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November 23rd, 2016, 01:24 AM  #1 
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3  Collatz Conjecture bounds
Cycle bounds Collatz Conjecture I have an update about what I was trying to find for so long. (I am still kinda stuck with finishing my proof on a few things.) I think I said when I will find a limit to x by y I will share it x = number of even values in a cycle y = number of odd values in a cycle x<8/9*floor(1.5^y)y I know it's really high but it works and is helpful on a few cases. Last edited by skipjack; November 23rd, 2016 at 07:34 PM. 
May 16th, 2017, 01:00 AM  #2 
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 
anyone with lower bounds??? i think this is the best results positive cycles: x = roundup ( y * ln3/ln2 ) negative cycles: x = rounddown ( y * ln3/ln2 ) (1,2) (1,2,4) (5,14,7,20,10) (−17,−50,−25,−74,−37,−110,−55,−16 4 ,−82,−41,−122,−61,−182,−91,−272,−1 36,−68,−34) 
June 30th, 2017, 09:54 PM  #3 
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 
i can't believe it after 2 years it was right under my nose d=2^x3^y if d=1 then you have 3 solutions (1,2) (1,2,4) (5,14,7,20,10) if d>1 then we know we have atlease 1 solution (−17,−50,−25,−74,−37,−110,−55,−16 4 ,−82,−41,−122,−61,−182,−91,−272,−1 36,−68,−34) if this is the only solution for d>1 then the collatz conjecture is true now the thing is i didnt know why this is for x=11 and y=7 and why this should be the only solution  the minimum value for any cycle is 12k+7 12k+11 or 96k+7 96k+31 96k+79 96k+91 96k+47 96k+59 96k+71 96k+95 etc ...  because of minimum (mod 12) = 7 minimum (mod 12) = 11 this force it to be d=2^113^7 i love Number Theory because of simple things like this  this not proving the collatz conjecture but this explain why the extra solution must be x=11 and y=7  its just like if d=1 then you have 3 solutions because 2^x3^y=1 gives x=1,y=1 x=2,y=1 x=3,y=2 Last edited by isaac; June 30th, 2017 at 10:10 PM. 
July 13th, 2017, 01:29 AM  #4 
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 
new update let y be the total odd values in a cycle let max be the cycle maximum value for a cycle with y odd values for y=5 max=13y*1.5^y for y=306 max=12.8y*1.5^y for y>=2966 max=1.6y*1.5^y nice ah? i proved that today Last edited by isaac; July 13th, 2017 at 01:38 AM. 

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