
Terence Tao Proves Result on the Collatz Conjecture - datashrimp
https://www.quantamagazine.org/mathematician-terence-tao-and-the-collatz-conjecture-20191211/
======
pg_bot
I got sucked into this problem as an undergraduate in mathematics. I similarly
wanted to limit my search space and my initial work lead me down the path of
prime factorization. If every prime number follows this pattern then logically
every number also follows this pattern. Then you get to the hard part which is
why or why not would a prime number follow this pattern. So then you go
looking for the reasons why prime cycles can or cannot exist, your brain
melts, and then you work on something else. Fun times though.

~~~
cyborgx7
As a computer science (Informatics) student, I shy away from any path in a
prove that makes me make some kind of statement about primes. They are the
death of every approach they are involved in, to me.

From my one hour of thinking about this problem, I think sums of finite
sequences are a much more promising approach. But obviously you'd know way
more about all of this than me.

~~~
Gene_Parmesan
Many of the smartest mathematicians who have ever lived have spent collective
man-decades thinking about this problem. Any idea any of us can come up with
in an hour of thought will have been considered and rejected a thousand times
already by others.

This is partly why it's such a trap. It's like trisecting an angle or squaring
a circle -- easy to state the problem, easy to start playing around with it.
In my amateur opinion, any proof related to Collatz would be similar in
complexity and non-obviousness as the proof of Fermat by Wiles, if such a
proof is even possible.

~~~
khawkins
For most, it's far more fruitful to jump on
[https://math.stackexchange.com/](https://math.stackexchange.com/) and work on
questions there. Low impact, but high probability it is within anyone's realm
of capabilities. Answering 100 questions people want to know is bigger impact
than making 0% progress towards a problem you have no realistic chance of
solving.

------
oefrha
Technical post from Terence Tao:
[https://terrytao.wordpress.com/2019/09/10/almost-all-
collatz...](https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-
orbits-attain-almost-bounded-values/)

The paper:
[https://arxiv.org/pdf/1909.03562.pdf](https://arxiv.org/pdf/1909.03562.pdf)

------
wil93
> Tao used this weighting technique to prove that almost all Collatz starting
> values — 99% or more — eventually reach a value that is quite close to 1.
> This allowed him to draw conclusions along the lines of 99% of starting
> values greater than 1 quadrillion eventually reach a value below 200.

Isn't this equivalent to saying that "99% of starting values greater than 1
quadrillion eventually reach 1"?

I mean, once you reach a value below 200 then you will continue and reach 1.
Not only below 200, but below any limit that was experimentally verified (i.e.
around 10^20)

~~~
soVeryTired
If Colmin(N) is the lowest value reached by iterating the process from initial
value N, Tao proved that

Colmin(N) < f(N)

for _arbitrary_ f provided f tends to infinity, for almost all N. Here,
"almost all" means something like "exceptions(N) / N, tends to 0 for large N",
where exceptions(N) is the count of values that do not obey the inequality [0]

So it's an asymptotic result. The first million integers could all be
exceptions - but _eventually_ the proportion of exceptions dies out.

The ability to pick arbitrary f is very powerful. Pick the slowest-growing
function you can think of. e.g. Tenfold-iterated logarithm. The inequality
says for all but a negligible fraction of integers, Colmin grows slower than
that function.

[0] Nitpick: I'm describing the _natural_ density, but Tao needs the
_logarithmic_ density, where each exception _n_ is weighted by _1 /n_.

~~~
pmiller2
Try the inverse Ackerman function instead:
[https://en.m.wikipedia.org/w/index.php?title=Ackermann_funct...](https://en.m.wikipedia.org/w/index.php?title=Ackermann_function)

If non-computable functions are on the table, define f(n) to be the smallest k
such that BB-k >= n, where BB-n is the n-th Busy Beaver number:
[https://en.wikipedia.org/wiki/Busy_beaver](https://en.wikipedia.org/wiki/Busy_beaver)

Edit: looks like @tromp managed to reply before I added the bit about BB-n.
:-)

~~~
tromp
Or the inverse TREE() function [1]. There's also the inverse Busy Beaver
function if you don't insist on computability...

