MySeniorSeminar

The Collatz Conjecture

Andrew StanleyUniversity of Virginia at Wise

Math 490 Presentationadvisor: Dr. Matt Harvey

November 1, 2008

Abstract

This paper examines different representations of the Collatz Conjecture. The Collatz Con-jucture,

Theorem 0.1. Let n ∈ Z+,Then the function,

f(n) =

{3n+ 1 if n is oddn/2 if n is even

converges to 1 for all n.

We examine the phenomena the conjecture exhibits over Z, R, C and a polynomial repre-sentation in a field with multiplication and addition modulo 2.

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Contents

1 Introduction 3

2 The Collatz Conjecture in Z+ 32.1 Using 3n+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Negative Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.2 Using 5n+ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 The Collatz Conjecture in reverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 The Collatz conjecture in R 7

4 The Collatz conjecture in C 8

5 The Collatz Conjecture over a field 105.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.1.1 First degree Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115.1.2 Second degree Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

6 Conclusion 15

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1 Introduction

The Collatz Conjecture has an interesting history. Since the conception of the problem it hastaken many different names, the 3n+ 1 problem, the Syracuse problem, Hasse’s algorithm and the”Hailstone” problem these are the most common names used to reference the original conjecture.Each name uses different methods for trying to prove the conjecture.

The problem was first posed by Lothar Collatz in 1937. Collatz was a German mathemati-cian who studied at the University of Berlin; Collatz published most of his papers in the area ofnumerical analysis. Collatz’s conjecture is what he is mostly remembered for, Collatz publishedmany important works, he contributed practical applications to boundary problems for ordinaryand partial differential equations. While his contributions to math are very important, Collatz ismostly recognized for his simple conjecture. [1]

Since the conception of the problem, various mathematicians have examined the problem usingvarious methods, but the problem remains unsolved to this day. Many different methods of proofhave been attempted but none have been successful. Different analogues of the problem have beenstated and some of these have been proven but none have led to any insight on how to prove theoriginal conjecture.

The Collatz conjecture is simple in structure, choose a number if that number is odd, use3n+ 1, if the number is even use n

2 . After either of the functions is applied, repeat the process. Itis theorized that all n ∈ Z+ will converge to 1. This is a simple process, but proving the conjectureto be true is difficult.

In this paper we will examine various ways to look at the problem, first using examples fromZ, we will look at the behavior the problem exhibits. After looking at the conjecture in Z, wewill transform the piecewise equation into a function that is continuous, we will be looking at thebehavior of the Collatz conjecture in R. Then we will proceed to C then we will look at a polynomialrepresentation. We will be looking at the graphs of these various analogues to the conjecture andlooking at how the translations create new ways of viewing the problem.

2 The Collatz Conjecture in Z+

2.1 Using 3n + 1

The Collatz equation actually stems from the field of numerical analysis. This is the area in whichthe creator of this problem, Lothar Collatz, worked in. Collatz’s examples are slightly differentthan the examples that are shown here. The original question that Collatz asked was whether, fora specific function f , the trajectory starting with 8 and the iterates of 8, contains 1 or not. [2]

For this research project we have used the following conjecture, which, is an extension of theoriginal question. We begin with the following:


f(n) =



It is theorized that no matter what integer we begin with, successive iterations will eventuallyend with the sequence, (4, 2, 1). Once 1 is the theorem converges to 1, a loop is created that goesfrom 1→ 4, 4→ 2, 2→ 1. Looking at a few examples we have:

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Integer Terms of the Iteration

10 (10→ [5])9 (9→ 28, 28→ 14, 14→ [7])8 (8→ 4, 4→ 2, 2→ 1)7 (7→ 22, 22→ 11, 11→ 34, 34→ 17, 17→ 52, 52→ 26,

26→ 13, 13→ 40, 40→ 20, 20→ 10, 10→ 5, 5→ 16,16→ 8, 8→ 4, 4→ 2, 2→ 1)

6 (6→ 3, 3→ 10, 10→ 5, 5→ 16, 16→ 8, 8→ 4, 4→ 2, 2→ 1)5 (5→ 16, 16→ 8, 8→ 4, 4→ 2, 2→ 1)4 (4→ 2, 2→ 1)3 (3→ 10, 10→ 5, 5→ 16, 16→ 8, 8→ 4, 4→ 2, 2→ 2, 2→ 1)2 (2→ 1)1 (1→ 4, 4→ 2, 2→ 1)

Table 1: Iterations of the Collatz Conjecture

The previous example shows that for the first 10 integers, do in fact, converge to 1. Also, theterms have converged to 1 and a loop began that cycled the numbers (4, 2, 1).

