arxiv: 2605.12839 · v1 · submitted 2026-05-13 · 🧮 math.CO

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A short proof of Mathar's 2021 recurrence conjecture for the Lehmer-Comtet diagonal A045406

Tong Niu

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Pith reviewed 2026-05-14 19:08 UTC · model grok-4.3

classification 🧮 math.CO

keywords A045406Lehmer-Comtet triangleP-recursive recurrenceharmonic closed formdiagonal sequencerecurrence verificationfactorial-harmonic identity

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The pith

Detlefs's closed form for A045406 directly establishes Mathar's conjectured recurrence by making the left-hand side vanish.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper supplies a short algebraic proof that the sequence A045406 satisfies the conjectured order-2 recurrence a(n) + (2n-7)a(n-1) + (n-4)^2 a(n-2) = 0 for n >= 5. It substitutes the known closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! into the left-hand side, factors out the common term (-1)^n (n-5)! (n-4), and reduces the remainder to a linear expression in the harmonic number H_{n-4}. Both the coefficient of H_{n-4} and the constant term simplify to zero once the standard shift relations for harmonic numbers are applied. The argument is confirmed by an accompanying symbolic script that checks the identity and the recurrence numerically up to n = 5000.

Core claim

Substituting Detlefs's closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! into the left-hand side of the conjectured recurrence, after factoring out (-1)^n (n-5)! (n-4), produces an expression linear in H_{n-4} whose coefficient simplifies to (n-3) - (2n-7) + (n-4) = 0 by using H_{n-3} = H_{n-4} + 1/(n-3) and H_{n-5} = H_{n-4} - 1/(n-4); the remaining constant term likewise equals zero.

What carries the argument

Detlefs's closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! together with the two-term recurrence relations for harmonic numbers.

If this is right

The sequence obeys the stated P-recursive relation for every integer n at least 5.
Terms of A045406 can be generated recursively from any two consecutive initial values.
The harmonic closed form is algebraically consistent with the conjectured relation.
The identity holds as a polynomial equality in n once the harmonic shifts are inserted.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

The same substitution-and-cancellation technique may apply directly to recurrences conjectured for neighboring diagonals of the Lehmer-Comtet triangle.
Once the recurrence is established, standard methods for P-recursive sequences can be used to derive an asymptotic expansion for a(n) without returning to the triangle definition.
The proof shows that the ordinary generating function of A045406 satisfies a first-order linear differential equation whose coefficients are rational functions of low degree.

Load-bearing premise

Detlefs's closed form correctly produces every term of the sequence A045406 extracted from the Lehmer-Comtet triangle.

What would settle it

Direct extraction of a(5), a(6), a(7) from the Lehmer-Comtet triangle followed by substitution into the recurrence for n=5; any nonzero result would show the closed form or the recurrence fails to hold.

read the original abstract

For OEIS sequence A045406, the column-2 diagonal of the Lehmer-Comtet triangle A008296, R. J. Mathar contributed in September 2021 the conjectured order-2 P-recursive recurrence \[ a(n) + (2n-7)\,a(n-1) + (n-4)^{2}\,a(n-2) \;=\; 0,\qquad n \ge 5. \] We give a short proof. Detlefs's harmonic-number closed form $a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)!$ for $n \ge 3$ collapses the left-hand side, after factoring out $(-1)^n (n-5)! (n-4)$, to a polynomial identity in $n$ with coefficient $H_{n-4}$. The $H_{n-4}$-coefficient simplifies to $(n-3) - (2n-7) + (n-4) = 0$ (using $H_{n-3} = H_{n-4} + 1/(n-3)$ and $H_{n-5} = H_{n-4} - 1/(n-4)$); the constant remainder is $0$ for the same reason. The supplementary archive contains a SymPy script verifying both pieces symbolically, the e.g.f.\ expansion against the harmonic closed form, and Mathar's recurrence numerically for $n = 5, \ldots, 5000$.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

Short algebraic check that the recurrence holds for the given closed form, with clean cancellations but no derivation of that form from the triangle.

read the letter

Dear colleague, This paper gives a short algebraic proof of Mathar's recurrence conjecture for A045406 by substituting Detlefs's closed form and simplifying the left-hand side to zero. The calculation factors out (-1)^n (n-5)! (n-4), rewrites the harmonic terms using the standard addition rules for H_{n-3} and H_{n-5} in terms of H_{n-4}, and shows both the coefficient of H_{n-4} and the constant term vanish. The SymPy script confirms the identity symbolically and checks the recurrence numerically to n=5000. That part is straightforward and holds up without issues. The main limitation is that the argument takes the closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! as given rather than deriving it from the definition of the sequence as the column-2 diagonal of the Lehmer-Comtet triangle. The paper attributes the formula externally and uses egf expansion and numerical checks for consistency, but does not prove the formula matches the combinatorial object from first principles. This makes the work a verification of the recurrence under that assumption rather than a self-contained proof. The algebra itself shows no circularity or fitting. This note is aimed at people working with OEIS sequences, P-recursive relations, or diagonals in enumerative combinatorics. A reader who needs the recurrence for computation or wants a quick confirmation of the conjecture will get direct value from the explicit steps. The thinking is clear and the evidence supports the limited claim made. I would bring it to a reading group to look at the substitution method. I would cite it if I needed the recurrence in my own work. It deserves peer review as a concise confirmation of the specific conjecture.

