{"id":"c865bebb-8835-4e8c-a2f4-97987a697ccd","arxiv_id":"1908.05594","paper_version":2,"verdict":"ACCEPT","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"The paper gives exact p-adic valuations or lower bounds for s(ap, ap-k) and derives a recursive pattern that partially proves Lengyel's conjecture for odd k.","lead":"This paper proves new formulas for how often a prime divides certain cycle-counting numbers, for inputs of the form ap and ap minus k. It partially confirms two open conjectures about such divisibility and about sums of reciprocals.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"No significant objection identified; the delicate step is congruence (4.21), which deserves an independent modular check.","rationale":"The paper's central claim is Theorem 1.1, and the reader correctly identifies the even-k part of (ii) as the most delicate. My stress-test read focused on the inductive proof of claim (4.21). I verified the range checks, the p-adic unit status of the coefficient, the deletion of terms in V_k, the treatment of i=p in the l=a case, and the use of Lemma 2.11 to pass from s_{p^n} to s. The only place where the text is terse is the sentence after (4.6) claiming v_p(sp((a−1)p,·))≥1 for all i∈V_k; the exceptional values k+i−p=1 and k+i−p=(a−1)(p−1)+1 are not literally instances of the induction hypothesis. However, they are covered by the elementary bound v_p(s(m,j))≥1 for (m+j) odd and by the already-proved part (iii), so this is a minor expositional compression, not a flaw. Since I could not substantiate a counterexample and the modular transfer of the equality condition is sound, I do not lower the reader's accept verdict. I still recommend an independent machine check of (4.21) at an irregular prime because the theorem's exact statement is more delicate than the regular-prime case and no formal verification is provided.","tokens_in":24182,"tokens_out":44819,"duration_ms":397913,"concrete_test":"Compute both sides of (4.21) for the first irregular prime p=37, a=2, k=32 (so ⟨k⟩=32 and v_37(B_32)>0): evaluate s(74,42) modulo 37^2 and compare it with 2*s(37,5) modulo 37^2; the congruence predicts equality of residues and v_37(s(74,42))≥2. If the residues differ, claim (4.21), and hence the exactness part of Theorem 1.1(ii), fails; if they agree, the most delicate step is supported. This check also tests the theorem's non-regular-prime prediction, since v_37(B_32)=1.","verdict_should_be":"UNCHANGED","load_bearing_attack":"I read the proof of Theorem 1.1 in full and found no gap that undermines the central claim. The exact equality in Theorem 1.1(ii) is read off the congruence (4.21), so that congruence is the load-bearing internal step. I checked the induction in Section 4: the coefficient a*C(a-1,l-1) is a p-adic unit for 1≤l≤a≤p−1; the terms indexed by V_k in (4.7) are killed modulo p^2 because both factors have positive valuation, and the edge cases not literally covered by the induction (k+i−p=1 or k+i−p=(a−1)(p−1)+1) are controlled by the direct valuation bound from Theorem 1.3 and by part (iii); the i=p term in Case 3 is shown to be 0 modulo p^2. A congruence modulo p^2 whose right side has valuation 1 transfers that valuation exactly, so the Bernoulli-number equality criterion follows. The only residual risk is that an omitted term in the decomposition (4.6) has valuation exactly 1 and changes the residue in (4.21); I could not construct such a term, and the proof appears to exclude all of them. Since the proofs are long and unformalized, this remains the point worth checking independently rather than a demonstrated error.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies the p-adic valuations of the unsigned Stirling numbers of the first kind s(n,k). The main theorem (Theorem 1.1) gives, for primes p≥5 and integers 1≤a≤p−1, 2≤k≤ap−2, an exact formula or lower bound for v_p(s(ap,ap−k)) in terms of v_p(k), the parity ε_k, the residue ⟨k⟩ modulo p−1, and the valuation of a Bernoulli number B_{2⌊⟨k⟩/2⌋}. The proof combines Washington's congruences for generalized harmonic numbers, the Newton-Girard identities, properties of the m-th Stirling numbers from the authors' previous work, and an induction on a. From Theorem 1.1 the authors derive Corollary 1.2, giving a negative logarithmic bound for v_p(H(ap−1,k)) for regular primes p≥7, which partially supports a conjecture of Leonetti and Sanna. Theorems 1.3 and 1.5 establish a recursion for odd offsets k, proving v_p(s(ap^{n+1},ap^{n+1}-k))=v_p(s(ap^n,ap^n-k))+2 under a largeness condition, thereby partially confirming a conjecture of Lengyel. Three conjectures on the valuations of s(ap^n,k) are stated in Section 5.","tokens_in":24442,"tokens_out":23175,"duration_ms":207307,"significance":"If the results are correct, the paper is a