$2^m - 1$ as the Integer Formulation to Govern the Dynamics of Collatz-Type Sequences

Preprint

Article

$2^m - 1$ as the Integer Formulation to Govern the Dynamics of Collatz-Type Sequences

Altmetrics

Downloads

283

Views

187

Comments

Gaurav Goyal^*

This version is not peer-reviewed

Submitted:

24 July 2024

Posted:

25 July 2024

You are already at the latest version

Alerts

Abstract

It has been discovered that representing odd integers as modified binary expressions of the form $\sum \limits_{n > m}2^n + 2^m - 1$ for $m \geq 1$ helps in understanding the dynamics of Collatz-type sequences. Starting with the original Collatz sequence $3x + 1$, it is found that when the odd step is applied to an odd integer $\sum \limits_{n > m}2^n + 2^m - 1$, an even integer $3\left(\sum \limits_{n > m}2^{n}\right) + 2^{m + 1} + 2^m - 2$ is obtained, which is exactly once divisible by 2, unless the lowest index reduces to zero. This implies that the sequence alternates between odd and even steps $m$ times. This governs the dynamics of the Collatz-type sequences because the value of $m$ determines the number of times the integer can be divided by 2 in each even step. A shortcut method is developed based on this dynamics that states that the even integer after $m$ odd-even steps are completed is $\left(\left(\frac{3}{2}\right)^m \sum \limits_{n > m}2^n\right) + 3^m - 1$. A shortcut method of this magnitude has never been utilized anywhere. The shortcut method for the modified Collatz sequence $5x + 1$ is also presented.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

The Collatz problem [1,2,3,4], defines the following set of rules: If n is odd, multiply it by 3, and add 1. If n is even, it is divided by 2.

The associated Collatz conjecture states that every integer ultimately reduces to unity. To prove this conjecture, it must be shown not only that the sequence eventually cycles through 1, 4, 2, 1, but also that no integer diverges to infinitely larger integers [5,6].

While a complete proof may be impossible, this article attempts to understand the working of Collatz-type sequences. For this, the odd integers are expressed as modified binary expressions ending in

2^{m} - 1

for

m \geq 1

. By examining the integers that result from this seed form, insights are gained into the patterns of the odd-even steps. The conditions that govern the progression of Collatz-type sequences are immediately made clear through the use of the modified binary form ending with

2^{m} - 1

. For this reason,

2^{m} - 1

is stated as the governing integer formulation for Collatz-type sequences.

2. Behavior of Collatz Sequence $3 n + 1$

Let x be an odd integer ending with

2^{m} - 1

. The term

2^{m} - 1

is crucial because it governs the sequence’s behavior. Although calculations will only explicitly show the evolution of this term, it is implied that the full odd integer, including higher index terms, is present but not explicitly written out.

Let

O {1}, O {2}, \dots

denote the resulting integer at the end of the odd step while

E {1}, E {2}, \dots

denote the resulting integer at the end of the even step. The modified binary expression of integers at the end of each step is:

\begin{matrix} x & = 2^{m} - 1 \\ O {1} & = (2 + 1) (2^{m} - 1) + 1 \\ = 2^{m + 1} + 2^{m} - 2 \\ E {1} & = 2^{m} + 2^{m - 1} - 1 \\ O {2} & = (2 + 1) (2^{m} - 1) + 1 + (2 + 1) (2^{m - 1}) \\ = 2^{m + 1} + 2^{m} - 2 + (2 + 1) (2^{m - 1}) \\ E {2} & = 2^{m} + 2^{m - 1} - 1 + (2 + 1) (2^{m - 2}) \\ O {3} & = 2^{m + 1} + 2^{m} - 2 + (2 + 1) (2^{m - 1}) + {(2 + 1)}^{2} (2^{m - 2}) \\ E {3} & = 2^{m} + 2^{m - 1} - 1 + (2 + 1) (2^{m - 2}) + {(2 + 1)}^{2} (2^{m - 3}) \end{matrix}

