{"id":"c1a04732-f88b-4788-b9c6-555cda5bd32f","arxiv_id":"2607.05823","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":6.0,"correctness_risk":"unknown","formal_verification":"none","parameter_count":1,"one_line_summary":"The author proves f(N) ≫ N exp(−(2√(log(24/7))+o(1))√(log log N)) for the largest harmonic-triple-free subset of [N], matching the form of the best 3-AP-free lower bound with log N replaced by log log N.","lead":"The paper constructs large sets of integers with no harmonic triples (distinct a,b,c with 2/a = 1/b + 1/c), proving a lower bound of N·exp(−(2√(log(24/7))+o(1))√(log log N)). This transfers the best known Behrend-type lower bound for progression-free sets to the reciprocal setting, giving the first construction of this size for harmonic-triple-free sets.","discovery_kind":"unclear","skeptic_critique":{"model":"glm-5.2","headline":"No significant objection identified. The argument is correct; the notation 'n−1' in B₀'s definition denotes modular inverse n⁻¹, not subtraction, as confirmed by the deletion argument's reference to 'inverse residues'.","rationale":"The reader correctly identified the two potential soft spots: dependence on the external r₃ bound of [4], and the tightness of the union bound in the deletion argument. Both are handled correctly in the paper. The external dependence is clearly stated and modular (improvements to [4] would directly improve Theorem 1.1). The deletion count U_q(Y) ≪ Y log Y / q is tight enough because q ≍ K log Y with K chosen large, making the deleted fraction O(1/K). The parametrization in Lemma 2.2 is complete: it captures all odd harmonic triples via the coprimality argument (xy, x+y) = 1, and the modular collapse q|d follows cleanly from cancelling the nonzero factor (h−d)m. The only issue I found is notational: 'n−1' in B₀'s definition should be read as n⁻¹ (modular inverse), which the context confirms but the typesetting obscures. This is a presentation concern, not a mathematical gap. The reader's ACCEPT verdict with HIGH confidence is appropriate.","tokens_in":4851,"tokens_out":9497,"duration_ms":462583,"concrete_test":"Verify the notation by checking the LaTeX source: confirm that the definition of B₀ uses $n^{-1} (mod q)$ (modular inverse) rather than $n - 1 (mod q)$ (subtraction). If it is indeed $n^{-1}$, no further action is needed. If it is $n-1$, re-derive the step 'the three residues must all be equal' to check whether a 3-AP in {b⁻¹, a⁻¹, c⁻¹} can be related to the filter on {b−1, a−1, c−1} via some additional argument not stated in the paper.","verdict_should_be":"UNCHANGED","load_bearing_attack":"After careful reading, the central argument is sound. The key step—showing that a harmonic triple surviving the random filter must collapse modulo q—requires the filter to be on n⁻¹ mod q (modular inverse), not n−1 mod q (subtraction). The text writes 'n−1modq' in the definition of B₀, which is ambiguous, but the surrounding context confirms the intended meaning is n⁻¹: (1) the modular reduction of 2/a = 1/b + 1/c gives 2a⁻¹ ≡ b⁻¹ + c⁻¹ (mod q), creating a 3-AP in {b⁻¹, a⁻¹, c⁻¹}; (2) the deletion argument explicitly states 'their inverse residues are also equal; the probability that this common inverse residue lies in R_{λ,μ} is exactly δ'; (3) the set V excludes multiples of q, which is necessary precisely for n⁻¹ to exist. Under this interpretation, the 3-AP-free property of R_{λ,μ} forces b⁻¹ = a⁻¹ = c⁻¹, hence a ≡ b ≡ c (mod q), and Lemma 2.2 gives q|d. The counting bound U_q(Y) ≪ Y log Y / q in (12) is also correct: since h+qj ≥ h for j ≥ 1, the inner sum Σ_{j<h/q} 1/(h(h+qj)) ≤ (h/q)·(1/h²) = 1/(qh), giving the stated bound. The deletion term δU_q(Y) ≪ δY/K is absorbable by choosing K large. No mathematical gap identified.","agreement_with_reader":"agree"},"referee_report":{"model":"glm-5.2","summary":"The paper proves that there exist subsets of [N] of size N exp(-(C_AP + o(1)) sqrt(log log N)) containing no harmonic triple (distinct a,b,c with 2/a = 1/b + 1/c), where C_AP = 2 sqrt(log(24/7)). The construction selects odd integers n <= N whose modular inverse n^{-1} mod q lies in a random affine image of a 3-AP-free set R in F_q (q ~ log N), then deletes the largest element of