{"id":"409ccafc-2316-43ca-b1f4-5839b6c665a9","arxiv_id":"2412.19866","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":4.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"The authors derive explicit asymptotic series for 1/li_k(x) whose coefficients come from a combinatorial recurrence, generalizing Panaitopol's expansion from primes to prime k-tuples.","lead":"This paper proves asymptotic expansions for the reciprocals of the integrals appearing in the first Hardy-Littlewood conjecture, generalizing Panaitopol's expansion for the prime counting function. The result connects a recurrence counting indecomposable permutations to higher-order prime number heuristics.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 2.7's proof substitutes the divergent series A(1/x) with x e^{-x} Ei(x) and then takes reciprocals without stating the standard Poincaré asymptotic-manipulation theorem; this unproved step is the load-bearing gap.","rationale":"The reader's weakest assumption identifies exactly the same load-bearing concern: Theorem 2.7 rests on an unproved algebraic-manipulation step for Poincaré asymptotic expansions, specifically replacing a divergent formal series by its asymptotic function and then taking a reciprocal. I agree that the claim is true in content and that a standard asymptotic-analysis theorem repairs the proof; the paper should state or cite that theorem. A secondary textual issue is the sign in Definition 2.5: as written, li_1(x) would equal -li(x), contradicting the appended note; Lemma 2.6 and Theorem 2.7 use the positive convention (li_k(x) = int_0^x dt/(log t)^k), so the minus sign should be removed. This is an error in the statement, not in the main argument. The proposed concrete test settles the concern by redoing the substitution with partial sums and explicitly tracking the remainder; if the remainder is beyond all orders, the conditional verdict stands. I therefore keep the reader's CONDITIONAL verdict unchanged.","tokens_in":5402,"tokens_out":14963,"duration_ms":125360,"concrete_test":"Run a truncation-level proof of the contested step in Theorem 2.7: fix N, write A(1/x) = sum_{n=0}^{N-1} n! x^{-n} + O(N! x^{-N}) and likewise x e^{-x} Ei(x) = same partial sum + O(N! x^{-N}); substitute both into the denominator, expand the reciprocal using (1+u)^{-1}, multiply through by x^k e^{-x}, and use Lemma 2.6 to compare with 1/li_k(e^x). If the remainder is o(x^{-M}) for every M, the manipulation is justified and the proof is repairable; if a non-negligible term survives, Theorem 2.7 requires correction.","verdict_should_be":"UNCHANGED","load_bearing_attack":"In the proof of Theorem 2.7, after the exact formal identity 1 - k/x - sum a_n x^{-n-1} = (k-1)! x^{1-k} / (A(1/x) - sum_{i=0}^{k-2} i! x^{-i}), the text invokes Proposition 2.4 to replace A(1/x) by x e^{-x} Ei(x) and immediately writes an asymptotic equivalence for the resulting reciprocal. This step requires a theorem: if f(x) - g(x) = o(x^{-N}) for every N, and if g(x) has a nonzero leading asymptotic term, then rational combinations of f and g, including the reciprocal, inherit asymptotic expansions from the formal expansion of g. Here the denominator has leading term (k-1)! x^{-(k-1)}, so a standard Poincaré reciprocal theorem applies, but the paper neither proves nor cites it. Without such a theorem, the line is a formal manipulation of divergent series rather than a consequence of Definition 1.1. The claimed expansion appears numerically correct (e.g., the k=2 case matches the direct integration-by-parts expansion), so this is a rigor gap rather than a counterexample, but it is the load-bearing point on which the proof rests.