{"id":"581180eb-c186-43ce-9d9d-16d80be09b92","arxiv_id":"2501.18048","paper_version":3,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":4,"one_line_summary":"For every n ≥ 1, the interval between n^2 and (n+1)^2 contains a number with at most 4 prime factors.","lead":"This paper proves that between every pair of consecutive square numbers, n^2 and (n+1)^2, there is always an integer with at most 4 prime factors, for every n from 1 upward. It is the first all-n result of this kind, combining an explicit weighted sieve with a large computational check.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 2.1 depends on Sorenson-Webster [17] verifying primes in each half of the square interval, but the cited title only promises Legendre's conjecture; if the half-interval strength is absent, the 4p extension to 1.98e28 collapses.","rationale":"The reader's weakest-assumption analysis is the same one I would single out. The central claim, Theorem 1.2, is established by combining a computational base case with an explicit weighted sieve. The sieve portion is internally coherent: the constants are tracked, the linear sieve lemma is quoted from [2], and the final positivity check, while numerically tight, appears to hold if the imported constants are correct. The genuinely exposed load-bearing input is Lemma 2.1's use of [17]. The title of [17] mentions Legendre's conjecture, which guarantees only one prime somewhere in (n^2,(n+1)^2), not necessarily in each half. The proof needs the half-interval assertion to obtain the multiplier 4p and push the computation from 4.97·10^27 up to 1.98·10^28. Without that, the computational range and sieve range do not overlap, and the theorem as written is unsupported. This is not a matter of internal consistency but of verifying an external computational claim; the cited paper may well contain the stronger statement, in which case the argument goes through. The final numerical margin deserves independent audit too, but it is not the primary load-bearing concern. The appropriate disposition remains conditional: the paper should be accepted only after confirming that [17] indeed verifies the half-interval prime assertions, or after the authors supply the needed computational evidence directly.","tokens_in":12712,"tokens_out":30781,"duration_ms":331175,"concrete_test":"Read Sorenson and Webster [17] (Res. Number Theory 11.4, 2025) and determine exactly what their theorem and algorithm certify. Specifically, check whether [17] proves or computationally establishes a prime in each of (n^2,n(n+1)) and (n(n+1),(n+1)^2) for all n up to the stated bound(s), or only a prime in the full interval (n^2,(n+1)^2). If only the full-interval statement is present, recompute Lemma 2.1's coverage and test whether Theorem 1.2 can still be recovered by lowering the sieve cutoff below 1.98·10^28; if the half-interval statement is present, record the exact upper bound and verify it exceeds the m-values used in the 4p construction.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Lemma 2.1 asserts that for n^2 ≤ 4.97·10^27, the computation of Sorenson and Webster [17] yields a prime in each half-interval (n^2,n(n+1)) and (n(n+1),(n+1)^2). The cited title, 'An algorithm to verify Legendre's conjecture up to 7·10^13,' advertises only a prime somewhere in the full interval (n^2,(n+1)^2). That weaker statement does not imply a prime in each half unless the algorithm records additional location data. This half-interval input is load-bearing: in Lemma 2.1 it supplies the prime p used to cover n^2 up to 1.98·10^28 via 4p—for even n, p in (m^2,m(m+1)), and for odd n, p in (m(m+1),(m+1)^2). If [17] is only a Legendre verification, the proof has no coverage between 4.97·10^27 and the sieve cutoff, and Theorem 1.2 is not established at that stage. The final numerical margin in the sieve step is also tight enough that this gap cannot be absorbed by slack elsewhere. Whether [17] actually proves the stronger half-interval statement must be settled before the proof can be accepted.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves that for every integer n ≥ 1, the interval (n^2, (n+1)^2) contains an integer with at most four prime factors (Theorem 1.2). The proof splits at n^2 ≈ 1.98·10^28: for smaller n, a computation of Sorenson and Webster is invoked; for larger n, an explicit version of the linear sieve due to Bordignon, Johnston, and Starichkova is combined with a new explicit version of Kuhn's weighted sieve. The authors derive explicit constant estimates for the sieve input and use them to show that the weighted sieve count is positive.","tokens_in":12928,"tokens_out":23196,"duration_ms":192520,"significance":"If the proof is correct, this is the first result of this kind valid for all n ≥ 1, giving an explicit analogue of Legendre's conjecture for almost primes. The paper also provides a self-contained explicit version