On a Family of Nested Recurrences and Their Arithmetical Solutions

Benoit Cloitre

REVIEW 3 major objections 4 minor 10 references

On a Family of Nested Recurrences and Their Arithmetical Solutions

T0 review · 3 major / 4 minor · reviewed 2026-08-07 · deepseek-v4-flash

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Pith's one-line read For each integer $m\ge1$, the nested recurrence $a(n+1)=n-a^{(m)}(n)+a^{(m+1)}(n)$ with $a(1)=1$ has, the paper claims, the unique solution $a(n)=n-h(n)$, where $h(n)$ is an explicit counting sequence.

desk verdict The main theorem is already in Iannucci and Mills-Taylor and the proof of the key identity is circular; only the small m=2 partial-sums lattice count is new and correct. read the letter →

arxiv 2506.00093 v1 pith:LLTICYJ7 submitted 2025-05-30 math.CO

Benoit Cloitre This is my paper

classification math.CO MSC 11B3705A1911B3911B83

keywords nestedrecurrencerelationsmeta-Fibonaccisequencesk-appearancesequenceConnellslowlygrowinggeneralizedpolygonalnumbersfloorformulacombinatorialinterpretation

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

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

The reading

The paper claims that for every integer $m\ge 1$, the nested recurrence $a(n+1)=n-a^{(m)}(n)+a^{(m+1)}(n)$ with initial condition $a(1)=1$ has exactly one solution, namely $a(n)=n-h(n)$. Here $h(n)$ is the counting sequence that starts with $h(1)=0$ and lets each non-negative integer $k$ appear exactly $mk+1$ times; equivalently, $h(n)$ counts how many generalized $m$-polygonal numbers $m\binom{j}{2}+j$ lie below $n$. The paper derives an explicit floor formula for this solution and proves the solution by reducing the recurrence to the identity $h(n+1)-1=h(a^{(m)}(n))$. If correct, this turns a self-referential meta-Fibonacci-type family into a simple conditional-increment sequence, and for $m=2$ it gives a lattice-point interpretation of the partial sums.

What carries the argument

The load-bearing object is the auxiliary sequence $h(n)$: the unique non-decreasing sequence with $h(1)=0$ in which each $k\ge0$ occurs $mk+1$ times. Its block boundaries are $T_k^* = 1 + m k(k-1)/2 + k$, and the candidate solution is $a_{\mathrm{sol}}(n)=n-h(n)$. The key identity $(4)$, $h(n+1)-1=h(a_{\mathrm{sol}}^{(m)}(n))$, is what converts the recurrence into a statement about $h$ alone. Lemma 4.2 supplies the two boundary properties that do most of the work: iterating $a_{\mathrm{sol}}$ exactly $m$ times from just before $T_{k+1}^*$ lands on $T_k^*$, and iterating from $T_k^*$ lands on $T_{k-1}^*$. Those properties, if they hold, force $h(a_{\mathrm{sol}}^{(m)}(n))=h(n)-1$ inside each constant block, which is precisely Identity (4).

What would settle it

Compute, in exact integer arithmetic for a fixed $m\ge2$ and all $k$ up to a few thousand, the value $a_{\mathrm{sol}}^{(m)}(n)$ for every $n$ in $T_k^* \le n < T_{k+1}^*-1$: if any such value equals $T_k^*$, then Identity (4) fails and Theorem 3.1 is false.

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Extended reading notes

Core claim

On the paper's own terms, Theorem 3.1 is the central claim: the sequence defined by the recurrence $a(n+1)=n-a^{(m)}(n)+a^{(m+1)}(n)$, $a(1)=1$, is uniquely $a_{\mathrm{sol}}(n)=n-h(n)$, with $h(n)=\left\lfloor \frac{m-2+\sqrt{(m-2)^2+8m(n-1)}}{2m}\right\rfloor$. The uniqueness follows by induction once the key identity $h(n+1)-1=h(a_{\mathrm{sol}}^{(m)}(n))$ is established for all $n\ge1$. The paper proves that identity by splitting $n$ into boundary cases and interior cases, using lemmas that pin down the values of $h$ after iterating $a_{\mathrm{sol}}$ at the boundary indices $T_k^*$. It also reports that $a(n)$ is slowly growing, that it increments exactly at the non-special integers, and that for $m=1,2,3,4$ the solution matches known tabulated sequences connected with triangular, square, pentagonal, and hexagonal numbers.

