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Almost all orbits of the Collatz map attain almost bounded values

Terence TaoForum of Mathematics, PiPublished 2022doi 10.1017/fmp.2022.8 Original︎ ↗Archived copy

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For any function f tending to infinity, almost all N in logarithmic density have Collatz orbit minimum below f(N).
Supports. This is the primary source: the refereed proof, published open access in Forum of Mathematics, Pi in 2022. The abstract states the bound with a weak inequality and the theorem with a strict one; since the function may be shifted by a constant, the two forms are equivalent. Sep 16, 2026

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Reduction to Fourier decay bound0 passages

Decay of Fourier coefficients0 passages

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Almost all orbits of the Collatz map attain almost bounded values1 passage

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Reduction to fine-scale mixing of the n-Syracuse offset map0 passages

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1 Introduction0 passages

Let19 passages

We have23 passages

Given a finite non-empty subset R of56 passages

Let8 passages

Thus, for instance, one has81 passages

Equivalently, one can write22 passages

This Syracuse orbit10 passages

We have16 passages

Let8 passages

Indeed, if Theorem 1.6 holds and22 passages

The iterates7 passages

A brief calculation shows that41 passages

The identity in equation ( 1.7 ) asserts that19 passages

If50 passages

We can make this heuristic precise as follows. Given two random variables6 passages

Note that2 passages

For any finite non-empty set R , let3 passages

Let18 passages

Then61 passages

In practice, the offset82 passages

The key proposition is then1 passage

For any y with16 passages

Informally, this theorem asserts that the Syracuse orbits of25 passages

It remains to establish Proposition 1.11 . Since the constant56 passages

We thus see that the7 passages

For any10 passages

Proof. Let22 passages

By Euler’s theorem, the quantity6 passages

Thus, for instance, we trivially have20 passages

Remark 1.13. One could view the Syracuse random variables41 passages

While the Syracuse random variables9 passages

For all19 passages

Informally, the above proposition asserts that the Syracuse random variable48 passages

Let71 passages

2 Notation and preliminaries0 passages

We use the asymptotic notation20 passages

If E is a set, we use26 passages

The following alternate description of the n -Syracuse valuation3 passages

Let12 passages

Proof. It is clear from equation ( 1.7 ) that16 passages

This implies that11 passages

Inserting this into equation ( 2.1 ) and using the fact that30 passages

Lemma 2.2 (Chernoff type bound(0 passages

Let41 passages

Thus, for instance for any26 passages

By contour shifting, we then have5 passages

From Taylor expansion and the non-degeneracy condition we have27 passages

If17 passages

3 Reduction to stabilisation of first passage2 passages

For7 passages

In particular, by equation ( 1.2 ), we have4 passages

In other words, for10 passages

Proof. We may assume that30 passages

Observe that if9 passages

In particular, the event10 passages

Let31 passages

If11 passages

By definition of13 passages

Now let20 passages

42 passages

In this section, we establish Proposition 1.9 . Let12 passages

We have1 passage

Proof. Write8 passages

From Lemma 2.1 and equation ( 1.3 ), we see that9 passages

Thus, if one conditions to the event17 passages

The tuples9 passages

Since7 passages

Bernoulli random variables2 passages

From Lemma 2.2 , we also have1 passage

From equation ( 1.9 ) and the triangle inequality, we therefore have1 passage

From Definition 1.7 , we have4 passages

By Lemma 2.1 , the event7 passages

Equivalently one has1 passage

This constrains10 passages

5 Reduction to fine-scale mixing of the n -Syracuse offset map16 passages

Since the random variable13 passages

We begin with the proof of equation ( 1.19 ). Let26 passages

Remark 5.1. This argument already establishes that5 passages

Indeed, the above arguments show that outside of an event of probability63 passages

Let52 passages

Proof. Fix E , and write5 passages

Hence, if6 passages

Suppose that8 passages

In particular, from equation ( 5.11 ), one has34 passages

As22 passages

Now suppose that n is an element of21 passages

We conclude that for any7 passages

With22 passages

In particular, the event2 passages

We conclude that we have the equality of events9 passages

Suppose that10 passages

Now we compute the right-hand side of equation ( 5.8 ). Let10 passages

Conversely, if equation ( 5.18 ) holds, then8 passages

Since13 passages

We conclude that1 passage

We will eventually establish the estimate7 passages

Since from equation ( 5.9 ), we have8 passages

It remains to establish equation ( 5.20 ). Fix12 passages

Lemma 5.3. We have6 passages

Proof. We can split3 passages

We now estimate14 passages

Also, from equation ( 1.7 ), we have that19 passages

If5 passages

Now suppose instead that7 passages

Applying equation ( 5.25 ) with14 passages

From equations ( 5.9 ) and ( 5.2 ), we have4 passages

6 Reduction to Fourier decay bound18 passages

Henceforth we assume equation ( 6.1 ). We also fix6 passages

Let30 passages

Now suppose that E holds. From equation ( 6.2 ), we have9 passages

From equation ( 6.2 ), we have1 passage

It thus suffices by the union bound to show that43 passages

Fix l . If we let6 passages

By Cauchy-Schwarz, it suffices to show that2 passages

By the Fourier inversion formula, we have7 passages

By equation ( 6.9 ), we can write1 passage

On the event3 passages

The key point here is that the random variable5 passages

For15 passages

Then by equation ( 1.23 ), we have12 passages

Remark 6.1. If we ignore the technical restriction to the events8 passages

For each natural number n , the n -Syracuse offset map2 passages

Proof. Suppose that7 passages

On the other hand, from equation ( 1.5 ), we have6 passages

We will need a more quantitative2 passages

Corollary 6.32 passages

Let18 passages

Proof. Suppose that3 passages

Applying equation ( 1.5 ) and multiplying by19 passages

Dividing by23 passages

7 Decay of Fourier coefficients1 passage

Let38 passages

Let16 passages

To extract some usable cancellation in the expression7 passages

For22 passages

For any30 passages

Clearly5 passages

From equation ( 7.4 ), we certainly have33 passages

Let16 passages

If3 passages

Proof. If17 passages

We specialise to the case5 passages

By hypothesis we have12 passages

As we shall see later, we can interpret the34 passages

The black set31 passages

Proof. We first observe some simple relations between adjacent values of14 passages

