REVIEW 2 major objections 4 minor 3 references
A simple recursive representation of the Faulhaber series
T0 review · 2 major / 4 minor · reviewed 2026-08-10 · deepseek-v4-flash
Pith's one-line read This paper presents a recurrence for the Faulhaber sums $\sum_{k=1}^n k^N$ that expresses each power sum in terms of the lower ones, using only the binomial theorem and no Bernoulli numbers.
desk verdict Correct elementary note whose central novelty claim is already in the paper's own cited reference. 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 central object is the recurrence of equation (3), which carries the argument. It comes from the binomial telescoping identity $$(k+1)^{N+1}-$k^{{N+1}}$=\sum_{j=0}^{N}\binom{N+1}{j}k^j,$$ summed over $k=1,\dots,n$. The cancellation leaves $(n+1)^{N+1}$ minus a weighted sum of lower power sums, and dividing by $N+1$ isolates $s(n,N)$. This identity is what makes the representation elementary and self-contained.
What would settle it
Compute $s(n,N)$ from the recurrence for any specific values, say $n=17$ and $N=12$, and compare the result with the direct sum $\sum_{k=1}^{17}k^{12}$; any mismatch would falsify the recurrence. For the factorization observation, test divisibility at $N=101$ or another value beyond the reported range, since the pattern is only asserted for $N\le 100$.
Extended reading notes
Core claim
Denote $s(n,N)=\sum_{k=1}^n k^N$. Starting from $s(n,0)=n+1$, the paper proves that for $N\ge 1$, $$s(n,N)=\frac{(n+1)^{N+1}-\sum_{j=0}^{N-1}\binom{N+1}{j}s(n,j)}{N+1}.$$ This is equation (3) of the paper. The proof expands $(k+1)^{N+1}-k^{N+1}$ by the binomial theorem, sums over $k$, and isolates $s(n,N)$. The paper also notes that $s(n,N)$ is a polynomial in $n$ of degree $N+1$ with leading term $n^{N+1}/(N+1)$, and reports an empirical factorization checked by computer algebra for $N\le 100$: for odd $N\ge 3$ the polynomial is divisible by $n^2(n+1)^2$, and for even $N\ge 2$ it is divisible by $n(n+1)(2n+1)$.
Load-bearing premise
The load-bearing premise is that the recurrence is a genuinely new presentation, not already contained in the earlier Faulhaber-sum literature the paper cites; the identity itself depends only on the binomial theorem.
Editorial extensions
If this is right
- For every integer $N\ge 1$, the exact polynomial formula for $\sum_{k=1}^n k^N$ can be generated from the single starting value $s(n,0)=n+1$.
- Each $s(n,N)$ is a rational-coefficient polynomial in $n$ of degree $N+1$ with leading term $n^{N+1}/(N+1)$.
- The recurrence turns evaluation of one high-degree power sum into a finite sequence of lower-degree computations, so it is easy to implement and to verify by direct summation.
- The paper's computer algebra output shows the reported divisibility pattern for even and odd $N$ holding for all computed cases $N=1,\dots,100$.
Reading between the lines
- An implication the paper leaves implicit is that this recurrence is algebraically the standard binomial telescoping derivation of power sums in a compact recursive form; its practical value is the explicit organization rather than a new number system or a new constant.
- The reported factorizations suggest a route to a fully general proof: for odd $N$, show $s(n,N)$ has double roots at $n=0$ and $n=-1$, and for even $N$, show it vanishes at the roots of $n(n+1)(2n+1)$.
- Comparing this recurrence coefficient by coefficient with the Bernoulli-number formula would let one recover Bernoulli numbers from the recursive representation, reversing the usual direction of use.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper defines s(n,N) = sum_{k=1}^n k^N and derives the recurrence s(n,N) = ((n+1)^(N+1) - sum_{j=0}^{N-1} binom(N+1,j) s(n,j)) / (N+1), Eq. (3), with the boundary convention s(n,0) = n+1. The proof is a telescoping argument over (k+1)^(N+1) - k^(N+1) followed by binomial expansion. The paper claims this is a simple recursive representation of the Faulhaber series without reference to Bernoulli numbers or polynomials. The appendix contains a MAPLE worksheet generating the polynomials s(n,N) for N = 1,...,100 and an empirical, explicitly hedged observation about factorization of these polynomials.
Significance. If Eq. (3) were a new representation, the paper would provide a clean, self-contained derivation and a reproducible MAPLE verification, which are genuine strengths. However, the derivation is a standard one-step consequence of the binomial theorem, and the same recurrence is already present in the cited Knuth paper [2], whose only difference is the boundary convention for s(n,0), an equivalent bookkeeping choice. The factorization observation in Section 2 is honestly labelled as empirical but is not proved and is not needed for the main identity. The mathematical content is correct, but the claimed novelty is not established, leaving only a pedagogical restatement of a known result.
major comments (2)
- [Section 1, Eq. (3)] The abstract and Section 1 present Eq. (3) as a new recursive representation, but the paper's own reference [2] (Knuth, 'Johann Faulhaber and Sums of Powers') derives the same telescoping recurrence for power sums; the only difference is the boundary convention s(n,0)=n+1 (used in the MAPLE appendix) versus s(n,0)=n, which is equivalent bookkeeping. The manuscript credits Knuth only with coining the term 'Faulhaber series' and gives no argument that the recurrence itself is absent from [2], so the central novelty claim is unsupported.
