My Math Forum collatz conjecture

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 March 11th, 2016, 06:39 AM #1 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 collatz conjecture x = the numbers of even values in a cycle y = the numbers of odd values in a cycle 2^x-3^y>0 x > y ln3/ln2 > y is there an upper limit to x in a way that i can write it with y somthing like: if ky > x then k=? can someone knows if and how i can find k? its seems that k can be above 2 or without limit ...
 March 12th, 2016, 04:37 AM #2 Senior Member   Joined: Mar 2012 Posts: 572 Thanks: 26 Off the top of my head, I don't see how you could prove an upper limit without finding an upper limit on how long a cycle could be.
 March 12th, 2016, 09:38 PM #3 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 You can get k in the Same way, you got the other direction of the inequality. Just notice 3n+1 < (3+1/n) n< (3+1/m) n wehre m is the spaltest number in the cycle. You can get a lower blind in m with the existing computational results. You wil get: Ln(3+1/m)/ln(2) y > x > ln 3/ln 2 y
March 13th, 2016, 12:09 AM   #4
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From: israel

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Quote:
 Originally Posted by Peter You can get k in the Same way, you got the other direction of the inequality. Just notice 3n+1 < (3+1/n) n< (3+1/m) n wehre m is the spaltest number in the cycle. You can get a lower blind in m with the existing computational results. You wil get: Ln(3+1/m)/ln(2) y > x > ln 3/ln 2 y
that's not even seem to be right .... what is m?
plus i want only y on the right side of the equation ...

the whole point was to write the upper bound with only y

March 13th, 2016, 03:52 AM   #5
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Sorry for the many typos. I have still to figure out how to switch between the auto corrects. But no need to get aggressive.

Quote:
 If ky > x then k=?
Of course not any formally correct k can be accepted as an answer but my reply was:

k=ln(3+ 1/m)/ln(2). Where m is a lower bound for the smallest element in your cycle.

For instance if you accept that people have checked that no number below 1000 is in a cycle (except for 1,2,4) you could take m = 1000. Giving you a factor k only 1 in 3000 bigger than the lower bound you stated yourselve.

March 14th, 2016, 12:03 AM   #6
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From: israel

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Quote:
 Originally Posted by Peter Sorry for the many typos. I have still to figure out how to switch between the auto corrects. But no need to get aggressive. Your Question was: Of course not any formally correct k can be accepted as an answer but my reply was: k=ln(3+ 1/m)/ln(2). Where m is a lower bound for the smallest element in your cycle. For instance if you accept that people have checked that no number below 1000 is in a cycle (except for 1,2,4) you could take m = 1000. Giving you a factor k only 1 in 3000 bigger than the lower bound you stated yourselve.
i am not aggressive all cool

but that is not what is asked

i wanted an upper limit (without m) only written formula by y (for every y)

x > y ln3/ln2 > y

i wanna replace x with a formula that uses y only and not another factor

 March 15th, 2016, 01:12 AM #7 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 nvm i found a way to do it problem solved you guys can close this thread

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