Collatz conjecture Puns

I took Real Analysis this semester so I had series and sequences on my mind. I had the random idea to look at the sum of the reciprocals for the Hailstone sequence of some numbers. So, taking 5 and applying the Collatz function iteratively you would get,

[5, 16, 8, 4, 2, 1]

Wherein I would stop. Then I would look at,

1/5 + 1/16 + 1/8 + 1/4 + 1/2 + 1/1 = 2.1375

I plotted this number with respect to its initial value (I started calling these seeds just to help me out).

Here are some of my findings.

Fig 1. Hailstone Sequences, Length of Sequence, Sum of Reciprocals

This first image was to get my head around everything. The first graph is of the hailstone sequences for the first 1000 integers. You can see how all the lines eventually fall down to 1.

The second graph plots the length of the sequences, from the starting seed to 1, against the seed's value. It seems to grow sort of logarithmically which is neat.

The third plot is my work. The sum of reciprocals for the sequences. The first thing to notice is that strange band at the bottom, followed by a gap, followed by a cluster of points. Very interesting. I'll comment more on the bottom band later. Second thing of note is how slowly it increases. This makes sense because the reciprocal of large numbers are small, so we have to go out very far to see any noticeable change. We know the harmonic series diverges and I am just taking different subsequences of the harmonic sequence. If there was some upper bound on these sums of sequences that could say something interesting. But I don't know.

Fig 2. Sum of Reciprocals up to 10,000

Here is a graph of just the sums of reciprocals but up to 10,000 integers. The interesting thing here is that you can see more pronounced bands forming. The lowest band around 2.0, a thin band about 2.3-2.4, a dense band around around 2.5, a thick band from 2.1 to 3.0, and then an upper band about 3.2. The bands themselves are interesting but what is even more interesting to me is the gaps between the bands.

Fig 3. Sum of Reciprocals up to 10E5

This figure is just to

continued from here

ty tnf for the run+assist

Calculator

next get is 664 (664 | 0) schedule

continued from here ty everyone whos counted the past month and /u/TheNitromeFan for the assist

The rule is that if you have an odd number, multiply 3 and add 1, and if it's even, divide by 2, until you hit 1.

The format is

[current number] ([starting number] | [steps])

Collatz Calculator

The next get is at 632 (632 - 0). Schedule

lol

Youre probably tired of hearing it, but I really think I solved it and I'm still in elementary school (I'm 8 years old). I tried around with it for 2 days and I finally have the proof. Where can I find someone to endorse me, so I can upload my paper to the arXiv?

It's simple, really.

We have:

-5=>-14=>-7=>-20=>-10=>-5

So we have a cycle starting from -5. QED.

Has anyone tried testing this yet? Just wondering.

Yesterday, the ongoing distributed computing project for Convergence verification of the Collatz problem by David Barina (u/lord_dabler) verified the convergence of all numbers below 2^(69). This is the same project that did it till 2^(68) last year.

The verification to numbers from 2^(68) to 2^(69) which is 2^(68) numbers took 19 months of computation which means more that 5.85 trillion (2^(42.4)) numbers per second.

LINKS:
Track the progress: collatz-problem.org

Path records: collatz-problem.org/table

Source code: github.com/xbarin02/collatz

Original paper: rdcu.be/b5nn1

En Español:

Uno de los problemas matemáticos tratados como insoluble, lo he resuelto con un método considerado básico por la comunidad, obviamente, no mostró el resultado sin tener testigos de que la solución la encontré yo.

Lo único que voy a revelar es que la hipótesis de Terry Tao es correcta, los posibles resultados de estos son ciclos positivos o negativos.

In English:

One of the mathematical problems treated as insoluble, I have solved it with a method considered basic by the community, obviously, it did not show the result without having witnesses that the solution was found by me.

The only thing I'm going to reveal is that Terry Tao's hypothesis is correct, the possible outcomes of these are positive or negative cycles.

In my proof of the Collatz conjecture, numbers of the format 36n+35, single dividers, appear at the end of the proof, as well as 36n+17, multiple dividers. See my proof for details.

I looked at the generated numbers, hoping to find some relationship among them, and when I tried the 3N+2 factor, there was a connection.

The factor 3N+2 appeared in the proof earlier.

