Comptes Rendus Mathématique

Hieu Thanh Nguyen

Phuong Le

Thanh Chi Vo

Monotonicity for quasilinear elliptic problems with a sign-changing nonlinearity in half-planes

Yong Hu

Yisheng Tian

Trivialité des groupes de Whitehead réduits avec applications à l’approximation faible et l’approximation forte

Petr Gurka

Daniel Hauer

Existence of an extremal function of critical Sobolev embedding with an $\alpha $-homogeneous weight

Nguyen Tien Dung

Le Vi

Pham Thi Phuong Thuy

Non-uniform Berry–Esseen bounds via Malliavin–Stein method

Karthik Elamvazhuthi

A linear test for global nonlinear controllability

Tatsuki Kuwagaki

Vivek Shende

Adjoints, wrapping, and morphisms at infinity

Aleksey Zinger

The (co)homology of the Deligne–Mumford moduli spaces of marked rational curves

Guillaume Rialland

Multi-solitary waves for the one-dimensional Zakharov system

Shalom Eliahou

Jean-Louis Verger-Gaugry

Comptes RendusMathématiqueShalom Eliahou and Jean-Louis Verger-GaugryThe number system in rational base 3/2 and the 3x +1 problemVolume 363 (2025), p. 329-336Online since: 20 May 2025https://doi.org/10.5802/crmath.662This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/C EN T R EMER S ENN EThe Comptes Rendus. Mathématique are a member of theMersenne Center for open scientific publishingwww.centre-mersenne.org — e-ISSN : 1778-3569Comptes Rendus. Mathématique2025, Vol. 363, p. 329-336https://doi.org/10.5802/crmath.662Research article / Article de rechercheCombinatorics, Number theory / Combinatoire, Théorie des nombresThe number system in rational base 3/2 andthe 3x +1 problemLe système de numération en base rationnelle 3/2 et leproblème 3x +1Shalom Eliahou ,a,b and Jean-Louis Verger-Gaugry ,ca Univ. Littoral Côte d’Opale, UR 2597 – LMPA – Laboratoire de Mathématiques Pures et AppliquéesJoseph Liouville, F-62100 Calais, Franceb CNRS FR 2037, Francec Université-Grenoble-Alpes, Université Savoie Mont Blanc, F-73000 Chambéry, FranceE-mails: eliahou@univ-littoral.fr, verger-gaugry.jean-louis@unsechat-math.frAbstract. The representation of numbers in rational base p/q was introduced in 2008 by Akiyama, Frougny& Sakarovitch, with a special focus on the case p/q = 3/2. Unnoticed since then, natural questions related torepresentations in that specific base turn out to intimately involve the Collatz 3x +1 function. Our purposein this note is to expose these links and motivate further research into them.Résumé. La représentation des nombres dans la base rationnelle p/q a été introduite en 2008 par Akiyama,Frougny & Sakarovitch, avec un accent particulier sur le cas p/q = 3/2. Des questions naturelles liées auxreprésentations dans cette base spécifique, passées inaperçues jusqu’ici, s’avèrent impliquer intimement lafonction de Collatz 3x + 1. Le but de cette note est d’exposer ces liens et de motiver des recherches plusapprofondies sur ceux-ci.Keywords. Collatz conjecture, iteration, numeration, odometer.Mots-clés. Conjecture de Collatz, itération, numération, odomètre.2020 Mathematics Subject Classification. 