almost all collatz orbits attain almost bounded valuesalmost all collatz orbits attain almost...

Almost all Collatz orbits attain almostbounded values

Terence Tao

University of California, Los Angeles

Apr 15, 2020

Terence Tao Almost all Collatz orbits attain almost bounded values

Let N = {0,1,2, . . . } be the natural numbers.Define the Collatz map Col : N + 1→ N + 1 by settingCol(N) := N/2 when N is even and Col(N) = 3N + 1 whenN is odd.Each positive integer N ∈ N + 1 then generates a Collatzorbit

ColN(N) := {N,Col(N),Col2(N), . . . }

Each Collatz orbit has a minimal value

Colmin(N) := inf ColN(N).


Examples:

ColN(1) = {1,4,2}.ColN(6) = {6,3,10,5,16,8,4,2,1}.ColN(27) = {27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214,107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412,206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263,790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167,502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638,319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858,2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051,6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433,1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23,70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}.


Collatz conjecture

We have Colmin(N) = 1 for all N ∈ N + 1.

Also known as the “3x + 1 problem” or “Syracuse problem”.This conjecture is a prime example of what Frank Harary(1969) called a mathematical disease, as it obeys the followingthree criteria:

Must be easy to stateMust seem to be accessible, even to an amateurMust also have been repeatedly “solved” (often by thesame person)


XKCD, Randall Monroe, March 5, 2010


Epidemiology1937: Lothar Collatz infected with Col (Collatz, 1986)1950: first reported person-to-person transmission(informal lecture, Collatz, ICM, Cambridge Mass.)1952: independent introduction by Thwaites1950s: multiple infections reported at Yale, U. Chicago,and Syracuse University1963: first detection of Col in a mathematical journal1970: Coxeter states the Collatz conjecture and offers $50for proof1983: Erdos offers $500 for a proof1996: Thwaites offers £1000 for a proof1999: 197 published papers on the conjecture (Lagarias,2011)2009: 341 published papers on the conjecture (Lagarias,2012)2012: Lagarias abandons effort to maintain Collatzbibliography


“For about a month everyone at Yale worked on it, with noresult. A similar phenomenon happened when I mentionedit at the University of Chicago. A joke was made that thisproblem was part of a conspiracy to slow downmathematical research in the U.S.” (Shizuo Kakutani, 1960)“Mathematics is not yet ripe enough for such questions.”,(Erdos 1983)“This is an extraordinarily difficult problem, completely outof reach of present day mathematics.” (Lagarias, 2010)


Selected partial results and obstructions

Colmin(N) = 1 for all 1 ≤ N ≤ 1020 (yoyo@home project,2017).Apart from the cycle 1,4,2, all other Collatz cycles musthave length at least 17,087,915 (Eliahou, 1993). Note thatthe Collatz conjecture rules out such cycles completely.{N ≤ x : Colmin(N) = 1} � x0.84 for all large x(Krasikov-Lagarias, 2003).There are variants Col of Col (piecewise affine onarithmetic progressions) and initial data N0 for which theassertion Colmin(N0) is undecidable (Conway, 1987).The Collatz conjecture implies that there are finitely manysolutions to 2n − 3m = k for each k (T., 2011). All knownunconditional proofs of the latter claim require somevariant of Baker’s theorem in transcendence theory.


One can work on the easier statistical version of the Collatzconjecture, where one only seeks to prove statements for“most” choices of initial data N.

One has Colmin(N) < N for almost all N (Terras, 1976)One has Colmin(N) < N0.869 for almost all N (Allouche,1979)One has Colmin(N) < N0.7925 for almost all N (Korec, 1994)

Here “almost all” is in the sense of natural density (one has∑N∈E ;N≤x 1 = ox→∞(x) for the set E of exceptions). As we

shall see, these results are analogous to “almost sure localwell-posedness” results in PDE.


Theorem (T., 2019)

If f : R+ → R+ is such that limN→∞ f (N) = +∞, then one hasColmin(N) < f (N) for almost all n.

Thus for instance Colmin(N) < log log log logN for almost all N.Here “almost all” is in the sense of logarithmic density (one has∑

N∈E ;N≤x1N = ox→∞(log x) for the set E of exceptions). This

result is analogous to an “almost sure almost globalwell-posedness” result in PDE.


As observed by Ben Green, an equivalent formulation is

TheoremFor every ε > 0 there exists C > 0 such that Colmin(N) ≤ C forN in a set of logarithmic density at least 1− ε.

Corollary

There exists C0 such that Colmin(N) = C0 for all N in a set ofpositive logarithmic density.

With more effort it may be possible to replaced “logarithmicdensity” by “natural density”, and in principle it may be able toperform a finite computation to allow one to take C0 = 1.