of Kuhn's weighted sieve for generic sifting sets, which may be useful for future applications. The constants are tracked carefully and the auxiliary computations (e.g., Mertens-type estimates) are documented, making the argument reproducible in principle.","major_comments":[{"comment":"The bound e^{2h(s)} ≤ e^2 · 3e^{-s}/s used in (4.9) is false for s ≥ 3. Since h(s) = 3e^{-s}/s for s ≥ 3, one has e^{2h(s)} = e^{6e^{-s}/s}; at s = 3.3 this is about 1.069, whereas e^2 · 3e^{-s}/s is about 0.248. With the correct value, the maximum of f(s) − εC2(ε)e^{2h(s)} on [3,4] is below 0.808, so the constant 8.8C(s) − 7.113 in (4.20) becomes negative and the claimed positivity of r4(A) is not obtained. This is a load-bearing numerical error and must be corrected.","section":"Section 4, Eq. (4.9)"},{"comment":"The lemma asserts that the computation in [17] yields a prime in each half-interval (n^2, n(n+1)) and (n(n+1), (n+1)^2) for all n with n^2 ≤ 4.97·10^27. The cited title only advertises verification of Legendre's conjecture, which guarantees a prime somewhere in the full interval (n^2, (n+1)^2), not necessarily in each half. The half-interval statement is load-bearing for the 4p extension up to n^2 ≤ 1.98·10^28. The authors should verify from [17] (or supply the computation) that the stronger half-interval assertion is indeed available.","section":"Lemma 2.1"},{"comment":"The factor (1 + k2/(2(log X)^2)) in the statement of Proposition 3.5 and in equation (3.13) is inconsistent with the derivation from Lemma 3.3. Since log z = log X / k1, the correction term from Lemma 3.3 is k1^2/(2(log X)^2), not k2/(2(log X)^2). The numerical evaluation in (4.15) uses 32 in the numerator, corresponding to k1^2/2 with k1 = 8, so the displayed formulas should read 1 + k1^2/(2(log X)^2).","section":"Proposition 3.5, Eq. (3.13)"}],"minor_comments":[{"comment":"The first sentence states that no element of A is divisible by a prime in P; in the application P is the set of all primes, which would make the hypothesis false. The intended condition is that no element of A is divisible by a prime in the complement of P, as in Lemma 2.2.","section":"Lemma 3.1"},{"comment":"The lower bound '> 107' for y should read '> 10^7' (or a more precise value). With the corrected value, the choice c2 = 0.07 > 1.1/log(10^7) is valid.","section":"Lemma 4.1"},{"comment":"In the displayed bound for the product, '2√z' should be '2/√z'; the subsequent numerical estimate uses the correct expression.","section":"Proposition A.2, Case 2"},{"comment":"The condition (2.4) is required for all 1 < u < z, but Proposition A.2 proves the estimate only for u ≥ 3. The remaining range u < 3 is easy to handle but should be explicitly mentioned.","section":"Lemma 2.2 / Proposition A.2"}],"recommendation":"major_revision","confidential_remarks":"The error in eq. (4.9) appears fatal for the numerical part of the proof as written; with the corrected bound the final inequality in (4.20) does not hold. I recommend the editor ask the authors to re-examine this step and to verify the half-interval claim from [17]. If these issues cannot be resolved, the main theorem would be unproven."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The paper proves that for every n ≥ 1, the interval (n^2, (n+1)^2) contains an integer with at most four prime factors. That is genuine progress: earlier explicit results were either asymptotic or allowed more prime factors, and this is the first all-n statement of its kind. The main technical contribution, an explicit version of Kuhn's weighted sieve for generic sifting sets, is cleanly stated and should be reusable. Constants are tracked carefully, and the self-cited linear sieve is a general tool rather than a conclusion-shaped assumption, so circularity is low. The appendix's explicit Mertens bound is also a useful small piece.\n\nThe soft spot is exactly where the stress-test points. Lemma 2.1 assumes that the Sorenson–Webster computation gives a prime in each half of (n^2, (n+1)^2). The title of [17] only advertises Legendre's conjecture, which is weaker. If [17] only certifies the full interval, the proof loses coverage between 4.97·10^27 and the sieve cutoff, and the 4p argument falls apart. The authors need to state which theorem in [17] supplies the half-interval primes, or provide the data directly. I don't think this is a dealbreaker — the stronger claim is plausible and likely extractable from the algorithm — but it is load-bearing and cannot be waved away.\n\nThe stress-test's secondary worry about the final numerical margin does not survive arithmetic. At N = 1.98·10^28, the positive term in (4.20) is about 5.8·10^11 and the negative term about 5.7·10^11, so the margin is small but real, and it widens as N grows.