11B83, 68R15.Manuscript received 24 April 2024, revised 22 May 2024 and 21 June 2024, accepted 29 June 2024.1. IntroductionThe representation of numbers in rational base p/q with p > q coprime positive integers wasintroduced in 2008 in [2], with a focus on the case p/q = 3/2 to study Mahler’s Z -number problemand the Josephus problem.Given n ∈N, its representation in rational base 3/2 is the unique expression of the formn =k∑i=0ai2(32)iISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/330 Shalom Eliahou and Jean-Louis Verger-Gaugrywith digits ai ∈ {0,1,2} for all i ≥ 0 and ak ̸= 0 if n ≥ 1. We then denote the corresponding word in{0,1,2}∗ by〈n〉 = ak ak−1 · · ·a0.We set 〈0〉 = ϵ, the empty word, and we identify any word 0w ∈ {0,1,2}∗ with w .Denoting L = {〈n〉 | n ∈N} ⊂ {0,1,2}∗, here are two natural questions about L :(1) If w ∈L , which extensions among w0, w1, w2 still belong to L ?(2) Given k ≥ 1, what is the lexicographically largest word of length k in L ? Equivalently,what is the largest n ∈N such that 〈n〉 is of length k?Apparently unnoticed so far, answers to these questions turn out to intimately involve the Collatz3x +1 function, namely T : N→N defined byT (n) ={n/2 if n ≡ 0 mod 2,(3n +1)/2 if n ≡ 1 mod 2. (1)The purpose of this note is to expose these links and motivate further research into them. Itscontents are as follows. In Section 2, we recall basic properties of the numeration system inrational base 3/2, in particular the recursive construction of the word 〈n〉 in Section 2.1 and theodometer allowing to construct 〈n + 1〉 from 〈n〉 in Section 2.2. We also propose a variant ofthe Collatz conjecture in Section 2.3. In Sections 3.1 and 3.2, we address the above questions (1)and (2), respectively. Section 4 concludes the note with two questions and two conjectures closelyrelated to the Collatz conjecture and possibly of the same order of difficulty.2. BasicsFollowing [2], every n ∈N has a unique expression of the formn = 12(ak (3/2)k +ak−1(3/2)k−1 +·· ·+a0)with ai ∈ {0,1,2} for all i and ak ̸= 0 if n ≥ 1. Accordingly, we associate to n the word〈n〉 = ak ak−1 · · ·a0 ∈ {0,1,2}∗,called the representation of n in rational base 3/2. We set 〈0〉 = ϵ, the empty word, and we identifyany word 0w ∈ {0,1,2}∗ with w .By [5, Lemma 4.1], the length of the word 〈n〉 is given by |〈n〉| = log3/2(n)+O(1).Definition 1. We say that a word w ∈ {0,1,2}∗ is admissible if w = 〈n〉 for some n ∈N. We denoteby L ⊂ {0,1,2}∗ the language of admissible words, i.e.L = {〈n〉 | n ∈N}.As observed in [2], the language L is prefix-closed, i.e. stable under taking prefixes. That is,every prefix w ′ of an admissible word w = w ′w ′′ is itself admissible.Conversely, given an admissible word w ∈ L , which extensions among w0, w1, w2 remainadmissible? The answer involves the Collatz 3x +1 function. See Proposition 7.2.1. Construction of 〈n〉Given n ∈ N, a recursive construction of 〈n〉 ∈ {0,1,2}∗ may be described as follows. Set 〈0〉 = ϵ,the empty word.Shalom Eliahou and Jean-Louis Verger-Gaugry 331Proposition 2. Let n0 ∈N\ {0}. Denote 〈n0〉 = ak · · ·a0 with ai ∈ {0,1,2} for all i . Then:{a0 ≡ 2n0 mod 3;〈n0〉 = 〈n1〉a0, where n1 = (2n0 −a0)/3.Moreover, we have a0 = 2n0 −3n1 and a0 ≡ n1 mod 2.Proof. As 〈n0〉 = ak · · ·a0, we haven0 = 12(ak (3/2)k +·· ·+a1(3/2)+a0)=(12(ak (3/2)k−1 +·· ·+a1))32+ a02.Hence a0 ≡ 2n0 mod 3, and setting n1 = (2n0 − a0)/3, we have 