ok, as some of the comments show this does not always work (n=3 and x=39). Checked that and found out that for odd n there should be transformation 3x+n instead of (n+2)x+n, then it works fine. I also checked the even n, for n=5 and x=35 it goes to infinity (or at least doesn't go down after 1000 steps (Python loop)). I changed the transformations: (x+1)/2 -> (x+3)/2 and (x/2)+n -> (x/2)+1. Now it always leads to n+1 or 3; not perfect, but couldn't find a better formula. I think it might lead somewhere if one finds the right formula for the even n.
I’m trying to show no four numbers in a row intersect before merging at one. Very tedious maths, but I’m working on that. I have shown other things in the conjecture and also used the Collatz sequence rules to prove that A002326 on OEIS =a(x) means that ((2^(a((q-1)/2)))-1)/q for odd q is always an integer. So if someone finds a way to show that for infinitely many composite odd q that on a((q-1)/2) there are primes, there are infinitely many Mersenne numbers with prime exponent, solving a big conjecture.
For n=3 the transformations are: Odd x: (3+2)x+3 = 5x+3 Even x: x/2 Sequence should always lead to 3. Starting with x=39 the process never leads to 3: 39 -> 198 -> 99 -> 498 -> 249 -> 1248 -> 624 -> 312 -> 156 -> 78 -> 39 And for n=4 the transformations are: Odd x: (x+1)/2 Even x: (x/2)+n = (x/2)+4 Sequence should always lead to n+1=4+1=5. Starting with x=8 the process never leads to 5: 8 -> 8
You're absolutely right; 39 won't converge to 3 under these transformation rules. Notably, when n = 1, the sequence converges to 1, but it passes through 38, which is very close to 39. This implies that a self-loop for n = 1 is also necessary for higher values of x. Furthermore, if the transformation breaks for n = 3 and 4, it's likely that the transformation for n = 1 will also break for higher values of x.
For n=3 and x = 13 , this will lead to 3.. According to my transformation, every transformation has one loop loop that's n for odd n ... For n=3 , x=13 has one 1 loop which is 3.. Use transformation (2+3)x+3 = 5x+3 for odd x For even x : x/2 13 => 67 => 34 =>17 =>88 => 11 => 36 =>9 =>30 =>15 => 48 =>3
I like your ideas, keep working in it.
Congrats!
ok, as some of the comments show this does not always work (n=3 and x=39). Checked that and found out that for odd n there should be transformation 3x+n instead of (n+2)x+n, then it works fine. I also checked the even n, for n=5 and x=35 it goes to infinity (or at least doesn't go down after 1000 steps (Python loop)). I changed the transformations: (x+1)/2 -> (x+3)/2 and (x/2)+n -> (x/2)+1. Now it always leads to n+1 or 3; not perfect, but couldn't find a better formula. I think it might lead somewhere if one finds the right formula for the even n.
thanks , i will upload new video on this topic soon
But where is the proof ?
Collatz conjuncture is a special case in collatz transformation when n=1
there is no proof its just clickbait
May be a new understanding of collatz conjuncture
@@log_menus_1 bot?
I’m trying to show no four numbers in a row intersect before merging at one. Very tedious maths, but I’m working on that. I have shown other things in the conjecture and also used the Collatz sequence rules to prove that A002326 on OEIS =a(x) means that ((2^(a((q-1)/2)))-1)/q for odd q is always an integer. So if someone finds a way to show that for infinitely many composite odd q that on a((q-1)/2) there are primes, there are infinitely many Mersenne numbers with prime exponent, solving a big conjecture.
For n=3 the transformations are:
Odd x: (3+2)x+3 = 5x+3
Even x: x/2
Sequence should always lead to 3.
Starting with x=39 the process never leads to 3:
39 -> 198 -> 99 -> 498 -> 249 -> 1248 -> 624 -> 312 -> 156 -> 78 -> 39
And for n=4 the transformations are:
Odd x: (x+1)/2
Even x: (x/2)+n = (x/2)+4
Sequence should always lead to n+1=4+1=5.
Starting with x=8 the process never leads to 5:
8 -> 8
You're absolutely right; 39 won't converge to 3 under these transformation rules. Notably, when n = 1, the sequence converges to 1, but it passes through 38, which is very close to 39. This implies that a self-loop for n = 1 is also necessary for higher values of x. Furthermore, if the transformation breaks for n = 3 and 4, it's likely that the transformation for n = 1 will also break for higher values of x.
If you put in "3" for n & 13 for the seed it will never reach a loop
For n=3 and x = 13 , this will lead to 3..
According to my transformation, every transformation has one loop loop that's n for odd n ...
For n=3 , x=13 has one 1 loop which is 3..
Use transformation (2+3)x+3 = 5x+3 for odd x
For even x : x/2
13 => 67 => 34 =>17 =>88 => 11 => 36 =>9 =>30 =>15 => 48 =>3
but how do you justify the transformation is equivalent to collatz?
n =1
uhh