The Proof of Collatz Conjecture - For Dummies

[Nobody Knows]

https://www.youtube.com/watch?v=J8kHjNPsFJY

[Nobody Knows]

2024-02-06_180619.png (426.26 KiB) Viewed 2575 times

https://youtu.be/J8kHjNPsFJY

[Guest]

I'll post it here since youtube (or you) deleted my comment:
This "proof" is even less convincing than the last one. Now you even fall into the probability argument 53:30 and your conclusion at 1:18:00 is even more laughable: "ALL number reaches 1 since probability says so".

[Nobody Knows]

Sure ... and just by an accident I was able to predict, using my method, the number of steps (odd numbers in a sequence) needed to reach 1 starting from 3^100000 - 1 with accuracy 0.07%. This video is just "for dummies" you are too smart and shouldn't bother yourself with it any more. It wasn't me who removed your comment, probably even youtube knows that your comment was not appropriate.

[Guest]

My comment is the one made above (even without the "laughable" term). Your accuracy is also laughable. Take 2^1000 and tell me what is your prediction for that.

[Nobody Knows]

There are always people who have sooo much to say .... Anyway thank you for you valuable insight.

[Guest]

no response? I guess you saw what was coming next...Anyway, you know what you need to prove, and still, you didn't even tried

[Nobody Knows]

Well, thank you immensely for your incredibly enlightening input.

[Guest]

You are welcome. Let me share this with the "community":
1) Your prediction on the number of odd iteration for 2^4150000000 would yield 10000000000 iteration predicted for an actual iteration of 0 ! As you know, there are infinitely large numbers with iteration 1, same for iteration 2,.... so YOU need to understand what your prediction means.
2) You say a lot of bulsh*t. e.g. 1:17:40 you say we can only influence a limited number of initial iterations: Totally untrue. You can set in a very predictable way any (yes any) initial number of iterations like it pleases you.
e.g. you want to have 10 increasing steps (of f(n)=(3n+1)/2) followed by 2 decreasing steps followed by 5 increasing steps? Easy: You represent that into a binary path 11111111110011111 which is 130975 in binary, and you use this little PARI/GP function bellow, findx(130975) which gives you n=54271 your starting number.

findx(v)=if(v==0,return(0));l=valuation(v,2)+1;v=(v/2^(l-1)-1)/2;c=l%2;N=(5^c*2^l-1)/3;S=l;while(v>0,l=valuation(v,2)+1;v=(v/2^(l-1)-1)/2;c=l%2;if(N%6==5^c,S=S+l;N=(N*2^l-1)/3,N=N+2*2^S;if(N%6==5^c,S=S+l;N=(N*2^l-1)/3,N=N+2*2^S;S=S+l;N=(N*2^l-1)/3)));return(N%2^S)

Now, your highest bits have a slope of 3. Your lowest bits generally start with a slope less than 3 (sometimes 2 for a number of steps like you showed) so you see the binary representation growing, then have a slope greater then 3 which means the binary representation is shrinking. That's only another representation of the Collatz rollercoaster (the number grow, then decrease, then grow, ...)
3) You didn't prove there is no cycles: You begin with a slope less then 3, then the slope tendency is to be larger then 3, and at some point one of your row contains the exact same binary representation as your last row or starting number (just shifted to the right). Then it indefinitely grow/shrink and your last row repeats every k rows (k=length of cycle).

4) You didn't prove there is no divergent trajectories: Mean slope staying smaller then 3 for eternity.

Your presentation is fine, but it does not prove Collatz at all. It just displays what happens on numbers reaching 1 (because no one knows any counter exemple), and yes you can do stats and averages on that, but that's just that....stats.

[Guest]

Mine was not deleted but simply shadow banned. This is certainly not Youtube. That's not pretty man....

https://imgur.com/RybXsvm

[Nobody Knows]

I told you already that this video is just "for dummies" I see now that you are definitely NOT .... in this category. I have no problem to address any questions on the subject, but I have a bit of problem to reply to aggressive and highly frustrated people (maybe because of some personal problems) quote "You say a lot of bulsh*t". I have to say it is a bit annoying. Anyways....

I would expect that you are able (at least) to listen carefully and understand (which I see is not the case here). So starting from your first point.
1. at 1:17:50 of the video it is stated "we can influence this process in two different ways either to shorten this number immediately to one ..." at your number 2^4150000000 is exactly this case. I can predict how many of odd numbers will be starting from a number like 2^4150000000-1 if you are interested ? This will be 19998192982 (odd numbers in the sequence) +/- 0.0000001%, you can check it.

2. Quote: "You say a lot of bulsh*t. e.g. 1:17:40 you say we can only influence a limited number of initial iterations: Totally untrue. You can set in a very predictable way any (yes any) initial number of iterations like it pleases you."

Yes you can set "any" length of the initial number and even set it in predictable way, which only support my statement that by doing this you will influence only LIMITED number of initial iterations. "Any length" of initial number is still "limited length". The same way as you can have any length of the sequence (if the number is big enough). You can have two opposite scenarios you can influence the number setting it in the way that it will collapse immediately to 1 or to make it the "worst case scenario" using only 1's in binary notation then it will be extended for some (limited) number of iterations. Everything between these two scenarios is kind of more or less chaotic setup. But by setting up limited number of digits you are only able to influence limited number of initial iterations!

