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REVIEW 3 major objections 4 minor 8 references

The 3n+1 problem: a partition of interest

T0 review · 3 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash

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Pith's one-line read The paper argues that, under a conjugate Collatz map, $\mathbb{N}\setminus\{1\}$ splits into finite strings running from $[2+3\mathbb{N}_0]$ to $[3+4\mathbb{N}_0]$, forcing every non-trivial trajectory through an odd number congruent to…

desk verdict An honest conjecture preprint with a novel structural claim about the Collatz tree, but the density argument does not establish the claim and one corollary is false as stated. read the letter →

arxiv 1908.01509 v1 pith:I6EINOHL submitted 2019-08-05 math.NT

Maarten J. Wensink This is my paper · ORCID
classification math.NT
keywords Collatzconjecture3n+1problemstringpartitionconjugatemap5mod8residue3n+pgeneralizationdensityargumentpigeonholeprinciple
topics The Collatz Conjecture
open problems The Collatz Conjecture
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

By renumbering the odd positive integers, the paper conjugates the accelerated Collatz map to a map $F$ on $\mathbb{N}$, and argues that under $F$ the set $\mathbb{N}\setminus\{1\}$ splits into finite 'strings': ordered subsets that start at a number $2+3k$ and end at a number $3+4k$. If this partition is right, every non-trivial Collatz trajectory must pass through an odd number congruent to $5$ mod $8$, a property that could be a step toward the full conjecture. The evidence is a counting argument: in every block of $3^m$ consecutive enumerated numbers, the strings started so far account for $3^m-2^m$ positions, and the remaining $2^m$ positions are exactly filled by the images of the $2^{m-1}$ live chain ends, so the pigeonhole balance is exact as $m\to\infty$. The paper also finds that, among the generalized maps $3n+p$ with odd $p$, only $p=1$ and $p=3$ appear to admit such a partition, and these are exactly the values for which the reduction-to-a-trivial-loop conjecture appears to hold. The paper presents this as a conjecture with supporting evidence rather than a complete proof.

What carries the argument

The carrying object is the conjugate map $F$ obtained by enumerating odd integers with $g(n)=(n+1)/2$, together with the equivalence map $E(x)=4x-1$, which marks all enumerated integers sharing the same image under the accelerated Collatz map. The 'lower part' $F_l$ restricts $F$ to the two classes $[2+2\mathbb{N}_0]$ and $[1+4\mathbb{N}_0]$; it is injective on its domain, and repeatedly applying $F_l$ from the starting class $[2+3\mathbb{N}_0]$ builds the strings, while applying $F_l^{-1}$ from the end class $[3+4\mathbb{N}_0]$ builds them from the other end. The counting argument uses the paper's 'z-proportionality' and 'y-proportionality' lemmas, which say that certain periodic subsets of $\mathbb{N}$ inherit the uniform distribution of map restrictions; this yields the exact identities $\sum_{k=0}^{m-1} 2^k 3^{m-k-1} = 3^m - 2^m$ inside $3^m$-blocks and the analogous $4^m - 3^m$ count inside $4^m$-blocks, so the number of positions not yet reached equals the number of later images produced by the live ends.

What would settle it

Find one starting value whose iterates under the accelerated Collatz map never visit an odd number congruent to $5$ mod $8$; equivalently, exhibit a non-trivial odd cycle whose elements all avoid that residue class.

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Extended reading notes

Core claim

The paper's central claim is that under the conjugate Collatz map $F$, $[\mathbb{N}\setminus\{1\}]$ is partitioned into strings running from $[2+3\mathbb{N}_0]$ to $[3+4\mathbb{N}_0]$, so every trajectory except the trivial loop passes through $[3+4\mathbb{N}_0]$; in the original odd-number formulation this says every non-trivial trajectory goes through an odd number congruent to $5$ mod $8$. The same construction applied to the family $3n+p$ with odd $p$ yields such a partition only for $p=1$ and $p=3$, and these are precisely the members for which all trajectories are suspected to reduce to the trivial loop; for $p=3$ the strings have a two-to-one structure. The paper gives two complementary recursive procedures, one applying the injective lower map $F_l$ forward from $[2+3\mathbb{N}_0]$ and one applying $F_l^{-1}$ backward from $[3+4\mathbb{N}_0]$, and uses density counts and a pigeonhole argument to argue that the ends meet.

Load-bearing premise

The load-bearing assumption is that the exact density match in finite blocks really forces every individual number into a string; the counting argument does not by itself rule out an element of $[\mathbb{N}\setminus\{1\}]$ being left out on an infinite chain or in a cycle that avoids $[3+4\mathbb{N}_0]$.

