A Brief Analysis of the Collatz Conjecture

Algorithm for the Collatz Conjecture.

Remark: If n = 1, the algorithm terminates all computations.

Crucial Idea: R+ is a dense set.

Remark: R+ is the set of all positive real numbers.

We claim

  1. n_{t} = (\frac{3}{2} )^{\frac{t}{2}} *  (\frac{3}{4} )^{\frac{t}{4}} *  (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} * r = 1

over the Collatz sequence of positive odd integers, from n_{0} > 1 to n_{t} = 1 where the index, t, is the number of trials it takes the Collatz sequence of odd integers to converge to one.

Remark: There are no infinite (nontrivial) cycles of any length ( (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} \rightarrow 0 and r \rightarrow \infty as t \rightarrow \infty) in the Collatz sequence since the index, t, is clearly finite in equation one.

In addition, our claim is fundamentally based on the distribution of maximum divisors, 2^{i_{max}}, for finite sets of consecutive positive even integers (e.g. {2, 4, 6, …, e_{max}}) where 2 is the minimum positive even integer and where e_{max} is the maximum positive even integer belonging to those sets…

The positive real number, r, is determined by the algorithm for the Collatz conjecture. And therefore, its computation is generally complicated because we cannot easily compute the maximum positive even integer, e_{max}, in the Collatz sequence (orbit) for many odd positive integers, n_{0} >  2^{68}.

Question: For any odd positive integer, n_{0} >  2^{68} , can we approximate the maximum positive even integer, e_{max}, in the Collatz sequence (orbit)?

The variable, k, is determined by the maximum positive even number, e_{max}, in the Collatz sequence: k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor.

Can we refute our analysis?

We assume maximum divisors, 2^{i}, for any positive even integer.

Remark: The Collatz conjecture is true for all positive integers, n_{0} \le 2^{68}.

For any positive odd integer, n_{0} > 2^{68} -1, we have

n_{1} =  \frac{3n_{0} +1}{2^{i_{1}}} = \frac{3}{2^{i_{1}}} * r_{1} where r_{1} = n_{0} +\frac{1}{3}.

n_{2} = \frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} * r_{2}

where r_{2} = \frac{1}{9}(2^{i_{1}} + 9n_{0} +3).

n_{3} = \frac{3(\frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}}) + 1}{2^{i_{3}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} *\frac{3}{2^{i_{3}}} *r_{3}

where r_{3} = \frac{2^{i_{1}}}{27}*(2^{i_{2}} + 3) + n_{0} +\frac{1}{3}.

n_{t} = r_{t} * 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) where r_{t} > 1 since 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) < 1 and since n_{t} \ge 1.

Question: Is 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) > 1 possible? No!

Why? Hint: n_{0} > 1.

Therefore, r_{t} \ge 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}

Thus, r_{t} = 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})} or r_{t} > 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})}

Moreover, n_{t} = r_{t} * 3^{t}\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}}) =  r * (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} \ge 1 where r_{t} > r.

We conclude r_{t} = 3^{-t}\frac{1}{\prod_{j=1}^{t } (\frac{1}{2^{i_{j}}})} since r_{t} is a positive rational number and since there are no infinite (nontrivial) cycles in any Collatz sequence. Hence n_{t} = 1.

Important Remark: The Collatz Conjecture is true! πŸ™‚

Example: If we let n_{0} = 57, then we compute e_{max} = 196, k = 7, and t = 10.

Therefore, r = r(57, 10) = 1/.0841  394 = 11.8850384.

Dave’s Conjecture: r = r(n_{0 }, t) = O(t) or r = c_{t} * t for some real number, c_{t}, such that either c_{t}  >  1 or 0 < c_{t} < 1.

Example: If we have t = 1 for some n_{0 }, then as k \rightarrow \infty,

r = \frac{1}{ \prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}}} \rightarrow \frac{4}{3 } = c_{1 }.

Therefore, r = r(n_{0 }, 1) \approx \frac{4}{3} for infinitely many positive odd integers, n_{0} > 1.

Remark: For our example, we compute n_{0 } \in {5, 21, 85, 341, …, \frac{2^{2j} - 1}{3}, …} for all positive integers, j > 1.

Questions: What are the values (convergent) for c_{2}, c_{3}, c_{4}, ...?

Example: We compute r = r(n_{0 }, 2) \approx 1 \frac{7}{9} for infinitely many positive odd integers, n_{0} > 1.

And therefore, c_{2} = \frac{8}{9} where

n_{0 } \in {7281, 29125, 116501, 466005,…, 4 *l_{j}+1, …} for all positive integers, j \ge 4.

Remark: For our example, we assume l_{1} = 7,281,  l_{2} = 29,125, l_{3} = 116,501, l_{4} = 466,005.

Old Example: If we let n_{0} = 57, then we compute e_{max} = 196, k = 7, and t = 10.

Therefore, r = r(57, 10) = 1/.0841  394 = 11.8850384. However,

r = r(n_{0}, 10) \approx \frac{1}{ \prod_{i=1}^{\infty }(\frac{3}{2^{i}})^{\frac{10}{2^{i}}}}  = \frac{1}{.0563135} \approx 17.75773127

for infinitely many positive odd integers, n_{0} > 1.

And therefore, c_{10}  \approx 1.775773127 where

n_{0 } \in {57, 229, 917, 3669,…, 4 *l_{j}+1, …} for all positive integers, j \ge 4.

Remark: For our old example, we assume l_{1} = 57, l_{2} = 229, l_{3} = 917, l_{4} = 3669.