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"This paper defines, for every positive integer k, an integer sequence (a_n^{(k)}) by the recurrence in Definition 2.1 (with a_0^{(k)} = 1) and proves Proposition 2.2, a formal power series identity generalizing Comtet's generating function for indecomposable permutations: 1 − (k−1)! (∑_{n≥0} (n+k−1)! x^n)^{-1} = kx + ∑_{n≥1} a_n^{(k)} x^{n+1}. Proposition 2.4 (the standard asymptotic expansion of the exponential integral) and Lemma 2.6 (a reduction of li_k to li) are then combined with this identity to prove Theorem 2.7, which asserts the complete asymptotic expansion 1/li_k(x) ∼ (log x)^k/x (1 − k/log x − ∑_{n≥1} a_n^{(k)}/(log x)^{n+1}) for the reciprocal of the k-th Hardy-Littlewood logarithmic integral. For k = 1 the expansion reduces to the known expansion of 1/li(x), and through the remark following Proposition 1.2 to Panaitopol's expansion for 1/π(x). The Hardy-Littlewood application is presented as motivation, with an explicit caveat that the result only provides auxiliary information for a possible strengthening of Conjecture 1.3.","tokens_in":5644,"tokens_out":30392,"duration_ms":227256,"significance":"If the proof is completed as suggested in Major Comment 1, the main theorem is a correct, parameter-free, complete asymptotic expansion whose coefficients come from an independently defined combinatorial recurrence; there is no circularity and no fitted constant. The k = 2 case matches the direct integration-by-parts expansion, the k = 1 case reproduces the Comtet/Panaitopol system, and the coefficient sequences are identified with OEIS entries; these are genuine strengths that make the expansion a concrete, checkable target for any future strengthening of the first Hardy-Littlewood conjecture. The significance of the result is moderate: it is a clean unification of a classical combinatorial generating function with the asymptotic theory of logarithmic integrals, and the paper's claims about the conjecture are appropriately hedged. The principal weakness is the unstated asymptotic-manipulation step in the proof of Theorem 2.7, which is the load-bearing point of the paper and needs a short lemma or a precise citation.","major_comments":[{"comment":"This step substitutes the divergent formal series A(1/x) = ∑_{n≥0} n!x^{−n} by the function xe^{−x}Ei(x) inside the rational expression (k−1)!x^{1−k}/(A(1/x) − ∑_{i=0}^{k−2} i!x^{−i}) and then takes a reciprocal, without any stated justification. As written, the line is a formal manipulation of a divergent series, not a consequence of Definition 1.1. What is needed is the standard theorem that Poincaré asymptotic expansions are preserved under rational combinations when the denominator's leading asymptotic term is nonzero — here the denominator has leading term (k−1)!/x^{k−1} — and this theorem is neither stated, nor proved, nor cited. This step is load-bearing, because it is what turns the formal identity of Proposition 2.2 into the asymptotic statement. I verified the conclusion independently: for k = 2 the theorem's expansion reproduces 1/li_2(x) ∼ (log x)^2/x (1 − 2/log x − 2/(log x)^2 − 8/(log x)^3 − 44/(log x)^4 − ···) from the direct expansion of li_2, so the gap is a rigor gap rather than an error. The fix is local: add a lemma (if f(x) ∼ ∑ c_n x^{−n} with c_0 ≠ 0, then 1/f(x) admits the formal reciprocal expansion, and rational expressions inherit expansions by substitution), with a short proof or a precise citation, and verify its hypotheses for the specific denominator used.","section":"Theorem 2.7, proof (step after 'Proposition 2.4 gives the relation A(1/x) ∼ xe^{−x}Ei(x); thus, ...')"}],"minor_comments":[{"comment":"The first displayed line reads '(k−1)!I_k(x) = k!x − ∑_{n≥1}(k−1)!a_n^{(k)}x^{n+1}', but since I_k(x) = kx + ∑_{n≥1}a_n^{(k)}x^{n+1} the sign before the sum should be '+'; the subsequent line is consistent with the corrected sign, so this is a typo, but it is very confusing.","section":"Proposition 2.2, proof"},{"comment":"The asymptotic sequence with respect to which the expansion holds is implicit; it should be stated explicitly (e.g., {x^{−1}(log x)^{k−m}}_{m≥0}) so that the statement conforms to Definition 1.1.","section":"Theorem 2.7, statement"},{"comment":"The notation 'A(1/x) ∼ xe^{−x}Ei(x)' is an abuse of notation, since A(1/x) diverges for every finite nonzero x; it would be clearer to write that xe^{−x}Ei(x) has the asymptotic expansion ∑_{n≥0} n!x^{−n}.","section":"Theorem 2.7, proof"},{"comment":"The sentence 'Proposition 1.2 may be quickly obtained ... by following the k = 1 case of the proof of Theorem 2.7' refers forward to a theorem proved later; consider rephrasing, for example, 'by following the proof of Theorem 2.7 in Section 2 with k = 1'.","section":"Introduction, remark