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arxiv: 2605.11351 · v1 · submitted 2026-05-12 · 🧮 math.CO

Recognition: no theorem link

A short proof of Mathar's 2020 recurrence conjecture for the generalized-Stirling sequence A001711

Tong Niu

Pith reviewed 2026-05-13 02:40 UTC · model grok-4.3

classification 🧮 math.CO
keywords generalized Stirling numbersharmonic numbersrecurrence relationsP-recursive sequencesOEIS A001711diagonal of Stirling triangleclosed-form identities
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The pith

The conjectured recurrence for sequence A001711 holds identically when the harmonic closed form is substituted.

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

The paper proves that the sequence A001711 obeys the order-2 recurrence a(n) minus (2n+5) times a(n-1) plus (n+2) squared times a(n-2) equals zero for all n at least 2. It reaches this by inserting the closed form 1/4 (n+3)! times (2 H_{n+3} minus 3), dividing out the common factorial prefactor (n+1)!/4, and showing that the remaining expression factors into a harmonic-number coefficient that simplifies to zero plus a constant term that is also zero. A sympathetic reader cares because the algebraic cancellation converts an empirical pattern observed in the OEIS into a proven identity that can be used to generate or analyze the sequence without repeated combinatorial counting.

Core claim

Direct substitution of the closed form a(n) = 1/4 (n+3)! (2 H_{n+3} - 3) into the left-hand side of the recurrence, followed by division through by (n+1)!/4, produces an expression whose coefficient of H_{n+2} is exactly (n+3) - (2n+5) + (n+2) and whose constant remainder is -3 times the same coefficient. Both pieces vanish because the defining shifts H_{n+3} = H_{n+2} + 1/(n+3) and H_{n+1} = H_{n+2} - 1/(n+2) make the linear combination of the three shifted harmonic numbers identically zero.

What carries the argument

Detlefs's closed-form expression a(n) = 1/4 (n+3)! (2 H_{n+3} - 3), which turns the recurrence verification into an explicit cancellation of harmonic-number coefficients.

If this is right

  • The recurrence holds for every integer n greater than or equal to 2.
  • Any two consecutive terms determine all later terms of the sequence via the relation.
  • The sequence satisfies a linear homogeneous recurrence with polynomial coefficients and is therefore P-recursive of order 2.

Where Pith is reading between the lines

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

  • The same substitution method could be tried on other sequences whose closed forms involve harmonic numbers to obtain their recurrences without generating-function machinery.
  • The recurrence supplies a practical way to compute large terms of A001711 once the first two values are known from the closed form.
  • The algebraic identity proved here is independent of any particular combinatorial interpretation of the sequence.

Load-bearing premise

The sequence A001711 is exactly equal to the closed form 1/4 (n+3)! (2 H_{n+3} - 3).

What would settle it

An integer n at least 2 where the value of a(n) computed from the OEIS definition of A001711 differs from the value given by the closed form, or where the left-hand side of the recurrence fails to evaluate to zero under that closed form.

read the original abstract

For the OEIS sequence A001711, contributed by N. J. A. Sloane long before the on-line era and identified there as the diagonal $T(n+4, 4)$ of a generalized-Stirling triangle, R. J. Mathar contributed in February 2020 the conjectured order-2 P-recursive recurrence \[ a(n) - (2n+5)\,a(n-1) + (n+2)^{2}\,a(n-2) \;=\; 0,\qquad n \ge 2. \] We give a one-page proof. Detlefs's harmonic-number closed form $a(n) = \tfrac{1}{4}(n+3)!\,(2 H_{n+3} - 3)$ collapses the left-hand side, after dividing through by $(n+1)!/4$, to a polynomial identity of $n$ with coefficient $H_{n+2}$. The harmonic-number coefficient simplifies to $(n+3) - (2n+5) + (n+2) = 0$ (using $H_{n+3} = H_{n+2} + \tfrac{1}{n+3}$ and $H_{n+1} = H_{n+2} - \tfrac{1}{n+2}$); the constant remainder is $-3 \cdot 0 = 0$ 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 $n = 2, \ldots, 5000$.

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.

Referee Report

0 major / 2 minor

Summary. The manuscript provides a one-page algebraic proof of Mathar's 2020 conjectured recurrence a(n) - (2n+5)a(n-1) + (n+2)^2 a(n-2) = 0 for n ≥ 2, where a(n) is the OEIS sequence A001711 (the diagonal T(n+4,4) of a generalized-Stirling triangle). It substitutes Detlefs's closed form a(n) = (1/4)(n+3)! (2 H_{n+3} - 3), divides the left-hand side by (n+1)!/4, and reduces the resulting expression to zero by applying the standard harmonic-number recurrences H_{n+3} = H_{n+2} + 1/(n+3) and H_{n+1} = H_{n+2} - 1/(n+2). A supplementary SymPy script verifies the identity symbolically, the exponential generating function, and the recurrence numerically up to n=5000.

Significance. If the result holds, the paper supplies a direct, elementary verification of the recurrence that avoids generating functions or combinatorial bijections, confirming an exact algebraic relation for this combinatorially defined sequence. The explicit link to harmonic numbers and the reproducible SymPy verification are strengths that facilitate further study of asymptotics or extensions to other diagonals of the generalized-Stirling triangle.

minor comments (2)
  1. The source of Detlefs's closed form (e.g., OEIS comment, personal communication, or prior publication) should be cited explicitly in the main text, as the substitution step depends on it.
  2. The abstract mentions an e.g.f. expansion match; if this is only in the supplement, a brief statement in the main text would clarify its role relative to the recurrence proof.

Simulated Author's Rebuttal

0 responses · 0 unresolved

We thank the referee for the positive assessment of the manuscript and the recommendation to accept. The report contains no major comments or requested revisions.

Circularity Check

0 steps flagged

No significant circularity identified

full rationale

The paper assumes the sequence equals Detlefs's independently contributed closed form a(n) = 1/4 (n+3)! (2 H_{n+3} - 3) and substitutes it into the conjectured recurrence to obtain an algebraic identity. This identity is reduced to zero using only the standard harmonic-number relations H_{n+3} = H_{n+2} + 1/(n+3) and H_{n+1} = H_{n+2} - 1/(n+2), with no parameter fitting, self-definition, or load-bearing self-citation by the present author. The closed form is treated as an external, verifiable input (confirmed numerically and via SymPy in the supplement), and the derivation steps are independent of the conjecture itself rather than reducing to it by construction.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The proof rests on the external closed-form expression for a(n); no free parameters are introduced and no new entities are postulated.

axioms (1)
  • domain assumption Detlefs's closed form a(n) = 1/4 (n+3)! (2 H_{n+3} - 3) equals the sequence A001711
    Invoked at the beginning of the proof to substitute into the recurrence left-hand side.

pith-pipeline@v0.9.0 · 5603 in / 1321 out tokens · 46340 ms · 2026-05-13T02:40:27.239601+00:00 · methodology

discussion (0)

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Reference graph

Works this paper leans on

11 extracted references · 11 canonical work pages · 2 internal anchors

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