[1]
[https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#TREE(...](https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#TREE\(3\))

------
vecter
One of the aspects of math that I find beautiful is the deep connection
between often seemingly unrelated topics. Connecting PDEs, which is a branch
of analysis and what I consider "continuous" mathematics to something that
seems to me like a discrete number theory problem is beautiful. The bit of
intuition for why this might be the case in the article is nice:

> With a PDE, you plug in some values, get other values out, and repeat the
> process — all to understand that future state of the system. For any given
> PDE, mathematicians want to know if some starting values eventually lead to
> infinite values as an output or whether an equation always yields finite
> values, regardless of the values you start with.

> For Tao, this goal had the same flavor as investigating whether you always
> eventually get the same number (1) from the Collatz process no matter what
> number you feed in. As a result, he recognized that techniques for studying
> PDEs could apply to the Collatz conjecture.

~~~
emmelaich
My favourite story on this is Feynman's use of PDEs in the Connection Machine
design.

[http://longnow.org/essays/richard-feynman-connection-
machine...](http://longnow.org/essays/richard-feynman-connection-machine/)

~~~
core-questions
I really want to buy the CM-1 T-Shirt, but with duty and shipping it's too
much for a shirt the comments describe as being of poor t-shirt quality....

[http://www.tamikothiel.com/cm/cm-
tshirt.html](http://www.tamikothiel.com/cm/cm-tshirt.html)

[https://shop.spreadshirt.com/mission-base-
creations/cm+1+log...](https://shop.spreadshirt.com/mission-base-
creations/cm+1+logo+flexprint?idea=5d89c895f937647d81fe06aa)

------
knzhou
Now that was a great example of an accessible, technically and culturally
accurate article. For stuff like this, Quanta magazine is generally the world
leader.

~~~
gdy
What does "culturally accurate' mean?

~~~
OJFord
I assume that the characterisation of the problem as 'dangerous', 'alluring',
tempting', etc. to mathematicians is accurate.

~~~
rsj_hn
meh. No, I think this is the sensationalization we expect from pop-sci
accounts.

------
cyborgx7
I can totally see how one could be sucked into this problem.

The solution seems immediately reachable. After giving it some thought, I can
prove that it is equivalent to "every number will at some point reach a power
of 2" because from a power of two you then go straight to 1 through divisions.

This result seems even more reachable since I no longer need to prove that
numbers eventually get smaller.

I'm sure this is not a new discovery to people who have worked on the problem.
But part of me really wants to spend the rest of the day puzzling around with
where that approach goes.

Edit: Thought about it more. The two ways it doesn't hit 1 would be for it to
find a loop, that never hits a power of 2 or to grow infinitely.

Thinking about how a loop with 3n+1 and n/2 would have to function, I think
it's probably pretty easy to prove that 4-2-1 is the only loop that is
possible.

That leaves us with the infinite grows, but without ever hitting a power of 2.

Edit2: Turns out 4-2-1 being the only loop that is possible is actually not
easy to prove at all and isn't proven. But no matter, that wasn't how I was
going to continue working on the problem anyway.

The most promising solution to me is an inductive proof. Since I know all
powers of 2 converge to 1 and only have /2 applied to them, my next step is to
find the properties of numbers n such that 3n+1 is a power of 2.

Only for every second power of 2 is this a whole number. The numbers for the
powers of 2: [4, 16, 64, 256, 1024] have the possible precedents: [1, 5, 21,
85, 341]. Interestingly every subsequent number can be built from the previous
one by 4n+1. Clearly it's 4 because it's every second power of 2. I feel like,
if you recognize the structure of every number in the tree, you can prove that
every natural number has one of those structures. And then you are done.

Once again, I'm completely aware none of this is original thought. But I enjoy
it anyway.

Edit3: Every even number will be divided by 2 until it is an odd number. Now
we only have to look at the structure of the odd numbers.

~~~
huffmsa
Had the same thought further down the comments.

Since you're getting even numbers at least 50% of the time, eventually you
should go on a run.

You also benefit from hitting a _y =2^n x N_ number because you get to greatly
reduced the size of _y_

Further, every time you make it down to 1, you can append the values you hit
on your way down to a list of seen values, as they're part of a convergence
chain, meaning if you see them in a future cycle, you can stop your iteration
early because you have a known outcome.

So this doesn't necessarily become a computationally heavy problem with large
numbers.

~~~
cyborgx7
Yeah, this is definitely at least a semi-decidable problem. By following the
process of the video in the beginning you can generate all values that
converge towards 1, and you will reach each value at some point. Now, all that
is missing is the other direction. How would you test if a number doesn't
converge towards 0 in an algorithmic way that always finishes?

------
RcouF1uZ4gsC
> The Collatz conjecture is quite possibly the simplest unsolved problem in
> mathematics — which is exactly what makes it so treacherously alluring.

IMO that distinction actually belongs to Goldbach's conjecture which is:

Every even integer greater than 2 can be expressed as the sum of two primes.

