Collatz conjecture

[RMLabrador]

Hi. some code

{p=11*13; a=p;v=vector(1500); \\
for(i=1,1500,if(lift(Mod(a,2)),v[i]=1;a=3*a+1,v[i] = 0;a=a/2);
if(a==1,z=i;break());
);print(z); \\real size of vector
for(i=1,z,if(v[i],y=3*y+1,y=y/2));
print(y);}

as easy to see, for every input p we got (if Collatz conjecture true)
(3^x)*p = 2^(z-x) - B(v(p)) (1)
where z - total number of cycles to 1, x - number of 3*y+1 point, z-x number of 2*y points,
v - some vector of 1 and 0 [1,0,...,0], where sum of 1 = x, sum of 0 = z-x, and v = v(p) only.

So, can we drop p, and treat vector v as variable for given z, and solve (1) for all integer p instead???
P.S. vector v have a lot of obvios restrictions, its zero-ended, have no more than 1 in row (no ..,1,1,...) and the number of 1 in v is limited by z, coz left side in (1) non-negative. I.e. if Collatz conjecture is true, (1) have at least one solution for p for any non-negative integer z.

[RMLabrador]

(1) have the solution in integers when

B = k*3^x+mod(2^(z-x),3^x) (2) => and p = (2^(z-x)-B)/3^x (3)

let z = 50. x can be any from 0 to some integer number around z/2.8; let x = 16, from (2),(3) and (1) we have p = 0...399. 399 -the maximum p, that may posess both z = 50 and x = 16. What about the lower bound??? The one exist ans safe to assume in this case p = 399 - 66 - 11 = 322 i.e. for other z and given x - max k value for x minus sum of all max k for x+1, x+2, till (1) have at least one solution for k > 0. I leave this unproven, lets say as assumption)
Now we have a 77 p values and some of them have the lenght of the Collatz cycle = 50. Here they are 329,338,339 and 359. Other is just a guests on our party from the other floors - they have different z and x values)

So, can we go to oversimple, and somehow remove the guest from the 77 or other number for other z and x???
(You may have been notice, that if for some z and x only guest in the room, then Collatz conjecture is wrong)

[RMLabrador]

... so if we fixed the number of cycles z and x (the number of 3*y+1 cases in p) we can
a) obtain the values of A and B, A<p<B, for the p that possess both z and x respectively
b) sieve an interval of B-A and find the values of p
the sieve is simple, just take some p and make the Collatz cycles till
1) run out of x - this is useful for small x
2) till z*/x*<=2 + epsilon, where z* and x* local values for some step, i.e. z = 50, x = 16 and p = 336, sieve =>
p* = 326/2/2/2/2 so z* = 50-4 = 46 and x* = x = 16; for next step in this sample z*=45 and x = 15 and so on
3) the value of epsilon = 0 is good, very safe for the all p, but overkill a little
4) as result, we have an explanation of behaivor of Collatz cycles as solution of (1) in integers (2), (3) and good practical method to find the values of p for fixed number of Collatz cycles

[RMLabrador]

meanwhile, (may be this is well nown as old broken drum) there is a some ordnung in Collatz cycles for mersenne numbers
i.e. for 2^n-1 the frist 2*n steps are decided, and they in the form of 2^a*3^b+-1 and a+b = n and after every next step value of a decreased by 1 and b incremented by 1. At 2*n cycles the number is 3^n-1, at 2*n+1 = 3^(n+1).
Interesting, the numbers of primes in form of 2^a*3^b-1, a or b is prime seems unusual high (for me).

After 3^(n+1), the card game begin in the House of Collatz; I have no energy to emphasize it clearly and put on the paper at least for now
P/S/
The Perfect numbers drop to Mersenne first after n iteration

[RMLabrador]

Everyone can drop a piles of it, sometime, dont You?)
I'm make a language typo mistake sometimes or continuously, this time is a bait to see a someone!
And what I got - А perfectionist!
I'm glad to see You)

Yes, the if we take the some number = (2^(n-1)*(2^n-1)) (i.e. Perfect Numbers for prime 2^n-1) and do the Collatz cycles, after n-1 of them we got Mersenne numbers 2^n-1.

[RMLabrador]

Leave the perfect numbers joke, look on the my hand creation)

if you use zoom, you may notice that that's joins on the bench is impossible.
Collatz conjecture is the same. And 2^n-1, and 3^n-1 is a key.
Something tells me that my explanations are f@cking incomprehensible???

[MPrimeFinder]

proof coming soon

hint: related to 3x+1 and x/2 never meeting each other

[mart_r]

Originally Posted by MPrimeFinder

proof coming soon

hint: related to 3x+1 and x/2 never meeting each other

Is this a part of the proof?

[MPrimeFinder]

Originally Posted by mart_r

Is this a part of the proof?

its not that but yes the stopping time is always 469 moves (above 300 quintillion)

[kriesel]

See https://www.collatzresearch.org/

[RMLabrador]

https://www.youtube.com/watch?v=dDZ9Mbf1GNo




For any other a value, there is such b in range from ceil(a/2.7) to 0, that number p have an exact numbers of Collatz cycles a&b.