1. The proof starts with single dividers 4n+3; after a Collatz transform (and removal of generated multiple dividers), the resulting single dividers are of the format 12n+11. The factor 3N+2 converts the starting 4n+3 group into the resulting, 12n+11 group.
2. N=4n+3 -> 3*(4n+3)+2 = 12n+11.
3. Likewise, upon a second application of the Collatz transform, 12n+11 turns into 36n+35.
4. N=12n+11 -> 3*(12n+11) + 2 = 36n+35.
5. The proof also generates multiple dividers of the format 36n+17.

Both 36n+35 and 36n+17 are also connected through the 3N+2 factor. This 3N+2 factor converts a 36n+35 number into another number of the same format.

N=36n+35 -> 3*(36n+35)+2 = 36*3n+107 = 36*3n + 36*2 + 35 = 36*(3n+2) + 35

and 36n+17 number into another 36n+17 number:

N=36n+17 -> 3*(36n+17) + 2 = 36*3n + 53 = 36*3n + 36 + 17 = 36*(3n+1) + 17

The factor 3N+2 offers some insight into odd numbers as well .

It converts a single divider into another single divider. The same goes for multiple dividers.

N=4n+3 -> 3*(4n+3) + 2 = 12n + 11 = 12n + 8 + 3 = 4(3n+2) + 3

N=4n+1 -> 3*(4n+1) + 2 = 12n + 5 = 12n + 4 + 1 = 4(3n+1) + 1

The odd numbers related through 3N+2 may have some other properties. If you find anything, post your results.

7->23; 5->17; 47->143...

In the proof, 36n+35 numbers turn into 36n+17 numbers after a number t of Collatz transforms. What is this number? It can be calculated using a parametric equation.

Let us consider a number 36n+35, a single divider. It turns into a 36n+17 number, a multiple divider, after t Collatz transforms.

A general equation is 3^t * n + (3^t - 2^(t-1)) = k*2^t where t,k,n are positive integers.

The first, lowest n = 2^(t-1) - 1, the step = 2^t, which means the next higher n1=n+step = n + 2^t, etc.

The n, in general, equals n = 2^(t-1) - 1 + k*2^t , or the lowest n=2^(t-1) - 1 plus a multiple k of step 2^t.

So, the 36n+35 numbers which go through t Collatz transforms to turn into a multiple divider have a step 2^t and its multiples (for n in the 36*n portion of 36n+35) between themselves, and the l

Warning: Zero practical applications ahead!

In a post I made a little over a month ago, I talked about different types of content that this sub could benefit from, one of which being art with the Collatz conjecture being used as its basis.

Here is a small art project, all generated programmatically and presented in Desmos. The rules are as follows:

1. Start with some integer and begin performing the collatz conjecture on it.

2. Draw a semicircle from the starting integer to the next. If the sequence is increasing, render it as red and above the x axis. If it is decreasing, render as blue and below the x axis.

3. Continue the sequence until you've reached the 4-2-1-4 loop.

Here is a small example using 5 as the starting point.

Quickly, these patterns get increasingly complex, here is 31 as a starting point.

Here is a more detailed view of the semicircles near the origin.

Here are the first 100 numbers and their semicircles. We get nice regular patterns this way. I'm curious to know what the lines tangent to both sets of semicircles are, if that piques your curiosity please let me know what they are as I'm curious too but far to busy to dedicate any more time to this tiny project.

If anyone is interested in playing with this, throw this into a .html file, change the number at the designated spot, and open in your browser to see what your own sequence looks like. This was thrown together fairly quickly and it was my first time working with the Desmos API, so the code may not be optimal.

Enjoy!

Hey guys, so I found this pretty recent proof that claims it proves the Collatz conjecture (also known as the 3n+1 problem), I took a look at it yesterday and as crazy as it sounds I couldn't find any flaw in it ( granted I was half asleep, and I'm not exactly Terrence Tao) So could you guys check it out, and point out where the mistake was made?

https://arxiv.org/pdf/2101.06107.pdf

Thanks in advance.

This is gonna be a real short post, since the user hasn't provided any math (good or bad) to debunk. Instead, he offers $10k to the person who can somehow read his mind and reconstruct his argument. But seeing as he doesn't even present his conclusion for consideration, it reeks of bad math to me.

If you want to take a crack at it, you can find his challenge here: https://www.reddit.com/r/Collatz/comments/q2fqzw/10k_reward_for_explaining_my_logic_here/?utm_medium=android_app&utm_source=share

Some suggestions:

• Take the collatz process and tweak it a little, see how the resulting system differs.
• Select a measurement, such as the average branch length over many many iterations and share your findings with us.
• Post visualizations of your findings or use the collatz conjecture as a basis for a piece of art.
• Hell, use the collatz conjecture in a fun, completely recreational way that might be unexpected or give rise to interesting results even if they are ultimately meaningless.