Load-bearing premise

The proof depends on the boundary fact that inside every block where $h(n)$ is constant, the $m$-times-iterated candidate sequence $a_{\mathrm{sol}}^{(m)}(n)$ never lands exactly on the next block boundary; the text's attempt to prove this substitutes the very identity that this fact is being used to prove, so the premise is effectively assumed rather than derived.

Editorial extensions

If this is right

The recurrence has a closed form: every term $a(n)$ is computable directly from the floor formula, so no iteration over previous values is needed.
The solution is slowly growing, and it increases by exactly $1$ when $n$ is not of the form $m\binom{j}{2}+j$; hence the nested recurrence is equivalent to a one-line conditional increment rule.
The solution is unique among sequences with $a(1)=1$: no other sequence can satisfy both the recurrence and this starting value.
For $m=2$, the partial sums $\sum_{i=1}^n a(i)$ count pairs $(x,y)$ with $y\le x\le y^2$ and $x\le n$.
For $m=1,2,3,4$, the solution reproduces known sequences associated with triangular, square, pentagonal, and hexagonal numbers, so the family is a generalization of the classical Connell sequence.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

Editorial inference: if the boundary fact used in Case 5.2 is actually false for some $m$, the uniqueness claim would fail even if the explicit floor formula still satisfied the recurrence numerically for long stretches; a small exact-arithmetic search would settle the paper's main theorem.
Editorial inference: the proof strategy suggests a broader family $a(n+1)=n-a^{(r)}(n)+a^{(s)}(n)$ might admit solutions built from similar $k$-appearance sequences, but the paper does not address that generalization.
Editorial inference: changing the multiplicity rule $N_k=mk+1$ to another arithmetic progression would likely destroy the exact floor formula; testing $N_k=mk+c$ would show which part of the structure is load-bearing.
Editorial inference: for $m=2$, the lattice-point interpretation of partial sums may generalize to higher $m$ by replacing the square bound $y^2$ with the appropriate $m$-gonal level sets, though the paper only works out $m=2$.

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the letter

Short version: this is an honest but derivative manuscript. Its main theorem and explicit floor formula appear in Iannucci and Mills-Taylor, with the corrected formula cited as [7] by the author himself, and the proof of the key identity is circular at exactly the boundary case that carries the weight. The one piece that is new and checks out is the m=2 partial-sums lattice count.

The paper is transparent about what it takes from the literature. Section 2 states the formula aligns with [7], and Section 5 says the proof structure follows [6, Theorem 1]. That candor is fine, but it means the paper’s contribution is supposed to be a self-contained proof of a known result. That proof fails. In Case 5.2 (T*_k ≤ n < T*_{k+1}-1, k ≥ 1), everything works until the strict upper bound a_sol^(m)(n) < T*_k. Monotonicity and Lemma 4.2(1) give only a_sol^(m)(n) ≤ T*_k. The paper then rules out equality by assuming a_sol^(m)(n0) = T*_k, substituting into Identity (4) — the identity under proof — and getting k-1 = k. That is circular. An independent arithmetical exclusion of equality is missing. Without it, Identity (4) is not established, and Theorem 3.1 is left unproved by this manuscript. The result itself is very likely true; it is in the cited literature. But this paper does not prove it.

The m=2 partial-sums observation is small but real: A2(n) = |{(x,y) ∈ Z_+^2 : y ≤ x ≤ y^2, x ≤ n}|, and the first-difference computation with ceil(sqrt(n))-1 = floor(sqrt(n-1)) is correct. The OEIS cross-references are mostly routine. The alternative-indexing remark in Section 7 is harmless.

Who is this for? A reader who wants a survey-style tour of the Connell-generalization literature with OEIS links, plus a small lattice count for the square case. Not a reader looking for a new theorem or a valid proof of the theorem already in [6,7].

The math is not incoherent; the author clearly knows the area and is honest about prior work. But the load-bearing argument is circular and the novelty is marginal. I would not send this to a serious referee; it deserves a desk reject, though perhaps an encouraging one pointing to [6,7] and the partial-sums note.

