20221024, 09:58  #1 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
65_{8} 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);} (3^x)*p = 2^(zx)  B(v(p)) (1) where z  total number of cycles to 1, x  number of 3*y+1 point, zx number of 2*y points, v  some vector of 1 and 0 [1,0,...,0], where sum of 1 = x, sum of 0 = zx, 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 zeroended, 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) nonnegative. I.e. if Collatz conjecture is true, (1) have at least one solution for p for any nonnegative integer z. Last fiddled with by RMLabrador on 20221024 at 10:45 Reason: *** 
20221026, 10:16  #2 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
53 Posts 
(1) have the solution in integers when
Code:
B = k*3^x+mod(2^(zx),3^x) (2) => and p = (2^(zx)B)/3^x (3) 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) 
20221108, 21:10  #3 
"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<B, for the p that possess both z and x respectively b) sieve an interval of BA 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* = 504 = 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 
20221128, 18:11  #4 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
110101_{2} 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^n1 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^n1, at 2*n+1 = 3^(n+1). Interesting, the numbers of primes in form of 2^a*3^b1, 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 20221128 at 18:37 Reason: inattention in those small mathematical signs 
20221129, 04:38  #5 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
65_{8} 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^(n1)*(2^n1)) (i.e. Perfect Numbers for prime 2^n1) and do the Collatz cycles, after n1 of them we got Mersenne numbers 2^n1. 
20221129, 20:19  #6 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
35_{16} 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^n1, and 3^n1 is a key. Something tells me that my explanations are f@cking incomprehensible??? 
20221202, 04:29  #7 
"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 20221202 at 04:33 Reason: hint 
20221209, 10:23  #10 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2×29×127 Posts 

20221212, 12:26  #11 
"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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