\n\nThis paper is for anyone working on explicit sieve methods or Legendre-adjacent problems. It deserves a serious referee. My recommendation: send it out, and have the referee verify the Sorenson–Webster half-interval claim before sign-off.","headline":"The result is new and the sieve work is solid; the half-interval reading of Sorenson–Webster needs to be checked before full confidence.","tokens_in":13520,"tokens_out":4891,"would_cite":true,"duration_ms":292588,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11N36","11N05"],"pacs":[],"model":"deepseek-v4-flash","headline":"The paper proves that for every $n \\ge 1$, the interval between $n^2$ and $(n+1)^2$ contains an integer with at most four prime factors.","keywords":["almost primes","Legendre's conjecture","sieve methods","linear sieve","weighted sieve","explicit results","prime gaps"],"falsifier":"Re-run or inspect the verification behind [17] and check, for every integer $n$ with $n^2 \\le 4.97\\cdot10^{27}$, whether both $(n^2,n(n+1))$ and $(n(n+1),(n+1)^2)$ contain a prime; if any such interval lacks a prime in one half, the small-$n$ lemma loses its force and the theorem would need another argument for that range.","tokens_in":12441,"feed_emoji":"🔢","tokens_out":11981,"duration_ms":96108,"temperature":0.7,"pith_summary":"This paper proves that between every pair of consecutive squares, $n^2$ and $(n+1)^2$, there is always an integer built from at most four primes. That is the first theorem of this kind to hold for every $n \\ge 1$; earlier work located such almost primes only for sufficiently large $n$. The proof combines a recent computational verification of Legendre's conjecture up to a large bound with an explicit version of Kuhn's weighted sieve. As a byproduct, the paper gives a fully explicit weighted-sieve framework for generic sifting sets. If the argument is correct, it provides an unconditional step toward Legendre's conjecture: the obstruction to finding a prime in every square gap is, in this sense, at worst four prime factors.","feed_headline":"Between every two squares sits a number with at most 4 prime factors","feed_subtitle":"No prime is guaranteed in these gaps, but a number made from at most 4 primes always is.","key_machinery":"The key machinery is an explicit version of Kuhn's weighted sieve, stated for generic sifting sets. The weight $w(a) = 1 - \\frac12 \\sum_{z \\le q < y,\\, q^\\ell \\parallel a} \\ell$ lets the count of integers with at most $k_2$ prime factors be bounded below by $S(\\mathcal{A},\\mathcal{P},z) - \\frac12 \\sum_{z \\le q<y} S(\\mathcal{A}_q,\\mathcal{P},z)$ minus explicit error terms, keeping elements that have at most one small prime divisor in $[z,y)$. Lower and upper bounds for the sifting functions come from an explicit linear sieve, and explicit Mertens-type estimates control the product and remainder terms. The specific choice $z = X^{1/8}$, $y = X^{1/4}$, with $k_1=8$, $k_2=4$ and $\\alpha = 0.07$, makes the positive main term dominate all errors once $N > 1.98\\cdot10^{28}$.","core_discovery":"The central claim is Theorem 1.2: for every positive integer $n$, the interval $(n^2,(n+1)^2)$ contains an integer $a$ with $\\Omega(a) \\le 4$, where $\\Omega(a)$ counts prime factors with multiplicity. The proof splits at $N = 1.98\\cdot10^{28}$. For smaller square endpoints, a cited computation gives primes in each half of the square interval, and the observation that $4p$ lies between $n^2$ and $(n+1)^2$ when $p$ lies in a suitably rescaled interval lifts the coverage up to the cutoff. For larger $N$, the authors sieve the set $\\mathcal{A} = \\mathbb{Z} \\cap (N, N+2\\sqrt{N})$ with $z = X^{1/8}$, $y = X^{1/4}$, where $X = \\lfloor N + 2\\sqrt{N}\\rfloor$, and apply an explicit weighted-sieve inequality to show that the count $r_4(\\mathcal{A})$ of elements with at most four prime factors is positive. This establishes the first unconditional result of this form valid for all $n$ rather than only for sufficiently large $n$.","pith_inferences":["In our reading, the small-range computation is the part most worth checking first: the argument needs a prime in each half of the square interval, which is stronger than the cited title's claim of verifying Legendre's conjecture, and a failure there would not be repaired by the sieve portion.","A natural next step is to optimize the sieve parameters $k_1$, $k_2$, and $\\alpha$; the paper's own rough estimate suggests that reaching $k=3$ for all $n$ would require the starting point to move to roughly $10^{50}$, so progress is more likely to come from extending the computational small-$n$ coverage.","Any future computation that certifies half-interval primes to a larger bound would directly extend the range of Theorem 1.2 and could reduce the number of prime factors in the small-$n$ regime."],"forward_implications":["For every integer $n \\ge 1$, the interval between consecutive squares contains an integer with at most four prime factors, so the almost-prime analogue of Legendre's conjecture with $k=4$ holds