**Idea 1**

Every (Collatz-type conjecture) contains an input-output loop for (where is odd). The twist is that this includes .

This is merely a factual statement because:

If then and .

It is only trivial for . Why? Because is the starting point and the endpoint of the loop.

**Idea 2**

The even numbers can be eliminated. Let me call this the function.

Since the next even term following is half, we know that it’s also latex $1$ minus the product of the previous odd and :

For example:

and , and , and

Each smaller is one-quarter of the previous one:

For the odds:

For the evens:

**Idea 3**

All odd-numbered sequences can be reduced to three operations of multiplication:

**a.** If is divisible by , multiply by , subtract

**b.** if is divisible by , multiply by , add

**c.** Else multiply by

We can call this the a, b, c congruence. It’s important for a number of reasons, including the fact that they are self-switching according the follow rules:

–For **a**, has a minimum exponent of for multiplication by . With each product of , the exponent decreases by latex$2$.

–For **b**, has a minimum prime factor exponent of $latex2^4$ for multiplication by . With each product of , the exponent decreases by .

–For **c**, which for always has prime factors , is a class that contains the odd terminating numbers, say for : **53**, 13, 3, **5**, **1**.

**Idea 4**

All complex loops are determined by the congruence of two successive numbers equaling .

For example, take sequence and :

**31→5**, 5→13, 13→25, 25→43, 43→35, 35→29, 29→49, 49→79, **79→17**, 17→31

There are two looping pairs: , and , because:

(mod )

(mod )

The amusing twist is that again this must be true for $latex $x = 1$. But unless the modulus is $latex $\geq 3$ this congruence-based looping cannot occur.

**Idea 5**

The regression to occurs because of exponents of , not powers of .

As we know, powers of terminate conventional sequences. For example, to take an extreme case, the entire sequence of is:

262144, 131072, 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

This becomes, in the sequence:

87381, 21845, 5461, 1365, 341, 85, 21, 5, 1

It’s the same pattern described above of division by , quartering in every step: .

There are no common odd numbers save one, .

(Excuse the long post.)

]]>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 :)