Editorial extensions

If this is right

  • If the partition holds, every accelerated Collatz trajectory other than the trivial loop hits $[3+4\mathbb{N}_0]$, i.e. an odd number congruent to $5$ mod $8$.
  • The partition would rule out any non-trivial cycle or infinite chain entirely contained in the complement of $[3+4\mathbb{N}_0]$.
  • For the family $3n+p$, the apparent coincidence between string partition and the reduction conjecture singles out $p=1$ and $p=3$ as candidate cases where a proof of the partition might directly yield the full conjecture.
  • The explicit intercept bounds in Lemmas 2 and 4 mean that the density count can be checked block-by-block from $[2]$ onward, making the argument computationally testable for larger and larger $m$.

Reading between the lines

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

  • Beyond the paper, the density equality could be turned into an algorithmic test: for increasing $m$, check whether every element of the initial block $[2,2+3^m)$ is eventually hit by forward iteration of $F_l$; a residue class that stays uncovered would falsify the density argument.
  • Beyond the paper, if the $5$ mod $8$ statement is true, a hypothetical non-trivial cycle would have to include an element of that residue class, which might allow cycle-length bounds from the first-return time to $5$ mod $8$.
  • Beyond the paper, the special status of $p=1$ and $p=3$ suggests that the string partition is a sharper invariant than mere cycle behavior for classifying $3n+p$ systems; one could test whether other $p$ values admit partial partitions on subsets of $\mathbb{N}$.
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Editorial analysis

A structured set of objections, weighed in public.

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

Referee Report

3 major / 4 minor

Summary. The paper studies the accelerated Collatz map on odd positive integers, enumerated as N via g(n)=(n+1)/2. It defines an explicit conjugate map F (Appendix 2, Lemmas 9-13) and considers a decomposition of N\{1} into 'strings' that start at elements of [2+3N0] and end at elements of [3+4N0]. The central claim is that N\{1} is partitioned into such strings, which would imply that every nontrivial Collatz trajectory passes through an odd number congruent to 5 mod 8 (i.e., an element of [3+4N0] in the enumerated space). The evidence is a counting argument in Sections 3.1 and 3.2: in blocks of length 3^m and 4^m, the number of elements hit by the first m forward (respectively backward) iterations tends to the full block size as m→∞. The paper also observes that among the generalized maps 3n+p, only p=1 and p=3 seem to admit such a partition, and links this to the apparent validity of the reduction-to-the-trivial-loop property for those p.

Significance. If the partition claim were proven, it would be a striking structural result about the Collatz map and would identify a nontrivial congruence condition (5 mod 8) that all Collatz trajectories (except the trivial loop) would have to satisfy. The explicit derivation of the conjugate map F and the partition of N into the domains of its restrictions (Lemma 13) are correct and clearly presented. The paper also includes an honest discussion of the limitations of the counting argument, which is commendable. However, the main result is not established: the density-one counting argument does not imply pointwise coverage, and the abstract's corollary about the original Collatz mapping is literally false for powers of two. The paper is therefore best viewed as a speculative and partially heuristic contribution, not as a proof of a theorem about the Collatz map.