In thinking of a method to prove the conjecture, The first method of proof one would think ofwould be that of the method of mathematical induction. This method in fact fails to work on thisparticular problem, because it does not help to assume that the first n cases are true to prove thecase for n+ 1.

This is what happens when we assume the theorem is true integers up to n. If n is odd, wemultiply it by three, then add one. Since we have done this, we have already went beyond ourinitial starting number, we cannot know for sure that any number greater than n is valid for theproof. So from one iteration of the function we have already jumped ahead to numbers that wecannot know are valid, so mathematical induction fails to prove the conjecture.[2]

Now we look at another way to examine the conjecture, that of a relation graph. With thegraph of the relation we can see the paths that each integer takes to 1. The relation graph lets uslook for any patterns that may be occurring in the conjecture. We can see from this graph thatstarting with some number and iterating the function, our numbers grow very quickly, but thenthe numbers eventually converge to some other number, once one path converges to another, wecan see the various paths the numbers take to reach 1.

This is called “hailstone” behavior.[4] The term hailstone is used because the conjecture emulatesthe process in which hailstones are created. Hail forms in clouds when supercooled water dropletsfreeze on contact with things in the air, such as dust. The storms updraft blows the hailstone intoan upper part of the cloud. When the updraft dissipates the hailstones fall back down and arelifted up again. As the hail continues this process it gets heavier, once the hail is to heavy to belifted back up again by the updraft, it falls to the ground.

The behavior of the collatz equation is much the same way. Our numbers grow larger, then it isshrunk again by n

2 , eventually the numbers reach a point where n2 drops our numbers to a sequence

that converge to 1.

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Figure 1: The Map of the Collatz Conjecture

2.1.1 Negative Integers

Now let us look at the Collatz conjecture using negative integers. Remember that our originaltheorem limits us to the positive integers. We ask here if all negative integers will end with thecycle (-4, -2, -1).


f(n) =



We have the following table:


-5 (−5→ −14,−14→ −7, 7→ −20,−20→ −10,−10→ −5)-4 (−4→ −2,−2→ −1)-3 (−3→ −8,−8→ −4,−4→ −2,−2→ −1)-2 (−2→ −1)-1 (−1→ −2,−2→ −1)

Table 2: Iterations of the collatz conjecture with n ∈ Z−

Here we see with -5 that not all n ∈ Z−, converge to -1, but, a loop is created.Now we begin looking at the collatz conjecture using different valued constants to see if the

conjecture holds true for different constants. The conjecture is initially stated for 3n+ 1, let’s lookat the behavior of when we change the coefficient 3 to another number.

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2.1.2 Using 5n+ 1

We begin by choosing another odd number for our coefficent for n. We have the following theorem:

Theorem 2.3. Let n ∈ Z+, then the function,

f(n) =



While working out the first few examples of the new theorem, it seemed like all the terms wouldreturn to 1. But using 5 as the starting point, the theorem no longer held. Iterating the functionfor 5 began a loop that never reached one, but began and ended with 26. At first the theoremseemed to follow the pattern that the 3n+ 1 problem ended with, (4,2,1).


