+\varepsilon}}. \] Hence \[ \frac{|B_j|}{2^j}\ \le\ \frac{2C}{(\log 2)^{1+\varepsilon}}\cdot \frac{1}{(j+1)^{1+\varepsilon}}. \] Let $u_j := \dfrac{2C}{(\log 2)^{1+\varepsilon}}(j+1)^{-(1+\varepsilon)}$. Then $\sum_j u_j<\infty$ because $1+\varepsilon>1$. Apply Lemma 4. ∎

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5.3 External input: Bloom–Sisask bound on $r_3(N)$

Theorem (Bloom–Sisask). There exists an absolute constant $c_0>0$ such that if $S\subseteq{1,\dots,N}$ contains no nontrivial 3‑term arithmetic progression, then \[ |S|\ \ll\ \frac{N}{(\log N)^{1+c_0}}. \] Equivalently, $r_3(N)\ll N/(\log N)^{1+c_0}$. ([arXiv][1])

I will use only the stated bound, not the proof.

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5.4 Full deduction of the $k=3$ Erdős conjecture

Theorem 6 (Erdős conjecture for $k=3$, with arbitrarily large 3‑APs). If $A\subseteq\mathbb N$ satisfies $\sum_{n\in A}\frac1n=\infty$, then (A) contains a nontrivial 3‑term arithmetic progression. In fact, for every (J) there exists a nontrivial 3‑term AP all of whose terms exceed $2^J$; hence (A) contains infinitely many 3‑APs with unbounded maximum element.

Proof (existence of some 3‑AP). Assume for contradiction that (A) contains no nontrivial 3‑term arithmetic progression. Then for every (N), \[ |A\cap[1,N]|\ \le\ r_3(N). \] By Bloom–Sisask, there exist constants $C_0>0$ and $N_1$ such that for all $N\ge N_1$, \[ |A\cap[1,N]|\ \le\ r_3(N)\ \le\ C_0\frac{N}{(\log N)^{1+c_0}}. \] Apply Corollary 5 with $\varepsilon=c_0$: this implies $\sum_{n\in A}\frac1n<\infty$, contradicting the hypothesis. Therefore (A) must contain a nontrivial 3‑term AP. ∎

Proof (arbitrarily large 3‑APs). Fix $J\ge 0$. Suppose for contradiction that (A) contains no 3‑term AP with all three terms $>2^J$. Then for every $j\ge J$, the block \[ B_j = A\cap(2^j,2^{j+1}] \] is 3‑AP‑free (any 3‑AP inside $B_j$ would have all terms $>2^j\ge 2^J$). Translation preserves 3‑APs, so $B_j$ is 3‑AP‑free as a subset of an interval of length $2^j$; hence \[ |B_j|\ \le\ r_3(2^j). \] Again by Bloom–Sisask, \[ \frac{|B_j|}{2^j}\ \le\ \frac{r_3(2^j)}{2^j} \ \ll\ \frac{1}{(\log 2^j)^{1+c_0}} \asymp \frac{1}{j^{1+c_0}}. \] Since $\sum_{j\ge J} j^{-(1+c_0)}<\infty$, Lemma 4 (with $u_j\asymp j^{-(1+c_0)}$) implies the tail $\sum_{n\in A\cap(2^J,\infty)}\frac1n$ converges, and adding finitely many initial terms gives $\sum_{n\in A}\frac1n<\infty$, contradiction.

Thus, for every (J), there exists some 3‑AP in (A) with all terms $>2^J$. In particular, there are infinitely many distinct 3‑APs and their maxima are unbounded. ∎

This completes the $k=3$ case in a fully rigorous way.

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5.5 A clean criterion reducing the (k)-case to a summability condition on $r_k$

This is a structural clarification not present in Round‑1.

Proposition 7 (dyadic summability criterion via $r_k$). Fix $k\ge 3$. If \[ \sum_{j=1}^\infty \frac{r_k(2^j)}{2^j}\ <\ \infty, \] then every $A\subseteq\mathbb N$ with $\sum_{n\in A}\frac1n=\infty$ must contain a nontrivial (k)-term arithmetic progression (indeed with arbitrarily large terms).