~~~
tlholaday
Is it your opinion that understanding the meaning of "two primes" is simpler
than understanding the meaning of even, odd, tripling, halving, and adding
on3?

~~~
Taek
In my opinion it's not simpler, primes are a difficult concept. Tripling and
halving is simpler.

------
btilly
See [https://terrytao.wordpress.com/2019/09/10/almost-all-
collatz...](https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-
orbits-attain-almost-bounded-values/) for a more detailed explanation.

His actual result is this.

Suppose that f(N) is a function from the natural numbers to the reals that has
infinity as a limit. Let X be the set of natural numbers who eventually go
below f(N). Then the logarithmic density of X is 1.

Here logarithmic density is the limit of the following ratio:

    
    
        sum(1/n for n in X and < N)
      / sum(1/n for n < N)
    

Now you just have to pick f that grows very, very slowly. For example f(N) =
log(log(log(x)))).

------
nickcw
A few years ago I got interested enough in this that I bought the book
mentioned in the article "The Ultimate Challenge: The 3x + 1 Problem."

If you want to see dead ends other people have gone down and read some actual
results and learn some history it is a good read for the mathematically
inclined. Some of the maths went over my head but I enjoyed it!

~~~
dmayle
I must admit to being curious about this book. I don't see how this hasn't
been proven already (though I can't write mathematical proofs myself)

If you look at this as a series of consecutive operations, every odd operation
grows by a little over three, and is guaranteed to be followed by an even
operation. Even operations shrink by a little more than three, so all that's
required is to compare the cardinality of those two offsets to see whether
continued operations are growing or shrinking. In the case of odd operations,
as the infinite series of operations continues, the offset gets smaller and
smaller. In contrast, the size of the offset for even operations doesn't
change as the series of operations continues, which should mean that numbers
shrink more than the grow, just ever so slightly.

(To understand the even operations, you have to look at expected values. 1/2
of all even operations will produce an odd operation after dividing by two,
1/4 after dividing by four, 1/8 after dividing by eight. The infinite sum
(1/2)^2n converges to 1/3, with an actual value of 1/3 - (1/3)*(1/2)^2n.

~~~
rmidthun
By using expected value, you are assuming that the distribution of the even
numbers is uniform. You will need to prove that first, and I suspect that will
be difficult.

------
jboggan
This was the problem that turned me from a music major to a math major and
eventually led me down the path to computer science. It's a beautiful and cold
problem and I love to see any progress made on it.

------
bentona
It is hilarious to me that the thread about this "Dangerous" problem is full
of approaches to solving it :)

------
jlarocco
It's an interesting puzzle, but is there any significance to it?

Are there an infinite number of sequences like this, but with different
numbers or different operations? Is there any value proving things about them?
Seems like a shot in the dark looking at an arbitrary sequence and proving
things about it.

~~~
baq
It isn’t about the result, it’s about the tools that have to be discovered to
get that result. It’s likely there’s an as yet unknown branch of number theory
required or perhaps a hidden link to something apparently unrelated.

------
atemerev
I have a stupid question. If binary representations of numbers in the
hailstone sequence for Collatz conjecture can be written as a 2-tag system (a
→ bc, b → a, c → aaa), and m-tag systems are Turing complete for m>1, and
Collatz conjecture is a sort of halting problem in this framework (there is a
halting criteria), and "deciding whether a particular algorithm for Turing
machine will/won't halt on every input" is a totality problem, which is also
undecidable — can a proof of undecidability of Collatz conjecture be
approached this way? What is the major problem with this line of thought?

~~~
inimino
If I understand your comment correctly, "deciding whether a particular
algorithm ..." means an arbitrary algorithm. For a specific algorithm, of
course sometimes it is possible to prove that it halts for every input. It's
only undecidable if the algorithm is regarded as an input.

------
scarejunba
That video at the top is very neat. Was it manually made or is there software
that makes it easy to do?

Also which is the comment they're talking about? I can't seem to find an
anonymous one talking about this on the original post
[https://terrytao.wordpress.com/2011/08/25/the-collatz-
conjec...](https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-
littlewood-offord-theory-and-powers-of-2-and-3/)

~~~
acqq
> That video at the top is very neat.