The above expressions are re-written in the expanded form as follow:

\begin{matrix} x & = 2^{m} - 1 \\ O {1} & = 2^{m + 1} + 2^{m} - 2 \\ E {1} & = 2^{m} + 2^{m - 1} - 1 \\ O {2} & = 2^{m + 2} + 2^{m - 1} - 2 \\ E {2} & = 2^{m + 1} + 2^{m - 2} - 1 \\ O {3} & = 2^{m + 2} + 2^{m + 1} + 2^{m - 1} + 2^{m - 2} - 2 \\ E {3} & = 2^{m + 1} + 2^{m} + 2^{m - 2} + 2^{m - 3} - 1 \end{matrix}

The following observations are made based on the expanded form:

Under normal conditions, each $O$ step is followed by exactly one $E$ step.
However, after each $OE$ step, the value of the lowest index decreases. Consequently, after m $OE$ steps, the lowest index becomes zero.
When the first $2^{0}$ term is reached, it cancels out the negative 1, resulting in an even integer. Additional $E$ steps are performed until the integer becomes odd again, meaning the next lowest index also becomes zero and a positive 1 is obtained.
The positive 1 is rewritten as $2^{1} - 1$ . An $O$ step is then performed, followed immediately by two $E$ steps.
This process repeats until m additional $E$ steps are completed. These m additional $E$ steps reduce the term $2^{m}$ to $2^{0}$ , causing all lower index terms to vanish. The value m is also subtracted from higher indices.

This demonstrates that, despite the overall growth of the integer, terms with progressively smaller indices are being generated. Once the smallest index is reduced to zero, it triggers a domino effect, causing the next smallest index to also become zero. This cascading process continues until all indices are reduced to zero.

Let the even integer obtained at the first even step after the

m^{t h}

odd step be

E^{(1)} {m}

\begin{matrix} E^{(1)} {m} & = 2^{m} + 2^{m - 1} + (2 + 1) (2^{m - 2}) + \dots + {(2 + 1)}^{m - 1} - 1 \\ = 3^{m} - 1 \end{matrix}

Suppose the actual integer is

\sum_{n > m} 2^{n} + 2^{m} - 1

, then the integer obtained at

E^{(1)} {m}

becomes

\begin{matrix} E^{(1)} {m} & = ({(\frac{3}{2})}^{m} \sum_{n > m} 2^{n}) + 3^{m} - 1 \end{matrix}

A few examples are given in Table 1.

3. Controlling the Collatz Sequence

Since the behavior of the Collatz sequence is understood, it is now possible to estimate the starting integer based on a given pattern of

OE

cycles, or to estimate the cycle pattern based on the integer. An example of estimating the integer based on a given

OE

cycle is presented.

3.1. Estimating Integer based on a Cycle Pattern

Suppose the following cycle pattern is desired:

︸_{α}^{O E E} ︸_{β}^{O E E} ︸_{γ}^{O E O E E}

The cycle is segmented into

α

β

, and

γ

blocks depending on termination by the extra

E

step. Let the integer be of the form

2^{γ} + 2^{β} + 2^{α} - 1

γ > β > α

. The value of

α

is visually determined as 1.

The residue of

2^{α} - 1

at the end of

α

block is one. The lowest index term at the end of

α

block determine the cycle pattern for the

β

block. The lowest index terms resulting from

2^{β}

and

2^{γ}

at the end of the

α

block are estimated using binary formulation in previous section.

\begin{matrix} O E E (x) & = \dots + 2^{γ - 1} + 2^{γ - 2} + 2^{β - 1} + 2^{β - 2} + 1 \end{matrix}

The higher index terms are ignored. The integer formulation should end in

2^{1} - 1

since there is one

O E

cycle. Therefore, the odd integer resulting from

2^{β - 2} + 1

should be equivalent to

2^{1} - 1

\begin{matrix} 2^{β - 2} + 1 & \equiv 2^{1} - 1 \\ β & \geq 4 \end{matrix}

In a similar manner, the lowest index at the end of

β

block are

\begin{matrix} O E E O E E (x) & = \dots + 2^{γ - 2} + 2^{γ - 3} + 2^{γ - 3} + 2^{γ - 4} + 2^{β - 2} + 2^{β - 3} + 2^{β - 3} + 2^{β - 4} + 1 \\ O E E O E E (x) & = \dots + 2^{γ - 1} + 2^{γ - 4} + 2^{β - 1} + 2^{β - 4} + 1 \\ O E E O E E (x) & = \dots + 2^{γ - 4} + 2^{β - 4} + 1 \end{matrix}

