REVIEW 2 minor 11 references
Three short proofs of Mathar's 2014 conjecture for OEIS A002627
T0 review · 0 major / 2 minor · reviewed 2026-05-19 · grok-4.3
Pith's one-line read The sequence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2.
desk verdict Niu gives three elementary proofs of the decade-old Mathar conjecture on A002627 and notes a simple homogenisation trick that may apply to similar OEIS cases. 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
Homogenisation by subtracting the first-order recurrence at adjacent indices, which cancels the inhomogeneous constant and yields the second-order homogeneous relation.
What would settle it
Compute a(3) step-by-step from a(0)=0, a(1)=1, a(2)=3, a(3)=10 and test whether 10 - 4*3 + 2*1 equals zero; the same direct check at any larger n would confirm or refute the relation.
Extended reading notes
Core claim
The sequence a(n) defined by the inhomogeneous recurrence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies the homogeneous relation a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for every n >= 2. The first proof subtracts the defining recurrence written at consecutive indices so the +1 terms cancel. The second extracts the same coefficient relation from the exponential generating function (e^x - 1)/(1 - x). The third applies the Pascal identity to telescope the explicit sum a(m) = sum_{k=0}^{m-1} k! binom(m, k).
Load-bearing premise
The first-order recurrence a(k) = k a(k-1) + 1 holds exactly at every integer k without boundary adjustments or remainder terms.
Editorial extensions
If this is right
- Later terms of the sequence can be generated from the two preceding values alone, without inserting the additive constant at each step.
- The same subtraction step immediately resolves any OEIS conjecture of the form a(n) = p(n) a(n-1) + q(n) whenever q(n) is constant or otherwise simple.
- The exponential generating function (e^x - 1)/(1 - x) encodes the recurrence coefficients so that differentiation or series expansion directly produces the second-order relation.
Reading between the lines
- Repeated differencing of the same kind could convert higher-order inhomogeneous recurrences into homogeneous ones of still higher order.
- The resulting linear relation may be solved explicitly to obtain a closed form or asymptotic expression for a(n) without summing the original binomial series.
- The technique supplies a uniform method for turning many listed OEIS conjectures into theorems without case-by-case generating-function work.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript provides three elementary proofs of the 2014 conjecture by R. J. Mathar for OEIS A002627: the sequence defined by the inhomogeneous recurrence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies the homogeneous second-order relation a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for n ≥ 2. The proofs consist of (i) direct subtraction of the defining relation at consecutive indices (homogenisation), (ii) extraction of the relation from the closed-form EGF F(x) = (e^x - 1)/(1 - x), and (iii) telescoping via the binomial-sum representation a(m) = ∑_{k=0}^{m-1} k! binom(m,k) using Pascal's identity.
Significance. The result resolves a decade-old OEIS conjecture with three independent, undergraduate-level derivations that require no advanced machinery. The homogenisation technique is highlighted as applicable to a broader class of sequences satisfying a(n) = p(n) a(n-1) + q(n) with simple q, potentially clearing multiple similar conjectures. The provision of multiple proofs (recurrence manipulation, generating functions, and combinatorial summation) supplies cross-verification and explicit credit to standard identities.
minor comments (2)
- The binomial-sum proof in §3 invokes Pascal's rule on the sum without explicitly stating the range adjustment for the k=0 term; adding one sentence clarifying that the k=0 contribution vanishes would improve readability for readers unfamiliar with the representation.
- The remark on the general homogenisation trick for OEIS conjectures is placed at the end; moving it to a dedicated short paragraph after the three proofs would better highlight its broader utility.
Simulated Author's Rebuttal
We thank the referee for the positive report, the accurate summary of the three proofs, and the recommendation to accept. No revisions are required.
Circularity Check
No circularity: all three proofs derive the claimed second-order relation as a direct algebraic or combinatorial consequence of the given first-order definition.
full rationale
The paper starts from the explicit first-order recurrence a(n) = n a(n-1) + 1 with a(0)=0 and shows that this forces the second-order homogeneous relation for n≥2. The first proof subtracts the defining equation written at n and at n-1, so the constant +1 cancels and the target identity appears immediately; this is a one-line algebraic identity, not a redefinition or fit. The EGF proof extracts the same coefficients from the closed form F(x)=(e^x-1)/(1-x) by differentiation or series manipulation, again without assuming the target relation. The binomial-sum proof uses Pascal's identity on the explicit sum representation, which is independently verifiable. None of the derivations feeds the conjectured relation back into itself, invokes self-citations for uniqueness, or renames a fitted parameter as a prediction. The derivation chain is therefore self-contained and non-circular.
Assumptions & free parameters
assumptions (2)
- domain assumption The sequence satisfies a(n) = n a(n-1) + 1 for every integer n >= 1 with a(0) = 0.
- domain assumption The exponential generating function of the sequence is F(x) = (e^x - 1)/(1 - x).
Cite this review
Pith. "Pith review of Three short proofs of Mathar's 2014 conjecture for OEIS A002627." pith.science (2026). https://pith.science/paper/DKBHILN5
@misc{pith2026260515500,
author = {Pith},
title = {Pith review of: Three short proofs of Mathar's 2014 conjecture for OEIS A002627},
year = {2026},
howpublished = {\url{https://pith.science/paper/DKBHILN5}},
note = {Machine review of arXiv:2605.15500}
}
abstract
For the OEIS sequence A002627, defined by the inhomogeneous first-order recurrence $a(n) = n\,a(n-1) + 1$ with $a(0) = 0$, R.~J.~Mathar recorded in February 2014 the conjectured second-order homogeneous recurrence \[ a(n) - (n+1)\,a(n-1) + (n-1)\,a(n-2) = 0, \qquad n \ge 2, \] which has remained marked as a conjecture on the OEIS for over a decade. We give three short proofs. The first is two lines: subtract the defining recurrence at adjacent indices and the constant cancels (we call this homogenisation). The second reads off the same relation from the exponential generating function $F(x) = (e^x-1)/(1-x)$. The third is a Pascal-rule telescoping on the binomial-sum form $a(m) = \sum_{k=0}^{m-1} k!\binom{m}{k}$. All three derivations are elementary, requiring nothing beyond undergraduate techniques. We remark that the same homogenisation trick clears an entire class of ``Conjecture: \dots'' entries on the OEIS, namely sequences satisfying $a(n) = p(n)\,a(n-1) + q(n)$ with simple $q$.
Reference graph
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N. J. A. Sloane (founder),The On-Line Encyclopedia of Integer Sequences, sequence A002627, https: //oeis.org/A002627. Acknowledgements Thanks to R. J. Mathar for the original conjecture, and to the OEIS maintainers for hosting an indispensable repository of integer sequences and the conjectured identities that go with them. AppendixA.Verification script:v...
work page 2014
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[8]
Compute a(0), ..., a(N) directly from the defining recurrence
Show all 11 references
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[9]
Cross-check against the OEIS b-file
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[10]
Verify the conjectured second-order recurrence to N >= 200
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[11]
"" from pathlib import Path ROOT = Path(__file__).resolve().parent.parent DATA_DIR = ROOT /
Save the extended data to data/A002627_terms.txt. """ from pathlib import Path ROOT = Path(__file__).resolve().parent.parent DATA_DIR = ROOT / "data" REFS_DIR = ROOT / "refs" DATA_DIR.mkdir(exist_ok=True) def compute_a(n_max: int) -> list[int]: """a(n) = n*a(n-1) + 1, a(0) = 0...
2014
Reviewed May 19, 2026 · model on record in the stance chip above.
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