Proof. Suppose (A) is (k)-AP‑free. Then each dyadic block $B_j$ is (k)-AP‑free, hence $|B_j|\le r_k(2^j)$. Round‑1 Lemma 1 gives \[ \sum_{n\in A\cap[2,\infty)}\frac1n\ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j} \ \le\ \sum_{j\ge 0}\frac{r_k(2^j)}{2^j}\ <\ \infty, \] contradicting divergence. The “arbitrarily large” refinement is proved exactly as in Theorem 6 by truncating below $2^J$. ∎

Remark. A sufficient (not necessary) condition for $\sum r_k(2^j)/2^j<\infty$ is an upper bound \[ r_k(N)\ \ll\ \frac{N}{(\log N)^{1+\varepsilon}} \quad\text{for some }\varepsilon>0, \] since then $r_k(2^j)/2^j \ll j^{-(1+\varepsilon)}$, summable.

For $k=3$ this holds by Bloom–Sisask, so Proposition 7 recovers Theorem 6. ([arXiv][1])

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5.6 Why the Round‑1 strategy still does not reach $k\ge 4$

The Erdős conjecture for all (k) remains listed as open in current problem databases. ([erdosproblems.com][2]) The precise reason, within the dyadic framework, is:

* For $k=4$, the best-known bounds are of the form $r_4(N)\ll N/(\log N)^c$ for some (small) $c>0$. ([erdosproblems.com][2]) Then $r_4(2^j)/2^j \ll 1/j^c$, and $\sum 1/j^c$ diverges whenever $c\le 1$, so Proposition 7 cannot be triggered.

* For general $k\ge 5$, the current best bounds (Leng–Sah–Sawhney) are \[ r_k(N)\ \ll\ \frac{N}{\exp!\big((\log\log N)^{c_k}\big)} \quad (c_k>0). \] ([erdosproblems.com][2]) Setting $N=2^j$ gives $r_k(2^j)/2^j\ll \exp!\big(-(\log j)^{c_k}\big)$, and the series $\sum_j \exp(-(\log j)^{c_k})$ still diverges (since $(\log j)^{c_k}\ll \log j$, so the terms are $\gg 1/j$ eventually). Therefore Proposition 7 still cannot be triggered for $k\ge 5$ using currently available $r_k$ bounds.

So the Round‑1 “compare to $r_k(N)$” mechanism is now fully closed for $k=3$ and cleanly pinpointed as failing for $k\ge 4$ because known bounds do not make $\sum r_k(2^j)/2^j$ summable.

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6) ADVERSARIAL VERIFICATION

I now try to break the new claims.

1. Did I illegitimately infer divergence of $\sum |B_j|/2^j$ from divergence of $\sum_{n\in A}1/n$? Round‑1 Lemma 1 gives $\sum_{n\in A\cap[2,\infty)}1/n \le \sum_j |B_j|/2^j$. If the left is $+\infty$, the right cannot be finite. So yes, the right diverges. No issue.

2. Translation step for dyadic blocks and $r_3(2^j)$: I used: if (B_j$\subset$$2^j,2^{j+1}]$ is 3‑AP‑free, then $|B_j|\le r_3(2^j)$. Justification: translate by $-2^j$: $T_j:={n-2^j:n\in B_j}\subset{1,\dots,2^j}$. A 3‑AP in $B_j$ corresponds bijectively to a 3‑AP in $T_j$. Hence $T_j$ is 3‑AP‑free, so $|T_j|\le r_3(2^j)$, and $|T_j|=|B_j|$. Solid.

3. Quantifier strength for “arbitrarily large 3‑APs”: I proved: for each (J), some block $B_j$ with $j\ge J$ contains a 3‑AP. That gives a 3‑AP with all terms $>2^j\ge 2^J$. This indeed gives unbounded maxima and infinitely many distinct progressions. No hidden reuse of the same small 3‑AP.

4. Any hidden dependence on monotonicity of logs / constants? Corollary 5 uses $\log 2^{j+1}=(j+1)\log 2$, valid for natural logs. The constants $(\log 2)^{-(1+\varepsilon)}$ are harmless.

5. Potential mismatch of constants in Bloom–Sisask $\ll$: In Theorem 6, I used only existence of some $C_0,N_1$ such that $r_3(N)\le C_0 N/(\log N)^{1+c_0}$ for all $N\ge N_1$. The arXiv abstract states $|A|\ll N/(\log N)^{1+c}$, which exactly provides such constants. ([arXiv][1]) No need for effective values.

All checks pass.