The higher index terms obtained from coalescing lower index terms are removed. The lowest index terms are

2^{γ - 4} + 2^{β - 4}

and this should be equivalent to

2^{2} - 1

since the

O E

continues for two cycle.

\begin{matrix} 2^{γ - 4} + 2^{β - 4} + 1 & \equiv 2^{2} - 1 \end{matrix}

β = 5

, the term

2^{β - 4} + 1

becomes equal to

2^{2} - 1

. Therefore, the value of

γ

is greater than 4. Some of the integers obtained for different values of

γ

along with their cycle are given.

3.1.1. $γ = 5$

The integer is

2^{5} + 2^{5} + 2^{1} - 1 = 65

. The Collatz cycle is

OEEOEEOEEE

. The cycle is correct till the

β

block but then differs for the

γ

block. It happens because the value of

γ

and

β

is taken equal that results in index coalescing.

3.1.2. $γ = 6$

The integer is

2^{6} + 2^{5} + 2^{1} - 1 = 97

. The Collatz cycle is

OEEOEEOEOEOEE

. The cycle is improvement over the previous one but still not correct. This occurs because

2^{γ - 4} + 2^{β - 4} + 1 = 2^{2} + 2^{2} - 1

. As one can see, the indices are same and coalesce to become

2^{3} - 1

for which the

OE

repeat three times.

3.1.3. $γ = 7$

The integer is

2^{7} + 2^{5} + 2^{1} - 1 = 161

and the Collatz cycle is

OEEOEEOEOEE

which is the desired cycle.

3.1.4. $γ = 8$

The integer is

2^{8} + 2^{5} + 2^{1} - 1 = 289

and the Collatz cycle is

OEEOEEOEOEE

which is also the desired cycle. Similarly, any integer

2^{γ} + 2^{5} + 2^{1} - 1

will follow the desired cycle for any

γ \geq 7

4. Constructing Integers That ‘Might’ Diverge

There are two ways the sequence can diverge:

m is infinite.
Alternatively, if every time the binary expression of an integer ends in positive 1 and is re-written as $2^{1} - 1$ , there is a sequence of $2^{1} + 2^{2} + 2^{3} + \dots$ that combines with $2^{1}$ to produce an index larger than m.

This discussion focuses on the second method of divergence. Let the seed integer be of the form

x = 2^{b} + 2^{a} - 1

As a starting point, let

a = 1

. The integer obtained after applying

OEE

is:

\begin{matrix} OEE (x) & = 2^{b - 1} + 2^{b - 2} + 2 - 1 \end{matrix}

If the terms

2^{b - 1} + 2^{b - 2}

are set to

4 + 2

for

b = 3

, the integer obtained after

OEE

becomes

OEE (x) = 2^{3} - 1

, and the seed integer is

x = 9

Next, let the seed integer be

x = 2^{c} + 2^{3} + 2^{1} - 1

. The integer obtained after applying

OEEOEOEOEE

is:

\begin{matrix} OEEOEOEOEE (x) & = OEEOEOEOEE (2^{a}) + 2^{4} + 2^{3} + 2^{1} \\ = 2^{a + 1} + 2^{a - 1} + 2^{a - 5} + 2^{4} + 2^{3} + 2^{1} \\ OEEOEOEOEEE (x) & = 2^{a} + 2^{a - 2} + 2^{a - 6} + 2^{3} + 2^{2} + 2 - 1 \end{matrix}

The integer is reduced to odd value in last step. For instance, let

a = 7

. The resulting integer is

2^{7} + 2^{5} + 2^{4} - 1

, giving a seed integer of

x = 137

. This shows that the value of the lowest index has increased once again.

