Collatz Conjecture: Some stuff I noticed

[Unity+]

So, working with my method, here is what I got so far:

 

[math]\frac{\mathrm{d} }{\mathrm{d} x}\left [ \left(3x+1\right)\left(\frac{x-1}{3}\right) \right ] = 2x-\frac{2}{3}[/math]

 

What I noticed was if you get the resulting function, you can do the same process with that function. This leads to the constant at the end of the result to eventually reach 0.

 

[math]f_{N}(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left [ f_{n}(x)f_{n}^{-1}(x) \right ] = 2x\pm \frac{1}{2^{n}}m[/math]

 

[math]\lim_{n\rightarrow \infty } 2x\pm \frac{1}{2^{n}}m = 2x[/math]

 

Some extra stuff, I decided to have the equation equal each function of the Collatz conjecture.

 

[math]\lim_{n\rightarrow \infty } 2x\pm \frac{1}{2^{n}}m = 3x+1[/math]

[math]2x\pm \lim_{n\rightarrow \infty } \frac{1}{2^{n}}m = 3x+1[/math]

[math]0 =x +1[/math]

 

[math]\lim_{n\rightarrow \infty } 2x\pm \frac{1}{2^{n}}m = x/2[/math]

[math]2x = x/2[/math]

[math]0 = -3x/2[/math]

I dont know where this will lead. Just thought it was interesting.

EDIT: Now the question is if [math]\left\{\begin{matrix} 3x+1 && x\mod2>0 \\\frac{x}{2}&& x\mod2=0 \end{matrix}\right.=\lim_{n\rightarrow \infty } 2x\pm \frac{1}{2^{n}}m[/math]

Edited February 21, 2015 by Unity+

[Daedalus]

Here's an interesting fact that I've noticed:

 

From the odd rule we know:

 

[math]\text{even}=3(\text{odd})+1[/math]

 

such that the inverse is:

 

[math]\text{odd}=\frac{\text{even}-1}{3}[/math]

 

Given that all even numbers are odd numbers that contain factors of 2:

 

[math]\text{even} = (\text{odd}) 2^m[/math]

 

we can show that odd numbers will never produce another odd number that contains a factor of 3:

 

[math]\text{odd numbers with factors of 3} = 3(2n-1) = 6n-3[/math]

[math]\text{even numbers that contain odd numbers with factors of three} = (6 n - 3) 2^m[/math]

 

[math]\text{odd} \ne \frac{\text{even}-1}{3} = \frac{\left ((6 n - 3) 2^m \right ) - 1}{3} = (2n-1)2^m - \frac{1}{3}[/math]

 

This proves that odd numbers of the form [math]6n-3[/math] for [math]n \ge 1[/math] can only start a hailstone sequence and will never occur in a sequence resulting from previous iterations. So, I've proposed another conjecture that all hailstone sequences start with an odd number that contains a factor of 3. This is because given any odd number, we can reverse the algorithm to determine all of the odd numbers that will reduce to the given odd number. Once you reach an odd number that contains a factor of three, we can go no further in reversing the algorithm. Thus, suggesting my conjecture is true.

[Unity+]

Here's an interesting fact that I've noticed:

 

From the odd rule we know:

 

[math]\text{even}=3(\text{odd})+1[/math]

 

such that the inverse is:

 

[math]\text{odd}=\frac{\text{even}-1}{3}[/math]

 

Given that all even numbers are odd numbers that contain factors of 2:

 

[math]\text{even} = (\text{odd}) 2^m[/math]

 

we can show that odd numbers will never produce another odd number that contains a factor of 3:

 

[math]\text{odd numbers with factors of 3} = 3(2n-1) = 6n-3[/math]

[math]\text{even numbers that contain odd numbers with factors of three} = (6 n - 3) 2^m[/math]

 

[math]\text{odd} \ne \frac{\text{even}-1}{3} = \frac{\left ((6 n - 3) 2^m \right ) - 1}{3} = (2n-1)2^m - \frac{1}{3}[/math]

 

This proves that odd numbers of the form [math]6n-3[/math] for [math]n \ge 1[/math] can only start a hailstone sequence and will never occur in a sequence resulting from previous iterations. So, I've proposed another conjecture that all hailstone sequences start with an odd number that contains a factor of 3. This is because given any odd number, we can reverse the algorithm to determine all of the odd numbers that will reduce to the given odd number. Once you reach an odd number that contains a factor of three, we can go no further in reversing the algorithm. Thus, suggesting my conjecture is true.