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7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

* The full Erdős conjecture (“divergent $\sum 1/n$ ⇒ (k)-APs for every (k)”) remains open. ([erdosproblems.com][2]) * Strict advance beyond Round‑1: I gave a complete, gap‑free proof of the $k=3$ case, including the stronger conclusion of arbitrarily large 3‑term APs, by:

* strengthening Round‑1 Lemma 2 to Corollary 5 / Lemma 4, and * plugging in Bloom–Sisask’s bound on $r_3(N)$. ([arXiv][1]) * I also isolated a sharp sufficient condition for general (k): the summability of $\sum r_k(2^j)/2^j$ (Proposition 7), and explained why known bounds for $k\ge 4$ do not (yet) meet it. ([erdosproblems.com][2])

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8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

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9) REFERENCES

* T. F. Bloom, O. Sisask, Breaking the logarithmic barrier in Roth’s theorem on arithmetic progressions, arXiv:2007.03528 $v2, 2021$. ([arXiv][1]) * Erdős Problems (Problem #3), status and summary of best known bounds on $r_k(N)$, last edited 28 Dec 2025. ([erdosproblems.com][2]) * J. Leng, A. Sah, M. Sawhney, Improved Bounds for Szemerédi’s Theorem, arXiv:2402.17995 $2024$. ([arXiv][3])

[1]: https://arxiv.org/abs/2007.03528 "https://arxiv.org/abs/2007.03528" [2]: https://www.erdosproblems.com/3 "https://www.erdosproblems.com/3" [3]: https://arxiv.org/pdf/2402.17995 "https://arxiv.org/pdf/2402.17995"

Round 3: Log-weighted resolution and minimal correction

1) ROUND-3 OBJECTIVE

Path (C): obstruction/correction. Round 2 established a complete proof of the Erdős conjecture for $k=3$ by combining a dyadic summability criterion with the Bloom--Sisask bound on $r_3(N)$. The original statement for general $k$ remains open. In this round I prove a minimal corrected statement that is currently unconditional for all $k\ge 3$:

If $A\subseteq\mathbb{N}$ satisfies $\sum_{n\in A,\ n\ge 3}\frac{1}{n\log n}=\infty$, then $A$ contains arithmetic progressions of every length. Moreover, for every $k\ge 3$ and every $J\ge 0$ there exists a $k$--term progression in $A$ whose terms are all $>2^J$.

This strengthens the Round 2 $k=3$ theorem in a different direction: it solves a log-weighted variant of the full conjecture using the current best bounds on $r_k(N)$ for all $k$.

2) ROUND-1/2 FOUNDATION USED

I will use the following Round 1/2 results as black boxes.

  • (Round 1 Lemma 1) Dyadic decomposition upper bound: with $B_j:=A\cap(2^j,2^{j+1}]$, \[\sum_{n\in A\cap[2,\infty)}\frac1n\ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j}.\]
  • (Round 1 Lemma 3) Definition link: if $|A\cap[1,N]|>r_k(N)$ then $A$ contains a nontrivial $k$--term AP in $[1,N]$.
  • (Round 2 Theorem 6) The unweighted $k=3$ case: if $\sum_{n\in A}1/n=\infty$ then $A$ contains infinitely many $3$--term APs with unbounded maximum element.
  • (Round 2 Proposition 7) Dyadic criterion via $r_k$: if $\sum_{j\ge 1} r_k(2^j)/2^j<\infty$ then any $A$ with $\sum_{n\in A}1/n=\infty$ contains a $k$--term AP (indeed arbitrarily large ones).

3) NEW INSIGHT / TOOL (ROUND-3)

New tool: weighted dyadic summability. Round 2 used dyadic decomposition to bound the harmonic weight $\sum_{n\in A}1/n$ by $\sum_j |B_j|/2^j$. Here I generalize this to weights of the form $1/(n\,\varphi(n))$ for monotone $\varphi$, and specialize to the threshold weight $\varphi(n)=\log n$.

The key point is that, with the current best bounds on $r_k(N)$ for $k\ge 4$, the series \[\sum_{j\ge 2}\frac{r_k(2^j)}{2^j\,j}\] converges, yielding a full theorem for the log-weighted conjecture.

