My Math Forum Collatz Conjecture bounds

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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^x-3^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=2^11-3^7 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^x-3^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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