Higher indices can be calculated in a similar manner. Changing the value of a will yield different seed integers.

5. Application to Collatz-Type $5 x + 1$ Sequence

Consider the modified binary expression of integers obtained when

5 x + 1

is applied to

2^{m} - 1

\begin{matrix} x & = 2^{m} - 1 \\ O {1} & = (2^{2} + 1) (2^{m} - 1) + 1 \\ = 2^{m + 2} + 2^{m} - 2^{2} \\ E^{(1)} {1} & = 2^{m + 1} + 2^{m - 1} - 2 \\ E^{(2)} {1} & = 2^{m} + 2^{m - 2} - 1 \\ O {2} & = (2^{2} + 1) (2^{m} - 1) + 1 + (2^{2} + 1) (2^{m - 2}) \\ = 2^{m + 2} + 2^{m} - 2^{2} + (2^{2} + 1) (2^{m - 2}) \\ E^{(1)} {2} & = 2^{m + 1} + 2^{m - 1} - 2 + (2^{2} + 1) (2^{m - 3}) \\ E^{(2)} {2} & = 2^{m} + 2^{m - 2} - 1 + (2^{2} + 1) (2^{m - 4}) \\ O {3} & = 2^{m + 2} + 2^{m} - 2^{2} + (2^{2} + 1) (2^{m - 2}) + {(2^{2} + 1)}^{2} (2^{m - 4}) \\ E^{(1)} {3} & = 2^{m + 1} + 2^{m - 1} - 2 + (2^{2} + 1) (2^{m - 3}) + {(2^{2} + 1)}^{2} (2^{m - 5}) \\ E^{(2)} {3} & = 2^{m} + 2^{m - 2} - 1 + (2^{2} + 1) (2^{m - 4}) + {(2^{2} + 1)}^{2} (2^{m - 6}) \end{matrix}

The following observations are made:

If m is odd then the pattern $O E^{(1)} E^{(2)}$ repeats for $(\frac{m - 1}{2})$ times. The integer obtained at $E^{(1)} {\frac{m + 1}{2}}$ is odd and a $O$ step follows, resulting in the pattern terminating with $O E^{(1)} O$ .
If m is even then the pattern $O E^{(1)} E^{(2)}$ repeats for $(\frac{m}{2} - 1)$ times. The integer obtained at $E^{(2)} {\frac{m}{2}}$ is even and another $E$ step follows, resulting in the pattern terminating with $O E^{(1)} E^{(2)} E^{(3)}$ . The actual number of $E$ steps depend on the integer.

5.1. Case 1: m is odd

Suppose m is odd and the

O {\frac{m + 1}{2} + 1}

step occurs after the

E^{(1)} {\frac{m + 1}{2}}

step. The even integer obtained at

O {\frac{m + 1}{2} + 1}

\begin{matrix} O \{\frac{m + 1}{2} + 1\} & = (2^{2} + 1) (2^{m + 1} - 2) + 1 \\ + (2^{2} + 1) (2^{m - 1}) + {(2^{2} + 1)}^{2} (2^{m - 3}) + \dots + {(2^{2} + 1)}^{\frac{m + 1}{2}} \\ = 2^{m + 3} + 2^{m + 1} \\ + (2^{2} + 1) (2^{m - 1}) + {(2^{2} + 1)}^{2} (2^{m - 3}) + \dots + {(2^{2} + 1)}^{\frac{m + 1}{2}} - 9 \\ = 2^{m + 3} + 2^{m + 1} + 5^{\frac{m + 3}{2}} - 5 \cdot 2^{m + 1} - 9 \\ = 5^{\frac{m + 3}{2}} - 9 \end{matrix}

As before, if the actual integer is

\sum_{n > m} 2^{n} + 2^{m} - 1

, then the integer obtained at

O \{\frac{m + 1}{2} + 1\}

becomes

\begin{matrix} O \{\frac{m + 1}{2} + 1\} & = (10 \cdot {(\frac{5}{2^{2}})}^{\frac{m + 1}{2}} \sum_{n > m} 2^{n}) + 5^{\frac{m + 3}{2}} - 9 \end{matrix}

A few examples are given in Table 2.