It doesn't contain number 9, so I somehow miss what it was supposed to
demonstrate.

~~~
_Microft
The animation is a bit strange. I think it's meant to show which numbers will
eventually end the cycle at 1. Having the numbers appear in the reverse order
compared to what the rules are does not help, in my opinion.

Take 16 and 5. While 16 is shown first and 5 after, the meaning is actually
that you can get from 5 to 16 by applying the odd rule (3*x+1) and from 10 to
5 by applying the even rule (x/2).

~~~
acqq
I understood the rule used. What I question is the:

\- usefulness of animation

\- the non obviousness of the numbers that are displayed, v.s. all other
remaining in the observed range.

At least, non animated version of end frame allows better comprehension of
what is shown.

------
odyssey7
Huh, neat. I just read about this after seeing it as an example in UPenn's CS
194 "Introduction to Haskell" notes.

    
    
      hailstone :: Integer -> Integer
      hailstone n
        | n `mod` 2 == 0 = n `div` 2
        | otherwise      = 3*n + 1
    

[https://www.seas.upenn.edu/~cis194/spring13/lectures/01-intr...](https://www.seas.upenn.edu/~cis194/spring13/lectures/01-intro.html)

~~~
bjoli
Now see how fast you can make it run! 22 year old me was the king of the
project Euler forum for that specific problem ("find the longest sequence
below 1000000") until my post got removed (which all newer posts for old
problems are).

Doing those stupid things actually made me learn a lot about how computers
work

------
archi42
My initial idea is to go the other way around: We can count natural numbers
using the successor function, starting at 1. E.g.: 1, s(1) = 2, s(s(1)) = s(2)
= 3, and so on. So if we see a number n, we know it's s(n-1), which is
s(s(n-2)) and so on, until we reach 1. So we have a nice, linear chain, and
it's trivial to construct an algorithm f(n) [or more mathematically, a
function f: N -> {s,ss,sss,...}] that (i) produces the necessary operations to
reach any given natural number n in that chain, starting at 1, and (ii)
terminates for every input n.

Now we don't use s(x), but instead two operations that are inverse to the
Collatz rules: a(x) = 2*x and b(x) = (x - 1)/3 (with b of course is not always
being possible). Starting at 1, this can now be used to construct a tree
containing "some" numbers, and we can use that it to describe a path to a
number from the root.

If the conjecture holds, the tree constructed from these functions has to
contain all natural numbers.

My nudge is then: If I have a number n, can I give an algorithm/function f': N
-> {a,b,aa,ab,ba,bb,...} that (i) finds a path to n, and (ii) always
terminates? Answer: I have no idea =)

E.g.: f'(5) = aaaab <=> b(a(a(a(a(1))))).

So we have f'(1)=no idea how to represent, f'(2)=a, f'(3)=aaaabab, f'(4)=aa,
f'(5)=aaaab, f'(6)=aaaababa, f'(7)=aaaabaaab[...], f'(8)=aaa,...

(However, maybe it is not possible to even construct f': I believe the target
space of f is of another infinity than the target space of f' \-- obviously
this isn't my area of expertise).

~~~
jcranmer
The algorithm is easy: enumerate all possible strings in increasing order of
length, and return the first string you found that returns n.

If there is such a string for n, then this will find it. It will terminate if
and only if there is such a string for every integer (i.e., the Collatz
conjecture).

~~~
archi42
If it only was that simple :) since we can not solve the halting problem, pure
breadth first search doesn't help. We need to find another algorithm that
makes use of the tree's structure.

If bfsearch could be proven to be the best algorithm, this would reduce the
conjecture to the halting problem.

------
devit
It's interesting that it seems that some very short propositions require very
long proofs.

Are there any results on the length (or maybe Kolmogorov complexity) of the
shortest proof for provable propositions of length N? (in a specific logic
system, I suppose)

I.e. how hard to prove is the hardest proposition of length of N and how does
it grow with N? how about the average random provable proposition?

------
huffmsa
The question as I, not a mathematician, read it is

> _For a given number, can using the operations 3n+1 and n /2 make it converge
> to a value of 2^i_

Because that's what you need. If you can get on the _2^i_ branch, you've won.

~~~
ReptileMan
Not a mathematician but that is too close to mesenne primes to make a quite
fun possible connection. Since messene you have points that 3n cannot hit.

~~~
huffmsa
But it's 3n+1, so you're always getting an even number 50% of the time.

And for every number you hit along the way, assuming you make it down to 1,
you can stop your iteration if you've previously seen the number

~~~
thaumasiotes
> And for every number you hit along the way, assuming you make it down to 1,
> you can stop your iteration if you've previously seen the number

You can stop your iteration if you come to a previously encountered value,
because the (n+1)th number in the sequence is strictly a function of the nth
number. Assuming that you make it down to 1 gives you nothing.

The whole point is to prove that you always make it to 1, or -- equivalently
-- to show that you can never reach a previously encountered number without
first reaching 1.

~~~
huffmsa
But you'll always reach a previously seen number because _3n+1_ will never be
a prime number.

~~~
Resbog
No. If you could prove that you'll always reach a previously seen number, then
you could simply say "Let n be the smallest number for which the Collatz
Conjecture is false, we have shown that we'll always reach a number < n for
which the Collatz Conjecture is true, hence n does not exist."

~~~
thaumasiotes
I don't follow this either. ((3n+1) / 2) is always greater than n. If all
Collatz sequences are periodic (or periodic after a finite prefix), how does
that prove that they all achieve a value less than some constant _k_?

(Or even that they all achieve a value which is less than their own first
element?)

~~~
huffmsa
(3n+1)/2 is always greater than n, yes, but you'd need to find a loop where
you're always alternating 3n+1 and n/2 to reach a rate of increase which could
not be overcome. Which I take to mean that if n/2 occurs > 50% of the time,
will eventually take the series to a number lower than your starting value and
eventually to 1.

Maybe the threshold isn't 50%, but the proof then becomes finding that
threshold

~~~
thaumasiotes
EDIT:

I am satisfied that, under the assumption that all Collatz sequences are
periodic[1], they must all achieve a value less than their own initial value.
(Unless that initial value is 1.)

Because the sequence is periodic[1], the ratio of (3n+1) applications to (n/2)
applications must be constant[1]. This implies that the value of a term of the
sequence will look like ((3^pn)k / (2^qn)), where p/q is approximately equal
to the ratio of (3n+1) applications to (n/2) applications. Since no integer
greater than 1 is simultaneously a power of 2 and a power of 3, this must
trend toward positive infinity or toward zero as n increases. But for the
sequence to increase without bound violates the assumption that it is
periodic[2] -- so the terms of the sequence must instead trend toward zero.

[1] I'll use the simplifying assumption that all Collatz sequences are
periodic after a prefix of 0 terms. For sequences that are periodic after a
prefix of more than 0 terms, just treat them as equivalent to the same
sequence with the prefix removed.

[2] A periodic sequence can take on only a finite number of values (equal to
the period). But if the sequence increases without bound, it must take on an
infinite number of distinct values.

------
tester89
> In the 1970s, mathematicians showed that almost all Collatz sequences — the
> list of numbers you get as you repeat the process — eventually reach a
> number that’s smaller than where you started

I don’t understand why this doesn’t prove the conjecture. Like if you start
with _x_ , and you reach _y_ : _y_ < _x_ , then couldn’t one reäpply the
quoted statement to show there exists _z_ : _z_ < _y_ < _x_ and wouldn’t
iterating of this statement eventually lead to 1 < … _z_ < _y_ < _x_

~~~
edjrage
Reread the text you pasted: It says they proved it for _almost_ all sequences.

~~~
tester89
Ohh I’m stupid lol

~~~
edjrage
Don't put yourself down. I (and I'm sure many others) have thought the same at
first. But maybe we're all stupid?

------
oneepic
Just reading the summary of Collatz reminds me a whole lot of the Josephus
problem. 3x+1 is a similar expression you'd see when you're trying to
enumerate indices in that problem. Dividing by two is similar to going once
around the circle killing every other index. I wonder if that's been explored?

------
AstralStorm
I suspect that about the only other class of numbers that could loop starts
from a prime. (Because otherwise after a series of steps you could essentially
run the procedure in parallel.) So you would have to find a specific class of
prime numbers. This is darn hard to find ever.

