# 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! π

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 $n_{0}$ β‘ $1$ (mod $4$) solutions for a given index, t. The solutions are not unique for a given index, t. We assume $n_{0} \gets 4n_{0} + 1$.

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, $n_{0} = 57, 855$.

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! πβ

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.

"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!" π P.S. Go Blue! π β

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