For n_{0} = 57, we compute the following Collatz sequence where t = 10:

172

86

#1: 43

130

#2: 65

196

98

#3: 49

148

74

#4: 37

112

56

28

14

#5: 7

22

#6: 11

34

#7: 17

52

26

#8: 13

40

20

10

#9: 5

16

4

2

#10: 1

For n_{0} = 917, we compute the following Collatz sequence where t = 10:

2752

1376

688

344

172

86

#1: 43

130

#2: 65

196

98

#3: 49

148

74

#4: 37

112

56

28

14

#5: 7

22

#6: 11

34

#7: 17

52

26

#8: 13

40

20

10

#9: 5

16

8

4

2

#10: 1

Remark: The common Collatz sequence of positive odd integers for our old example is

{43, 65, 49, 37, 7, 11, 17, 13, 5, 1}.

Final Remark (Hmm): The 3n_{0} + 1 Problem (Collatz Conjecture) has n_{0} ≑ 1 (mod 4) solutions for a given index, t. The solutions are not unique for a given index, t. We assume n_{0} \gets 4n_{0} + 1.

Reference Link:

1. Wolfram Cloud (Mathematica Online);

The function, nvalue[t], computes a random positive odd integer, 21 \le n_{0} < 100,000 (initially), for a small index, 1 \le t \le 10 (number of trials for a Collatz sequence of odd integers to converge to one from the computed n_{0}).

Source Code:

nvalue[t_] := (
tt = t;
icnt = 0;
n = 2 * RandomInteger[{10, 50000}] + 1;
While[icnt != tt,
While[n != 1,
If[icnt == 0, nstart = n];
n = 3n + 1;
While[EvenQ[n], n = n/2];
icnt = icnt + 1;
If[icnt > tt, n = nstart + 2 *RandomInteger[{1,10}]]
If[icnt > tt, icnt = 0]]];
Return[{tt, nstart}])

__________________________________________________________

Some Examples:

nvalue[2] computes for t = 2 a random value, n_{0} = 54, 613;

nvalue[10] computes for t = 10 a random value, n_{0} = 67, 077;

nvalue[5] computes for t = 5 a random value, n_{0} = 7,885.

nvalue[6] computes for t = 6 a random value, n_{0} = 82,485.

nvalue[4] computes for t = 9 a random value, n_{0} = 69,965.

nvalue[1] computes for t = 1 a random value, n_{0} = 87, 381.

nvalue[41] computes for t = 41 a random value, n_{0} = 29, 137.

nvalue[75] computes for t = 75 a random value, n_{0} = 57, 855.

Graph of Collatz Sequence for n_{0} = 57, 855.

nvalue[75] computes for t = 75 a random value, n_{0} = 15, 051.

nvalue[150] computes for t = 150 a random value, n_{0} = 2, 139, 623.

nvalue[150] computes for t = 150 a random value, n_{0} = 2, 143, 079.

nvalue[200] computes for t = 200 a random value, n_{0} = 6, 899, 327.

nvalue[211] computes for t = 211 a random value, n_{0} = 21, 069, 775.

nvalue[250] computes for t = 250 a random value, n_{0} = 46, 912, 975.

nvalue[300] computes for t = 300 a random value, n_{0} =  330, 552, 319.

nvalue[351] computes for t = 351 a random value, n_{0} = 1, 505, 110, 398, 721, 395.

nvalue[351] computes for t = 351 a random value, n_{0} = 1, 514, 992, 426, 990, 283.

nvalue[408] computes for t = 408 a random value, n_{0} = 319, 762, 236, 301, 419, 029.

Go Blue! πŸ‘βœŒ

Relevant Reference Links:

Wolfram Alpha Computation (r3);

Collatz Conjecture Calculator;

Our_Response_to_4.2_A_probabilistic_heuristic;

Proof of Collatz Conjecture;

The Collatz Equation that supports the Collatz Conjecture;

An Analysis of the Collatz Conjecture.

“Counting and ordering stuff (objects, sets, numbers, spaces, etc.) are fundamental.”

P.S. FIGHT RACISM IN THE SCIENCES INCLUDING MATHEMATICS! THANK YOU!

Oops! The Proceedings of the London Mathematical Society rejected the paper,  “A Brief Analysis of the Collatz Conjecture, for publication! Why?

We are confident our work is valid, and we suspect our work was rejected because of political reasons… It happens…

Reference Link: Two Important Properties of Convergent Collatz Sequences.

Bakuage Offers Prize of 120 Million JPY to Whoever Solves Collatz Conjecture, Math Problem Unsolved for 84 Years.

Dave.

Published by Dave

"May all beings be happy and peaceful. Thank you, Lord GOD! Praise you, Lord GOD! Love you, Lord GOD! Trust you, Lord GOD! Bless you, Lord GOD! Amen!" πŸ˜‚ P.S. Go Blue! πŸ‘ ✌

10 thoughts on “A Brief Analysis of the Collatz Conjecture

  1. Wow that was unusual. I just wrote an very long comment but after I clicked submit my comment didn’t appear. Grrrr… well I’m not writing all that over again. Anyhow, just wanted to say excellent blog!

    Like

  2. Hello I am so happy I found your weblog, I really found you by error, while I was researching on Google for something else, Anyhow I am here now and would just like to say many thanks for a marvelous post and a all round interesting blog (I also love the theme/design), I don’t have time to browse it all at the moment but I have saved it and also included your RSS feeds, so when I have time I will be back to read more, Please do keep up the great job.

    Like

Leave a Reply to Victor Kimel Cancel reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: