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n1 = 3n0 + 1 where n0 is any odd integer > 1;
l1 = Log( n1 / greatest odd factor > 1 of numerator) / Log(2);
n2 = 3((3n0 + 1) / 2^ (l1) ) + 1 = 3n1 / 2^ (l1) + 1;
l2 = Log( n2 / greatest odd factor > 1 of numerator) / Log(2);
...
nm = 3*nsub(m-1) / 2^ (lsub(m-1)) + 1;
lm = Log( nm / greatest odd factor > 1 of numerator) / Log(2);
Note: nm = 2^lm * (greatest odd factor > 1 of nm);
Probability( n1 is a multiple of 2) = 1;
Probability( n1 is a multiple of 2^2 | n1 is a multiple of 2) = 1/2;
Probability( n1 is a multiple of 2^3 | n1 is a multiple of 2^2) = 1/2;
...
Probability( n1 is a multiple of 2^2) = Probability( n1 is a multiple of 2^2 | n1 is a multiple of 2)
* Probability( n1 is a multiple of 2) = (1/2) * 1 = 1/2;
Probability( n1 is a multiple of 2^3) = Probability( n1 is a multiple of 2^3 | n1 is a multiple of 2^2) * Probability( n1 is a multiple of 2^2) = (1/2) * (1/2) = 1/4;...
So, Probability(n1 is at least a multiple of 2^2) = Sum[i = 0 to l1 - 1 of (1/2)^(i+1) ]
...
Probability(nm is at least a multiple of 2^2) = Sum[i = 0 to lm - 1 of (1/2)^(i+1) ]
Hence,
Probability(Collatz sequence does not converge to 1)
= Product[ m=1 to infinity of Sum[ i=0 to lm - 1 of (1/2)^(i+1) ] ] --> 0.
So, the Collatz conjecture is true!
Note: 'sub' indicates a long subscript of a variable.
Please Be A Good Sport! Take any detection of mistakes as learning exercises...
And please let me know too. Thank you!
P.S. There's more analysis that I did include here for brevity sake...
FYI: The Last Calculation In My Proof Of The Collatz Conjecture Revisited
Note: I assumed there exists a Collatz sequence which does not converge to 1.
Recall Final Result: Probability(Collatz sequence does not converge to 1)
= Product[ j=1 to infinity of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ] ] --> 0. Why?
Let N0 be the event that 3*n0 + 1 is a multiple of 2 for any odd integer > 1;
Let N1 be the event that n1 = 3*n0 + 1 is at least a multiple of 2^2;
Let N2 be the event that n2 = 3*n1/(2^(l1)) + 1 is at least a multiple of 2^2;
...
Let Nm be the event that nm = 3*nsub(m-1)/(2^(lsub(m-1)) ) +1 is at least a multiple of 2^2.
Note: 'sub' indicates a long subscript of a variable.
Note: lj = Log( nj / greatest odd factor > 1 of numerator) / Log(2).
Note: Probability(N0) = Prob(N0) = 1.
Prob(N1) = Prob(N1 | N0) * Prob(N0) = Prob(N1 | N0) = Sum[ i=0 to l1 - 1 of (1/2)^(i+1) ];
Prob(N2) = Prob(N2 | N1) * Prob(N1) = Prob(N2 | N1) * Prob(N1 | N0)
= Sum[ i=0 to l2 - 1 of (1/2)^(i+1) ] * Sum[ i=0 to l1 - 1 of (1/2)^(i+1) ] ;
...
...
Prob(Nm) = Product[ j = 1 to m of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ];
Therefore, I have the following result:
Probability(Collatz sequence does not converge to 1)
= Prob(Nm as m --> infinity)
= Product[ j =1 to infinity of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ] ] --> 0. Q.E.D.
P.S. Please share your comments. Thank you!
Recall: Basics of Conditional Probability
Prob(N2|N1)P(N1) = Prob(N2 ∩ N1) = Prob(N2).
In context of my proof of the Collatz conjecture, N2 ∩ N1 means that the greatest odd factor of events, N2 and N1 is greater one, and the events, N2 and N1 are at least multiples of 4.
And since the prior existence of event N1 in the Collatz sequence indicates the existence of the succeeding event N2 ... all according to my prior assumption that there exists a Collatz sequence that does not converge one. And thus, N2 ∩ N1 = N2.
I hope this explanation helps you to follow the previous post about my proof of the Collatz conjecture.
Please share your comments too. Thanks!
David Cole
aka primework123
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Thank you! Thank Lord GOD!
