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Probably many of you have heard of the infamous and unsolved 3x+1 problem. Many others have not, I'm sure.

Here it is, in my words:

*****

Take any integer greater than zero. Call this number x.

If it is even, halve it. In other words, x becomes x/2

If it is odd, x becomes 3x+1

The sequence, for all numbers tested so far, ends up in the endlessly repeating loop of 4,2,1,4,2,1,4,2,1...

While many numbers have been tested by computers, this doesn't prove or disprove the conjecture that the sequence will fall into this loop. Can you provide a counterexample (it would have to be BIG), or give a proof as to why it always iterates to 1, or a proof why it sometimes doesn't? Will you be the first?

a few obvious things:

* it makes sense that to enter the endless loop of 4,2,1, a number would have to become a power of 2, from where it would quickly divide down to 1

* if x is odd, than 3x is also odd and thus 3x+1 is even

* a number, if even, will divide by the biggest power of 2 it is divisible by, and the result would be an odd number (if the odd numbers is 1, your even number was a power of 2, and have entered the endless loop

Just curious to see what you guys find. :P

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You mispelled some of those:

Feramt should be Fermat

Pythagaros should be Pythagoras

Wiles should be Willes (I think)

anyway, yeah. I would also like to be a coauthor ;D lol. I was thinking we could discuss it or something. I dunno. Somebody will be the first to solve it, if it's solvable :P which I hope it is

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You mispelled some of those:

Feramt should be Fermat

Pythagaros should be Pythagoras

Wiles should be Willes (I think)

anyway, yeah. I would also like to be a coauthor ;D lol. I was thinking we could discuss it or something. I dunno. Somebody will be the first to solve it, if it's solvable :P which I hope it is

shouldn't it end up as only 1 or 2, because 4/2 is 2...

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You mispelled some of those:

Feramt should be Fermat

Pythagaros should be Pythagoras

Wiles should be Willes (I think)

anyway, yeah. I would also like to be a coauthor ;D lol. I was thinking we could discuss it or something. I dunno. Somebody will be the first to solve it, if it's solvable :P which I hope it is

Probably true am digitaly dyslexic when i'm tired am ko this mroning htough!

Good luck with it!

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Another question I would have...how fast does a number reach the repeating sequence?

10-5-16-8-4-2-1-4-2-1

11-34-17-52-26-13-40-20-10-5-16-8-4-2-1-4-2-1

12-6-3-10-5-16-8-4-2-1-4-2-1

13-40-20-10-5-16-8-4-2-1-4-2-1

14-7-22-11-34-17-52-26-13-40-20-10-5-16-8-4-2-1-4-2-1

is there a relationship between the starting x value and the length of the sequence?

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no, at least not that's been found, I think. The number 27 takes like a lot of steps. Like over a hundred, I think. And, let's say you have 2^100. A huge number, I'm sure. But it only takes 100 steps to reach 1. So I think it jumps around

Wait I found the list for 27, it takes 111 steps:

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

here's a graph of the numbers 2 to 9999 and their total stopping time:

300px-Totalstoppingtime.png

a fractal made from the sequences:

500px-CollatzFractal.png

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