any surviving harmonic triple. The key observation is that a surviving triple must collapse to a 3-AP in R, forcing a = b = c mod q; a parametrization then shows q | d, limiting the number of such triples to O(Y log Y / q), which is absorbable.","tokens_in":5192,"tokens_out":1177,"duration_ms":109877,"significance":"The result cleanly transfers the best known 3-AP-free lower bound (Elsholtz–Hunter–Proske–Sauermann) to the harmonic-triple setting, with log N replaced by log log N reflecting the reciprocal map. The construction is explicit and the proof is short and self-contained apart from the cited external bound. The constant C_AP is inherited parameter-free from [4]. The open Question 1.2 (whether f(N) = o(N)) is a natural and well-motivated complement.","major_comments":[{"comment":"The notation 'n^{-1} mod q' is written as 'n-1modq' in the definition of B_0 = {n in V : n-1 mod q in R_{lambda,mu}}. This is ambiguous and could be read as subtraction rather than modular inverse. The surrounding text (reduction of (1) mod q giving 2a^{-1} = b^{-1} + c^{-1}, the set V excluding multiples of q, and the phrase 'inverse residues') confirms the intended meaning is n^{-1} mod q. This should be stated unambiguously, as it is load-bearing for the entire argument: the 3-AP-free property of R_{lambda,mu} forces b^{-1} = a^{-1} = c^{-1}, hence a = b = c mod q, which is the key step.","section":"Lemma 3.1, definition of B_0"},{"comment":"The inner sum bound in (12) writes sum_{1 <= j < h/q} 1/(h(h+qj)) << 1/(qh), using h+qj >= h. This is correct but the step should be slightly more explicit: since h+qj >= h for j >= 1, each summand is <= 1/h^2, and there are at most h/q terms, giving <= (h/q)(1/h^2) = 1/(qh). As written the inequality is valid but a reader might want this one-line justification, especially since the summability of 1/(qh) over h is what makes the deletion term O(delta Y log Y / q) = O(delta Y / K).","section":"Lemma 3.1, equation (12)"}],"minor_comments":[{"comment":"The wrap-around argument states |x+z-2y| < q for x,y,z in [1, floor(q/3)]. Strictly, x+z-2y ranges in (-q/3, q), so |x+z-2y| < q holds. This is correct but could be stated more precisely.","section":"Lemma 2.1"},{"comment":"The parametrization requires b < a < c, but the converse direction notes that not all parameter choices yield odd solutions. The construction only uses odd integers (via V), so a brief remark that the parametrization still applies to the odd triples that arise in the deletion argument would improve clarity.","section":"Lemma 2.2"},{"comment":"The constant C_AP = 2 sqrt(log(24/7)) is described as 'the natural-logarithm form of the base-two constant appearing in [4]'. A brief note on the conversion (base-2 to natural log) would help readers verify the constant.","section":"Introduction, equation (2)"},{"comment":"The set V = {n <= Y : n odd, q does not divide n} has |V| >> Y, but the implicit constant depends on q being prime and odd. Since q >= K log Y, the fraction of multiples of q up to Y is 1/q = o(1), so |V| ~ Y/2. Making this explicit would be helpful.","section":"Lemma 3.1"},{"comment":"Reference [4] is cited as a preprint (arXiv:2406.12290, 2024). If it has since been published or refereed, the citation should be updated.","section":"References"}],"recommendation":"minor_revision","confidential_remarks":"The paper is short (6 pages) and the argument is clean. The only substantive issue is the notation ambiguity in B_0's definition, which is a presentation problem rather than a mathematical gap. I see no reason this cannot be resolved quickly. The result is a natural and correct transfer of the Elsholtz et al. bound to the harmonic setting."