The proof of the result proceeds in several stages:

1 Passing to the Syracuse formulation of the problem2 Reduction to locating an “almost invariant” (or more

precisely “almost self-similar”) measure3 Reduction to establishing mixing of a “Syracuse probability

measure” Syrac(Z/3nZ)4 Reduction to controlling Fourier coefficients of

Syrac(Z/3nZ)5 Reformulation in terms of random walks hitting triangles6 Exploiting a key separation property of triangles


Syracuse formulation

The dynamics of the Collatz map Col can be clarified byskipping the even numbers and looking only at the actionon the odd numbers 2N + 1 = {1,3,5, . . . }.Define the Syracuse map Syr : 2N + 1→ 2N + 1 by

Syr(N) :=3N + 1

2ν2(3N+1)

where ν2(N) is the 2-valuation of N (the largest j such that2j divides N).It is easy to see that for any odd N and k ≥ 0,SyrN(N) ⊂ ColN(2kN) and Syrmin(N) = Colmin(N). Hencethe Collatz conjecture is equivalent to asserting thatSyrmin(N) = 1 for all N ∈ 2N + 1, and the main theoremequivalent to asserting that Syrmin(N) ≤ f (N) for almost allN ∈ 2N + 1.


A heuristic proof of the (statistical) Collatz conjecture

If N is a “random” odd number, then 3N + 1 is even, will bedivisible by 4 “with probability 1/2”, divisible by 8 “withprobability 1/4”, and so forth.This leads to the heuristic probabilistic distributionν2(3N + 1) ≡ Geom(2), where Geom(2) ∈ N + 1 is arandom variable with the geometric distribution of mean 2:

P(Geom(2) = k) = 2−k

We therefore morally have Syr(N) ≈ 32Geom(2) N, or

log Syr(N) ≈ logN + log 3−Geom(2) log 2. Thus wepredict the dynamics log Syrn(N),n = 0,1,2, . . . to behavelike a random walk with drift

E(log 3−Geom(2) log 2) = log34< 0.

The law of large numbers then predicts the orbit shouldalmost surely drift down to 1.


A plot of n 7→ log Syrn(N), compared against the linear driftn 7→ logN + n log 3

4 , with N = 27382591.


Can this heuristic be made rigorous?

One can calculate

Syrn(N) =3n

2a1+···+anN + Fn(a1, . . . ,an)

where a1, . . . ,an are the valuations

ai := ν2(3Syri−1(N) + 1)

and Fn(a1, . . . ,an) is the offset

Fn(a1, . . . ,an) := 3n−12−a1−···−an−1+3n−22−a2−···−an+· · ·+2−an

For short times n ≤ c logN, c small, and N uniformlydistributed from 1 to some large x , it is not difficult to showthat a1, . . . ,an behave like independent copies ofGeom(2). The law of large numbers givesa1 + · · ·+ an ≈ 2n for almost all N.


Thus we can basically show that

Syrn(N) ≈ 3n

22n N

for n ≤ c logN and almost all N. This sort of “short time”control lets us show results of the form Syrmin(N) ≤ N1−c

for almost all N and some small c > 0, in the spirit ofprevious results of Allouche and Korec.However, for long times n > c logN, we lose control on thedistribution of the valuations an.


What if we iterate?

The above arguments show that for almost all N, we havecan find an element M = Syrn(N) of the Syracuse orbit ofN with M ≤ N1−c . It is then tempting to iterate and claimthat we can find an element M ′ = Syrn′(M) of the Syracuseorbit of M with M ′ ≤ M1−c ≤ N(1−c)2

. This would imply thatSyrmin(N) ≤ N(1−c)2

, and one could hope to use furtheriteration to improve the bound even more.But there is an obstruction: if N is uniformly distributed onsome interval [1, x ], the output M is not going to beuniformly distributed (on, say, [1, x1−c]). In particular it isconceivable that the Syracuse dynamics forces the outputM to lie in the small exceptional set of the short timealmost sure result, thus preventing us from iterating.


The problem here is that the Syracuse mapSyr(N) = 3N+1

2ν2(3N+1) is biased with respect to the uniformdistribution: if N is a uniformly distributed random variable(on some interval), then Syr(N) will not be uniformlydistributed.For instance, it is obvious that Syr(N) is never a multiple of3. Furthermore, for n ≥ 1, Syrn(N) will be biased to betwice as likely to equal 2 mod 3 than 1 mod 3.For n ≥ 2, Syrn(N) acquires further biases modulo 32; forn ≥ 3, there are biases modulo 33; and so forth.


Histogram of N 7→ Syrn(N) mod 3 for odd N ≤ 104 and n = 0,1,together with the limiting Syracuse measure.


Histogram of N 7→ Syrn(N) mod 32 for odd N ≤ 104 andn = 0,1,2, together with the limiting Syracuse measure.


Histogram of N 7→ Syrn(N) mod 33 for odd N ≤ 104 andn = 2,3, together with the limiting Syracuse measure.