Referee Report

3 major / 4 minor

Summary. The paper studies the family of nested recurrence relations a(n+1)=n-a^{(m)}(n)+a^{(m+1)}(n) with initial condition a(1)=1, parameterized by m≥1. The author claims that the unique solution is a(n)=n-h(n), where h(n) is the k-appearance sequence in which each nonnegative integer k appears exactly mk+1 times, and derives an explicit floor formula for h(n) (Eq. (2)). The proof reduces the recurrence to a key identity h(n+1)-1=h(a_sol^{(m)}(n)) (Identity (4)) and attempts to prove this identity by arithmetical lemmas about the boundary behavior of the iterated function a_sol^{(m)}. The paper also includes combinatorial interpretations of a(n) and its partial sums for m=2, and discusses connections to OEIS sequences, including generalizations of Connell's sequence.

Significance. If the main theorem were established, it would provide a clean arithmetic description of solutions to a nontrivial family of nested recurrences, unifying several known sequences (e.g., triangular, square, pentagonal number counting) and connecting them to the Connell sequence literature. The explicit floor formula and the combinatorial interpretations are potentially useful. However, the proof of the central identity is circular in a key case, so the main theorem is not established by this manuscript. The paper's value is therefore contingent on a repair of the proof gap.

major comments (3)

[5, Case 5.2] The proof of the strict upper bound a_sol^{(m)}(n) < T_k^* is circular. After Lemma 4.1 and Lemma 4.2(1) yield only a_sol^{(m)}(n) ≤ T_k^*, the author supposes equality a_sol^{(m)}(n0)=T_k^* and derives a contradiction by substituting into Identity (4), the very identity being proved. This only shows that equality is incompatible with the statement under proof; it does not provide an independent arithmetical reason to exclude equality. Consequently, the claim h(a_sol^{(m)}(n))=k-1 in Case 5.2 is not established.
[Theorem 3.1 and Section 5] The Main Theorem depends entirely on Identity (4). Since Case 5.2 is the only place that handles the interior of each block T_k^*≤n<T_{k+1}^*-1 and it fails to prove the required inequality, Identity (4) remains unproved for all n≥1. Without Identity (4), the induction step in Theorem 3.1 does not go through, and the theorem is unproved.
[5 (opening sentence)] The proof of Identity (4) is not self-contained: the author explicitly states that the proof structure follows Iannucci and Mills-Taylor [6, Proof of Theorem 1], and the explicit formula in Section 2 is imported from the corrected formula in [7]. While importing a known proof structure is not itself an error, it means the manuscript does not provide an independent derivation of the key identity. The circular dependency in Case 5.2 is therefore not mitigated by an alternative argument elsewhere in the text.

minor comments (4)

[Throughout] The iterated notation a^{(m)}(n) is used in the abstract and in Sections 3–5 but never formally defined. It would be helpful to state explicitly that a^{(m)} denotes the m-fold iterate of the sequence a.
[Section 2] The phrase 'This aligns with the corrected formula in Iannucci and Mills-Taylor [7]' appears without a precise citation to the errata page; the reference list should include the full bibliographic details of [7].
[Section 6.2] The identity ceil(sqrt(n))-1 = floor(sqrt(n-1)) is used without proof; a one-line justification (e.g., by squaring) would improve readability.
[Section 7] The definition of h_m^{(0)}(x) says 'each integer k≥0 appears mk+1 times' but does not specify the starting index for the first appearance; it would be clearer to state h_m^{(0)}(0)=0 and h_m^{(0)}(x)=k for T_k^* -1 ≤ x < T_{k+1}^*-1, consistent with the 0-indexed convention.

Circularity Check

1 steps flagged · score 7.0 of 10

Case 5.2 proves the key strict inequality by assuming Identity (4), the identity under proof; the Main Theorem is therefore not established by the manuscript.

other [Section 5, Case 5.2 (proof of Identity (4))]
"Suppose, for contradiction, that a_sol^(m)(n0) = T*_k for some n0 in the range T*_k ≤ n0 < T*_{k+1}−1. Then h(a_sol^(m)(n0)) = h(T*_k) = k. Substituting this into the identity we are trying to prove (h(n0+1)−1 = h(a_sol^(m)(n0))): Since n0 < T*_{k+1}−1, we have h(n0+1)=k. So the identity would read k−1=k, which is a contradiction."