major comments (3)
  1. [Section 3.1, Eqs. (13)-(15) and Section 3.1.1] The counting argument establishes only that the density of the union of A_k for k<m in blocks of length 3^m tends to 1 as m→∞. It does not show that every element of [N\1] is eventually covered. A density-zero set, such as an infinite chain that never reaches [3+4N0] or a nontrivial cycle, would be invisible to this limit. The paper itself concedes this in Section 3.1.1 ('it is unsure whether such a counting process could just by itself constitute proof') and even gives a concrete counterexample to the method for 3n-1 numbers, where the counting misses the cycle {3,4}. Therefore the statement in Section 3.3 that 'it follows that under the conjugate Collatz map F, [N\1] is partitioned in strings' is not justified by the preceding arguments.
  2. [Abstract and Section 4] The abstract claims that the partition result implies 'all trajectories except for the trivial loop go through an element of {3+4N0} ({5+8N0} for the original mapping).' This is literally false for the original Collatz mapping: the trajectory of 8 is 8→4→2→1, which never visits an odd number congruent to 5 mod 8. The statement can at most hold for the accelerated map restricted to odd numbers, as the enumerated map F only tracks odd numbers. The paper should either remove this corollary or qualify it precisely; as written, it is an overstatement of what the (putative) partition would imply.
  3. [Section 3.2, Eqs. (20)-(22)] The reverse counting argument in Section 3.2 suffers from the same gap as the forward argument. The identity lim_{m→∞} ∑_{k=0}^m 3^k·4^{m-k-1} = 4^m shows that, asymptotically, the union of the first m sets B_k has the same density as the whole space, but it does not rule out a measure-zero exceptional set that is never reached by backward iteration. The 'pigeonhole principle' invoked here is only a statement about counts in finite blocks; it does not imply that the open positions are eventually filled. The paper's own discussion of spillover between bins (Section 3.1.1) underscores that the finite-block counts are only averages and cannot certify pointwise coverage.
minor comments (4)
  1. [Lemma 4 statement] Lemma 4 is mis-stated: it refers to 'some z-proportional subset of [A_k]', but the context is the backward iteration and the sets [B_k]; it should say 'y-proportional subset of [B_k]'. Similarly, the variables [D_k] and [W_k] are not used consistently with the earlier notation.
  2. [Section 3.3 and Abstract] The paper oscillates between conjectural language ('seems', 'I give reasons for this conjecture') in the abstract and definite assertions ('it follows', 'the finding ... means') in Section 3.3. The authors should decide whether the partition claim is a theorem or a conjecture and use consistent language throughout, especially in the abstract and the concluding section.
  3. [Section 4, simulation paragraph] The sentence 'I have succesfully tested this in a simulation up to element [159902416]' is a numerical check, not a proof. The paper should explicitly label this as computational evidence and avoid implying that a test up to a finite bound supports the universal claim.
  4. [Throughout] There are numerous typos and formatting issues (e.g., 'N0 = 0 ∪ N' should be 'N0 = N ∪ {0}' or similar; missing spaces after commas in formulas; inconsistent use of 'F−1 l'). A careful proofreading pass is needed.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: the conjugate-map analysis is self-contained; the acknowledged gaps are matters of proof strength, not definitional circularity.

full rationale

The paper's derivation chain does not reduce to its own inputs by construction. The conjugate map F is derived explicitly in Appendix 2 from the accelerated Collatz map via g(n)=(n+1)/2, with lemmas proving the range and domain; no target result is assumed. The equivalence E([x])=4[x]-1 is introduced with a lemma proving F(E([x]))=F([x]), so it is not a self-definitional shortcut. The central string-partition claim is supported by density counting in Sections 3.1-3.2: equations (13)-(15) and (20)-(22) are self-contained identities showing that the union of finitely many iterates fills blocks of size 3^m or 4^m in density. No parameter is fitted to data and then renamed a prediction; the claimed 5+8N0 corollary is a direct translation of [3+4N0] back through the enumeration. The author explicitly concedes in Section 3.1.1 that the limit/counting process may not constitute proof and that the same method misses the 3n-1 cycle, which is an acknowledged proof gap rather than a circular step. The false literal reading of the 5+8N0 statement for original Collatz trajectories such as 8,4,2,1 is a correctness issue, not circularity. Citations to Lagarias, Pickover, and Wirsching are background or standard references and are not load-bearing. Accordingly, no circular step is present.

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

The paper introduces the concept of 'strings' as ordered subsets, but these are defined sets, not postulates of new entities. The main load-bearing assumptions are the pointwise-coverage step of the density argument and the intercept bounds; both are flagged by the author as uncertain. No free parameters are fitted.

assumptions (3)
  • ad hoc to paper Density-one coverage in finite blocks implies pointwise coverage of all of [N\1] by the strings.
    The pigeonhole argument in Section 3.1 shows that in any block of 3^m consecutive elements exactly 3^m elements are eventually included in the union of A_k, but this only establishes density 1, not pointwise coverage. The paper itself flags this uncertainty in Section 3.1.1.
  • domain assumption The intercepts of all z-proportional subsets of A_k and y-proportional subsets of B_k remain below their intervals for all k (Lemmas 2 and 4).
    Lemma 2's proof assumes C_k <= V_k to derive C_{k+1} < V_{k+1} and only checks the base case A_1; it does not handle all orderings. Lemma 4 is stated with a copied typo and has the same inductive gap. These bounds are load-bearing for the 'identical sections start at 2' claims.
  • standard math The accelerated Collatz map and the enumerated conjugate F capture all Collatz trajectories on natural numbers.
    The conjugacy g(n)=(n+1)/2 and Lemma 13 partition the domain, so this is proven in Appendix 2; included as a standard background fact.

