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On a Conjecture of Cusick on a sum of Cantor sets

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Pith's one-line read The paper proves that for every integer $k \geq 2$, the sumset $S(k)+S(k)$ contains the interval $[0, 1/(k-1)]$, which gives Lebesgue measure at least $1/(k-1)$ and disproves Cusick's conjecture that this measure is zero for every $k \geq…

desk verdict Cusick's conjecture is false: Shulga's constructive algorithm shows S(k)+S(k) contains [0,1/(k-1)], and the proof checks out despite some compressed parity cases. read the letter →

arxiv 2411.17379 v2 pith:TXJ2RCPL submitted 2024-11-26 math.NT

Nikita Shulga This is my paper · ORCID
classification math.NT MSC 11A5511J7028A80
keywords continuedfractionsCantorsetsarithmeticsumsofCusick'sconjecturepartialquotientsLebesguemeasuregapsinsumset
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

Cusick conjectured in 1971 that the set of reals in $[0,1]$ expressible as a sum of two continued fractions whose partial quotients avoid $1,\dots,k-1$ has Lebesgue measure zero for every $k \geq 3$. This paper proves the opposite: for every $k \geq 2$ the sumset $S(k)+S(k)$ contains the whole interval $[0, 1/(k-1)]$, so its measure is at least $1/(k-1)$ and Conjecture 1.1 is false. The proof is constructive: it gives an algorithm that, for any $x$ in that interval, produces two continued fractions with all partial quotients at least $k$ and whose sum converges to $x$. The same machinery yields positive-measure results for sums $S(m)+S(n)$ of two different sets, an exact interval in the case $S(m)+S(m^2)$, and a countable family of genuine gaps in $S(k)+S(k)$. A sympathetic reader should care because the result overturns a fifty-year-old expectation and replaces a dimension-only lower bound with an explicit interval containment.

What carries the argument

The engine is an explicit iterative algorithm. Starting with $c_1 = a_1(x)+1$, it defines $b_n = a_n(x - p_n/q_n)$ and $c_{n+1} = a_{n+1}(x - s_n/t_n)+1$, where $p_n/q_n$ and $s_n/t_n$ are the convergents of the two fractions being built; the construction alternates between the two summands so that each is a good approximation to $x$ minus the other. The proof that the algorithm is well-defined, converges, and produces partial quotients $\geq k$ reduces to a system of inequalities comparing the continuant denominators $q_n$ and $t_n$ (Lemmas 3.3–3.8), with the key amplification estimate (Lemma 3.10) showing that once $c_1 \geq 3$, all later $c_{n+1} \geq (c_1-1)^2$. This denominator comparison is what carries the argument; it replaces the thickness and critical-exponent machinery of earlier work with a direct construction.

What would settle it

Implement the algorithm from equations (2)–(3) for $k=3$ on a fine grid of points in $(0, 1/2)$, say $x=1/3$, and check whether every produced partial quotient is at least $3$ and whether the residuals stay inside the predicted cylinders; the first failure, or a computed $c_n$ or $b_n$ below $3$, would refute Theorem 1.2. A more targeted check is to verify the even-index case of inequality (16) numerically, for example $n=2$ with $c_1=3$, $b_1=6$, $c_2=5$, since the paper only spells out the odd case.

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Extended reading notes

Core claim

The central discovery is that the addition of two copies of the continued-fraction Cantor set $S(k)$ is much larger than previously believed: for any $k \geq 2$, $S(k)+S(k)$ contains $[0, 1/(k-1)]$. Since $S(k)+S(k)$ is trivially contained in $[0, 2/k]$, this does not reach the whole maximal interval, but it gives Lebesgue measure at least $1/(k-1)$, which for $k \geq 3$ is positive, contradicting Cusick's Conjecture 1.1. In addition, the paper shows that $S(k)+S(k)$ is not the full maximal interval by exhibiting countably many open gaps $G_{k,n}$ whose endpoints lie in the sumset but whose interiors do not, and it extends the interval-containment result to unequal sets $S(m)+S(n)$ under the condition $n \leq (m-1)^2$, with an exact equality $S(m)+S(m^2) = [0, (m+1)/m^2]$.

Load-bearing premise

The construction works only if the denominator estimates that control the algorithm's induction hold for even indices exactly as claimed for odd indices; the paper writes out the odd-index cases of Lemmas 3.4 and 3.7 in detail and asserts the even-index cases follow by the same argument with signs reversed, so a hidden sign error there would break the decomposition for some $x$.

