1. Notice as I mentioned, in the weak Collatz conjecture, 2^(ak+1) > 3^k. Thus, if we keep repeating this cycle, the ratio (3^k)/(2^(ak+1)) keeps decreasing. Could we use this fact to somehow prove that the cycle also would have to be decreasing (or close), leaving the only option n = 1?

2. Would it be possible to find the maximum of (3^(k-1) + 3^(k-2)*2^(a2) + … + 2^(ak))/(2^(ak+1) – 3^(k)) using calculus or advanced analysis? This is beyond my knowledge, but if we could find this value, and show that every value of n below it converges to 1, we will have proven that the only possible value of n can be 1.

Chris Smith

]]>And you say “with three moves”, …Yes!! It might be the same thing to you what I see in there. but I don’t think the conjecture is a probability game, though!

Your point of veiw and the data is very exact, Thank you very much for sharing your study with us!

]]>If it is to me, thanks! if not, sorry I’m embarassing..

Although I’ve found these ideas in my own way, however I found a paper by Prof. Hellekalek just yesterday. and found in there that my notion for the conjecture is almost the same to his notion. I don’t understand the paper completely because I’m an amateur, but I would like to support his study.

Here is the paper by Professor Perter Hellekalek.

https://arxiv.org/pdf/1605.02634.pdf

According to his paper, these above inverse mapping and shortcut fuction are not new. such function is called “the accelerated Collatz function” T.

I’m just an amature, so my result for the conjecture is more specific…well, what I would like to say is, there is more straightfoward formulea with numbers.

Anyway, thanks for your comment! (..if it is to me)

]]>I am the ultimate amateur. It may be heresy: I don’t see this as a difficult problem. That doesn’t mean a formal proof would be easy or that there is not a lot of interesting work to be done, like yours, but from my POV I feel confident the game has only one solution.

]]>I can grasp a part of it but not completely a whole thing.

I’m tackling the conjecture with another way of yours, although each of them seem to describe something like the same thing, I think.

Forgive me my lack of understanding. lack of explanation, English and my rudeness.

Thanks for sharing your ideas!

]]>I’ve done the generalisation of the convergence of the sets and the representration of ID for all number. So there is no randomness.

*but it’s very elementary way, so mathematicians will laugh at it when they look at it.

Those are not enough for solving the conjecture, I know. But the result of the structure is simple and interesting.

Our approaches seem to be different, but very similar I think.

so good luck, looking forward to read your paper in near future :)

If so, It would of course generates every odd number. But there is the certain rules about the indices of 2. I mean, because the generator of every odd number is the Collatz odd (multivalued) inverse mapping.

That’s why I use the shortcut function ” f^1+j= 3x+1 /2^j ” for x≔odd≢0(mod.3). I don’t deal with 3x+1 and /2^j separately.

And even is just the (multivalued) inverse mapping of odd. So even can be ignored.

Your findings:

1. If n+1 is divisible by 4, add 0.5n to n.

2. If n-1 is divisible by 8, subtract 0.25n from n.

3. If neither case applies, subtract 0.75n from n.

Actually I still could not understand cleary(..sorry that’s why I am an amatrue!) But I think it might be related the rules of the indices of 2 what I explained above.

And the mod. 4 list of 27–111, there is {27,41,31,47,71…}≡{-1,+1,-1,-1,-1…(mod.4)} Does it have any regularity of the ordering of -,+,-.. ?

Personally, I think this randomness IS undefinable.

My mind image of the 3x+1 structure, there is actually the symmetrical or fractal hierarchy (which represents the depths of the subsets of odd), but it’s not so important. There is just the permutation of the terms of odd number.

I have only four formulea which represent about the sets: Two parent sets(sueprset) and Two child sets(which converge thier parent sets). But the member of the parent sets is also being a member of the child sets… I can not explain cleary anymore.

Anyway, that sets can show almost all x≻1 is converge the number which less than x.

So, the final problem of the conjecture will be able to done If all such number; x<f^1+j(x) can be mapped to x'≻f^1+j(x'). It is just my personal opinion. ..And it can.

Our approachs are different, but I hope they will reach the same conclusion :) …I can not write the paper, though..

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