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by Guest » Fri Oct 21, 2016 1:03 pm
Keywords: Modified Collatz Conjecture, Undecidability, Halting Problem, Gödel's Incompleteness Theorems; Proof of Collatz Conjecture.
In the Collatz conjecture, what will happen if instead of multiplication by three in the Collatz process, we choose to multiply by five or more?
If the odd multiplier is greater than three, the Collatz sequence may never converge to one in some cases because division by [tex]2^{1 + k}[/tex] where integer, k>1, only occurs with a collective probability of twentyfive percent according to the Law of Large Numbers.
Example: Suppose we replace the odd multiplier, three, with five in the Collatz Conjecture. We compute the Collatz sequence of odd integers given the first number, seven.
We have the following Collatz sequence of odd integers:
{7, 9, 23 , 29, 73, 183, 229, 573, 1433, 3583, 4479, 5599, 6999, 8749, 21873, 54683, 34177, 85443, 26701, 66753, 166883, 52151, 65189, 162973, 407433, 1018583, 1273229, 3183073, 7957683, 310847, 388559, 485699, 151781, 379453, 948633, 2371583, 2964479, 3705599, 4631999, 5789999, 7237499, 4523437, 11308593, 28271483, 17669677, 44174193, 110435483, 69022177, 172555443, 6740447, 8425559, 10531949, 26329873, 65824683, 41140427, 25712767, 32140959, 40176199, 50220249, 125550623, 156938279, 196172849, 490432123, 306520077, …}
Unless there is some odd integer, n, such that 5n + 1 = [tex]2^{ k}[/tex] for the above Collatz sequence where k is a positive integer, the sequence will never converge to one. That is a very interesting observation because the equation, 5n + 1 = [tex]2^{ k}[/tex], has infinitely many solutions!
Will the above sequence converge to one or diverge? That question may be undecidable!
However, it seems the sequence diverges (never halts) when you begin with n =7.
For example, the 2280th odd integer of above sequence is greater than 10^200.
Mathematica Software 11 function, f[], code:
f[l_,m_]:=(
n = l;i=0;
While[n>1,n = m*n +1;
x = n;
While[
EvenQ[x],x = x/2];
n=x;
i = i + 1;
];
Return[i]
);
f[7,5] = ? (Undecidable?) Note: 5 is the multiplier and 7 is the starting value.
When i =2280, n_i = n > 10^{200}.
 https://www.quora.com/IstheCollatzConjecturesolvable/answer/DavidCole146

Guest

by Guest » Fri Oct 21, 2016 1:06 pm
Guest wrote:Keywords: Modified Collatz Conjecture, Undecidability, Halting Problem, Gödel's Incompleteness Theorems; Proof of Collatz Conjecture.
In the Collatz conjecture, what will happen if instead of multiplication by three in the Collatz process, we choose to multiply by five or more?
If the odd multiplier is greater than three, the Collatz sequence may never converge to one in some cases because division by [tex]2^{1 + k}[/tex] where integer, k>1, only occurs with a collective probability of twentyfive percent according to the Law of Large Numbers.
Example: Suppose we replace the odd multiplier, three, with five in the Collatz Conjecture. We compute the Collatz sequence of odd integers given the first number, seven.
We have the following Collatz sequence of odd integers:
{7, 9, 23 , 29, 73, 183, 229, 573, 1433, 3583, 4479, 5599, 6999, 8749, 21873, 54683, 34177, 85443, 26701, 66753, 166883, 52151, 65189, 162973, 407433, 1018583, 1273229, 3183073, 7957683, 310847, 388559, 485699, 151781, 379453, 948633, 2371583, 2964479, 3705599, 4631999, 5789999, 7237499, 4523437, 11308593, 28271483, 17669677, 44174193, 110435483, 69022177, 172555443, 6740447, 8425559, 10531949, 26329873, 65824683, 41140427, 25712767, 32140959, 40176199, 50220249, 125550623, 156938279, 196172849, 490432123, 306520077, …}
Unless there is some odd integer, n, such that 5n + 1 = [tex]2^{ k}[/tex] for the above Collatz sequence where k is a positive integer, the sequence will never converge to one. That is a very interesting observation because the equation, 5n + 1 = [tex]2^{ k}[/tex], has infinitely many solutions!
Will the above sequence converge to one or diverge? That question may be undecidable!
However, it seems the sequence diverges (never halts) when you begin with n =7.
For example, the 2280th odd integer of above sequence is greater than 10^200.
Mathematica Software 11 function, f[], code:
f[l_,m_]:=(
n = l;i=0;
While[n>1,n = m*n +1;
x = n;
While[
EvenQ[x],x = x/2];
n=x;
i = i + 1;
];
Return[i]
);
f[7,5] = ? (Undecidable?) Note: 5 is the multiplier and 7 is the starting value.
When i =2280, [tex]n_{i }[/tex]= n > [tex]10^{200}[/tex].
 https://www.quora.com/IstheCollatzConjecturesolvable/answer/DavidCole146

Guest

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