mersenneforum.org Collatz conjecture
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 2022-10-24, 09:58 #1 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 658 Posts 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: ***
 2022-10-26, 10:16 #2 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 53 Posts (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)
 2022-11-08, 21:10 #3 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 53 Posts ... 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* = 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
 2022-11-28, 18:11 #4 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 1101012 Posts 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
 2022-11-29, 04:38 #5 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 658 Posts 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.
 2022-11-29, 20:19 #6 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 3516 Posts 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???
 2022-12-02, 04:29 #7 MPrimeFinder   "Anonymous" Sep 2022 finding m52 3×7 Posts 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
2022-12-02, 16:45   #8
mart_r

Dec 2008
you know...around...

23·37 Posts

Quote:
 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?

2022-12-09, 04:15   #9
MPrimeFinder

"Anonymous"
Sep 2022
finding m52

3×7 Posts

Quote:
 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)

 2022-12-09, 10:23 #10 kriesel     "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 2×29×127 Posts
 2022-12-12, 12:26 #11 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 53 Posts Default I may not have a brain gentleman, but I have an idea)) 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.

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