REVIEW 2 major objections 4 minor 36 references
On the $p$-adic properties of Stirling numbers of the first kind
T0 review · 2 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read 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.
desk verdict 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. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
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.
What would settle it
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.
Extended reading notes
Core claim
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.
Load-bearing premise
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.
Editorial extensions
If this is right
- 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.
Reading between the lines
- 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.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
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.
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 (2)
- [Section 4, proof of claim (4.21), Case 3 (paragraph containing Eq. (4.39))] 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 4, Eq. (4.21) and the passage from Eq. (4.6)] 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.
minor comments (4)
- [Abstract] 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 2, proof of Lemma 2.8] After equation (2.5), the text says "Hence Lemma 2.9 is true when r=2"; the intended reference is Lemma 2.8.
- [Section 4, Eq. (4.29)] 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 3, proof of Theorem 1.5, final display] 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.
Circularity Check
No significant circularity; the main theorems are derived from external congruences and in-paper induction, not from the claims themselves.
full rationale
The paper's central results are not assumed in their own proofs. Theorem 1.1 is proved by induction on a, and the load-bearing congruence (4.21) is explicitly established by induction: 'We claim that if (l−1)(p−1)+2≤k≤l(p−1)−2 for any integer l with 1≤l≤a, then s(ap,ap−k)≡a(a−1 choose l−1)s(p,1)^{l−1}s(p,p−⟨k⟩) (mod p^2).' The claim is then proved case by case, not assumed. The valuation equalities in Theorem 1.1 are read off from this congruence together with Corollary 2.10, which is itself derived from Washington's congruence (Lemma 2.5), an external classical result. Theorem 1.3 is proved independently from Lemma 2.1 and elementary p-adic estimates, and Theorem 1.5 follows from Theorems 1.3 and 1.4, where Theorem 1.4 is Lengyel's published result. The only self-citations are Lemmas 2.3 and 2.4 from the authors' earlier paper [27]; these are standard identities for m-th Stirling numbers with published proofs, and they are not used as a substitute for the main result. Lemma 2.11 invokes Theorem 1.3, but Theorem 1.3 is already proved earlier in the paper. No fitted parameter is renamed as a prediction, no uniqueness theorem is imported from the authors' prior work to forbid alternatives, and no ansatz is smuggled in by citation. The regularity assumption merely specializes the Bernoulli-number condition v_p(B_{2⌊⟨k⟩/2⌋})=0; it is not an input that forces the valuation formula. The delicate part, congruence (4.21), may merit independent modular checking as a correctness issue, but it is not circular.
Assumptions & free parameters
assumptions (10)
- standard math Identity s(n+1,k+1)=n!H(n,k) and basic Stirling number recurrences from Comtet [5].
- standard math Lemma 2.1 of Adamchik [1]: if n+k is odd, s(n,k)=1/2 * sum_{i=k+1}^n (-1)^{n-i} n^{i-k} binom(i-1,i-k) s(n,i).
- standard math Lemma 2.2 of Boyd [3]: H_{np} ≡ H_n / p (mod p^2) and H_{np+k} ≡ H_{np} + H_k (mod p).
- standard math Lemma 2.5 of Washington [32]: congruences for generalized harmonic numbers H_{p-1}^{(r)} modulo p^2 or p^3 involving Bernoulli numbers.
- standard math Lemma 2.7 Newton-Girard identities [29].
- standard math Lemmas 2.3 and 2.4 from Qiu-Hong [27] on the m-th Stirling numbers of the first kind.
- standard math Theorem 1.4 of Lengyel [19]: for even k satisfying a valuation condition, v_p(s(ap^{n+1}, ap^{n+1}-k)) = v_p(s(ap^n, ap^n-k)) + 1.
- standard math Von Staudt-Clausen theorem, giving v_p(B_n) >= 0 for even n with (p-1) not dividing n.
- standard math Wilson's theorem s(p,1)=(p-1)! ≡ -1 (mod p).
- standard math p-adic isosceles triangle principle (Koblitz [15]).
Cite this review
Pith. "Pith review of On the $p$-adic properties of Stirling numbers of the first kind." pith.science (2026). https://pith.science/paper/2VFDZG7K
@misc{pith2026190805594,
author = {Pith},
title = {Pith review of: On the $p$-adic properties of Stirling numbers of the first kind},
year = {2026},
howpublished = {\url{https://pith.science/paper/2VFDZG7K}},
note = {Machine review of arXiv:1908.05594}
}
abstract
Let $n, k$ and $a$ be positive integers. The Stirling numbers of the first kind, denoted by $s(n,k)$, count the number of permutations of $n$ elements with $k$ disjoint cycles. Let $p$ be a prime. In recent years, Lengyel, Komatsu and Young, Leonetti and Sanna, Adelberg, Hong and Qiu made some progress in the study of the $p$-adic valuations of $s(n,k)$. In this paper, by using Washington's congruence on the generalized harmonic number and the $n$-th Bernoulli number $B_n$ and the properties of $m$-th Stirling numbers of the first kind obtained recently by the authors, we arrive at an exact expression or a lower bound of $v_p(s(ap, k))$ with $a$ and $k$ being integers such that $1\le a\le p-1$ and $1\le k\le ap$. This infers that for any regular prime $p\ge 7$ and for arbitrary integers $a$ and $k$ with $5\le a\le p-1$ and $a-2\le k\le ap-1$, one has $v_p(H(ap-1,k))<-\frac{\log{(ap-1)}}{2\log p}$ with $H(ap-1, k)$ being the $k$-th elementary symmetric function of $1, \frac{1}{2}, ..., \frac{1}{ap-1}$. This gives a partial support to a conjecture of Leonetti and Sanna raised in 2017. We also present results on $v_p(s(ap^n,ap^n-k))$ from which one can derive that under certain condition, for any prime $p\ge 5$, any odd number $k\ge 3$ and any sufficiently large integer $n$, if $(a,p)=1$, then $v_p(s(ap^{n+1},ap^{n+1}-))=v_p(s(ap^n,ap^n-k))+2$. It confirms partially Lengyel's conjecture proposed in 2015.
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