after Proposition 1.2"},{"comment":"Please confirm that the k ≥ 2 cases of Theorem 2.7 do not already appear in [5] or in the literature following [11]; if they do, the abstract's novelty claim should be adjusted accordingly.","section":"References / novelty"},{"comment":"There are several typographical artifacts (e.g., 'C onsequently' in the abstract, 's eries' in the first paragraph of Section 1); a careful proofread is advised.","section":"General"}],"recommendation":"major_revision","confidential_remarks":"The core mathematics checks out: I verified the algebra of Proposition 2.2, Lemma 2.6, and the k = 2 case of Theorem 2.7 against the direct integration-by-parts expansion of li_2, and the coefficients from Definition 2.1 are consistent with the table. The single substantive defect is the missing justification for algebraic manipulation (including reciprocation) of Poincaré asymptotic expansions in the proof of Theorem 2.7; this is standard and easily supplied, so I expect a straightforward revision rather than a reworking of the results. I would also ask the author to double-check that the k ≥ 2 expansion is not already implicit in [5], which is cited for closely related material, or elsewhere in the post-Panaitopol literature. The Hardy-Littlewood discussion is motivational and properly hedged; it should not be used as a criterion for acceptance. Overall the paper is modest but sound, and after the lemma is added it should be publishable in a specialist journal."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Quick take: this is a small but legitimate result. The paper generalizes Comtet's generating-function identity to the series with factorial coefficients shifted by k, and uses it to write down the full asymptotic expansion of 1/li_k(x). For k>1 that expansion is new, and the k=2 case checks out against direct integration by parts. The combinatorial connection to indecomposable permutations only really bites at k=1, but the generalization is natural.\n\nWhat the paper does well: the formal step in Proposition 2.2 is clean and self-contained, and Theorem 2.7 is clearly stated. The coefficients a_n^(k) come from an independent recurrence, not from fitting the target integral, so there is no circularity. The relation to the first Hardy-Littlewood conjecture is stated honestly as auxiliary information, not as a stronger theorem.\n\nThe soft spots are minor but real. The proof of Theorem 2.7 substitutes the divergent formal series A(1/x) with its Poincaré asymptotic expansion x e^{-x} Ei(x) and then manipulates the resulting asymptotic series, including taking reciprocals, without citing or proving the standard theorem that rational operations preserve Poincaré expansions when the denominator's leading term is nonzero. That is exactly the step the stress-test note flags. It is a rigor gap rather than a counterexample: the expansion is correct, and the k=2 check confirms it. The fix is a one-line citation or a short argument.\n\nDefinition 2.5 also has a sign typo: the minus sign makes li_1(x) equal -li(x), contradicting the text and Lemma 2.6. This is clearly a typo, but it needs fixing.\n\nA minor clarity point: the paper could state explicitly that Theorem 2.7 is the reciprocal of the standard asymptotic expansion of li_k. That would make the relationship to existing work plainer.\n\nWho should read this: people who work with li_k in asymptotics or in computations for prime k-tuples. It is not a breakthrough, but it is a solid technical note. If it is submitted, I would send it to a referee; with the manipulation step cited and the typo corrected, it is acceptable. I wouldn't cite it in my own work unless I needed that specific expansion, but it deserves a place in the literature.","headline":"A correct, modest generalization of Panaitopol's reciprocal li expansion; the proof has a repairable rigor gap and a sign typo, but the main result stands.","tokens_in":6163,"tokens_out":5113,"would_cite":false,"duration_ms":41898,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["05A15","11N05","41A60","05A05"],"pacs":[],"model":"deepseek-v4-flash","headline":"For every positive integer k, the reciprocal of the k-th Hardy-Littlewood logarithmic integral has an