I never thought of that before, with the function and inverse. Therefore, I decided to continue your idea and add on my idea:

 

[math]\frac{\mathrm{d} }{\mathrm{d} x}\left [ (3(odd)+1)(\frac{even-1}{3}) \right ]=\frac{\mathrm{d} }{\mathrm{d} x}\left [ (3(6x-3)+1)(\frac{(6x-3)2^{m}-1}{3}) \right ]=[/math]

[math]\frac{\mathrm{d} }{\mathrm{d} x}\left [ (3(6x-3)+1)(\frac{(6x-3)2^{m}-1}{3}) \right ]=2^{m+1}(36x-17)-6[/math]

 

Therefore, the result is an even number multiplied by an even/odd number, subtracting an even number.

Edited February 21, 2015 by Unity+

[Daedalus]

I never thought of that before, with the function and inverse.

 

And, that is precisely why I wanted to explore my equation and methods myself before releasing my work. It's not that I'm worried that someone will steal my ideas for themselves, but that no one has really approached the problem from this angle. Because if they did, there would be references to the proofs my method has already produced. As is the case for proving that no odd number will iterate to an odd number that has a factor of three, or the Collatz coefficient as I define it. So, once again, I'm going to give you the method that I used to reverse the algorithm:

 

Let's define the 3 in the rule for odd numbers in the Collatz conjecture as the Collatz coefficient, [math]C[/math] where [math]\left \{C \in \mathbb{N}, C\,\text{is odd}\right \}[/math], and let's define the 1 as the Collatz constant, [math]S[/math] where [math]\left \{S \in \mathbb{N}, S\,\text{is odd}\right \}[/math].

 

From the odd rule we know:

 

[math]\text{even}=C(\text{odd}_1)+S=\text{odd}\times\text{odd}+\text{odd}[/math] [1]

 

such that the inverse is:

 

[math]\text{odd}_1=\frac{\text{even}-S}{C}[/math] [2]

 

Given that all even numbers are odd numbers that contain factors of 2, we can redefine the equation for even numbers as:

 

[math]\text{even} = (\text{odd}_2) 2^m[/math] [3]

 

Now that we have an equation for all even numbers [3] and an equation for the odd rule [1], we set them equal to each other:

 

[math](\text{odd}_2) 2^m = C(\text{odd}_1)+S[/math]

 

and solve for [math]\text{odd}_1[/math] and values of [math]m[/math] that produce integer / natural number results:

 

[math]\text{odd}_1 = \frac{(\text{odd}_2) 2^m-S}{C}[/math] [4]

 

Because we are interested in odd numbers that reduce to one, we simply set [math]\text{odd}_2=1[/math]:

 

[math]\text{odd}_1 = \frac{2^m-1}{3}[/math]

 

and check for integer / natural number results:

 

[math]{\frac{1}{3}, 1, \frac{7}{3}, 5, \frac{31}{3}, 21, \frac{127}{3}, 85, \frac{511}{3}, 341, ...}[/math]

 

We can see that every other number is an integer / natural number so the equation that predicts all odd number that reduce to one in one iteration of the odd rule in the Collatz conjecture is:

 

[math]\text{odd}_1 = \frac{2^{2m}-1}{3}[/math] [5]

 

We can use induction to prove equation [5] is true, but I'll leave that as an exercise for the reader. From here, we want to know all of the odd numbers that reduce to one of the odd numbers predicted by equation [5]. That will give us every odd number that reduces to one in two iterations of the rule for odd numbers. By using equation [5] as the source for [math]\text{odd}_2[/math] in equation [4] , we can reverse the algorithm once again:

 

[math]\text{odd}_1 = \frac{(\text{odd}_2) 2^m-S}{C} = \frac{(\frac{2^{2 m_1}-1}{3}) 2^{m_2}-1}{3}[/math]

 

Again, after verifying integer results, we get an equation that predicts every single odd number that reduces to one given two iterations of the rule for odd numbers. Because of the cyclic nature of the odd number 1, all of the odd numbers that reduce to one using one iteration of the odd rule are included in the results for the odd numbers that reduce to one using two iterations of the odd rule:

 