- [Section 2, factorization remark] The statement that s(n,N) factors as n^2(n+1)P(n,N) for odd N >= 3 and as n(n+1)(2n+1)P(n,N) for even N >= 2 is verified only empirically for N = 2,...,100 and is explicitly hedged; if this observation is retained, it needs a proof or a reference, and it should be clearly marked as a separate conjecture rather than presented alongside the main theorem.
minor comments (4)
- [Section 2, proof] The displayed proof of Eq. (3) is heavily corrupted in the typeset version, with broken lines and missing equation numbers; the telescoping argument is recognizable and valid, but it should be rewritten with clear line breaks and labelled displays so that it can be read without reconstruction.
- [Section 2, equation numbering] Equation numbers (1) and (4) are referenced in the text, but no equation number appears next to the displayed formula for s(n,0) or the factorization statement; please number all displays consistently.
- [Appendix] The MAPLE output is truncated mid-expression at several places, and the final output for N = 100 is incomplete; the appendix should either be complete or replaced by a short, self-contained verification script.
- [References] Reference [3] is cited for the polynomial form of s(n,N), but the phrase 'cf. [3], p.3' should be expanded to state the precise Bernoulli-polynomial formula so that the reader can directly compare it with Eq. (3).
Circularity Check
No significant circularity: Equation (3) is proved from the binomial theorem and telescoping, with no fitted inputs or self-citation chain.
full rationale
The central claim, Eq. (3), is obtained by expanding (k+1)^(N+1) - k^(N+1) via the binomial theorem, summing over k, and telescoping. The derivation starts from the definition s(n,N)=sum_{k=1}^n k^N and the binomial expansion; it does not assume Eq. (3) or any Bernoulli-based formula. The boundary condition s(n,0)=n+1 is a consistent convention for the empty coefficient sum, not a fitted parameter. The MAPLE appendix generates examples by applying the already-proven recurrence, which is a legitimate use rather than circular verification. The factorization remark is explicitly hedged as 'Seemingly' and 'can be shown empirically at least for N=2,...,100'; it is not a predicted result and is not used to prove Eq. (3). Any concern that the same recurrence appears in Knuth [2] is a novelty or priority issue, not circularity of the derivation. The paper is self-contained against the binomial theorem, so no circular step is present.
Assumptions & free parameters
assumptions (3)
- standard math Binomial theorem for integer exponents: (a+b)^N = sum_j binom(N,j) a^j b^(N-j).
- standard math Telescoping sums: sum_{k=1}^n (f(k+1) - f(k)) = f(n+1) - f(1).
- standard math Iterated substitution into a proven recurrence is a valid way to compute s(n,N) for consecutive N.
Cite this review
Pith. "Pith review of A simple recursive representation of the Faulhaber series." pith.science (2026). https://pith.science/paper/E5GIKDP4
@misc{pith2026250113548,
author = {Pith},
title = {Pith review of: A simple recursive representation of the Faulhaber series},
year = {2026},
howpublished = {\url{https://pith.science/paper/E5GIKDP4}},
note = {Machine review of arXiv:2501.13548}
}
abstract
We present a simple elementary recursive representation of the so called Faulhaber series $\sum_{k=1}^n k^N$ for integer $n$ and $N$, without reference to Bernoulli numbers or polynomials.
Reference graph
Works this paper leans on
-
[2]
Knuth (1993): Johann Faulhaber and Sums of Powers
D.E. Knuth (1993): Johann Faulhaber and Sums of Powers. Mathematics of Computation (61), no. 203, 277 294.-
work page 1993
-
[1]
J. H. Conway and R. K. Guy (1996): The Book of Numbers. Springer, N.Y
work page 1996
-
[3]
H. Richter and B. Schiekel (2004): Potenzsummen, Bernoulli-Zahlen und Eulersche Summenformel. Universität Ulm, doi:10.18725/OPARU-1819. Appendix MAPLE Worksheet > restart: > s[n,0]:=n+1; := s ,n 0 n 1 > for N from 1 to 100 do s[n,N]:=sort(factor(simplify(((n+1)^(N+1)-sum(binomial(N+1,j)*s[ n,j],j=0..N-1))/(N+1)))) od; := s ,n 1 () n 1 n 2 := s ,n 2 ...
Reviewed August 10, 2026 · model on record in the stance chip above.
Discussion (0). Continue with ORCID to comment.