explicit full asymptotic expansion in powers of 1/log x, with coefficients given by a recurrence that generalizes the indecomposable…","keywords":["asymptotic expansions","Hardy-Littlewood logarithmic integrals","reciprocal logarithmic integral","indecomposable permutations","formal power series","prime counting function","Hardy-Littlewood conjecture","exponential integral"],"falsifier":"Evaluate $1/\\mathrm{li}_2(x)$ numerically to high precision at $x=10^j$ for a fixed truncation order $N$ and check whether $\\frac{(\\log x)^{N+2}}{x}\\left(1/\\mathrm{li}_2(x) - \\text{partial sum}_N(x)\\right)$ stays bounded as $j$ grows. If it does not, the asserted asymptotic expansion for $k=2$ is false; checking the first coefficients against Proposition 2.2 would catch algebraic errors.","tokens_in":5194,"feed_emoji":"🔢","tokens_out":11003,"duration_ms":90629,"temperature":0.7,"pith_summary":"The paper defines a family of integer sequences by a recurrence and uses them to write down a full asymptotic expansion, for each positive integer k, of the reciprocal of the k-th Hardy-Littlewood logarithmic integral $\\mathrm{li}_k(x)$. These integrals are the ones appearing in the first Hardy-Littlewood conjecture on prime k-tuples, so the expansion supplies higher-order corrections beyond the leading term. In the case k=1 the result reproduces the known expansion for the reciprocal of the prime counting function, since $\\pi(x)$ and $\\mathrm{li}(x)$ have the same asymptotic behavior. A sympathetic reader would care because the proof ties a combinatorial counting problem, the enumeration of indecomposable permutations and its generalizations, to analytic number theory.","feed_headline":"Full expansion found for reciprocal Hardy-Littlewood integrals","feed_subtitle":"A recurrence gives all correction terms for reciprocal logarithmic integrals, from prime counts to every k.","key_machinery":"The load-bearing object is the coefficient sequence $a_n^{(k)}$ together with the generating-function identity of Proposition 2.2: $$1 - (k-1)!\\left(\\sum_{n\\ge 0}(n+k-1)!\\,x^n\\right)^{-1} = kx + \\sum_{n\\ge 1} $a_n^{{(k)}}$ $x^{{n+1}}$$$ as formal power series. This identity connects the recurrence to the reciprocal of a factorial series; when $x$ is replaced by $1/\\log x$ and the factorial series is read through the asymptotic expansion of the exponential integral, it becomes the correction series in the expansion of $1/\\mathrm{li}_k(x)$. The sequence's $k=1$ case counts indecomposable permutations, and the paper shows the higher-$k$ sequences play the same combinatorial role for the higher logarithmic integrals.","core_discovery":"The main result, Theorem 2.7, states that for each $k \\in \\mathbb{Z}^+$, as $x \\to \\infty$, $$\\frac{1}{\\mathrm{li}_k(x)} \\sim \\frac{(\\log x)^k}{x}\\left(1 - \\frac{k}{\\log x} - \\sum_{n\\ge 1}\\frac{$a_n^{{(k)}}$}{(\\log x)^{n+1}}\\right),$$ where the coefficients $a_n^{(k)}$ are fixed by the recurrence $$(k-1)!\\,$a_n^{{(k)}}$ = (n+k)! - k(n+k-1)! - \\sum_{m=0}^{n-2} (n-m+k-2)!\\,a_{m+1}^{(k)},$$ with $a_0^{(k)}=1$. The proof starts from a formal power-series identity for the generating function of these coefficients and transfers it into an asymptotic statement via the standard asymptotic expansion of the exponential integral and a change of variables. When $k=1$, the coefficients count indecomposable permutations and the expansion reduces to the known reciprocal prime-counting expansion.","pith_inferences":["If the first Hardy-Littlewood conjecture is ever promoted to a full asymptotic expansion for prime $k$-tuple counts, the coefficients $a_n^{(k+1)}$, up to the singular-series constant of the tuple, are the natural correction coefficients; this is an extrapolation, since the paper only says the expansion is 'salient' to the conjecture.","The $k=1$ equivalence between $1/\\mathrm{li}(x)$ and $1/\\pi(x)$ uses an error bound much larger than the individual terms of the new series, so comparing the expansion with numerical values of $\\mathrm{li}_k(x)$ directly would be a sharper test than comparing with prime counts.","The recurrence likely admits a combinatorial interpretation for every $k$, since its $k=1$ case counts indecomposable permutations; such