4) ATTACK PLAN (ROUND-3)

  1. Prove a weighted dyadic inequality: for nondecreasing $\varphi$, \[\sum_{n\in A\cap[2,\infty)}\frac{1}{n\varphi(n)}\ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j\,\varphi(2^j)}.\]
  2. Combine this with the translation bound $|B_j|\le r_k(2^j)$ for $k$--AP--free sets to obtain \[\sum_{n\in A\cap[2,\infty)}\frac{1}{n\varphi(n)}\ \le\ \sum_{j\ge 0}\frac{r_k(2^j)}{2^j\,\varphi(2^j)}\qquad (A\text{ $k$--AP--free}).\]
  3. Insert the best known upper bounds on $r_k(N)$:
    • $k=3$: $r_3(N)\ll N/(\log N)^{1+c}$ (Bloom--Sisask).
    • $k=4$: $r_4(N)\ll N/(\log N)^{c}$ (Green--Tao).
    • $k\ge 5$: $r_k(N)\ll N\exp(-(\log\log N)^{c_k})$ (Leng--Sah--Sawhney).
    Taking $\varphi(n)=\log n$ makes the dyadic series converge for each $k$. Contraposition gives the desired corrected statement.

5) WORK (ROUND-3)

5.1 Weighted dyadic decomposition

Throughout, $\log$ denotes the natural logarithm. For $A\subseteq\mathbb{N}$ and $j\ge 0$ set $B_j:=A\cap(2^j,2^{j+1}]$.

Lemma (Weighted dyadic bound): Let $\varphi:[2,\infty)\to(0,\infty)$ be nondecreasing. Then \[ \sum_{n\in A\cap[2,\infty)}\frac{1}{n\varphi(n)} \ \le\ \sum_{j\ge 0}\frac{|B_j|}{2^j\,\varphi(2^j)}. \]
Proof: Fix $j\ge 0$ and $n\in(2^j,2^{j+1}]$. Then $n\ge 2^j$ and $\varphi(n)\ge \varphi(2^j)$, so \[ \frac{1}{n\varphi(n)}\ \le\ \frac{1}{2^j\,\varphi(2^j)}. \] Summing over $n\in B_j$ gives \[ \sum_{n\in B_j}\frac{1}{n\varphi(n)}\ \le\ \frac{|B_j|}{2^j\,\varphi(2^j)}. \] Finally sum over $j\ge 0$ and use that the blocks $B_j$ partition $A\cap[2,\infty)$. ∎

5.2 From $k$--AP--freeness to dyadic upper bounds

Lemma (Translation to $[2^j]$): Fix $k\ge 3$. If $A$ contains no nontrivial $k$--term arithmetic progression, then for every $j\ge 0$, \[ |B_j|\ \le\ r_k(2^j). \]
Proof: Let $T_j:=\{n-2^j:\ n\in B_j\}$. Then $T_j\subseteq\{1,\dots,2^j\}$ and translation preserves $k$--term arithmetic progressions, so $T_j$ is $k$--AP--free. Hence $|T_j|\le r_k(2^j)$ by definition of $r_k$, and $|T_j|=|B_j|$. ∎
Proposition (Weighted $r_k$ criterion): Fix $k\ge 3$ and let $\varphi:[2,\infty)\to(0,\infty)$ be nondecreasing. If $A\subseteq\mathbb{N}$ is $k$--AP--free then \[ \sum_{n\in A\cap[2,\infty)}\frac{1}{n\varphi(n)} \ \le\ \sum_{j\ge 0}\frac{r_k(2^j)}{2^j\,\varphi(2^j)}. \] Consequently, if the series on the right converges then every set $A$ with \[\sum_{n\in A,\ n\ge 2}\frac{1}{n\varphi(n)}=\infty\] must contain a nontrivial $k$--term arithmetic progression.
Proof: Combine Lemma Lemma (Weighted dyadic bound) with Lemma lem:translation. The final statement is the contrapositive. ∎

5.3 Convergence for $\varphi(n)=\log n$ using current $r_k$ bounds

Set $\varphi(n):=\log n$ for $n\ge 3$ (and extend it arbitrarily to $[2,3]$). Then $\varphi$ is nondecreasing.