The goal of Case 5.2 is to prove the strict upper bound a_sol^(m)(n) < T*_k, which is needed to conclude h(a_sol^(m)(n)) = k−1 and hence Identity (4). Lemmas 4.1 and 4.2(1) only give a_sol^(m)(n) ≤ T*_k. The proof rules out equality by substituting a_sol^(m)(n0)=T*_k into 'the identity we are trying to prove' and deriving k−1=k. That contradiction is not an independent arithmetical property; it is just the negation of the statement under proof. Thus the only proof of the upper bound assumes the target identity, so Identity (4) is not established by the manuscript.

full rationale

The paper's central derivation of the Main Theorem depends entirely on Identity (4), h(n+1)−1 = h(a_sol^(m)(n)). In Case 5.2, the only missing ingredient is the strict upper bound a_sol^(m)(n) < T*_k; Lemmas 4.1 and 4.2(1) supply only the non-strict bound. The text explicitly excludes equality by substituting into 'the identity we are trying to prove', making the proof of the identity circular at the decisive point. This is a genuine logical circularity, not a data-fitting or self-citation issue. The external references [6,7] (Iannucci and Mills-Taylor, with errata) are prior work by other authors and do provide independent support for the explicit formula, so this is not a self-citation-chain case; however, as written, the manuscript's own proof of the key identity and therefore of Theorem 3.1 is not self-contained. I assign 7: substantial partial circularity in the central proof, short of the 8-10 range reserved for derivation-by-definition or forced self-citation chains.

Assumptions & free parameters 0 free parameters · 3 assumptions · 0 invented entities

No data-fitting parameters appear. The central proof rests on the ansatz a(n)=n-h(n) with h(1)=0 and appearance counts N_k=mk+1 (Definition 2.1), on standard monotonicity of a_sol, and, because the new proof is incomplete in Case 5.2, on the earlier Iannucci-Mills-Taylor result [6,7] for the correctness of the closed form.

assumptions (3)

domain assumption Solution ansatz a(n)=n-h(n) with h(1)=0 and appearance counts N_k=mk+1.
Definition 2.1 fixes h by fiat rather than deriving it from the recurrence; the counts are chosen to match the known solution.
standard math Monotonicity and 0/1 increment behavior of a_sol(n).
Proven in Lemma 4.1 as a consequence of the definition; used throughout Sections 4 and 5 to control iterates.
domain assumption The prior Iannucci-Mills-Taylor theorem [6,7] that the recurrence has this closed-form solution.
Section 2 states the formula aligns with the corrected formula in [7]; Section 5 says the proof structure follows [6, Proof of Theorem 1]. Because the new proof is incomplete, the truth of the theorem rests on this cited background.

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Cite this review

Pith. "Pith review of On a Family of Nested Recurrences and Their Arithmetical Solutions." pith.science (2026). https://pith.science/paper/LLTICYJ7

@misc{pith2026250600093,
  author       = {Pith},
  title        = {Pith review of: On a Family of Nested Recurrences and Their Arithmetical Solutions},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/LLTICYJ7}},
  note         = {Machine review of arXiv:2506.00093}
}

abstract

A family of nested recurrence relations $a(n+1) = n - a^{(m)}(n) + a^{(m+1)}(n)$, parameterized by an integer $m \ge 1$ with initial condition $a(1)=1$, is studied. We prove that $a(n)=n-h(n)$ is the unique solution satisfying this condition, where $h(n)$ is an arithmetical sequence in which each non-negative integer $k$ appears $mk+1$ times, with $h(n)$ 1-indexed such that $h(1)=0$. An explicit floor formula for $h(n)$ (and thus for $a(n)$) is derived. The proof of the main theorem involves establishing a key identity for $h(n)$ that arises from the recurrence; this identity is then proved using arithmetical properties of $h(n)$ and the iterated function $a^{(m)}(n)$ at critical boundary points. Combinatorial interpretations for $a(n)$ and its partial sums (for $m=2$), and connections to The On-Line Encyclopedia of Integer Sequences (OEIS), including generalizations of Connell's sequence, are also discussed.