The equations for the exponents of two, [math]m_1[/math] and [math]m_2[/math], are:

 

[math]m_1 = \frac{6 {m_1} - (-1)^{m_1}-3}{2}[/math]

[math]m_2 = \frac{4 {m_2} - (-1)^{m_1}-1}{2}[/math]

 

which when plugged into the variables for our exponents of 2 gives us:

 

[math]\text{odd}_2 = \frac{(\frac{2^{\frac{6 {m_1} - (-1)^{m_1}-3}{2}}-1}{3}) 2^{\frac{4 {m_2} - (-1)^{m_1}-1}{2}}-1}{3}[/math] [6]

 

Of course, I'm close to having the general solution that predicts the equations for any value of N iterations. Then by using equation [4], I hope to map all odd numbers to families of odd numbers, and show that the odd numbers that are predicted by my equations are complete. As it turns out, it seems that the equation that predicts all of the equations for odd numbers that reduce to one in N iterations will be an infinite series containing an infinite number of variables where the results of the previous iteration are contained within the next iteration.

 

As a bonus, I have attached the software that I wrote to explore the indices for the exponents of the twos for anyone who wishes to explore the Collatz conjecture with me. If you have any problems with the software, please PM me the log file it produces in the location you installed the .exe.

Collatz Solver.zip

Edited February 21, 2015 by Daedalus

[Unity+]

 

And, that is precisely why I wanted to explore my equation and methods myself before releasing my work. It's not that I'm worried that someone will steal my ideas for themselves, but that no one has really approached the problem from this angle. Because if they did, there would be references to the proofs my method has already produced. As is the case for proving that no odd number will iterate to an odd number that has a factor of three, or the Collatz coefficient as I define it. So, once again, I'm going to give you the method that I used to reverse the algorithm:

 

Let's define the 3 in the rule for odd numbers in the Collatz conjecture as the Collatz coefficient, [math]C[/math] where [math]\left \{C \in \mathbb{N}, C\,\text{is odd}\right \}[/math], and let's define the 1 as the Collatz constant, [math]S[/math] where [math]\left \{S \in \mathbb{N}, S\,\text{is odd}\right \}[/math].

 

From the odd rule we know:

 

[math]\text{even}=C(\text{odd}_1)+S=\text{odd}\times\text{odd}+\text{odd}[/math] [1]

 

such that the inverse is:

 

[math]\text{odd}_1=\frac{\text{even}-S}{C}[/math] [2]

 

Given that all even numbers are odd numbers that contain factors of 2, we can redefine the equation for even numbers as:

 

[math]\text{even} = (\text{odd}_2) 2^m[/math] [3]

 

Now that we have an equation for all even numbers [3] and an equation for the odd rule [1], we set them equal to each other:

 

[math](\text{odd}_2) 2^m = C(\text{odd}_1)+S[/math]

 

and solve for [math]\text{odd}_1[/math] and values of [math]m[/math] that produce integer / natural number results:

 

[math]\text{odd}_1 = \frac{(\text{odd}_2) 2^m-S}{C}[/math] [4]

 

Because we are interested in odd numbers that reduce to one, we simply set [math]\text{odd}_2=1[/math]:

 

[math]\text{odd}_1 = \frac{2^m-1}{3}[/math]

 

and check for integer / natural number results:

 

[math]{\frac{1}{3}, 1, \frac{7}{3}, 5, \frac{31}{3}, 21, \frac{127}{3}, 85, \frac{511}{3}, 341, ...}[/math]

 

We can see that every other number is an integer / natural number so the equation that predicts all odd number that reduce to one in one iteration of the odd rule in the Collatz conjecture is:

 

[math]\text{odd}_1 = \frac{2^{2m}-1}{3}[/math] [5]

 

We can use induction to prove equation [5] is true, but I'll leave that as an exercise for the reader. From here, we want to know all of the odd numbers that reduce to one of the odd numbers predicted by equation [5]. That will give us every odd number that reduces to one in two iterations of the rule for odd numbers. By using equation [5] as the source for [math]\text{odd}_2[/math] in equation [4] , we can reverse the algorithm once again:

 

[math]\text{odd}_1 = \frac{(\text{odd}_2) 2^m-S}{C} = \frac{(\frac{2^{2 m_1}-1}{3}) 2^{m_2}-1}{3}[/math]

 