〈n1〉 = ak · · ·a1. The claimsfollow. □2.2. The odometerThe odometer constructs the word 〈n+1〉 from the word 〈n〉. It is described as a finite automatonin [2]. Here is an equivalent description.Let σ = (2 1 0) be the cyclic permutation 2 7→ 1 7→ 0 7→ 2. For a word u ∈ {0,1,2}∗, let uσ ∈{0,1,2}∗ denote the word obtained by applyingσ to each letter of u. For instance, 12012σ = 01201,then identified to 1201.Proposition 3. Let n ∈N. If 〈n〉 = u0v ∈ {0,1,2}∗ with v ∈ {1,2}∗, then 〈n +1〉 = u2vσ.Proof. See [2, Section 3.2.3, pp. 66–67]. □The rule applies even if u or v is the empty word. In particular, if 〈n〉 = v ∈ {1,2}∗, thenidentifying v with 0v , we get 〈n + 1〉 = 2vσ. Table 1, listing 〈n〉 for 1 ≤ n ≤ 27, was constructedby hand using this description of the odometer. The interested reader will effortlessly extend itmuch farther.Table 1. The words 〈n〉 for 1 ≤ n ≤ 27.n 〈n〉1 22 213 2104 2125 21016 21207 21228 210119 21200n 〈n〉10 2120211 2122112 21011013 21011214 21200115 21202016 21202217 21221118 2101100n 〈n〉19 210110220 210112121 212001022 212001223 212020124 212022025 212022226 212211127 210110002.3. The Collatz conjectureDating back to the 1930’s, this conjecture deals with iterates T (k) of the Collatz function T : N→Ndefined in (1).Conjecture 4. Let n0 ∈N, n0 ≥ 1. Then there exists k ≥ 1 such that T (k)(n0) = 1.332 Shalom Eliahou and Jean-Louis Verger-GaugryHere is an equivalent version, as shown below.Conjecture 5. Let n0 ∈N, n0 ≥ 1. Then the proportion of even numbers in the sequence of iteratesT (k)(n0) tends to 1/2 as k tends to infinity.Proposition 6. Conjectures 4 and 5 are equivalent.Proof. If Conjecture 4 holds, then at some point the sequence of iterates T (k)(n0) collapses to1,2,1,2,1,2, . . . , whence Conjecture 5.Conversely, assume Conjecture 5 holds. Given an integer n ≥ 1, letΩ(n) = (n,T (n),T (2)(n), . . .)denote the trajectory of n under iteration of T . Moreover, given k ≥ 1, letΩk (n) =(n,T (n),T (2)(n), . . . ,T (k−1)(n)),and let k1 denote the number of odd integers in Ωk (n). Assume for a contradiction that Conjec-ture 4 fails. Then there is an integer m ≥ 10 such that minΩ(m) = m. (Note that minΩ(n) = 1 forn ≤ 9.) We claim that for all k ≥ 1, we haveT (k)(m) ≤ (3.1)k12km. (2)The proof is straightforward, by induction on k. Indeed, (2) holds for k = 1, since m is oddby minimality in Ω(m) and T (m) = (3m + 1)/2 = (3 + 1/m)m/2 ≤ (3.1)m/2 since m ≥ 10. Letnow k ≥ 1 and assume that T (k)(m) ≤ (3.1)k1 /2k m. Denote n = T (k)(m). If n is even, thenT (k+1)(m) = T (n) = n/2 ≤ (3.1)k1 /2k+1m, as desired. If n is odd, thenT (k+1)(m) = T (n) = (3+1/n)n/2 ≤ (3+1/m)n/2 ≤ (3.1)n/2since n ≥ m ≥ 10. Hence T (k+1)(m) ≤ (3.1)k1+1/2k+1m, as desired. The claim is proved.Now (3.1)k1 /2k ≥ 1 if and only if k1/k ≥ ln(2)/ln(3.1), and ln(2)/ln(3.1) > 0.6. By Conjecture 5,for k large enough, the proportion k1/k will fall below 0.6. Hence T (k)(m) ≤ (3.1)k1 /2k m < m,contradicting the minimality of m. This completes the proof of the proposition. □At the time of writing, Conjecture 4 has been verified up to n0 ≤ 268. See [3]. With this bound,the results of [4] and its Table 2 on page 54 imply that, besides the trivial cycle {1,2} under T , noother cycle, if any, can have