substantial contribution to the p-adic theory of Stirling numbers of the first kind. The explicit Bernoulli-number criterion for equality in Theorem 1.1(ii) is a strong and falsifiable statement, and the partial confirmations of the Leonetti-Sanna and Lengyel conjectures are significant. The proofs are detailed and internally consistent for the most part, and the paper makes good use of known congruences rather than introducing ad-hoc assumptions. The main weaknesses are that the proofs are long and not machine-checked, and that the exact equality in Theorem 1.1(ii) is read off a single delicate congruence, Eq. (4.21). I found one missing case in the proof of that congruence, but it is easily repairable.","major_comments":[{"comment":"The case split for even k in the subcase p≤ap−k≤p+a−3 omits k=(a−1)p−1. This value occurs, for example, when a=4 and p=5, where k=14 is even and satisfies (a−1)(p−1)+2≤k≤a(p−1)−2. The text states that for even k one has (a−1)(p−1)+2≤k≤(a−1)p−2 or k=(a−1)p, but k=(a−1)p−1 is even whenever a is even and is not covered. The subsequent appeal to part (iii) does not apply to s((a−1)p,1), since that case has second argument 1 and, for a=4, the required parameter a−1 is only 3. The desired congruence mod p^2 is nevertheless true because v_p(s((a−1)p,1))=v_p(((a−1)p−1)!)=a−2≥2, so the gap is local and fixable. This case must be added explicitly, since the exact equality condition in Theorem 1.1(ii) is read from congruence (4.21).","section":"Section 4, proof of claim (4.21), Case 3 (paragraph containing Eq. (4.39))"},{"comment":"The proof of claim (4.21) suppresses many terms from the decomposition (4.6) using the V_k condition (4.7), and the authors then read off the exact valuation equality in Theorem 1.1(ii) from the resulting congruence modulo p^2. Because this is the load-bearing step for the exact equality, I recommend adding an explicit verification, perhaps as a short lemma or a displayed paragraph, that every discarded term in each of the cases of the induction is divisible by p^2. In particular, the edge values k+i−p=1 and k+i−p=(a−1)(p−1)+1 are only handled indirectly in the current text. I did not find a counterexample, but this step is delicate enough that a dedicated check would make the proof auditable.","section":"Section 4, Eq. (4.21) and the passage from Eq. (4.6)"}],"minor_comments":[{"comment":"The formula in the abstract reads \"v_p(s(ap^{n+1},ap^{n+1}-))\" and is missing the k; it should be \"v_p(s(ap^{n+1},ap^{n+1}-k))\".","section":"Abstract"},{"comment":"After equation (2.5), the text says \"Hence Lemma 2.9 is true when r=2\"; the intended reference is Lemma 2.8.","section":"Section 2, proof of Lemma 2.8"},{"comment":"In the displayed congruence (4.29), the summand should be s(p,i)sp((a−1)p,ap−k−i), not s(p,i)sp(p,2p−k−i), to match equation (4.27) in the general a case; the typo appears only in the display and does not affect the argument.","section":"Section 4, Eq. (4.29)"},{"comment":"The final formula should be v_p(s(ap^n,ap^n−k)) = v_p(s(ap^{n1},ap^{n1}−k)) + 2(n−n1); the text omits the \"−k\" in the first term on the right-hand side.","section":"Section 3, proof of Theorem 1.5, final display"}],"recommendation":"major_revision","confidential_remarks":"The paper is a solid contribution and the missing case in the proof of (4.21) is easily repairable, so I do not see grounds for rejection. I would ask the authors to fix the missing case, add the explicit verification of the discarded terms in (4.21), and correct the minor typos listed above. After these changes, I expect the paper to be suitable for publication."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague —\n\nQuick verdict: accept, subject to minor revision. The paper does more than extend earlier formulas: Theorem 1.1 gives exact valuations for v_p(s(ap, ap-k)) over the broad range 1≤a≤p−1, and it is genuinely new compared with Lengyel, Komatsu-Young, and the authors' own p=2,3 papers. Corollary 1.2 gives a partial confirmation of Leonetti-Sanna; Theorem 1.5 conditionally confirms the +2 recursion in Lengyel's odd-k conjecture. I read the proof of Theorem 1.1 in full and did not find a gap. The induction is long, but the structure is coherent: terms indexed by V_k are killed modulo p^2 by valuation bounds, edge cases are handled by Theorem 1.3 and part (iii), and the i=p term is shown to vanish modulo p^2. The equality conditions tied to Bernoulli valuations follow from Corollary 2.10 and the congruence (4.21).\n\nThe strongest feature is that the main theorem is load-bearing and plausible: it reduces to a clean p^2 congruence, not to an external conjecture. The use of Washington's congruences and Newton-Girard is standard, and the two lemmas imported from [27] are published with independent proofs, so self-citation is not a problem here.