Editorial extensions

If this is right

  • For every $k \geq 2$, the Lebesgue measure of $S(k)+S(k)$ is at least $1/(k-1)$; in particular $S(3)+S(3)$ has measure at least $1/2$, so the zero-measure conjecture fails already at $k=3$.
  • By monotonicity of the nested sets $S(k)$, the containment $S(k)+S(k) \supseteq [0, 1/(k-1)]$ gives the lower bound $\lambda(S(m)+S(n)) \geq 1/(\max(m,n)-1)$ for any $m,n \geq 2$ (Corollary 1.4).
  • Under the condition $3 \leq m < n \leq (m-1)^2$, the same algorithm shows $S(m)+S(n) \supseteq [0, 1/(m-1)]$, and for the special pair $(m, m^2)$ the sum is exactly the interval $[0, (m+1)/m^2]$.
  • The gap theorem shows that $S(k)+S(k) \neq [0, 2/k]$ for every $k \geq 3$: the countably many open intervals $G_{k,n}$ lie inside the maximal interval but are disjoint from the sumset, with endpoints in the sumset.

Reading between the lines

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

  • Because the algorithm is explicit and converges quickly in the paper's worked examples, it may serve as a practical tool for representing real numbers as sums of two restricted continued fractions, with possible use in Diophantine approximation or digit-expansion problems where simultaneous restrictions are wanted.
  • The amplification estimate $c_{n+1} \geq (c_1-1)^2$ hints that the construction is quite robust: once the first partial quotient exceeds $k$, later ones grow at least quadratically, so the interval $[0, 1/(k-1)]$ might be far from the largest interval contained in $S(k)+S(k)$, a question the paper does not attempt to optimize.
  • The gap pattern around $2/S_k$, twice the reciprocal metallic mean, suggests that $S(k)+S(k)$ has a periodic, self-similar boundary; characterizing the full complement inside $[0, 2/k]$ is an open direction the paper only begins.
  • If, as the numerical data suggest, the merged partial-quotient sequence $c_1, b_1, c_2, b_2, \dots$ is non-decreasing (open Problem 6.1), then every real number in $(0,1]$ would be a sum of two elements from Good's set of numbers whose partial quotients tend to infinity, resolving the paper's Problem 6.2.
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Editorial analysis

A structured set of objections, weighed in public.

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

Referee Report

0 major / 7 minor

Summary. The paper proves that Cusick's conjecture is false by showing that for every k≥2 the sumset S(k)+S(k) contains the interval [0,1/(k−1)], so its Lebesgue measure is at least 1/(k−1). The proof is constructive: an iterative algorithm, defined from the continued fraction digits of x and of the residual x−s_n/t_n, produces two numbers whose partial quotients are all at least k and whose sum converges to x. The paper also constructs countably many gaps in S(k)+S(k), proves containment results for S(m)+S(n), and gives an exact interval statement S(m)+S(m^2)=[0,(m+1)/m^2].

Significance. If the proof is correct, this is a significant resolution of a long-standing conjecture: the sum of two copies of S(k) has positive measure for every k≥3, contrary to Conjecture 1.1. The proof is self-contained and algorithmic, and it does not depend on any numerical fitting or on assuming the conjecture. The lower-bound chain b_n≥c_n and c_{n+1}≥(c_1−1)^2 is the engine of the argument, and the apparent circularity between Lemmas 3.6 and 3.8 is resolved by an induction on the index. The paper also recovers Cusick's S(2)+S(2)=[0,1] as a corollary, which is a useful consistency check.

minor comments (7)
  1. [Section 3, Lemmas 3.4 and 3.7] The even-index cases are not written out: Lemma 3.4 says that for even indices the process is exactly the same except for inequality signs, and Lemma 3.7 says the same. Since these inequalities are used for every index in the induction, please include the even-index verification or at least a precise statement of which inequalities reverse. I checked the even case of Lemma 3.4 and it does reduce to (16), so this is a completeness issue rather than a mathematical error.
  2. [Section 5, proof of Theorem 1.3] The proof of the even-n case is deferred with the sentence "The case of even n can be done similarly with minor changes." Because Theorem 1.3 is one of the main results, the even-n argument should be written out or explicitly reduced to the odd-n case by a clear symmetry.
  3. [Lemma 3.4] The statement says that inequality (16) holds for all 1≤n≤k, but for n=1 the right-hand side contains t_{n−2}=t_{−1}, which is undefined; the intended range appears to be 2≤n≤k. In the same proof, the phrase "all partial quotients b_i with indices smaller or equal to 2k" should read "2m".
  4. [Section 6, numerical data] In the third numerical example, the displayed error should be |2(√2−1) − [2,51,139299,23380586] − [3,2143,8527219,38512412]|; a minus sign is missing before the second continued fraction.
  5. [Proof of Theorem 1.6] The line "by (25) for k=1 we get c_2 > 4 + 1/2" does not follow directly from (25), which with c_1=2 and b_1≥4 gives c_2 > b_1^2/4; the displayed constant appears to be a typo, although the conclusion c_2≥5 remains valid.
  6. [Proof of Theorem 1.5] The symbol n is used both for the parameter in S(n) and as the running index in expressions such as c_{n+1} ≥ (c_1−1)^2; this should be disambiguated, for example by writing c_{r+1} for the running index.
  7. [Section 3] The initialization of the sequences uses q_0, t_0, s_0, and p_0 without an explicit definition; please state the standard conventions p_0=0, q_0=1, s_0=0, t_0=1 before the first use.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity found: the proof is self-contained and constructive, with no fitted parameters or self-citation carrying the argument.