This data suggests that the Syracuse dynamics, when viewedmodulo 3n, converges to an invariant measure in Z/3nZ. Thiscan in fact be proven, and the measure is the probabilitydistribution of the Syracuse random variable

Syrac(Z/3nZ) := 3n−12−a1−···−an+3n−22−a2−···−an+· · ·+2−an mod 3n

where a1, . . . ,an are iid copies of Geom(2).


One can use these measures to propose an “invariantmeasure” µx on [1, x ]. Roughly speaking, this is thelogarithmic measure

∑n≤x

1nδn, weighted by the density of

the Syracuse random variable for some n = bc log xc, andthen normalised to be a probability measure.If one can show that some (variable length) iteration of theSyracuse map pushes forward µx to µx1−c with small error(e.g., O(log−c x) in total variation norm), then on iterationwe can push forward µx to be less than f (x) withprobability 1− o(1) for any f that goes to infinity as x →∞.This idea was motivated by arguments of Bourgain (1994)exploiting an invariant measure for the nonlinearSchrödinger equation to boost a almost sure localwell-posedness result to an almost sure globalwell-posedness result.


After some routine calculations, we can check that thisstrategy works as long as the Syracuse random variablesSyrac(Z/3nZ) stabilise as n→∞. Specifically, it willsuffice to show that

dTV (Syrac(Z/3nZ),ExtZ/3nZ(Syrac(Z/3mZ)))�A m−A

for all 1 ≤ m ≤ n and A > 0, whereExtZ/3nZ(Syrac(Z/3mZ)) is an element of Z/3nZ thatreduces to Syrac(Z/3mZ) modulo 3m, uniformly chosenamongst all such choices.


The Syracuse random variables obey a “skew-convolution”relation that roughly speaking takes the form

Syrac(Z/3nZ) = Syrac(Z/3mZ)+3m2−a1−···−amSyrac(Z/3n−mZ)

where a1, . . . ,am are the independent geometric variablesinvolved in defining Syrac(Z/3mZ).Using (a rigorous version of) this relation and some Fourieranalysis, one can obtain the desired stabilisation propertyif one can establish the Fourier uniformity property

E exp(2πiξSyrac(Z/3nZ)/3n)�A n−A

whenever ξ ∈ Z/3nZ is not divisible by 3.


The Syracuse random variable

Syrac(Z/3nZ) := 3n−12−a1−···−an + 3n−22−a2−···−an

+ · · ·+ 2−an mod 3n

is not quite a sum of independent random variables, so thecharacteristic function

E exp(2πiξSyrac(Z/3nZ)/3n)

does not easily factor.


However, if we group the terms into pairs, and introducethe sums

b1 := a1 + a2, b3 := a3 + a4, . . .

one can rewrite the above random variable as

3n−22−b2+···+bn/2(3× 2−b1 + 2−a2)

+ 3n−42−b3+···+bn/2(3× 2−b2 + 2−a4)

+ · · ·+ (3× 2−bn/2 + 2−an).

The key point is that once one conditions on the sumsb1, . . . ,bn/2, the summands are independent and now thecharacteristic function factorises!


3n−22−b2+···+bn/2(3× 2−b1 + 2−a2)

+ 3n−42−b3+···+bn/2(3× 2−b2 + 2−a4)

+ · · ·+ (3× 2−bn/2 + 2−an).

Modulo a minor technicality (ignored here), the problem nowboils down to this: uniformly for all ξ ∈ Z/3nZ not divisible by 3,show that with high probability a large number of the quantities

3n−22−b2+···+bn/2ξ/3n,3n−42−b3+···+bn/2ξ/3n, . . . , ξ/3n

are not close to an integer.


This problem can be phrased geometrically as follows. Color apoint (j , k) in the lattice Z2 “black” if 32j2−kξ/3n is close to aninteger, and white otherwise. The task is then to show with highprobability, the two-dimensional random walk

(0,0), (1,b1), (2,b1 + b2), . . .

passes through a lot of white points. (Due to a technicality, weactually work with a subset of this random walk known as atwo-dimensional renewal process, but we ignore this subtletyhere.)


We are coloring points (j , k) black when 32j2−kξ/3n isclose to an integer.Using elementary number theory, one can show that nomatter what ξ is, the set of black points organise intodisjoint right angled “triangles” (of slope log 9/ log 2); thehypothesis that ξ is not a multiple of three places someupper limit on the size of the triangles. The question thenboils down to showing that a certain random walk with driftwill encounter a large number of non-black points.


Small black triangles are not much of a problem, becausethey happen to be surrounded by a “moat” of white points.The enemy is the large black triangles, and in particular ifthe random walk hops from one large triangle to anotherwithout much of a gap of white points between thetriangles.A key geometric observation saves us: if a large trianglelies just beneath one or more other large triangles, thelower vertices of the latter triangles must be well separatedfrom each other, so a random walk passing through thefirst large triangle is unlikely to immediately land on one ofthe latter triangles. Making this observation rigorous andquantitative concludes the argument.


Thanks for listening!


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