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Old 2022-10-24, 09:58   #1
RMLabrador
 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.

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Default Collatz conjecture

Hi. some code
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.

Last fiddled with by RMLabrador on 2022-10-24 at 10:45 Reason: ***
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Old 2022-10-26, 10:16   #2
RMLabrador
 
"Chereztynnoguzakidai"
Oct 2022
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Default

(1) have the solution in integers when
Code:
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)
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Old 2022-11-08, 21:10   #3
RMLabrador
 
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... 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
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Old 2022-11-28, 18:11   #4
RMLabrador
 
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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

Last fiddled with by RMLabrador on 2022-11-28 at 18:37 Reason: inattention in those small mathematical signs
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Old 2022-11-29, 04:38   #5
RMLabrador
 
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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.
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Old 2022-11-29, 20:19   #6
RMLabrador
 
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Leave the perfect numbers joke, look on the my hand creation)
Click image for larger version

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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???
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Old 2022-12-02, 04:29   #7
MPrimeFinder
 
"Anonymous"
Sep 2022
finding m52

2110 Posts
Talking i'm about to solve it

proof coming soon

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

Last fiddled with by MPrimeFinder on 2022-12-02 at 04:33 Reason: hint
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Old 2022-12-02, 16:45   #8
mart_r
 
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you know...around...

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Quote:
Originally Posted by MPrimeFinder View Post
proof coming soon

hint: related to 3x+1 and x/2 never meeting each other
Is this a part of the proof?
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Old 2022-12-09, 04:15   #9
MPrimeFinder
 
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finding m52

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Quote:
Originally Posted by mart_r View Post
Is this a part of the proof?
its not that but yes the stopping time is always 469 moves (above 300 quintillion)
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Old 2022-12-09, 10:23   #10
kriesel
 
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See https://www.collatzresearch.org/
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Old 2022-12-12, 12:26   #11
RMLabrador
 
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Default Default I may not have a brain gentleman, but I have an idea))

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


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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.
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