full rationale

The paper's central theorem, Theorem 1.2, is proved by an explicit algorithm that, for every x in [0, 1/(k-1)], produces two continued fractions c and b whose partial quotients are all at least k. The construction defines c1 = a1(x)+1, b_n = a_n(x - p_n/q_n), and c_{n+1} = a_{n+1}(x - s_n/t_n)+1, and the proof verifies well-definedness, convergence, and lower bounds on partial quotients through Lemmas 3.3-3.10. These lemmas are proved from classical continued-fraction facts (cylinder intervals, continuant inequalities, Farey neighbor properties) and from prior inductive steps. No statistical fit, no normalization forcing, and no parameter calibrated to a target output appears. Cusick's earlier theorem S(2)+S(2)=[0,1] is recovered as Corollary 3.9, but it is derived from the algorithm, not assumed. The only external references used for load-bearing claims are standard continued-fraction facts (e.g., Proposition 2.1, Lemma 2.2) which are proved or cited to Khintchine, and they do not contain the target result. The occasional statements that even-index cases follow 'with inequality signs reversed' are expositional compression, not circularity; the underlying arguments are symmetric and do not import the conclusion. The gap theorem (Theorem 1.3) is also proved by direct comparison with convergents. No step in the derivation reduces by definition to the conjecture or to a fitted input. The paper is therefore self-contained with respect to the circularity concerns in scope.

Assumptions & free parameters 0 free parameters · 4 assumptions · 0 invented entities

The proof uses only standard facts about continued fractions, the standing definition of S(k), and a termination convention for rational residuals. No free parameters are fitted and no new entities are postulated.

assumptions (4)
  • standard math Standard continued fraction facts, including cylinder interval descriptions, continuant identities, and Farey neighbor estimates
    Used throughout Section 3; taken from Khintchine and classical continued fraction theory.
  • standard math Rational numbers use the continued fraction representation whose last partial quotient is at least 2
    Stated in Section 2; required so membership in S(k) is well-defined for rationals.
  • domain assumption 0 is treated as having a partial quotient +infinity, so 0 belongs to S(k)
    Stated in the introduction; needed to include the left endpoint of [0,1/(k-1)].
  • domain assumption The algorithm terminates at exact rational identities when a residual has fewer than the required number of partial quotients
    Remarks 3.1 and 3.5 define this convention; finite decompositions are treated as valid end states.

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Pith. "Pith review of On a Conjecture of Cusick on a sum of Cantor sets." pith.science (2026). https://pith.science/paper/TXJ2RCPL

@misc{pith2026241117379,
  author       = {Pith},
  title        = {Pith review of: On a Conjecture of Cusick on a sum of Cantor sets},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/TXJ2RCPL}},
  note         = {Machine review of arXiv:2411.17379}
}
abstract

In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as a sum of two continued fractions with no partial quotients equal to $1$. In other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{ for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also conjectured that this result is unique in the sense that if you exclude partial quotients from $1$ to $k-1$ with $k\geq3$, then the Lebesgue measure $\lambda$ of the set of numbers which can be expressed as a sum of two continued fractions with no partial quotients from $\{1,\ldots,k-1\}$ is equal to $0$, that is $$\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3.$$ In this paper, we disprove the conjecture of Cusick by showing that $$ S(k)+S(k) \supseteq \left[0,\frac{1}{k-1}\right]. $$ The proof is constructive and does not rely on ideas from previous works on the topic. We also show the existence of countably many 'gaps' in $S(k)+S(k)$, that is intervals, for which the endpoints lie in $S(k)+S(k)$, while none of the elements in the interior do so. Finally, we prove several results on the sums $$ S(m)+S(n) $$ for $m\neq n$.

Figures

Figures reproduced from arXiv: 2411.17379 by the authors.

Figure 1
Figure 1. Known structure of S(k) + S(k). Proposition 5.2. For any fixed k ≥ 3 the intervals Gk,n are disjoint. Proof. Fix k ≥ 3. From Proposition 5.1 we know that G odd k \ G even k = . Thus, it remains to show that intervals in the collection {Gk,2n−1}∞ n=1 are disjoint and intervals in the collection {Gk,2m}∞ m=1 are disjoint, too. Let us only prove the claim for {Gk,2n−1}∞ n=1 as the other one will follow similar lines wi… view at source ↗

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Forward citations

Cited by 1 Pith paper

Reviewed papers in the Pith corpus that reference this work. Sorted by Pith novelty score. Full citation record

  1. Every real number is a sum of two real numbers with diverging partial quotients

    math.NT 2025-07 conditional novelty 7.0 of 10

    Every real number is a sum of two numbers whose continued fraction partial quotients tend to infinity.

Reference graph

Works this paper leans on

18 extracted references · 17 canonical work pages · cited by 1 Pith paper

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