Lemma (A calculus fact): For every $c>0$ the series \[ \sum_{j\ge 3}\frac{\exp(- (\log j)^c)}{j} \] converges.
Proof: Consider the integral \[ \int_3^{\infty}\frac{\exp(- (\log x)^c)}{x}\,dx. \] With the substitution $t=\log x$ we have $dt=dx/x$, so the integral becomes \[ \int_{\log 3}^{\infty} \exp(-t^c)\,dt, \] which converges for every $c>0$. The integral test gives the claim. ∎
Theorem (Log-weighted finiteness for $k$--AP--free sets): Fix $k\ge 3$. If $A\subseteq\mathbb{N}$ contains no nontrivial $k$--term arithmetic progression, then \[ \sum_{n\in A,\ n\ge 3}\frac{1}{n\log n}\;<\;\infty. \]
Proof: By Proposition prop:weighted-rk with $\varphi(n)=\log n$ it suffices to show that \[ \sum_{j\ge 2}\frac{r_k(2^j)}{2^j\,j} \] converges.

Case $k=3$. Bloom--Sisask proved $r_3(N)\ll N/(\log N)^{1+c_3}$ for some $c_3>0$. Taking $N=2^j$ gives \[ \frac{r_3(2^j)}{2^j\,j}\ \ll\ \frac{1}{j\,(\log 2^j)^{1+c_3}}\ \asymp\ \frac{1}{j^{2+c_3}}, \] which is summable.

Case $k=4$. Green--Tao proved $r_4(N)\ll N/(\log N)^{c_4}$ for some $c_4>0$. Again with $N=2^j$, \[ \frac{r_4(2^j)}{2^j\,j}\ \ll\ \frac{1}{j\,(\log 2^j)^{c_4}}\ \asymp\ \frac{1}{j^{1+c_4}}, \] which is summable.

Case $k\ge 5$. Leng--Sah--Sawhney proved that for each fixed $k\ge 5$ there exists $c_k>0$ such that \[ r_k(N)\ll N\exp(- (\log\log N)^{c_k}). \] With $N=2^j$ this becomes \[ \frac{r_k(2^j)}{2^j\,j} \ \ll\ \frac{\exp(- (\log\log 2^j)^{c_k})}{j} \ \asymp\ \frac{\exp(- (\log j)^{c_k})}{j}, \] which is summable by Lemma Lemma (A calculus fact).

In all cases the dyadic series converges, proving the claim. ∎

5.4 The corrected statement: divergence of $\sum_{n\in A}1/(n\log n)$ forces arbitrarily long progressions

Theorem (Log-weighted Erdős conjecture (now unconditional)): Let $A\subseteq\mathbb{N}$. If \[ \sum_{n\in A,\ n\ge 3}\frac{1}{n\log n}=\infty, \] then $A$ contains arithmetic progressions of every length. More precisely: for every integer $k\ge 3$ and every $J\ge 0$ there exist integers $a\ge 1$ and $d\ge 1$ such that \[ \{a,a+d,\dots,a+(k-1)d\}\subseteq A\cap(2^J,\infty). \]
Proof: Fix $k\ge 3$. Suppose for contradiction that $A$ contains no nontrivial $k$--term arithmetic progression. Then Theorem thm:log-weighted-finite implies \[ \sum_{n\in A,\ n\ge 3}\frac{1}{n\log n}<\infty, \] contradicting the hypothesis. Thus $A$ must contain at least one $k$--term progression.

For the strengthened statement with all terms $>2^J$, fix $J\ge 0$ and assume instead that $A\cap(2^J,\infty)$ is $k$--AP--free. Apply Theorem thm:log-weighted-finite to the set $A\cap(2^J,\infty)$ to deduce \[ \sum_{n\in A,\ n>2^J}\frac{1}{n\log n}<\infty. \] Adding the finitely many terms with $n\le 2^J$ gives $\sum_{n\in A,\ n\ge 3}1/(n\log n)<\infty$, again contradicting the hypothesis. Therefore for each $J$ the set $A\cap(2^J,\infty)$ contains a nontrivial $k$--term progression, as claimed. ∎

5.5 A sharp barrier for this method

Proposition prop:weighted-rk shows that to force $k$--term progressions from divergence of a weight $1/(n\varphi(n))$ it suffices that \[ \sum_{j\ge 0}\frac{r_k(2^j)}{2^j\varphi(2^j)}<\infty. \] For the original (unweighted) conjecture $\varphi\equiv 1$, this becomes the dyadic summability condition from Round 2, namely $\sum_j r_k(2^j)/2^j<\infty$. The current best bounds for $k\ge 4$ do not yield this convergence.