Among other things, this implies that4 passages

These identities have the following consequences. Call a point29 passages

Now we begin the proof of the lemma. Suppose5 passages

On the other hand, since19 passages

Suppose that67 passages

Now let43 passages

Let31 passages

Applying the equality case of equation ( 7.18 ), we conclude that21 passages

If25 passages

Now suppose that26 passages

Suppose for contradiction that6 passages

If14 passages

Now suppose that84 passages

We now describe the random set50 passages

The random variable10 passages

Proof. From the definition of45 passages

From equation ( 7.29 ) and dominated convergence, we have6 passages

Since19 passages

We conclude that12 passages

Let20 passages

Informally, this lemma asserts that as a rough first approximation one has2 passages

Proof. Note that by construction of13 passages

Writing3 passages

We can restrict to the region4 passages

This is in turn implied by27 passages

From Lemma 7.6 and Lemma 2.2 , one has18 passages

Observe that for each22 passages

One can think of10 passages

Suppose that we can prove a bound of the form24 passages

Clearly we have12 passages

We have14 passages

It remains to establish Proposition ( 7.8 ). Let36 passages

For any17 passages

We can bound20 passages

For m large enough depending on16 passages

We turn to the details. Set14 passages

This is a finite random variable since the22 passages

Since11 passages

Summing in84 passages

We will need a large constant P (much larger than A or14 passages

We again let9 passages

Applying equation ( 7.35 ), we then have2 passages

Applying equation ( 7.38 ) to12 passages

Let us first consider the event that13 passages

Thus the contribution of this case to equation ( 7.54 ) is5 passages

Since m is large compared to4 passages

Since the left-hand side of equation ( 7.55 ) is at most29 passages

Given an initial point32 passages

The key estimate relating3 passages

Let11 passages

Informally the estimate in equation ( 7.57 ) asserts that when21 passages

Suppose that25 passages

We thus obtain the inequality3 passages

Since the left-hand side is equal to5 passages

For each20 passages

We may of course assume that the event12 passages

To use this bound we need to show that the renewal process5 passages

Let28 passages

Proof. We can assume that9 passages

Similarly, from Lemma 2.2 , one has3 passages

In a similar spirit, from equation ( 7.48 ) and summing in11 passages

Thus, if12 passages

Suppose now that we are outside the event14 passages

Suppose that the lower tip of5 passages

Then by equation ( 7.63 ), we can find an integer9 passages

From equation ( 7.11 ), we have7 passages

Thus17 passages

On the other hand, from equation ( 7.11 ), we have8 passages

Thus, outside the event12 passages

Now suppose we have two distinct triangles21 passages

Applying equation ( 7.66 ) and the definition of5 passages

We conclude that for the triangles25 passages

We conclude that1 passage

From equation ( 7.48 ), we see that1 passage

Summing and using the6 passages

From Lemma 7.10 , we have11 passages

Next, we apply Lemma 7.9 with13 passages

Outside of the event5 passages

In particular, if we set9 passages

Now suppose we lie outside of both6 passages

We argue by contradiction. Suppose that equation ( 7.68 ) fails; thus1 passage

Then the point17 passages

Thus by equation ( 7.11 ), for26 passages

Acknowledgments1 passage

Conflicts of interest1 passage

Financial support1 passage

Footnotes11 notes

2 Indeed, if the latter assertion failed, then there exists a11 passages

4 As an alternative to reversing the order of the tuple8 passages

5 This Markov process may possibly be related to the5 passages

7 Note that we have reversed the order of variables4 passages

References11 references

Chamberland , M. , A9 passages

Everett , C. J. , Iteration of the number-theoretic function9 passages

Kontorovich , A. , Lagarias , J. , Stochastic models for the35 passages

Roosendaal , E. , www.ericr.nl/wondrous . Google Scholar15 passages

Wirsching , G. , The Dynamical System Generated by the2 passages

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Let f : N+1 → R be any function with lim_{N→∞} f(N) = +∞. Then one has Col_min(N) < f(N) for almost all N ∈ N+1 (in the sense of logarithmic density).

For any function f tending to infinity, almost all N in logarithmic density have Collatz orbit minimum below f(N).VerifiedReading · this claim’s Steward

Supports. This is the primary source: the refereed proof, published open access in Forum of Mathematics, Pi in 2022. The abstract states the bound with a weak inequality and the theorem with a strict one; since the function may be shifted by a constant, the two forms are equivalent.

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journal article

Authors

Terence Tao

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Forum of Mathematics, Pi

Published

2022

DOI

10.1017/fmp.2022.8

Link

https://www.cambridge.org/core/journals/forum-of-mathematics-pi/article/almost-all-orbits-of-the-collatz-map-attain-almost-bounded-values/1008CC2DF91AF87F66D190C5E01C907F

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https://web.archive.org/web/20261005071026/https://www.cambridge.org/core/journals/forum-of-mathematics-pi/article/almost-all-orbits-of-the-collatz-map-attain-almost-bounded-values/1008CC2DF91AF87F66D190C5E01C907F

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