\n\nSoft spots, in proportion: the proof of (4.21) is the point I would want an independent modular check. The exact equality in Theorem 1.1(ii) is read off that congruence; if any term in the decomposition (4.6) had valuation exactly 1 and had been overlooked, the Bernoulli-number criterion could shift. I could not construct such a term, and the paper appears to exclude all of them, but the proof is unformalized and long enough that this is a genuine verification burden rather than a demonstrated error. Minor typographical slips—the abstract is missing a k in one formula, and there is a Lemma 2.8/2.9 mix-up in Section 2—do not affect the mathematics. No data or code are involved, so the evidentiary standard is proof-checking, and by that standard the paper is solid.\n\nWho is this for? People working on p-adic valuations of combinatorial sequences or on harmonic sums. The regular-prime corollary is nice but conditional on regularity, so the reach is modest. General number theorists can skip; specialists should read.\n\nRecommendation: send to a referee comfortable with p-adic congruences and Stirling numbers. I would accept after minor revision, asking the referee specifically to double-check (4.21) and the equality condition. Desk rejection would be wrong; this is a serious contribution.","headline":"A real advance on p-adic valuations of Stirling numbers: exact formulas via a long induction, one delicate p^2 congruence worth an independent check, otherwise solid.","tokens_in":24978,"tokens_out":2778,"would_cite":true,"duration_ms":28907,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11B73","11A07"],"pacs":[],"model":"deepseek-v4-flash","headline":"The paper derives exact p-adic valuations of Stirling numbers of the first kind s(ap, ap−k) for primes p≥5 and small a, with equality governed by whether a Bernoulli number is divisible by p.","keywords":["p-adic valuation","Stirling numbers of the first kind","Bernoulli numbers","regular primes","generalized harmonic numbers","elementary symmetric functions","permutation cycles","p-adic analysis"],"falsifier":"Compute v_p(s(ap,ap−k)) directly for p=37 or p=59, the first irregular primes, for a=1 or 2 and an even k in the range 2≤k≤a(p−1) with k not divisible by p−1, and compare with the predicted value (v_p(k)+1)ε_k+1. Since the Bernoulli condition is known for these primes, a single mismatch between the computed valuation and the Bernoulli-predicted equality would disprove the theorem.","tokens_in":24000,"feed_emoji":"🔢","tokens_out":5897,"duration_ms":55659,"temperature":0.7,"pith_summary":"The paper aims to determine the p-adic valuation of Stirling numbers of the first kind s(ap, ap−k) for primes p≥5 and 1≤a≤p−1. It proves an exact formula when k is congruent to its parity modulo p−1, and a lower bound with a precise equality condition otherwise. These valuations control the denominators of harmonic elementary symmetric functions, so the results give partial confirmation of two standing conjectures about how large those p-adic valuations can be. The main tool is a family of generalized Stirling numbers that splits s(ap, ap−k) into manageable product terms.","feed_headline":"Valuation formula ties Stirling numbers to Bernoulli divisibility","feed_subtitle":"For primes p≥5, the p-adic order of s(ap,ap−k) is Bernoulli-determined, partially proving two conjectures.","key_machinery":"The carrying object is the family of m-th Stirling numbers of the first kind, defined as the coefficients of (x+m)(x+m+1)⋯(x+m+n−1), together with a convolution identity expressing ordinary Stirling numbers as sums of products s(m,i)s_m(n,k−i). This decomposition splits s(ap,ap−k) into terms s(p,i)s_p((a−1)p, ap−k−i); most terms vanish modulo $p^{2}$, leaving a binomial coefficient and powers of s(p,1), and the irreducible core s(p,p−⟨k⟩) whose p-adic valuation is controlled by the Bernoulli number condition.","core_discovery":"The central result is Theorem 1.1: for every prime p≥5, every 1≤a≤p−1 and every 2≤k≤ap−2, if k≡ε_k (mod p−1) then v_p(s(ap,ap−k))=(v_p(k)+1)ε_k; otherwise the valuation is at least (v_p(k)+1)ε_k+1, with equality exactly when the Bernoulli number B_{2⌊⟨k⟩/2⌋} is not divisible by p. For regular primes this becomes an exact equality in all cases. The proof builds the valuation from a p-adic expansion of s(ap,ap−k) in which only one or two product terms survive modulo $p^{2}$; the surviving term contains s(p,1)^{l−1}s(p,p−⟨k⟩), and the valuation of s(p,p−⟨k⟩) is controlled by a Bernoulli number through a congruence for generalized harmonic numbers.","pith_inferences":["If the same mod-p^2 expansion can be iterated to n≥1, it would yield the exact formulas proposed in the paper's final conjectures for s(ap^n, ap^n−k), reducing them