One can also quantify the near-optimality of taking $\varphi(n)=\log n$ for this approach when the $k\ge 5$ bounds have exponent at most $1$.

Lemma (No improvement to $1/(n(\log n)^\alpha)$ for $\alpha<1$ when $0<c\le 1$): Fix $0<c\le 1$ and $\alpha<1$. Then \[ \sum_{j\ge 3}\frac{\exp(- (\log j)^c)}{j^{\alpha}}=\infty. \]
Proof: Choose $\varepsilon>0$ such that $\alpha+\varepsilon\le 1$. Since $c\le 1$, we have $(\log j)^c/\log j=(\log j)^{c-1}\to 0$. Hence for all sufficiently large $j$, \[ (\log j)^c\le \varepsilon\log j. \] Exponentiating gives $\exp(- (\log j)^c)\ge \exp(-\varepsilon\log j)=j^{-\varepsilon}$ for all sufficiently large $j$, and therefore \[ \frac{\exp(- (\log j)^c)}{j^{\alpha}}\ge \frac{1}{j^{\alpha+\varepsilon}}. \] Since $\alpha+\varepsilon\le 1$, the comparison series $\sum_j 1/j^{\alpha+\varepsilon}$ diverges, so the original series diverges. ∎

Interpretation. In the range $0<c_k\le 1$, Lemma lem:alpha-barrier shows that the current $k\ge 5$ bound $r_k(2^j)/2^j\ll \exp(- (\log j)^{c_k})$ cannot, via Proposition prop:weighted-rk, force finiteness of $\sum_{n\in A}1/(n(\log n)^\alpha)$ for any $\alpha<1$. Thus within this dyadic+$r_k$ strategy, the threshold weight $1/(n\log n)$ is essentially best possible unless one obtains substantially stronger savings in $r_k(N)$ (or proves that the available exponent $c_k$ is $>1$).

6) ADVERSARIAL VERIFICATION

  • Monotonicity requirements. Lemma Lemma (Weighted dyadic bound) requires $\varphi$ nondecreasing. For $\varphi(n)=\log n$ this holds on $[3,\infty)$, and the finitely many terms $n=2$ (and $n=1$) are irrelevant to divergence/convergence.
  • Translation step. Lemma lem:translation uses the bijection $B_j\leftrightarrow T_j$ by shifting by $2^j$. This preserves $k$--APs because adding a constant to all terms of a progression keeps equal differences.
  • Summability in the $k\ge 5$ case. Lemma Lemma (A calculus fact) proves convergence of $\sum_j \exp(- (\log j)^c)/j$ for all $c>0$ by a direct integral substitution; no asymptotics are used.
  • Quantifiers for arbitrarily large progressions. In Theorem Theorem (Log-weighted Erdős conjecture (now unconditional)), the refinement "all terms $>2^J$" is verified by applying Theorem thm:log-weighted-finite to the tail set $A\cap(2^J,\infty)$; if that tail were $k$--AP--free then the weighted tail-sum would be finite, contradicting global divergence.
  • Compatibility with Round 2. Theorem Theorem (Log-weighted Erdős conjecture (now unconditional)) does not assume $\sum_{n\in A}1/n=\infty$ and does not imply it; it is a genuinely different (stronger) hypothesis. For $k=3$, Round 2 already proves much more from the weaker harmonic divergence.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). The original Erdős conjecture (harmonic divergence $\sum_{n\in A}1/n=\infty$ implies arithmetic progressions of every length) remains open for $k\ge 4$. However, Round 3 gives a complete proof of a minimal corrected statement: divergence of the log-weighted sum $\sum_{n\in A}1/(n\log n)$ forces progressions of every length (indeed with arbitrarily large terms), using current best bounds for $r_k(N)$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 60%

9) REFERENCES

References

  • [BlSi20] T. F. Bloom and O. Sisask, Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions, arXiv:2007.03528.
  • [GrTa17] B. Green and T. Tao, New bounds for Szemerédi's theorem, III: A polylogarithmic bound for $r_4(N)$, arXiv:1705.01703 (accepted in Mathematika, 2017).
  • [LSS24] J. Leng, A. Sah, and M. Sawhney, Improved bounds for Szemerédi's theorem, arXiv:2402.17995.
  • [ErdosProblems3] T. F. Bloom, Erdős Problem #3, https://www.erdosproblems.com/3 (accessed 2026-01-20).