I am all for mathematical discovery, but walls of numbers with no context and declarations that the conjecture is solved based on the most basic unfounded statements make this sub a drag to browse.

I love the collatz conjecture, but to me it becomes a whole lot more interesting when I realize that, on my own, I’m not going to solve it… and that can make toying with it a lot more fun.

Edit: Oh yeah, and as a challenge, try to make your results as interpretable to others as possible. I understand not knowing the mathematical terms and I’m not the most well versed person myself, but clarity in an explanation can make up for that.

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a tier

> If the number is odd, ×3 +1
> If the number is even, ×0.5 > > Whenever a sequence reaches 1, set the beginning integer for the next sequence on +1:
> 5 (5 | 0)
> 16 (5 | 1)
> 8 (5 | 2)
> 4 (5 | 3)
> 2 (5 | 4)
> 1 (5 | 5)
> 6 (6 | 0)
> 3 (6 | 1) > > Formatting will be: x (y | z)
> x = current number
> y = beginning of current sequence
> z = number of steps since the beginning of sequence > > Collatz Calculator

the next get (schedule) is at 599 (599 | 0)

This has been answered and what was written is either wrong or trivial. Thank You.

Hello,

I am merely able to consider myself a beginner at math and I have come across the Collatz conjecture.

After not having thought about it too much, trying to learn more about it and failing to understand the math used to proof this problem I decided to post my approach to the problem.

I hope you can give me some advice or point out mistakes. I am in no way trying to prove anything and am only posting casual notes of mine.

The Collatz conjecture basically is about whether or not every positive integer results in 1 if used in the following sequence:

Number: n

If Number is odd: 3n + 1 → always = even

If Number is even: n/2

So, what I thought about was to try and prove that there is no positive integer for which the sequence does not eventually return 1.

If there is a number that does not return 1 if applied to the sequence it would have its own sequence.

For that to happen we would have to find a number n for which the sequence eventually returns n.

As a result, I came up with this equation:

• (3n+1) /2 = n

We know that the sequence with the number n has to have an odd number, because if not it would just be divided until it reaches an odd number or 1. We also know that every odd number times 3 returns an odd number, add 1 and it becomes even. This means that there has to be a sequence with at least two numbers.

Since 3n /2 is always greater than n I basically now know that there is no sequence with two positive integers that does not eventually reach 1.

But since there might be a sequence of more than two numbers for which the result does not eventually reach 1, I can not be sure whether or not there exists one.

I sadly am not skilled enough to go further into it as I neither know how to express the continuous sequence mathematically nor do I know how to use this approach with a continuous sequence. I appreciate everyone pointing out any mistakes I made or linking posts that have explained this approach.

Thanks for your patience and thank you for reading.

I wish everyone a merry Christmas and hope you all enjoy the holiday.

Side note: I thought of (3n+1)/2 = n being the same as 1,5n+0.5 = n which would not make any sense at all as it results in different solutions and seems to have no connection with the Collatz conjecture anymore. I am interested in what I am getting wrong here.

I haven't received a single scheduled task on any of my machines in over a month. The project website makes no mention of this and I have tried syncing the project multiple times on different PCs.

I use GRC as an account manger for several other projects for years, I have never had a problem like this. Whenever I try and press the "update" button for the new projects it goes back to saying "communication deferred". Right now it is at the 18 hour mark for this which is probably when I last restarted the PC.

Anyone else able to test and confirm if they have been getting Collatz projects on their end or know what is going on with the project? It seemed poorly ran. But it would be bizarre if it went ghost like this without warning after over a decade of reliable runtime.

I stumbled across a video on the Collatz conjecture, which is described as 'an extraordinarily difficult problem, completely out of reach of present day mathematics'. From what I've read, the problem is still unsolved but many mathematicians consider the conjecture to be true.

To save people clicking the link, excerpt from Wikipedia of the conjecture is below;

>start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half of the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

What is the benefit to solving problems such as these?

How are problems such as these solved?

Continued from here. Thanks to /u/atomicimploder for the finish!

The next get is at 650 (650 | 0).

Calculator

Schedule

Below is an analytical proof of the Collatz conjecture. The conjecture is proven true for all positive integers.

I got new ideas while I was watching (for the 3rd time) the movie "Proof" where Gwyneth's character proves some theorem using new methods she had devised.