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Cite this review

Pith. "Pith review of The 3n+1 problem: a partition of interest." pith.science (2026). https://pith.science/paper/I6EINOHL

@misc{pith2026190801509,
  author       = {Pith},
  title        = {Pith review of: The 3n+1 problem: a partition of interest},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/I6EINOHL}},
  note         = {Machine review of arXiv:1908.01509}
}
abstract

A mapping conjugate to the Collatz mapping seems to imply that $\N=\{1,2,3,\ldots\}$ is partitioned in a trivial loop $\{1\}$ and `strings' that are ordered subsets of $\{\N \setminus 1\}$ that run from an element of $\{2+3\0\}$ to an element of $\{3+4\0\}$ ($\0=0 \cup \N$). In particular, this means that all trajectories except for the trivial loop go through an element of $\{3+4\0\}$ ($\{5+8\0\}$ for the original mapping). I give reasons for this conjecture. Next, I note that the 3n+1 numbers and the 3n+3 numbers are the only numbers from the generalization $3n+p, p \in \{\ldots,-3,-1,1,3,\ldots\}$ for which such a partition seems to exist. Suspiciously, these are also the only members for which the conjecture (reduction to the trivial loop) seems to hold.

Figures

Figures reproduced from arXiv: 1908.01509 by the authors.

Figure 1
Figure 1. A. [N\1] is partitioned in strings. Elements of [3+4N0], which have a first lower equivalent, are colored red. Elements of [2 + 3N0], that are not in the domain of F −1 l , are partially transparent. Internal vertices are just black. The cardinality of the strings differs. One string has only one element, which is true for [{2 + 3N0} ∩ {3 + 4N0}], and is therefore transparent red. Other strings have various cardinal… view at source ↗
Figure 2
Figure 2. Strings for 3n + 3 numbers: every position that is no a head has a co-tail, exactly half or double, while every position that is mapped to is mapped to exactly twice. And yet, it seems that “a Collatz conjecture for 3n+3 numbers” holds: every trajectory ends up in the ‘trivial-loop-plus-one’ for p = 3 (as simulated on Klaas IJntema’s Collatz Calculation Center). Simultaneously, only the 3n + 1 and 3n + 3 numbers see… view at source ↗

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Reference graph

Works this paper leans on

8 extracted references · 8 canonical work pages

  1. [1]

    When an element of[3 + 4N0] is hit, a string ends, as[3 + 4N0] is not in the domain ofFl: these elements generate no image in the next iteration

    work page

  2. [2]

    All elements hit through this procedure are indeed in[3 + 3N0∪4 + 3N0], since this is the range ofFl if [1↦→1] is ignored

    work page

  3. [3]

    It remains to be shown thatall of [3 + 3N0∪4 + 3N0] is indeed hit

    All elements of [N] that are hit through this procedure, are hit exactly once, sinceFl is one-to-one and [2 + 3N0] (the starting point) is not in the range ofFl. It remains to be shown thatall of [3 + 3N0∪4 + 3N0] is indeed hit. 6 Similarly, the inverse ofFl, F−1 l : [3 + 3 N0∪1 + 3N0→2 + 2N0∪1 + 4N0], such that F−1 l ([3 + 3m]) = [2 + 2m]|m∈N0, F−1 l ([1...

    work page

  4. [4]

    When an element of[2 + 3 N0] is hit, a string ends, as[2 + 3 N0] is not in the domain of F−1 l : these elements generate no image in the next iteration

    work page

  5. [5]

    All elements hit through this procedure are indeed in[2 + 2N0∪5 + 4N0 = N\1\3 + 4N0], since this is the range ofF−1 l if [1↦→1] is ignored

    work page

  6. [6]

    of any and allN consecutive elements of [N\1], exactlyN are included in the strings

    All elements of[N] that are hit through this procedure, are hit exactly once, sinceF−1 l is one-to-one and [3 + 4N0] (the starting point) is not in the range ofF−1 l . It remains to be shown thatall of [2 + 2N0∪5 + 4N0] is indeed hit. If it could be shown that recursive application ofFl on [2 + 3N0] hits all of[3 + 3N0∪4 + 3N0], which in union with[2+3N0]...

    work page

  7. [7]

    of any and all 4m consecutive elements of [N\1], exactly so many are in the strings

    and [4]. Thus, we verify manually that[4], [5], [6], and [8] are in strings, which is the case. Together with Lemma (4), which assures that the intercepts remain below the interval for all y-proportional subsets that make up all[Bk], we are thus sure for some large enoughm that the identical sections of4m consecutive elements of[N\1] start at 2: all the i...

    work page 2010

  8. [8]

    If this preimage is 3 (mod 4), take the preimage of it

    has a unique preimage underE(·). If this preimage is 3 (mod 4), take the preimage of it. This can be continued until an element not 3 (mod 4) is reached. Since the position decreases at each step, halting will happen. This completes the proof. 26

    work page

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