5.2. Case 2: m is even

Let z additional

E

steps occur after

E^{(2)} {\frac{m}{2}}

. The resulting odd integer will be

\begin{matrix} E^{(2 + z)} \{\frac{m}{2}\} & = \frac{1}{2^{z}} (2^{m} + 2^{m - 2} - 1 + (2^{2} + 1) (2^{m - 4}) + \dots + {(2^{2} + 1)}^{\frac{m}{2} - 1}) \end{matrix}

The odd step produces an even integer given by

\begin{matrix} O \{\frac{m}{2} + 1\} & = \frac{1}{2^{z}} ((2^{2} + 1) (2^{m} - 1)) + 1 \\ + \frac{1}{2^{z}} ((2^{2} + 1) (2^{m - 2}) + {(2^{2} + 1)}^{2} (2^{m - 4}) + \dots + {(2^{2} + 1)}^{\frac{m}{2}}) \\ = \frac{1}{2^{z}} (5^{\frac{m}{2} + 1} - 5) + 1 \end{matrix}

As before, if the actual integer is

\sum_{n > m} 2^{n} + 2^{m} - 1

, then the integer obtained at

O \{\frac{m}{2} + 1\}

becomes

\begin{matrix} O \{\frac{m}{2} + 1\} & = (\frac{5}{2^{z}} \cdot {(\frac{5}{2^{2}})}^{\frac{m}{2}} \sum_{n > m} 2^{n}) + \frac{1}{2^{z}} (5^{\frac{m}{2} + 1} - 5) + 1 \end{matrix}

The variable z is chosen such that

O \{\frac{m}{2} + 1\}

is even.

A few examples are given in Table 3.

6. Conclusions

It has been recently discovered that expressing odd integers in the alternate binary form

2^{m} - 1

aids in understanding the workings of Collatz-type sequences. The understanding leads to formulations that generate the even integer obtained after m odd-even steps are completed. For the original Collatz sequence

3 x + 1

, the said even integer is given by

({(\frac{3}{2})}^{m} \sum_{n > m} 2^{n}) + 3^{m} - 1

. This formulation allows for construction on odd integers that follow a specific odd-even step pattern.

For the modified Collatz-type sequence

5 x + 1

, the formulation for the even integer depends on m. If m is odd, the even integer is

(10 \cdot {(\frac{5}{2^{2}})}^{\frac{m + 1}{2}} \sum_{n > m} 2^{n}) + 5^{\frac{m + 3}{2}} - 9

. However, when m is even, the even integer is

(\frac{5}{2^{z}} \cdot {(\frac{5}{2^{2}})}^{\frac{m}{2}} \sum_{n > m} 2^{n}) + \frac{1}{2^{z}} (5^{\frac{m}{2} + 1} - 5) + 1

. Here, the maximum value of z is such that the integer remains even.

Data Availability Statement

Data availability is not applicable to this article as no new data were created or analysed in this study.

References

Lagarias, J.C. The 3x+ 1 problem: An annotated bibliography (1963–1999). The ultimate challenge: the 3x 2003, 1, 267–341. [Google Scholar]
Lagarias, J.C. The 3x+ 1 problem: An annotated bibliography. preprint 2004. [Google Scholar]
Lagarias, J.C. The 3x+ 1 problem: An annotated bibliography, II (2000-2009). arXiv preprint math/0608208 2006. [Google Scholar]
Lagarias, J.C. The ultimate challenge: The 3x+ 1 problem; American Mathematical Soc., 2010.
Terras, R. A stopping time problem on the positive integers. Acta Arithmetica 1976, 3, 241–252. [Google Scholar] [CrossRef]
Tao, T. Almost all orbits of the Collatz map attain almost bounded values. Forum of Mathematics, Pi. Cambridge University Press, 2022, Vol. 10, p. e12.

Table 1. Examples of the even integer obtained at

E^{(1)} {m}

for various seed integers.

Table 1. Examples of the even integer obtained at

E^{(1)} {m}

for various seed integers.

