TakenItSeriously

Proof for the Collatz Conjecture

Recommended Posts

The collatz conjecture 
If for any given positive integer n, a series is created such that:
  • If the number is even, divide it by two.
  • If the number is odd, triple it and add one.
Then the Collatz Conjecture states that if the operation repeats often enough, the resulting series of integers created will always converge to 1 regardless of the size of n.
 
Proof:

If you tripple any odd number and add one the result will always be an even number.

If you divide any even number by two, then half of the results will be odd while the other half will be even. 

Since every odd number will be changed to an even number by 3n+1 then we know a divide by 2 will always follow. ⇒ for any odd number we can instead apply:

  • (3n+1)/2,

i.e. half of the results for all operations will be odd and the other half will be even.

⇒ half of the time the number will expand to slightly more than 3n/2, while the other half of the time, the number will contract to n/2.

⇒ 3n/2·n/2 = 3n/4

or for the long term average the number will be reduced by 25%.per operation on the series.

For any given positive integer, n, such that all integers < n have already been proven to converge to 1, then any series that drops below n will then converge to one.

Since the long term average is -EV to about 3n/4, then all series must drop below n eventually.

 
Edited by TakenItSeriously

Share this post


Link to post
Share on other sites

Edit to add:

I forgot to mention that the distribution of numbers that expand or contract are distributed evenly by the periodicity of prime factors: two and three, which produces a random like distribution similar to prime number distributions, which are infact goverened by growing cyclical distributions between Mersenne Numbers thereby forcing the -EV long term average assertion to be true.

 

CEE1AC8C-39DA-47EB-99E6-D2D822220717.thumb.png.fe3a0da72e4df45fd48abfd6f5373244.png

Figure 1: Reveals the cyclical pattern based on 2n where for each column the repeating cycle doubles in size.

Edited by TakenItSeriously

Share this post


Link to post
Share on other sites

How, exactly, is this supposed to prove the Collatz conjecture? You have a "probabilistic" reason to think that most things get down to 1; that simply is not enough. The conjecture must be true not just "probabilistically", but for all integers.

Edited by uncool
  • Upvote 1

Share this post


Link to post
Share on other sites
17 hours ago, uncool said:

How, exactly, is this supposed to prove the Collatz conjecture? You have a "probabilistic" reason to think that most things get down to 1; that simply is not enough. The conjecture must be true not just "probabilistically", but for all integers.

You’re right.

I thought that there was a bounded symmetry within regions related to Mersenne and perfect numbers where the number of odd and even numbers must eventually be equal. While that’s true for the numbers in the vertical direction, it’s not necessarily true for the series in the x direction.

It’s not a valid proof as I’ve stated it.

It does reveal some interesting relationships though like any Mersenne number will begin with an odd series equal to the exponential. while perfect squares, of course, converge to 1 in the same number of series.

Also it seems like there should be an exponential function that could predict the point where the series drops below n, based on a persistant pattern I see that seems to relate to an exponential curve.

Share this post


Link to post
Share on other sites
On 11/18/2017 at 6:32 PM, TakenItSeriously said:

You’re right.

I thought that there was a bounded symmetry within regions related to Mersenne and perfect numbers where the number of odd and even numbers must eventually be equal. While that’s true for the numbers in the vertical direction, it’s not necessarily true for the series in the x direction.

It’s not a valid proof as I’ve stated it.

It does reveal some interesting relationships though like any Mersenne number will begin with an odd series equal to the exponential. while perfect squares, of course, converge to 1 in the same number of series.

Also it seems like there should be an exponential function that could predict the point where the series drops below n, based on a persistant pattern I see that seems to relate to an exponential curve.

As I left this proof in the last post, I had believed that there was an even distribution of odd to even numbers within a Mersenne range that must eventually force a series to converge only my mistake was that I could only demonstrate that distribution existed in the y axis (positive integers) and not necessarily in the x axis (the series of integers created by the conjecture) as shown in Fig 1 in the second post.

However I noted that mersenne numbers and perfect squares showed a direct correlation such that given a mersenne number (2ᵐ-1) or aperfect square (2ᵐ) the series would initially expand with m+1 odd numbers in a row for the mersenne numbers but would contract directly to 1, with m even numbers in a row 

e.g. for m = 5, then  

MN = 31 and the PS = 32

their series would respectively begin as:

31, 47, 71, 107, 161, 245...

or 6 odd numbers

32, 16, 8, 4, 2, 1.

or 5 even numbers plus 1

showing that there must be something that linked the x axes to the y axes in terms of their odd or even results with respect to their exponential integers.

After following up on those clues:

I believe that I made a profound discovery that may provide the missing step to finishing this proof which is based on showing that a hidden symmetry exists between the x & y axes. However, I’m still uncertain of if or how it prooves that the distribution of odd to even numbers must force a convergence at least as far as I can understand it, at the moment, at least until I have had time to digest this rather odd discovery a little more.

Perhaps a mathematician can interpret it better.

Since the pattern itself clearly shows that an x to y symmetry exists without yet fully understanding how it works right now I will first just show the pattern I discovered:

BEB2CF60-5353-4940-B8E1-2B0737757476.thumb.png.fd84918a6c66fd5538e898f5b6ac84a6.png

Figure 2: the first 30 of 64 numbers in a series for n = 27 are shown.for a series produced by the collatz conjecture, bottom row.