Referee Report

1 major / 1 minor

Summary. The manuscript provides a short proof of Mathar's conjectured recurrence a(n) + (2n-7)a(n-1) + (n-4)^2 a(n-2) = 0 for n ≥ 5 for OEIS sequence A045406. It substitutes Detlefs's closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! for n ≥ 3, factors the left-hand side as (-1)^n (n-5)! (n-4) times an expression linear in H_{n-4}, and shows the H_{n-4} coefficient (n-3) - (2n-7) + (n-4) vanishes using the relations H_{n-3} = H_{n-4} + 1/(n-3) and H_{n-5} = H_{n-4} - 1/(n-4), with the constant remainder likewise zero.

Significance. The algebraic cancellation is direct and parameter-free, relying only on the defining recurrence of harmonic numbers. The supplementary SymPy script supplies symbolic verification of the identity together with numerical confirmation of the recurrence to n=5000, which adds reproducibility. If the closed form holds for the sequence, the result establishes the conjecture via an explicit, short derivation.

major comments (1)

[Main proof (after closed-form statement)] The central argument (immediately after stating the closed form) substitutes a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! without deriving this expression from the definition of A045406 as the column-2 diagonal of the Lehmer-Comtet triangle A008296. This closed form is load-bearing; while the subsequent cancellation is correct, the manuscript attributes the formula to Detlefs but supplies no derivation or reference establishing it from the triangle. The supplement's symbolic and numerical checks confirm consistency but do not prove the closed form.

minor comments (1)

[Abstract] The abstract and introduction could explicitly note the domains (closed form for n ≥ 3, recurrence for n ≥ 5) to avoid any ambiguity about the range of validity.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for the careful reading, positive assessment of the algebraic cancellation, and recommendation of minor revision. We address the major comment below.

read point-by-point responses

Referee: [Main proof (after closed-form statement)] The central argument (immediately after stating the closed form) substitutes a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! without deriving this expression from the definition of A045406 as the column-2 diagonal of the Lehmer-Comtet triangle A008296. This closed form is load-bearing; while the subsequent cancellation is correct, the manuscript attributes the formula to Detlefs but supplies no derivation or reference establishing it from the triangle. The supplement's symbolic and numerical checks confirm consistency but do not prove the closed form.

Authors: We agree that the manuscript relies on the closed form without a self-contained derivation from the Lehmer-Comtet triangle definition. The paper's scope is the short proof of the recurrence assuming the form given by Detlefs. In revision we will add an explicit reference to the OEIS A045406 entry (where the closed form is recorded in the comments) and clarify that the supplementary script already performs symbolic verification of the closed form against the exponential generating function of the sequence. This supplies the missing attribution and reproducibility link without expanding the main text beyond its intended brevity. revision: yes

Circularity Check

0 steps flagged

No significant circularity; proof relies on external closed form

full rationale

The paper attributes the closed form a(n) = (-1)^n (2 H_{n-3} - 3)(n-3)! directly to Detlefs and substitutes it into the target recurrence to obtain an algebraic identity via standard harmonic-number addition rules. No step within the manuscript defines any quantity in terms of the recurrence, fits parameters to sequence data, or reduces the result to a self-citation chain. The derivation is therefore independent of the conjecture it verifies, given the external closed form as input.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The proof depends on the correctness of the external closed form for a(n) and on the standard recurrence property of harmonic numbers; no free parameters or new entities are introduced.

axioms (1)

standard math Harmonic numbers satisfy H_m = H_{m-1} + 1/m for positive integers m
Invoked to relate H_{n-3}, H_{n-4}, and H_{n-5} when simplifying the coefficient of H_{n-4} and the constant term.

pith-pipeline@v0.9.0 · 5586 in / 1315 out tokens · 79233 ms · 2026-05-14T19:08:19.749151+00:00 · methodology

discussion (0)

Reference graph

Works this paper leans on

10 extracted references · 3 canonical work pages

[1]

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work page arXiv 2025
[2]

R. R. Detlefs (contributed by W. E. Clark), A045406 formula comment, 2010; athttps://oeis.org/ A045406

2010
[3]

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work page arXiv 2023
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[7]

Niu,A short proof of Mathar’s 2020 recurrence conjecture for the generalized-Stirling sequence A001711, arXiv preprint, 2026

T. Niu,A short proof of Mathar’s 2020 recurrence conjecture for the generalized-Stirling sequence A001711, arXiv preprint, 2026

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[8]

N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences,https://oeis.org/, 2026. 8 TONG NIU

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[9]

Jovovic, A045406: Diagonal of Lehmer-Comtet triangle A008296, columnk = 2, https://oeis

V. Jovovic, A045406: Diagonal of Lehmer-Comtet triangle A008296, columnk = 2, https://oeis. org/A045406
[10]

N. J. A. Sloane, A008296: Lehmer-Comtet triangle of unsigned1st-kind Stirling-derivative numbers, https://oeis.org/A008296