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I noticed that there is a linear increase of +1 in the stopping times of a sequence of numbers with the seeds being the sum of the previous number in the sequence added to itself.

I tried this on Excel. The hailstone sequences converge to 1, as expected.

for a sequence starting with 3

I also tried it with different numbers as the starting point (ex. 5 10 20 40...)

Does this have any significance?

Rodrigo de Azevedo
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PassylA
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    That’s just the fact that the stopping time of $2n$ is one plus the stopping time of $n$, no? – Aphelli Mar 09 '22 at 08:52
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    Please use embedded ASCII instead of images. – Rodrigo de Azevedo Mar 09 '22 at 09:08
  • Do you mean that every new record (in the number of steps) is only one better than the current record ? – Peter Mar 09 '22 at 10:02
  • I'm not exactly sure what I mean either. hehe – PassylA Mar 09 '22 at 10:06
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    The depth in which the Collatz conjecture has been studied is unimaginable. Small start values exhibit some notable patterns which have almost surely be detected by someone which does not mean that this observation was also published. Unfortunately, all patterns eventually vanish because of the dynamic of the sequence. A proof of this conjecture in , say , the next $20$ years would be more than sensational. – Peter Mar 09 '22 at 10:08

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That is a very trivial observation. If the stopping time of a number $x$ is $n$, then the stopping time for the double of $x$: $2x$ is $n + 1$ because it takes one step for $2x$ to drop down to $x$ by division of 2.

machine_1
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  • hmmm does this mean that since all natural numbers follow this observation, that the collatz conjecture is true? – PassylA Mar 10 '22 at 01:48
  • or that you only need to prove that the conjecture is true for odd and prime numbers, to prove that the conjecture is true for all natural numbers? – PassylA Mar 10 '22 at 02:44
  • If you prove the conjecture true for odd numbers, then it's true for all numbers – machine_1 Mar 10 '22 at 09:03
  • @PassylA it means that if you first prove that all odd numbers $x$ terminate, you have proven that all even numbers of the form $2^nx$ also terminate. But to prove the conjecture, you must first prove every odd number terminates. – it's a hire car baby Mar 11 '22 at 20:44