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Message 5496 - Posted: 12 Jan 2010, 17:31:05 UTC - in response to Message 5494.

Hello. I'm a new user. I came from Poland and my english is weak, so probably I'll not understand every things what you'll try to tell me and you'll probably not understand every things what I'll try to tell you :) Whatever...

On this page:

http://boinc.thesonntags.com/collatz/highest_steps.php

I was found "Collatz Best Results". I understand if the starting value n = 2,362,032,211,856,451,015,323 is chosen, the sequence takes 2514 steps (before descending to 1).

I have a question. Do you know the number who total stopping time is longer than 2514 steps? If you don't know and there are realy Collatz Best Results may I have something what can interest you...


Because I was found a number who total stopping time is 59071 (but it's a big number).

If this interests you and you would know that number - write.

The problem is not finding numbers with with a lot of steps. The problem is checking all numbers to see if the sequence converges all the time to 1. And to do that, one starts at the bottom.

Just coming up with numbers having an arbitrary step count is easy. I can tell you a number with exactly 59072 steps. It's 2^59072. And 2^59072 + 1 has even more steps. So that isn't the question of this project.

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Message 5591 - Posted: 16 Jan 2010, 11:46:45 UTC - in response to Message 5587.

Just coming up with numbers having an arbitrary step count is easy. I can tell you a number with exactly 59072 steps. It's 2^59072. And 2^59072 + 1 has even more steps. So that isn't the question of this project.

Right, but you know that your example (2^59072) is trivial and we don't know do 2^59072 + 1 converges to 1?

Of course this was a trivial example. But the numbers this project uses are also unchecked so far and we don't know if they converge.

Whatever... I understand that this project try to disprove Collatz conjecture and you try to find a number who aren't converges to 1 (you try to find a cycle). So I have another question.

The problem is checking all numbers to see if the sequence converges all the time to 1.

Do you really checking all numbers?

Yes.

I'm asking because:

First: we know that numbers 2^n converges to 1, so we don't need checking it.

What do you think, how many 2^n numbers do we have in an average WU?
The answer is quite close to zero. Only every some billionth WU has a 2^n number in it. Hardly worth the effort excluding it.

Second: there's a quite simple prove that if starting number is 3n we don't need checking that number too. There's no "3n cycle".

Could it diverge? In principle there are 3 possibilities for a number, convergence, cycle, or divergence.

Indeed, to check only for convergence could be sped up. But we would loose the possibility to check for delay records. For the latter we are using a brute force approach (that is only really feasable with GPUs) and optimized the hell out of it. So it is really fast in what it is doing. Looking at some special cases would slow it down, even just the check if it is a 3n number or not).

If the project runs without issues and Slicker has nothing else to do, we could look into developing a second application for this project in the future, maybe doing other things like searching path records (highest number that occurs in the sequence) or glide records (amount of steps necessary for the numbers in the sequence to get smaller than the starting number). As said in another thread, there are quite a few different metrics for the Collatz problem one can play with.


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