},"author_rebuttal":{"model":"glm-5.2","summary":"We thank the referee for a careful reading and for identifying two notational/presentational issues in the proof of Lemma 3.1. Both points are well-taken and will be addressed in the revised manuscript. The mathematical content is unaffected; the revisions concern clarity of exposition only.","responses":[{"response":"The referee is correct. In the definition of B_0 = {n in V : n^{-1} mod q in R_{lambda,mu}}, the superscript on n^{-1} was not properly typeset, making the modular inverse ambiguous with subtraction. This is indeed load-bearing for the argument: the 3-AP-free property of R_{lambda,mu} forces b^{-1} = a^{-1} = c^{-1} mod q, hence a = b = c mod q, which is the key step. We will rewrite the definition of B_0 to state unambiguously that n^{-1} mod q denotes the modular multiplicative inverse of n modulo q, and will add a brief sentence clarifying this convention when B_0 is first introduced.","revision_made":"yes","referee_comment":"The notation 'n^{-1} mod q' is written as 'n-1 mod q' in the definition of B_0, which is ambiguous and could be read as subtraction rather than modular inverse."},{"response":"The referee's suggested one-line justification is exactly right. Since h + qj >= h for j >= 1, each summand satisfies 1/(h(h+qj)) <= 1/h^2, and there are at most h/q terms in the sum, giving the bound (h/q)(1/h^2) = 1/(qh). We will insert this one-line justification into the display between the second-to-last and last lines of (12), so that the reader can directly verify the step leading to the O(Y log Y / q) bound.","revision_made":"yes","referee_comment":"The inner sum bound in (12) writes sum_{1<=j<h/q} 1/(h(h+qj)) << 1/(qh), using h+qj >= h. This is correct but the step should be slightly more explicit."}],"tokens_in":4656,"tokens_out":464,"duration_ms":27892,"standing_objections":[]},"desk_editor":{"model":"glm-5.2","letter":"This is a short, well-written paper that proves the first lower bound of the form f(N) ≫ N exp(−C√(log log N)) for sets free of harmonic triples (distinct a,b,c with 2/a = 1/b + 1/c). The key idea is to filter odd integers by a random affine image of a 3-AP-free set in F_q with q ≍ log N, then delete collapsed triples. The reciprocal map turns the harmonic condition into a 3-AP condition modulo q, and the log N → log log N compression falls out naturally from the fact that q is logarithmic in N. It is a real new result and the proof is correct as far as I can tell. The two main ingredients are both new: the parametrization of odd harmonic triples (Lemma 2.2), which shows that collapsed triples have their difference parameter divisible by q, and the deletion argument (Lemma 3.1), which bounds the expected deletions via a union bound and absorbs them by choosing the constant K in q ≍ K log N large enough. The counting bound U_q(Y) ≪ Y log Y / q in (12) checks out: the inner sum is bounded by (h/q)·(1/h²) = 1/(qh), and summing over h gives the log factor. The wrap-around argument in Lemma 2.1 is also correct since |x+z−2y| < q for elements in [1, ⌊q/3⌋]. The paper relies externally on the Elsholtz–Hunter–Proske–Sauermann lower bound for r_3(X), which is a citation to a different group and not a circular dependency. If that bound is later improved, the constant here improves automatically. One notational issue: the definition of B_0 writes 'n−1 mod q' where the intended meaning is n⁻¹ mod q (modular inverse). The surrounding context makes this unambiguous — the set V excludes multiples of q precisely so inverses exist, the modular reduction gives 2a⁻¹ ≡ b⁻¹ + c⁻¹, and the deletion argument explicitly references 'inverse residues.' Still, a referee should ask the author to fix the notation. This is a minor typesetting issue, not a mathematical gap. The paper is appropriate for any number theory or combinatorics journal. It is short (6 pages), self-contained modulo the external citation, and poses a natural open question (Question 1.2: is f(N) = o(N)?). It deserves a serious referee. I agree with the reader's assessment on all axes.","headline":"Clean, correct, and short. The paper proves the first Behrend-quality lower bound for harmonic-triple-free sets. The argument is sound and deserves a serious referee.","tokens_in":5707,"tokens_out":626,"would_cite":true,"duration_ms":110486,"reading_group":"no","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11B25","11B75"],"pacs":[],"model":"glm-5.2","headline":"Large harmonic-triple-free sets match Behrend-type bound","keywords":["harmonic triples","arithmetic progressions","Behrend construction","additive combinatorics","progression-free sets","reciprocal map","finite fields"],"falsifier":"If the count of collapsed triples U_q(Y) were substantially larger than Y