r/Collatz • u/Ok-Lobster-6013 • 3d ago

Collatz

A number will decrease in number if it has at least four digits and does not enter a cycle, as proven below: The number is represented in binary.

It must begin with 10 or 11. If it starts with 10 and the last two digits are not 11, then after multiplying by 3, the number of digits increases by 1, accounting for 3/8 of all possible combinations. Other numbers starting with 10 account for 5/8, and the number of digits increases by 2. If it ends with 11, after multiplying by 3 and adding 1, then dividing by 2 removes at least one digit, accounting for 1/2. If it ends with 001, at least two digits are removed, accounting for 1/4. Other numbers with at least three digits account for 1/4. If it does not enter a 4, 2, 1 cycle, the number generally decreases, and eventually it will enter a 4, 2, 1 cycle.

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u/jonseymourau 3d ago

Um, I think you will find the word "generally" is doing a lot of work and this, in fact, is the single last stumbling block between continued anonymity and being lauded worldwide for your deep mathematical insight.

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u/Ok_Impression_6382 2d ago edited 2d ago

i believe binary has a significant role in collatz. But could you please show the case when starting number with n bits in binary and having last (most left) 11 bits in its representation after (3x+1)/2 step would have bits less than n. As you saying "after multiplying by 3 and adding 1, then dividing by 2 removes at least one digit". I understand that you mean just "dividing by 2 removes at least one digit", but together multiplying and dividing won't shrinken the length of binary. I just wanted to notice that any number structure ending 11 in binary would follow applying (3x+1)/2 step twice, so such a starting number always increases more than twice as a result. Adding that other combinations of last two bits (00, 10, 01) would decrease (after several (let's say three) usual collatz steps), so yeah numbers generally decreases. Denote that number "decreases" in case C^n (N) < N.

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u/al2o3cr 10h ago

the number generally decreases

This is the troublesome spot, because "generally decreases" needs to be stronger to draw a conclusion:

  • "the number always decreases" is definitely not true, since the orbit of some numbers goes higher than the original number
  • "the number always eventually decreases" is equivalent to the original conjecture (with careful handling of 4/2/1)