unconditionally.","Because the weighted-sieve inequalities are proved for generic sifting sets, other interval problems can reuse the explicit constants without re-deriving the sieve bounds.","With adjusted parameters, the authors note, the same method should give at most three prime factors between consecutive cubes and at most two between consecutive fourth powers for all $n$.","The effective range begins at $N > 1.98\\cdot10^{28}$, unlike the earlier prime-between-cubes result, which only starts at a doubly exponential scale."],"supporting_citations":[{"why":"Supplies the half-interval prime check that covers $n^2 \\le 4.97\\cdot10^{27}$ and, through the multiples $4p$, extends the small-range case to $1.98\\cdot10^{28}$.","marker":"[17]"},{"why":"Provides the explicit linear sieve upper and lower bounds used to estimate the sifting functions in Lemmas 2.2 and 2.3.","marker":"[2]"},{"why":"Introduces the weighted-sieve weight that lets the proof keep elements with one small prime divisor.","marker":"[12]"},{"why":"Supplies the explicit Mertens estimates behind the product $V(z)$ and the numerical sieve constants.","marker":"[16]"},{"why":"Supplies explicit bounds on sums of reciprocals of primes and on Mertens products used in the error terms and the appendix.","marker":"[18]"},{"why":"Supplies the explicit bound on the number of squarefree integers used to control the sieve remainder sum.","marker":"[15]"},{"why":"Gives the summation identity used to evaluate the weighted sum over $S(\\mathcal{A}_q,\\mathcal{P},z)$ in Proposition 3.5.","marker":"[9]"}],"fun_headline_variants":["Every square gap holds a number with ≤4 prime factors","4-prime numbers appear in every square interval","Each square pair contains a near-prime (≤4 factors)","All square gaps contain a 4-almost prime","Numbers with ≤4 primes found in every square gap"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing premise is that the cited computation [17] verifies a prime in each half of the square interval up to about $4.97\\cdot10^{27}$, not just a prime somewhere in the whole interval, since that half-interval strength is what lets the small cases be extended to $1.98\\cdot10^{28}$.","fun_headline_variants_meta":{"raw":{"variants":["Every square gap holds a number with ≤4 prime factors","4-prime numbers appear in every square interval","Each square pair contains a near-prime (≤4 factors)","All square gaps contain a 4-almost prime","Numbers with ≤4 primes found in every square gap"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000261,"raw_usage":{"total_tokens":1561,"prompt_tokens":882,"completion_tokens":679,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":498,"completion_tokens_details":{"reasoning_tokens":611}},"tokens_in":498,"tokens_out":679,"duration_ms":6034,"temperature":1.0,"reasoning_tokens":611,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-10T00:54:59.001960+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Re-run or inspect the verification behind [17] and check, for every integer $n$ with $n^2 \\le 4.97\\cdot10^{27}$, whether both $(n^2,n(n+1))$ and $(n(n+1),(n+1)^2)$ contain a prime; if any such interval lacks a prime in one half, the small-$n$ lemma loses its force and the theorem would need another argument for that range.","supporting_citations":[{"cited_title":"An algorithm to verify Legendre’s conjecture up to 7 · 1013","cited_arxiv_id":null,"evidence_quote":"Supplies the half-interval prime check that covers $n^2 \\le 4.97\\cdot10^{27}$ and, through the multiples $4p$, extends the small-range case to $1.98\\cdot10^{28}$."},{"cited_title":"An explicit version of Chen’s theorem and the linear sieve","cited_arxiv_id":null,"evidence_quote":"Provides the explicit linear sieve upper and lower bounds used to estimate the sifting functions in Lemmas 2.2 and 2.3."},{"cited_title":"Neue Absch¨ atzungen auf Gund der Viggo Brunschen Siebmethode","cited_arxiv_id":null,"evidence_quote":"Introduces the weighted-sieve weight that lets the proof keep elements with one small prime divisor."},{"cited_title":"Approximate formulas for some functions of prime numbers","cited_arxiv_id":null,"evidence_quote":"Supplies the explicit Mertens estimates behind the product $V(z)$ and the numerical sieve constants."},{"cited_title":"Explicit Mertens Sums","cited_arxiv_id":null,"evidence_quote":"Supplies explicit bounds on sums of reciprocals of primes and on Mertens products used in the error terms and the appendix."},{"cited_title":"From explicit estimates for primes to explicit estimates for the M¨ obius function","cited_arxiv_id":null,"evidence_quote":"Supplies the explicit bound on the number of squarefree integers used to control the sieve remainder sum."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Gives the summation identity used to evaluate the weighted sum over $S(\\mathcal{A}_q,\\mathcal{P},z)$ in Proposition 3.5."}],"review_version":1}