The 30 x 30 matrix above represents 30 numbers in the y axis that each produce their own series in the x axis (first 30 only are displayed).

The function that produces the numbers for the y axis is:

*f(y) = 2ᵐ+n

for y = (1:30)

where

n is some positive integer

m is position in the series and also the exponential in the function.

* note: I’m sure I expressed that improperly, but I was in too much of a hurry to post to look up the proper symbology form which I can never seem to retain due to my disfunctional short term memory)

Note that one important consequence of this function is that for any position (m) in a series for any given number (n), you can calculate which numbers will produce the identical pattern of odd and even results up to that parcicular position, after which, the series will always diverge, taking independent paths.

eg.

for n = 31 and the 6th position in the series:

n = 31: 

31, 47, 71, 107, 161, 242, 121, 182

 

y = 2⁶+31 = 95

n = 95: 

95, 143, 215,  323, 485, 728, 364, 182...

e.i. both series begin with 5 even numbers and 1 odd but they diverge ofter that.

o,o,o,o,o,e,o,e

o,o,o,o,o,e,e,e

if we chose m = 7 then the pattern would be identical to 7 positions.

I hope to follw up  with a more detailed, and clearly stated explanation shortly.

 

Edit to add:

After reading through it again, the reason I found the math confusing is probably due to the fact that its not using known math that I’m familiar with. The function which typically describes a curve or relationship between x and y axes is actually describing a scale of the y axes.

I’m not sure this operation or methodology falls into any particular field of math so it may require some new form  math, though I’m certainly no expert in terms of math.

 

Edited by TakenItSeriously

Share this post


Link to post
Share on other sites
7 hours ago, TakenItSeriously said:

As I left this proof in the last post, I had believed that there was an even distribution of odd to even numbers within a Mersenne range that must eventually force a series to converge only my mistake was that I could only demonstrate that distribution existed in the y axis (positive integers) and not necessarily in the x axis (the series of integers created by the conjecture) as shown in Fig 1 in the second post.

However I noted that mersenne numbers and perfect squares showed a direct correlation such that given a mersenne number (2ᵐ-1) or aperfect square (2ᵐ) the series would initially expand with m+1 odd numbers in a row for the mersenne numbers but would contract directly to 1, with m even numbers in a row 

e.g. for m = 5, then  

MN = 31 and the PS = 32

their series would respectively begin as:

31, 47, 71, 107, 161, 245...

or 6 odd numbers

32, 16, 8, 4, 2, 1.

or 5 even numbers plus 1

showing that there must be something that linked the x axes to the y axes in terms of their odd or even results with respect to their exponential integers.

After following up on those clues:

I believe that I made a profound discovery that may provide the missing step to finishing this proof which is based on showing that a hidden symmetry exists between the x & y axes. However, I’m still uncertain of if or how it prooves that the distribution of odd to even numbers must force a convergence at least as far as I can understand it, at the moment, at least until I have had time to digest this rather odd discovery a little more.

Perhaps a mathematician can interpret it better.

Since the pattern itself clearly shows that an x to y symmetry exists without yet fully understanding how it works right now I will first just show the pattern I discovered:

BEB2CF60-5353-4940-B8E1-2B0737757476.thumb.png.fd84918a6c66fd5538e898f5b6ac84a6.png

Figure 2: the first 30 of 64 numbers in a series for n = 27 are shown.for a series produced by the collatz conjecture, bottom row.

The 30 x 30 matrix above represents 30 numbers in the y axis that each produce their own series in the x axis (first 30 only are displayed).

The function that produces the numbers for the y axis is:

*f(y) = 2ᵐ+n

for y = (1:30)

where

n is some positive integer

m is position in the series and also the exponential in the function.

* note: I’m sure I expressed that improperly, but I was in too much of a hurry to post to look up the proper symbology form which I can never seem to retain due to my disfunctional short term memory)

Note that one important consequence of this function is that for any position (m) in a series for any given number (n), you can calculate which numbers will produce the identical pattern of odd and even results up to that parcicular position, after which, the series will always diverge, taking independent paths.

eg.

for n = 31 and the 6th position in the series:

n = 31: 

31, 47, 71, 107, 161, 242, 121, 182

 

y = 2⁶+31 = 95

n = 95: 

95, 143, 215,  323, 485, 728, 364, 182...

e.i. both series begin with 5 even numbers and 1 odd but they diverge ofter that.

o,o,o,o,o,e,o,e

o,o,o,o,o,e,e,e

if we chose m = 7 then the pattern would be identical to 7 positions.

I hope to follw up  with a more detailed, and clearly stated explanation shortly.

 

Edit to add:

After reading through it again, the reason I found the math confusing is probably due to the fact that its not using known math that I’m familiar with. The function which typically describes a curve or relationship between x and y axes is actually describing a scale of the y axes.

I’m not sure this operation or methodology falls into any particular field of math so it may require some new form  math, though I’m certainly no expert in terms of math.

 

 Correction:

In case anyone tries to verify the results shown in the pic, I mislabled the odd and even numbers. the light blue cells should be even numbers while the peach colored cells would be odd.

Edited by TakenItSeriously

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now