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March 11th, 2016, 07:39 AM   #1
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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 ...
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March 12th, 2016, 05:37 AM   #2
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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.
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March 12th, 2016, 10:38 PM   #3
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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
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March 13th, 2016, 01:09 AM   #4
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Quote:
Originally Posted by Peter View Post
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
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March 13th, 2016, 04:52 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.

Your Question was:
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.
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March 14th, 2016, 01:03 AM   #6
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Quote:
Originally Posted by Peter View Post
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
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March 15th, 2016, 02:12 AM   #7
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nvm i found a way to do it

problem solved

you guys can close this thread
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