Your example is not correct (not as you described it) there are not "10 increasing steps followed by 2 decreasing steps followed by 5 increasing steps" it is actually 10 incr,1 decr, 6 incr steps. As you can see it is not as easy as you think. Anyways still you were able ONLY influence the limited number of initial iterations then CHAOS WON !

If you think that it is so easy to control give me an example to have 5 inc, 7 dec (slope 4) , 3 inc, 8 dec (slope 4), then 10 dec (slope 16=2^4) and 3 inc. You've said it is easy ...

Your example on the picture.

2024-02-07_194319.png (111.69 KiB) Viewed 2497 times

3. It was not the goal of this video to prove anything, only explain a main idea of my paper which contains the proof. It was presented that positive bits have a slope of 3, the smallest negative bits you can control only for a while and then they tend to have a slope of 4, so contraction of the length is a must.

4. To achieve this, your initial number has to be infinitely long; otherwise, CHAOS will WIN!

[Guest]

Check again, 54271 as 10 increasing steps, 2 decreasing and 5 increasing (of course for the last part, it is "at least 5 increasing" since the next steps are free to grow or shrink).
You want an example of 5 inc, 7 dec, 3 inc, 8 dec, (no inc here? I will put 2 inc), 10 dec, 3 inc? I gave you a simple PARI GP code to do so you didn't try? it gives 241249015647 (with the 2 inc I added).
1:18:48 "we showed that we can construct. ... for any number" then "we can conclude that the conjecture is true", so when you say that the goal of your video was not to prove anything...

You didn't fully responded to my points:
1) It was an example, and I added that for any (small or large) number of iterations, there are an infinity of numbers with any number of bits so your prediction only work for an insignificant number of numbers. how many numbers have 3 iterations? an infinity having as many bit length as you want. How many have 50? same.
2) Yes you can control the exact number of iterations on any length you want, and my function will give you the smallest n (there are infinitely many n starting with those iterations). You are right, you can't predict what will come next down to 1 (it can be abrupt or very long and chaotic). But again, hiding behind chaos is not a proof.
3) What you say does not prohibit cycles as I described it
4) Again, you talk about chaos and use it as a predictable (chaos??predictable??) way to show your number will reach 1

Concerning your paper and first video, it was already discussed in another thread

[Nobody Knows]

You have your example 54271 on the picture check on your own. I can accept 6 instead of 5, but it was 1 decreasing instead of 2.

In this new example you are again wrong, check the picture below.

2024-02-07_212040.png (109.22 KiB) Viewed 2494 times

I said at the beginning of the video that my proof is in the paper, in the video it is just a presentation of the idea of the proof. It is not a strict mathematical proof for sure.

When I will have time I will reply to your questions, unfortunately I have also some other things to do. (I will do my best to address all substantive questions.)
4. If you have chaotic behavior controlled by external limits or rules, that is what i meant. Boiling water in a jar, chaos but in controlled environment and you can be sure it will evaporate all. Is it clear ?
Have a look at 1:16:32 this straight line is a jar ... but there is a chaos inside and you can see it when you zoom.

[Nobody Knows]

I see now that we use different terminology ... your decreasing means something else then mine ... i use term like "factor of increase" which is 2 (only one increasing the length of the number) then 4,8,16 are decreasing for me.

[Guest]

so you can't have more than 1 decrease ?

[Guest]

increase in the "Collatz sense" meaning f(n)=(3n+1)/2 yield another odd, and decrease is f(n)=n/2
54271, 81407, 122111, 183167, 274751, 412127, 618191, 927287, 1390931, 2086397, 3129596, 1564798, 782399, 1173599, 1760399, 2640599, 3960899, 5941349

[Nobody Knows]

My process is going only through "odd" numbers. There are no deviations by two (it is nothing more then a shift to the left). For me what is important is an increase in length of a binary number. You have an increase when the next odd number in a sequence is longer (in binary notation) then the previous odd number. When you have a sequence of 1's like 111111 you will have an increase of the length of the binary number in first 6 odd numbers in the sequence. Positive bit at the top increases by a factor of 3 always! When the smallest negative bit increases by a factor of 2 then the length of the binary number increases as well. When the smallest bit increases by a factor of 4 or 8 or 16 or more ... then the length of the binary number decreases.

[Nobody Knows]

2) Yes you can control the exact number of iterations on any length you want, and my function will give you the smallest n (there are infinitely many n starting with those iterations). You are right, you can't predict what will come next down to 1 (it can be abrupt or very long and chaotic). But again, hiding behind chaos is not a proof.

I said in my video that it can be shorter and it is fine but it can not be prolonged too much in a sense that chaotic changes can slightly extend the number or you can setup starting number in a way that it will initially increase but you can influence ONLY a limited number of initial iterations. Then chaos as described in the video controls the process. It is nicely presented starting from 1:16:30.

[Guest]

Yes, you work only with odd numbers, but controlling the number of decrease (n/2) is also important for your initial slope. In your last XLS, you can see that a decrease of 7 after the increase of 5 have some effect (long and almost horizontal slope before the next increase).

[Guest]

This is math's comedy hour or whatever. It's quite hilarious.

This guy, 'Nobody Knows', and his guest who has posted more nonsense here, are probably the same person. What a waste of time!