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Reference graph

Works this paper leans on

10 extracted references · 9 canonical work pages

[7]

On Generalizing the Connell Sequence

D. E. Iannucci and D. Mills-Taylor, Errata for "On Generalizing the Connell Sequence",J. Integer Seq.2(1999), Article 99.1.7. Available at [Errata Page]

work page 1999
[1]

Connell, Problem E1350, Amer

I. Connell, Problem E1350, Amer. Math. Monthly 66 (1959), p. 413; Solution by J. H. McKay and G. W. Walker, Amer. Math. Monthly 67 (1960), pp. 85-86. 8

work page 1959
[2]

Deugau and F

C. Deugau and F. Ruskey, The Combinatorics of Certain k-ary Meta-Fibonacci Sequences,J. Integer Seq.12(2009), Article 09.4.3. [Online]

work page 2009
[3]

Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences

N. Fox, Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences, arXiv:2203.09340v2 [math.CO], 2022. [arXiv]

work page Pith review arXiv 2022
[4]

R. K. Guy,Unsolved Problems in Number Theory, 3rd ed., Springer, 2004

2004
[5]

D. R. Hofstadter,Gödel, Escher, Bach: an Eternal Golden Braid, Basic Books, 1979

work page 1979
[6]

D. E. Iannucci and D. Mills-Taylor, On Generalizing the Connell Sequence,J. Integer Seq.2 (1999), Article 99.1.7. [Online]

work page 1999
[8]

Kimberling and K

C. Kimberling and K. B. Stolarsky, Slow Beatty sequences, devious convergence, and partitional divergence,Amer. Math. Monthly,123(No. 2, 2016), 267-273. DOI:10.4169/amer.math.monthly.123.3.267

work page doi:10.4169/amer.math.monthly.123.3.267 2016

Show all 10 references

[9]

C. L. Mallows, Conway’s challenge sequence,Amer. Math. Monthly98(1991), no. 1, 5-20. DOI:10.2307/2324028

1991 doi
[10]

(2024), The On-Line Encyclopedia of Integer Sequences, https: //oeis.org

OEIS Foundation Inc. (2024), The On-Line Encyclopedia of Integer Sequences, https: //oeis.org. 9

2024

Pith tools

Reviewed August 7, 2026 · model on record in the stance chip above.

[1] [7]

On Generalizing the Connell Sequence

D. E. Iannucci and D. Mills-Taylor, Errata for "On Generalizing the Connell Sequence",J. Integer Seq.2(1999), Article 99.1.7. Available at [Errata Page]

work page 1999

[2] [1]

Connell, Problem E1350, Amer

I. Connell, Problem E1350, Amer. Math. Monthly 66 (1959), p. 413; Solution by J. H. McKay and G. W. Walker, Amer. Math. Monthly 67 (1960), pp. 85-86. 8

work page 1959

[3] [2]

Deugau and F

C. Deugau and F. Ruskey, The Combinatorics of Certain k-ary Meta-Fibonacci Sequences,J. Integer Seq.12(2009), Article 09.4.3. [Online]

work page 2009

[4] [3]

Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences

N. Fox, Connecting Slow Solutions to Nested Recurrences with Linear Recurrent Sequences, arXiv:2203.09340v2 [math.CO], 2022. [arXiv]

work page Pith review arXiv 2022

[5] [4]

R. K. Guy,Unsolved Problems in Number Theory, 3rd ed., Springer, 2004

2004

[6] [5]

D. R. Hofstadter,Gödel, Escher, Bach: an Eternal Golden Braid, Basic Books, 1979

work page 1979

[7] [6]

D. E. Iannucci and D. Mills-Taylor, On Generalizing the Connell Sequence,J. Integer Seq.2 (1999), Article 99.1.7. [Online]

work page 1999

[8] [8]

Kimberling and K

C. Kimberling and K. B. Stolarsky, Slow Beatty sequences, devious convergence, and partitional divergence,Amer. Math. Monthly,123(No. 2, 2016), 267-273. DOI:10.4169/amer.math.monthly.123.3.267

work page doi:10.4169/amer.math.monthly.123.3.267 2016