length less than 114,208,327,604. See [1] for a general reference onthe conjecture, and [6] for recent strong results about it.3. Links with the Collatz functionWe now address the two questions about L = {〈n〉 | n ∈ N} asked in the Introduction, and showthat in both cases, the answer naturally involves the Collatz function T .3.1. Extending admissible wordsHere is the first question. Given w ∈L , which extensions among w0, w1, w2 still belong to L ?Proposition 7. Let w ∈L , say w = 〈n〉 for some n ∈N. Then:{w1 ∈L ⇐⇒ n ≡ 1 mod 2, and then w1 = 〈(3n +1)/2〉 = 〈T (n)〉;w0 ∈L ⇐⇒ n ≡ 0 mod 2 ⇐⇒ w2 ∈L , and then w0 = 〈3n/2〉 and w2 = 〈(3n +2)/2〉.Proof. If 〈n〉1 ∈L , Proposition 2 implies 1 ≡ n mod 2 and w1 = 〈(3n+1)/2〉. Similarly, if 〈n〉0 ∈Lthen 0 ≡ n mod 2 and w0 = 〈3n/2〉. Finally, if 〈n〉2 ∈ L then 2 ≡ n mod 2 and w2 = 〈(3n +2)/2〉.The stated equivalences follow. □Corollary 8. Let n ∈N be odd. Then 〈T (n)〉 = 〈n〉1.Shalom Eliahou and Jean-Louis Verger-Gaugry 333Proof. Immediate from the above proposition. □Remark 9. The map n 7→ n/2 for n even is straightforward in the base 2 representation of n:just remove the last digit, 0. Symmetrically, Corollary 8 shows that the map n 7→ (3n + 1)/2 forn odd is straightforward in the base 3/2 representation of n: just add 1 as rightmost digit. Byany chance, could it happen that considering the map n 7→ T (n) simultaneously in base 2 and inrational base 3/2 may lead to new insights on the Collatz conjecture?3.2. Saturated wordsGiven a number system B over a finite set of digits, we say that a positive integer n ≥ 1 is saturatedif its representation 〈n〉B in base B is of length strictly less than that of 〈m〉B for all m ≥ n +1.Accordingly, we say that 〈n〉B is a saturated word. For instance in base 10, the saturated integersand corresponding words are the n = 10ℓ−1 = 9 · · ·9.The digits 1,2 of the saturated words in rational base 3/2 turn out to be unpredictable in somesense and closely related to the 3x + 1 problem. Recall that |w | denotes the length of the finiteword w .Notation 10. Given k ≥ 1, we denote by sat(k) the largest positive integer whose representationin base 3/2 is of length k. That is,sat(k) = max{n ∈N, |〈n〉| = k}.Proposition 11. For all k ≥ 1, 〈sat(k)〉 is the unique admissible word of length k with digits in{1,2}.Proof. By the odometer as described in Proposition 3, if 〈n〉 is of length k and contains a 0 digit,then 〈n+1〉 is of the same length k. This shows that the inequality |〈n+1〉| > |〈n〉| happens exactlywhen 〈n〉 contains no 0 digit except possibly as an irrelevant prefix. □For instance, a look at 〈n〉 for 1 ≤ n ≤ 27 in Table 1 yields sat(k) and the corresponding wordfor k ≤ 7 in Table 2. Observe in Table 2 that 〈sat(k −1)〉 is a prefix of 〈sat(k)〉. Indeed, this is truefor all k ≥ 1.Proposition 12. For all k ≥ 2, we have 〈sat(k)〉 = 〈sat(k − 1)〉a0 where a0 ∈ {1,2} and a0 ≡sat(k −1) mod 2.Proof. Since L is prefix-closed, it follows from Proposition 11 that the length k − 1 prefix of〈sat(k)〉 equals 〈sat(k − 1)〉. Thus 〈sat(k)〉 = 〈sat(k − 1)〉a0 where a0 ∈ {1,2}, and a0 ≡ sat(k −1) mod 2 by Proposition 2. □Table 2. The first seven saturated numbers and words.k sat(k) 〈sat(k)〉1 1 22 2 213 4 2124 7 21225 11 212216 17 2122117 26 2122111334 