an interpretation would give an independent, purely finite check of the analytic coefficients."],"forward_implications":["For every $k$, the reciprocal $1/\\mathrm{li}_k(x)$ now has a complete, explicitly computable asymptotic series in powers of $1/\\log x$, with integer coefficients fixed by an elementary recurrence.","Setting $k=1$ recovers the known expansion for the reciprocal prime counting function, since $\\mathrm{li}(x)-\\pi(x)$ is much smaller than any term in the asymptotic sequence.","The coefficients $a_n^{(k)}$ generalize the indecomposable permutation counts, giving sequences originally studied combinatorially a role in the analytic theory of prime distributions.","The expansion supplies the correction terms needed if the first Hardy-Littlewood conjecture is strengthened from a leading-order asymptotic to a full asymptotic expansion for prime $k$-tuple counts."],"supporting_citations":[{"why":"Supplies the original generating-function identity for the reciprocal of $\\sum n!x^n$, which Proposition 2.2 generalizes to the family $\\sum (n+k-1)!x^n$.","marker":"[2]"},{"why":"Gives the classical estimate $\\pi(x)=\\mathrm{li}(x)+O(x e^{-a\\sqrt{\\log x}})$ used to identify the $k=1$ expansion with the prime-counting expansion.","marker":"[3]"},{"why":"Provides the standard asymptotic expansion of the exponential integral $\\mathrm{Ei}(x)$ that the proof substitutes into the formal power series.","marker":"[4]"},{"why":"States the $k=1$ expansion for the reciprocal prime counting function that Theorem 2.7 generalizes.","marker":"[11]"}],"fun_headline_variants":["Recurrence fixes every coefficient in reciprocal Hardy-Littlewood expansions","All correction terms for reciprocal Hardy-Littlewood integrals via recurrence","Complete reciprocal log integral asymptotics from a single recurrence","Explicit asymptotic expansion for reciprocal li_k from recurrence"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing assumption is that substituting the divergent series for the exponential integral by its asymptotic expansion, and then reciprocating and re-expanding termwise, preserves the final asymptotic expansion; this operation is used in the proof of Theorem 2.7 without being stated or proved.","fun_headline_variants_meta":{"raw":{"variants":["Recurrence fixes every coefficient in reciprocal Hardy-Littlewood expansions","All correction terms for reciprocal Hardy-Littlewood integrals via recurrence","Complete reciprocal log integral asymptotics from a single recurrence","Explicit asymptotic expansion for reciprocal li_k from recurrence"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.001452,"raw_usage":{"total_tokens":5774,"prompt_tokens":800,"completion_tokens":4974,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":416,"completion_tokens_details":{"reasoning_tokens":4906}},"tokens_in":416,"tokens_out":4974,"duration_ms":34769,"temperature":1.0,"reasoning_tokens":4906,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-11T00:35:09.766702+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Evaluate $1/\\mathrm{li}_2(x)$ numerically to high precision at $x=10^j$ for a fixed truncation order $N$ and check whether $\\frac{(\\log x)^{N+2}}{x}\\left(1/\\mathrm{li}_2(x) - \\text{partial sum}_N(x)\\right)$ stays bounded as $j$ grows. If it does not, the asserted asymptotic expansion for $k=2$ is false; checking the first coefficients against Proposition 2.2 would catch algebraic errors.","supporting_citations":[{"cited_title":"Comtet, Sur les coefficients de l’inverse de la s´ erie formelle ∑ n!xn, C","cited_arxiv_id":null,"evidence_quote":"Supplies the original generating-function identity for the reciprocal of $\\sum n!x^n$, which Proposition 2.2 generalizes to the family $\\sum (n+k-1)!x^n$."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Gives the classical estimate $\\pi(x)=\\mathrm{li}(x)+O(x e^{-a\\sqrt{\\log x}})$ used to identify the $k=1$ expansion with the prime-counting expansion."},{"cited_title":"Panaitopol, A formula for π (x) applied to a result of Koninck-Ivi´ c, Nieuw Arch","cited_arxiv_id":null,"evidence_quote":"States the $k=1$ expansion for the reciprocal prime counting function that Theorem 2.7 generalizes."}],"review_version":1}