https://hubertschaetzel.wixsite.com/website Terras sheet

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I suspect if such a set existed, the minimum set would be pretty trivial and not unique. Yet, a brute force search in 40 bit space (each Ki checked for up to 10 bit for 2, 3, 4 terms) brought no results.

]]>If we consider as a single step multiplying the previous number by 3, adding 1, and then discarding all least significant zeros, such a step will (on logarithmic average) multiply the number by 3/4.

On average, each step discards two zeros, which is equivalent to dividing by 4. (one zero is discarded with probability 1/2, two zeros with probability 1/4, 3 zeros with probability 1/8, and such series 1/2+2/4+3/8+4/16… sums to 2 zeros on average).

Thus, any long enough sequence, even though it may wander up and down, would inevitably come below its starting number.

Note that all starting numbers expressible as 4K+1 reach a smaller number after a single step. Only the starting numbers expressible as 4K+3 will produce a non-trivial sequence before reaching a smaller number.

I address (im)possibility of loops in my previous comment.

]]>To avoid confusion:

Of course, the canonical formula for Collatz (from 2Ki+1) would be:

(6Ki+4)/((2Ki+1)*2^Ni)

but we can just drop that extra 2, so it becomes 3Ki+2 in the numerator.

]]>Mr Li Jiang,

What are your thoughts on my comment below?

Regards,

Alex

If it’s possible to prove that the product:

П((3Ki+2)/(2Ki+1))

can never be a power of 2 (means its factors can never fully cancel each other, leaving only 2s from the numerator), then this will prove that a Collatz sequence can never loop.

Context:

A Collatz sequence of odd numbers 2Ki+1 is equivalent to series of multiplications by:

(3Ki+2)/((2Ki+1)*2^Ni),

where division by 2^Ni means discarding all least significant zeros from the intermediate result.

If the sequence loops to the initial number, this means the product of such terms becomes equal exactly to 1, or the product of terms without 2^Ni in the denominator becomes equal to exact power of 2.

It seems that, without even considering Ki belonging to a Collatz sequence, it’s not possible to find such a set that the product above will become a power of 2.

The only such trivial value is K=0, which makes the term equal to 2, which corresponds to the only known loop starting from 1.

]]>Aditionally, in the following sentence appears theta which is not defined before.

Thank you for your cool blog.

Fabian

*[Corrected, thanks – T.]*

For the collatz conjecture, we define an iterative formula of odd integers

according to the basic theorem of arithmetic, and give the concept of iterative exponent. On this basis, a continuous iterative general formula for odd numbers is derived. With the formula, the equation of cyclic iteration is deduced and get the result of the equation without a positive integer solution except 1. On the other hand, the general formula can be converted to linear indefinite equation. The solution process of this equation reveals that odd numbers are impossible to tend to infinity through iterative operations. Extending the result to even numbers, it can be determined that all positive integers can return 1 by a limited number iterations.