Integer	Modified binary	$E^{(1)} {m}$	Integer value
7	$2^{3} - 1$	$3^{3} - 1$	26
19	$2^{4} + 2^{2} - 1$	${(\frac{3}{2})}^{2} 2^{4} + 3^{2} - 1$	44
34603007	$2^{25} + 2^{20} - 1$	${(\frac{3}{2})}^{20} 2^{25} + 3^{20} - 1$	115063885232
57343	$2^{15} + 2^{14} + 2^{13} - 1$	${(\frac{3}{2})}^{13} (2^{15} + 2^{14}) + 3^{13} - 1$	11160260

Table 2. Values of even integer obtained at

O \{\frac{m + 1}{2} + 1\}

for various seed integers with odd m.

Table 2. Values of even integer obtained at

O \{\frac{m + 1}{2} + 1\}

for various seed integers with odd m.

Integer	Modified binary	$O \{\frac{m + 1}{2} + 1\}$	Integer value
7	$2^{3} - 1$	$5^{\frac{3 + 3}{2}} - 9$	116
95	$2^{6} + 2^{5} - 1$	$\begin{matrix} c 10 \cdot {(\frac{5}{2^{2}})}^{\frac{5 + 1}{2}} 2^{6} + 5^{\frac{5 + 3}{2}} - 9 \end{matrix}$	1866
57343	$2^{15} + 2^{14} + 2^{13} - 1$	$\begin{matrix} c 10 \cdot {(\frac{5}{2^{2}})}^{\frac{13 + 1}{2}} (2^{15} + 2^{14}) + 5^{\frac{13 + 3}{2}} - 9 \end{matrix}$	2734366

Table 3. Values of even integer obtained at

O \{\frac{m}{2} + 1\}

for various seed integers with even m. (*6 is obtained when

O

is applied to 1.)

Table 3. Values of even integer obtained at

O \{\frac{m}{2} + 1\}

for various seed integers with even m. (*6 is obtained when

O

is applied to 1.)

Integer	Modified binary	$O \{\frac{m}{2} + 1\}$	z	Integer value
3	$2^{2} - 1$	$\frac{1}{2^{z}} (5^{\frac{2}{2} + 1} - 5) + 1$	2	6*
79	$2^{6} + 2^{4} - 1$	$\begin{matrix} c (\frac{5}{2^{z}} \cdot {(\frac{5}{2^{2}})}^{\frac{4}{2}} 2^{6}) + \frac{1}{2^{z}} (5^{\frac{4}{2} + 1} - 5) + 1 \end{matrix}$	2	156
53247	$2^{15} + 2^{14} + 2^{12} - 1$	$\begin{matrix} c (\frac{5}{2^{z}} \cdot {(\frac{5}{2^{2}})}^{\frac{12}{2}} (2^{15} + 2^{14})) + \frac{1}{2^{z}} (5^{\frac{12}{2} + 1} - 5) + 1 \end{matrix}$	2	253906

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

MDPI Initiatives

Important Links

Choose an area of interest and we will send you notifications of new preprints at your preferred frequency.

Disclaimer

$2^m - 1$ as the Integer Formulation to Govern the Dynamics of Collatz-Type Sequences

Abstract

1. Introduction

2. Behavior of Collatz Sequence 3 n + 1

3. Controlling the Collatz Sequence

3.1. Estimating Integer based on a Cycle Pattern

3.1.1. γ = 5

3.1.2. γ = 6

3.1.3. γ = 7

3.1.4. γ = 8

4. Constructing Integers That ‘Might’ Diverge

5. Application to Collatz-Type 5 x + 1 Sequence

5.1. Case 1: m is odd

5.2. Case 2: m is even

6. Conclusions

Data Availability Statement

References

MDPI Initiatives

Important Links

Subscribe

2. Behavior of Collatz Sequence $3 n + 1$

3.1.1. $γ = 5$

3.1.2. $γ = 6$

3.1.3. $γ = 7$

3.1.4. $γ = 8$

5. Application to Collatz-Type $5 x + 1$ Sequence