Again, after verifying integer results, we get an equation that predicts every single odd number that reduces to one given two iterations of the rule for odd numbers. Because of the cyclic nature of the odd number 1, all of the odd numbers that reduce to one using one iteration of the odd rule are included in the results for the odd numbers that reduce to one using two iterations of the odd rule:

 

The equations for the exponents of two, [math]m_1[/math] and [math]m_2[/math], are:

 

[math]m_1 = \frac{6 {m_1} - (-1)^{m_1}-3}{2}[/math]

[math]m_2 = \frac{4 {m_2} - (-1)^{m_1}-1}{2}[/math]

 

which when plugged into the variables for our exponents of 2 gives us:

 

[math]\text{odd}_2 = \frac{(\frac{2^{\frac{6 {m_1} - (-1)^{m_1}-3}{2}}-1}{3}) 2^{\frac{4 {m_2} - (-1)^{m_1}-1}{2}}-1}{3}[/math] [6]

 

Of course, I'm close to having the general solution that predicts the equations for any value of N iterations. Then by using equation [4], I hope to map all odd numbers to families of odd numbers, and show that the odd numbers that are predicted by my equations are complete. As it turns out, it seems that the equation that predicts all of the equations for odd numbers that reduce to one in N iterations will be an infinite series containing an infinite number of variables where the results of the previous iteration are contained within the next iteration.

 

As a bonus, I have attached the software that I wrote to explore the indices for the exponents of the twos for anyone who wishes to explore the Collatz conjecture with me. If you have any problems with the software, please PM me the log file it produces in the location you installed the .exe.

Sorry for the late reply, finally got back to my studies.

 

I also found the same pattern that you got when working with the method I proposed:

 

 

I wonder if there is some connection between what we are both doing.

 

EDIT: Here is the pattern I have seen between our work.

 

The formula I get from taking the product of the function and its inverse, I get:

 

Now, I take your and my pattern and subtract them from each other:

 

p2-p1

 

And I get the following sequence:

 

[math]1,-\frac{1}{3},-1,-\frac{1}{3},3,\frac{35}{3},31,\frac{215}{3},155,\frac{971}{3},...[/math]

Edited February 22, 2015 by Unity+

[John]

So, I've proposed another conjecture that all hailstone sequences start with an odd number that contains a factor of 3. This is because given any odd number, we can reverse the algorithm to determine all of the odd numbers that will reduce to the given odd number. Once you reach an odd number that contains a factor of three, we can go no further in reversing the algorithm. Thus, suggesting my conjecture is true.

 

It is certainly true, though usually we allow the term "hailstone sequence" to apply with any starting number, e.g. {6, 3, 10, 5, 16, 8, 4, 2, 1} is a hailstone sequence.

 

 

Anyway, consider that any odd natural number n is congruent to 0, 1 or 2 (mod 3). In order to show that your conjecture holds, we simply must show for each case whether (n * 2^x) - 1 = 0 (mod 3) (where x is, of course, a natural number).

 

Case 0: Since n = 0 (mod 3), we have 2n = 0 (mod 3), 2(2n) = 0 (mod 3), etc. That is, subtracting 1 always yields 2 (mod 3), i.e. for all x, 2^xn - 1 = 2 (mod 3) and as such is not divisible by 3.

 

Case 1: Since n = 1 (mod 3), we have 2n = 2 (mod 3), 2(2n) = 1 (mod 3), etc. That is, repeated doublings cycle between 1 and 2 (mod 3), with the former corresponding with even powers of 2 and the latter corresponding with odd powers of 2. Thus, in this case, 2^xn - 1 is divisible by 3 only if x is even.

 

Case 2: Since n = 2 (mod 3), we have 2n = 1 (mod 3), 2(2n) = 2 (mod 3), etc. This is similar to Case 1, except in this case, 2^xn - 1 is divisible by 3 only if x is odd.

 

In Cases 1 and 2, we have shown that another odd natural number can precede n in a Collatz hailstone sequence, whereas in Case 0, n can only be preceded by even natural numbers.

[craniumonempty]

Yes, I noticed all odd only sequences end in a number n=3*c... I created a reverse Collatz approach (on reddit https://www.reddit.com/r/math/comments/4r01xn/density_of_a_subset_of_numbers_compared_to_the/d4xo29q ), but other than things like this, I'm not sure it's good for much.