log Y / q (for instance, if the parametrization missed a family of triples), the deletion step could remove too many elements and the construction would fail.","tokens_in":4984,"feed_emoji":"🔢","tokens_out":688,"duration_ms":62725,"temperature":0.7,"pith_summary":"A harmonic triple is three distinct integers whose reciprocals form an arithmetic progression: 2/a = 1/b + 1/c. The paper proves that one can find a subset of {1,...,N} of size at least N exp(-(C+o(1)) sqrt(log log N)), with C = 2 sqrt(log(24/7)), containing no such triple. The construction takes a dense three-term-arithmetic-progression-free set in a prime field of size comparable to log N, uses a random affine image of it to filter the odd integers up to N, and then deletes elements that participate in surviving harmonic triples. The key observation is that any harmonic triple surviving the filter must collapse modulo the prime q, forcing its difference parameter to be divisible by q, which limits the number of such triples to O(N log N / q). A union bound over these collapsed triples shows that the deletions are absorbable, leaving a set of the claimed size. The bound has the same shape as the best known lower bound for progression-free sets, with log N replaced by log log N, reflecting the compression induced by the reciprocal map.","feed_headline":"Harmonic-Triple-Free Sets Reach Behrend-Scale Lower Bound","feed_subtitle":"New construction transfers progression-free sets through the reciprocal map, matching the best known 3-AP bound with log N compressed to log","key_machinery":"Random affine images of progression-free sets in F_q, parametrization of odd harmonic triples via b=(h-d)hm, a=(h-d)(h+d)m, c=(h+d)hm, union bound over collapsed triples with q|d","core_discovery":"The transfer mechanism works because the reciprocal map converts the additive structure of arithmetic progressions into the multiplicative structure of harmonic triples, but at the cost of compressing the effective scale from N to roughly log N. By working in a prime field of size q ~ log N, the author embeds a progression-free set into the residue classes of odd integers. The parametrization of odd harmonic triples (Lemma 2.2) shows that triples surviving the modular filter must have their difference parameter d divisible by q, making them rare enough to delete. The constant C_AP = 2 sqrt(log(24/7)) is inherited directly from the best known progression-free construction, and the log log N取代","pith_inferences":[],"forward_implications":[],"fun_headline_variants":["Behrend-scale integer sets avoiding harmonic triples","Progression-free sets transfer to harmonic-triple-free construction","Reciprocal map links 3-AP-free sets to harmonic-triple avoidance","Dense harmonic-triple-free sets built via modular filter","Near-Behrend lower bound for sets with no harmonic triples"],"cache_read_input_tokens":0,"weakest_assumption_plain":"The construction relies on the external lower bound for progression-free sets r_3(X) >= X exp(-(C_AP+o(1)) sqrt(log X)) and on the union bound over collapsed triples being tight enough that the deletion term O(N log N / q) is absorbable when q is chosen large enough relative to log N.","fun_headline_variants_meta":{"raw":{"variants":["Behrend-scale integer sets avoiding harmonic triples","Progression-free sets transfer to harmonic-triple-free construction","Reciprocal map links 3-AP-free sets to harmonic-triple avoidance","Dense harmonic-triple-free sets built via modular filter","Near-Behrend lower bound for sets with no harmonic triples"]},"model":"glm-5.2","effort":"high","cost_usd":0.0,"raw_usage":{"total_tokens":639,"prompt_tokens":556,"completion_tokens":83,"prompt_tokens_details":null},"tokens_in":556,"tokens_out":83,"duration_ms":15870,"temperature":1.0,"reasoning_tokens":null,"cache_read_input_tokens":0,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-07-08T23:07:18.680478+00:00","model_set":{"reader":"glm-5.2"},"falsifier":"If the count of collapsed triples U_q(Y) were substantially larger than Y log Y / q (for instance, if the parametrization missed a family of triples), the deletion step could remove too many elements and the construction would fail.","supporting_citations":[],"review_version":1}