I realized I should devise a new method, at the same time preserving, for as long as possible, the only symmetry I could see; that is every other number is a single divider: when Collatz transform is applied to it, the resulting even number is divisible by 2 only once.

Collatz transform = take an odd number n. Calculate 3n+1, an even number. Divide the number by 2 one or more times until you get an odd number.

1. Let's consider a set of odd numbers 2n+1, n=0,1,2,3....

   1,3,5,7,9,11,13,15,17,19...

   We can subdivide it into 2 subsets:

   A. a subset of single dividers, or numbers divisible by 2 only once upon using the Collatz transform. Their format is 4n+3. Example:

   3,7,11,15,19,23,27,31,35,39,43... and

   B. a subset of multiple dividers, or numbers divisible by 2 two or more times, format 4n+1. Example:

   1,5,9,13,17,21,25,29,33,37,41,45...

2. 4n+1 numbers (multiple dividers) convert to 1 or 4n+3 numbers (single dividers) when a Collatz transform is applied (one or several times), so only 4n+3 numbers have to be proved.

3. The Collatz transform is applied to 4n+3 numbers only. This yields a mix of single and multiple dividers. Example:

   3, 7,11,15,19,23,27,31,35,39,43,47,51,55,59... after a Collatz transform turn into

   5,11,17,23,29,35,41,47,53,59,65,71,77,83...

   Multiple dividers are removed because we handled them in step 2. This yields the format 12n+11. Example:

   5, 11,17,23,29,35,41,47, 53, 59, 65, 71,77,83... after removing multiple dividers turn into

   11,23,35,47,59,71,83,95,107,119,131,143...

4. Another Collatz transform is applied. Example:

   11,23,35,47,59, 71, 83, 95,107,119,131,143... after a Collatz transform turn into

   (17),35,(53),71,(89),107,(125),143,(161),179,(197),215... Multiple dividers are enclosed in parentheses.

   The multiple dividers removed in step 4. are: 17,53,89,125,161,197,233,269,305,341,377,413,449,485,521,557,593,629,665,701,737,773,809,845,881,917,953,989,1025... Their format is 36n+17.

   All these numbers have the format 18n+17.

   Multiple dividers have the format 36n+17, or 4(9n+4)+1.

   Single dividers have the f

Continued from here. Thanks to /u/fourteensquares and /u/Cox_1920 for the finish!

The rule is that if you have an odd number, multiply 3 and add 1, and if it's even, divide by 2, until you hit 1.

The format is as

[current number] ([starting number] | [steps])

Collatz Calculator

The next get is at 619 (619 | 0). Schedule

I took Real Analysis this semester so I had series and sequences on my mind. I had the random idea to look at the sum of the reciprocals for the Hailstone sequence of some numbers. So, taking 5 and applying the Collatz function iteratively you would get,

[5, 16, 8, 4, 2, 1]

Wherein I would stop. Then I would look at,

1/5 + 1/16 + 1/8 + 1/4 + 1/2 + 1/1 = 2.1375

I plotted this number with respect to its initial value (I started calling these seeds just to help me out).

Here are some of my findings.

Fig 1. Hailstone Sequence, Length of Sequences, Sum of Reciprocals

This first image was to get my head around everything. The first graph is of the hailstone sequences for the first 1000 integers. You can see how all the lines eventually fall down to 1.

The second graph plots the length of the sequences, from the starting seed to 1, against the seed's value. It seems to grow sort of logarithmically which is neat.

The third plot is my work. The sum of reciprocals for the sequences. The first thing to notice is that strange band at the bottom, followed by a gap, followed by a cluster of points. Very interesting. I'll comment more on the bottom band later. Second thing of note is how slowly it increases. This makes sense because the reciprocal of large numbers are small, so we have to go out very far to see any noticeable change. We know the harmonic series diverges and I am just taking different subsequences of the harmonic sequence. If there was some upper bound on these sums of sequences that could say something interesting. But I don't know.

Fig 2. Sum of Reciprocals up to 10,000

Here is a graph of just the sums of reciprocals but up to 10,000 integers. The interesting thing here is that you can see more pronounced bands forming. The lowest band around 2.0, a thin band about 2.3-2.4, a dense band around around 2.5, a thick band from 2.1 to 3.0, and then an upper band about 3.2. The bands themselves are interesting but what is even more interesting to me is the gaps between the bands.

Fig 3. Sum of Reciprocals up to 10E5

This figure is just