------
ptah
> Then this past August an anonymous reader left a comment on Tao’s blog. The
> commenter suggested trying to solve the Collatz conjecture for “almost all”
> numbers, rather than trying to solve it completely.

maybe a time traveller nudging a result that will be significant for
humanity's future?

~~~
Grue3
Well, studying a conjecture for "almost all" numbers doesn't actually prove
anything about the conjecture because just a single exception would disprove
it. There are a few conjectures that had particularly large exceptions [1].
It's possible that Collatz exception will be so large that it's effectively
uncomputable or it's existence is unprovable in ZFC [2].

Somebody probably tried it, but I wonder if describing particularly long
sequences where r_(n+1) = 3*r_n+1 is always odd would help. There are results
for arbitrarily large arithmetic progressions having some specific property
[3], so there's no reason why other progressions couldn't be studied in that
way.

[1]
[https://en.wikipedia.org/wiki/P%C3%B3lya_conjecture](https://en.wikipedia.org/wiki/P%C3%B3lya_conjecture)

[2]
[https://en.wikipedia.org/wiki/Zero_sharp](https://en.wikipedia.org/wiki/Zero_sharp)

[3]
[https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem](https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem)

~~~
mkl
The rule is that _if r_n is odd_ , r_{n+1} = 3 * r_n+1, so because of that
condition, 3 * r_n+1 will always be even.

~~~
Grue3
Yeah, I forgot that little fact. It should be (3*r_n + 1) / 2 then.

------
FrozenVoid
What don't they instead work from the opposite direction: Create an abstract
model of a Collatz loop that doesn't end with 1(instead looping back to same
number) and prove it can't exist(Reductio ad absurdum).

~~~
cevi
This approach has been explored. It has only been helpful for showing that
certain types of (short-ish) cycles can't occur [1]. Showing that the sequence
can't get larger and larger forever (without repeating) with this approach has
been a non-starter.

[1] See
[https://en.wikipedia.org/wiki/Collatz_conjecture#Cycle_lengt...](https://en.wikipedia.org/wiki/Collatz_conjecture#Cycle_length)
for a few references. The most recent appears to be the paper "Theoretical and
computational bounds for m-cycles of the 3n + 1 problem" by Simons and de
Weger.

------
SubiculumCode
I know this is a fluff comment and meta, but I know many of us came here from
/. years ago. This article was put on /. and comparing the comments of /. and
HN on this article really highlights how far /. has fallen.

~~~
gefh
Is slashdot full of people who think they can solve it in the comment thread?

~~~
SubiculumCode
No it is filled with people who think that the mathematician should do
something more useful with their time, like make targeted ad algorithms, or
something.

------
idclip
I want to know who made that comment, what a unit.

------
jorgenveisdal
Tao will be remembered 500 years from now.

------
paulpauper
the proof if it exists will involve math is the very complicated and abstract
and cutting-edge, similar to the proof of Fermat's Last Thereon . It will
involve some sort of result, I am guessing, from complex analysis an number
theory and then generalized in such a way as to prove this result.

------
wyatt777
Approximating almost up-to, is not rigorous proof in math, but looking forward
to positive conjectures.

------
sidcool
The title seems to have been changed. Did he prove it for good or for most of
the values?

------
dclowd9901
What would it mean if they found a number — one number — that didn’t adhere to
this rule?

~~~
JoeAltmaier
Of course that's one way of proving this wrong. But you have to start above 18
quadrillion. Because all the numbers less than that have been tried.

~~~
3pt14159
Has anyone tried really, really large numbers picked at random?

~~~
JoeAltmaier
One approach. But I'd suggest it has less than a one-in-18-quadrillion chance
of working. From the statistics so far.

------
jeromebaek
Also known as hailstone problem from GEB

~~~
aroberge
The Collatz conjecture predates GEB by many years. Renaming it to something
else by the author of GEB was a way to misappropriate someone else's work.

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stan_rogers
Um, no. Collatz's conjecture was that the sequence will always reach one, and
that conjecture was entirely avoided in GEB. Any number that would result in
the 1-4-2-1 closed loop was referred to as "a wondrous number" in the GEB
dialogs, and the concentration was on the idea that while you could test any
number for the property of wondrousness, nobody knew how to prove that any
given number would be wondrous without testing it. It was used as a soft
introduction to the halting problem.

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johnrob
1 is the only number for which 3n + 1 = 4n. So a number that didn’t reduce to
1 would have to get stuck in a loop of multiple repeated values. That seems
unlikely.

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thetanil
any even number can be divided by 2 until it reaches 1. any odd number ends in
1,3,5,7,9. Any of those numbers multiplied by 3 and added to 1 ends in an even
number. I don't get why this is hard?

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bhaak
> any even number can be divided by 2 until it reaches 1.

6 wants to have a word with you.

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thetanil
thank you