to the same Bernoulli valuation condition.","The role of Bernoulli valuations suggests a testable dichotomy: for a fixed irregular prime, the floors v_p(s(p,p−k)) for even k should mirror the irregular pairs of that prime, so direct computation of these values for the first irregular primes would provide an independent cross-check.","Because Corollary 1.2 only gives v_p(H(ap−1,k))≤4−a, the paper's method cannot by itself prove the conjectured logarithmic bound for all k; bridging the gap would require controlling valuations in the middle range between a(p−1) and ap, where only the linear lower bound applies."],"forward_implications":["For regular primes p≥7, the inequality for H(ap−1,k) in Corollary 1.2 follows, giving the claimed partial support to a 2017 conjecture on harmonic symmetric functions.","For odd k satisfying the theorem's condition, v_p(s(ap^{n+1},ap^{n+1}−k))=v_p(s(ap^n,ap^n−k))+2 for all large n, so the valuation increments by exactly 2 along the p-power ladder.","In the exceptional range a≥4 and a(p−1)+2≤k≤ap−2, the linear lower bound v_p≥a+k−ap holds, bounding how small these valuations can be.","Since the equality condition is governed by whether a certain Bernoulli number is divisible by p, irregular primes are exactly where the simple lower bound can be strict."],"supporting_citations":[{"why":"Supplies the congruences on generalized harmonic numbers modulo p^2 and p^3 that are the source of the Bernoulli-number valuation condition.","marker":"[32]"},{"why":"Introduced the m-th Stirling numbers of the first kind and the convolution and expansion identities used throughout the proof of Theorem 1.1.","marker":"[27]"},{"why":"Established the even-k recursion theorem and proposed the odd-k +2 conjecture that Theorem 1.5 partially proves.","marker":"[19]"},{"why":"Provided exact p-adic valuations of Stirling numbers via Newton polygons, used as a baseline for the paper's concluding conjectures.","marker":"[16]"},{"why":"Supplies the identity s(n+1,k+1)=n!H(n,k) and the conjecture on logarithmic upper bounds that Corollary 1.2 partially supports.","marker":"[20]"},{"why":"Gives the 3-adic valuation formula used to verify the paper's conjectures in the p=3 case.","marker":"[26]"},{"why":"Supplies congruences for partial sums of harmonic series modulo p, used in the proof of part (iii) of Theorem 1.1.","marker":"[3]"},{"why":"Provides the parity identity for Stirling numbers that starts the proof of Theorem 1.3.","marker":"[1]"}],"fun_headline_variants":["Bernoulli numbers determine p-adic order of Stirling s(ap,ap-k)","p-adic valuation of Stirling numbers tied to Bernoulli B_n","Stirling s(ap,ap-k) valuations from Bernoulli numbers","New p-adic formula for Stirling numbers, partial conjectures","Partial proof of two Stirling-number conjectures via p-adics"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof's exact equality condition in part (ii) depends on a congruence modulo $p^{2}$ that assumes all but one product term in a certain decomposition vanish sufficiently fast; if that cancellation fails for some k, the Bernoulli-number equality condition could change.","fun_headline_variants_meta":{"raw":{"variants":["Bernoulli numbers determine p-adic order of Stirling s(ap,ap-k)","p-adic valuation of Stirling numbers tied to Bernoulli B_n","Stirling s(ap,ap-k) valuations from Bernoulli numbers","New p-adic formula for Stirling numbers, partial conjectures","Partial proof of two Stirling-number conjectures via p-adics"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000939,"raw_usage":{"total_tokens":4125,"prompt_tokens":1170,"completion_tokens":2955,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":786,"completion_tokens_details":{"reasoning_tokens":2864}},"tokens_in":786,"tokens_out":2955,"duration_ms":20223,"temperature":1.0,"reasoning_tokens":2864,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T13:09:18.198658+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute v_p(s(ap,ap−k)) directly for p=37 or p=59, the first irregular primes, for a=1 or 2 and an even k in the range 2≤k≤a(p−1) with k not divisible by p−1, and compare with the predicted value (v_p(k)+1)ε_k+1. Since the Bernoulli condition is known for these primes, a single mismatch between the computed valuation and the Bernoulli-predicted equality would disprove the theorem.","supporting_citations":[{"cited_title":"W ashington, p-Adic L-function and sums of powers, J","cited_arxiv_id":null,"evidence_quote":"Supplies the congruences on generalized harmonic numbers modulo p^2 and p^3 that are the source of the Bernoulli-number valuation condition."},{"cited_title":"Qiu, Y.L","cited_arxiv_id":null,"evidence_quote":"Gives the 3-adic valuation formula used to verify the paper's conjectures in the p=3 case."}],"review_version":1}