Shalom Eliahou and Jean-Louis Verger-Gaugry3.3. The function UIt turns out that the sequence(sat(k))k≥1 coincides with the sequence of iterates U(k)(0) of acertain function U closely related to the Collatz function T . See Proposition 16 below.Notation 13. Let U : N→N be defined asU (n) ={(3n +2)/2 = T (n)+n +1 if n ≡ 0 mod 2,(3n +1)/2 = T (n) if n ≡ 1 mod 2.Remark 14. We have U (n) ≡ T (n)+n +1 mod 2 for all n ∈N.Starting from any n0 ∈ N, let us consider the sequence of iterates U (k)(n0) for k ≥ 1. Thefollowing proposition shows that, for the sequence of corresponding words 〈U (k)(n0)〉 in rationalbase 3/2, each term is a prefix of the next one and the added digit is either 1 or 2.Proposition 15. Let n ∈N. Then:〈U (n)〉 ={〈n〉1 if n ≡ 1 mod 2,〈n〉2 if n ≡ 0 mod 2.Proof. Straightforward from Proposition 7. □A particularly interesting sequence of iterates U (k)(n0) is for n0 = 0.Proposition 16. For all k ≥ 1, we have sat(k) =U (k)(0).Proof. For k = 1, we have sat(1) = 2 and U (1) = T (1) = 2. For k ≥ 2, we only need showsat(k) =U (sat(k −1)). Indeed, it follows from Proposition 12 thatsat(k) = 3sat(k −1)+a02where a0 ∈ {1,2} and a0 ≡ sat(k − 1) mod 2. Hence sat(k) = U(sat(k − 1)) by definition of U , asdesired. □Consequently, the growth factor of the sequence sat(k) tends to 3/2.4. Open questionsWe conclude this note with two questions and two conjectures.Question 17. Is the Collatz conjecture equivalent to some behavior of the iteration of the func-tion U ? For instance, starting with any n0 ∈ N, is it true that the proportion of 1’s in 〈U (k)(n0)〉tends to 1/2 as k goes to infinity, as supported by numerical evidence? Note that by Proposition 15,the number of digits “0” remains constant along the trajectory of n0.More specifically, for n0 = 0, we have U (k)(0) = sat(k) for all k ≥ 1 by Proposition 16. Havingcomputed sat(k) up to k = 220 = 1,048,576, we find that the word w = 〈sat(220)〉, of length k = 220,counts exactly 525,067 digits “1” and 523,509 digits “2”. Moreover, for all 1 ≤ ℓ < 17, the word wcontains all 2ℓ words in {1,2}ℓ as subwords. See Table 3 for the word 〈sat(200)〉, with exactly 100digits “1”. This numerical evidence leads us to the following conjecture.Conjecture 18. The proportion of digits “1” in the word 〈sat(k)〉 tends to 1/2 as k tends to infinity.Shalom Eliahou and Jean-Louis Verger-Gaugry 335Is Conjecture 18 equivalent to the Collatz conjecture? The question arises from the closerelationship between the functions T and U , Remark 14 and the analogy with Conjecture 5. Atany rate, even if not equivalent to Conjecture 5, Conjecture 18 might well be of the same order ofdifficulty.In fact, the numerical evidence below Question 17 leads to an even stronger conjecture. Since〈sat(k −1)〉 is a prefix of 〈sat(k)〉 for all k ≥ 2, we may define the limit〈sat(∞)〉 = limk→∞〈sat(k)〉,an infinite word over {1,2}. Informally, recall that an infinite word w over a finite alphabet B ofcardinality |B| = b is said to be normal if for all ℓ ≥ 1, all bℓ words in Bℓ occur as subwords inw[n], the length n prefix of w , with frequency tending to 1/bℓ as n →∞. Our stronger conjectureruns as follows.Conjecture 19. The word 〈sat(∞)〉 is normal.Table 3. The word 〈sat(200)〉 = 〈U 200(0)〉.21221112212112212121122121121211111221211111112122222112212122121211211222212212222221211111221221212211211222211122111222122221211212122122212122212121211111111212122222111112211121212222122111112112Our second question is about fully understanding the Collatz map T in rational base 3/2.Question 20. For n odd, we have seen that 〈T (n)〉 = 〈n〉1. For n even, how does one determine〈T (n)〉 from 〈n〉? Is any answer susceptible to lead to new insights on the Collatz conjecture?Note that determining the parity of n ∈ N from the digits ai of the word 〈n〉 = ak · · ·a0 ∈ L ismore cumbersome than in base 2 or 10. Indeed, the following criterion is straightforward but notquite practical for calculations by hand.Proposition 21. Let n ∈N\ {0} and 〈n〉 = ak · · ·a0 ∈L . Thenn ≡ 0 mod 2 ⇐⇒k∑i=02k−i 3i ai ≡ 0 mod 2k+2.Proof. We have n = 12(∑ki=0(3/2)i ai) = 2−(k+1)(∑ki=0 2k−i 3i ai ). Hence n ≡ 0 mod 2 ⇐⇒ 2k+1n ≡0 mod 2k+2 ⇐⇒ ∑ki=0 2k−i 3i ai ≡ 0 mod 2k+2. □AcknowledgmentsWe thank the anonymous referee for their careful reading and constructive comments, in partic-ular for insisting to formally state Proposition 6 and carefully prove it, as opposed to just a remarkwith informal arguments as in the original version of this article.Declaration of interestsThe authors do not work for, advise, own shares in, or receive funds from any organizationthat could benefit from this article, and have declared no affiliations other than their researchorganizations.336 Shalom Eliahou and Jean-Louis Verger-GaugryReferences[1] J. C. Lagarias (ed.), The ultimate challenge: the 3x + 1 problem, American MathematicalSociety, 2010, pp. xiv+344.[2] S. Akiyama, C. Frougny and J. Sakarovitch, “Powers of rationals modulo 1 and rational basenumber systems”, Isr. J. Math. 168 (2008), pp. 53–91.[3] D. Barina, “Convergence verification of the Collatz problem”, J. Supercomput. 77 (2021),pp. 2681–2688.[4] S. Eliahou, “The 3x + 1 problem: new lower bounds on nontrivial cycle lengths”, DiscreteMath. 118 (1993), no. 1-3, pp. 45–56.[5] J. F. Morgenbesser, W. Steiner and J. M. Thuswaldner, “Patterns in rational base numbersystems”, J. Fourier Anal. Appl. 19 (2013), no. 2, pp. 225–250.[6] T. Tao, “Almost all orbits of the Collatz map attain almost bounded values”, Forum Math. Pi10 (2022), article no. E12 (56 pages).

The number system in rational base $3/2$ and the $3x+1$ problem

Ioannis Iakovoglou

Markov partitions for non-transitive expansive flows

Not a member yet

Monotonicity for quasilinear elliptic problems with a sign-changing nonlinearity in half-planes

Trivialité des groupes de Whitehead réduits avec applications à l’approximation faible et l’approximation forte

Existence of an extremal function of critical Sobolev embedding with an $\alpha$ -homogeneous weight

Non-uniform Berry–Esseen bounds via Malliavin–Stein method

A linear test for global nonlinear controllability

Adjoints, wrapping, and morphisms at infinity

The (co)homology of the Deligne–Mumford moduli spaces of marked rational curves

Multi-solitary waves for the one-dimensional Zakharov system

The number system in rational base $3/2$ and the $3x+1$ problem

Markov partitions for non-transitive expansive flows

5,302

5,993