5 (5→ 26, 26→ 13, 13→ 66, 66→ 33, 33→ 166, 166→ 83, 83→ 416,416→ 208, 208→ 104, 104→ 52, 52→ 26)

4 (4→ 2, 2→ 1, )3 (3→ 16, 16→ 8, 8→ 4, 4→ 2, 2→ 1)2 (2→ 11, 11→ 56, 56→ 23, 23→ 116, 116→ 58,

58→ 24, 24→ 12, 12→ 6, 6→ 3, 3→ 16, [16])1 (1→ 6, 6→ 3, 3→ 16, [16])

Table 3: Iterations of the Collatz Conjecture

Notice that the [16] means that the sequence has arrived at a previous numbers sequence andthe rest of the sequence is included in another numbers sequence.We see here again a pattern that emerged in one of our earlier examples. Initially the modifiedtheorem converged to one and created a loop of (4, 2, 1), but using 5, another cycle was createdthat never converged. It seems that it is easier to create a repeating loop than it is to find functionsthat will converge to 1.

2.2 The Collatz Conjecture in reverse

Now we want to see what happens when we begin with 1 and go in reverse. The collatz conjecturestates that if we pick some number n then it will go to 1. We begin here by looking at 1 andseeing what numbers it goes to. We want to see if the numbers between 1 and n are included in thesequence. The behavior that arises from this is interesting. The behavior is infrequent but when

3n+ 1 =n

2(1)

the paths “fork”.We have at n = 4 a trivial case, since when n = 4 the path begins to loop. The nontrivial cases

are at (10, 16, 22,...). There are branches that seem to have many ”forks” in it. One ”fork” richbranch is that of (1, 2, 3, 8, 2k), branches that seem to be “fork” poor are those that are multiplesof 3.Here we look at a radial drawing of the reverse Collatz conjecture. A radial drawing is a repre-sentation of a graph in which the vertices’s are constrained to lie on concentric circles of finiteradius.[6]

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Figure 2: Radial drawing of the reverse Collatz path

Figure 3: Layered drawing of the reverse Collatz path

3 The Collatz conjecture in R

We wanted to see what the conjecture looked like over C. To look at the behavior of the functionwe had to transform the piecewise function into an equation that was continuous. The first thingthat we wanted was the terms of the sequence to be bounded between 3x+ 1 and x

2 . Since we arein R we knew that to make the function continuous we needed to use trigonometric functions. The

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equation needed to take the form of:

f(x) = g(x) · (3x+ 1) + (1− g(x)) · x2

(2)

The equation needs to run between the values of 0 and 1. Using the graphs of sinx and cosx andshifting the graphs to fit what we needed. The values of sinx and cosx the we need are,

12

sin(πx− π

2

)and

12

cos(πx)

With these values for Sine and Cosine, we plug them into our previous equation:

f(x) =(

12

sin(πx− π

2

)+

12

)· (3x+ 1) +

(12

cos(πx)) +12

)·(x

2

)(3)

Which is bounded by the 3x+1 and x2 . Now that the function is continuous we look at the graph

of the equation over R+. Our graph covers the entire range of R+. Since our graph is continuouswe know that all of R+ is included in the domain of the function.

Figure 4: The Collatz function in R

4 The Collatz conjecture in C

Since the Collatz conjecture seemingly exhibits random and chaotic behavior. Looking at theconjecture over C seems like a natural place to examine the function. We want to graph thefunction and observe what is taking place. Fractals are a way to graph and recognize patterns inchaotic systems. Since the collatz problem is an iterative problem and displays chaotic behavior,we converted the collatz conjecture into a complex valued function to observe its behavior.

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f(z) =(−1

2cos(πz) +

12

(3z + 1))

+(

12

sin(πz) +12

(z2

))

Figure 5: The fractal associated with the Collatz Conjecture

An interesting property of fractals is that of self-similarity, meaning that if we zoom in to seethe details of a small subset, a replica of the original pattern appears.[5] One of the best ways toobserve fractals is through the use of software. While exploring the Mandelbrot set, zooming in onthe map at around -1.63, the image below was noticed. This is interesting because it is part of theequation z2 + c and not part of our equation for the collatz conjecture in the complex plane.

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Figure 6: A zoomed in view of the Mandelbrot set

5 The Collatz Conjecture over a field

The next thing we look at is the collatz conjecture using polynomial representations of the function.We limit this discussion to polynomials in field. This is a field with coefficients of 1 and 0 on ourpolynomials. We also have multiplication and addition modulo 2.

Theorem 5.1. The function

g(n) =

{f(n)

n if n divides f(n)(n+ 1)(f(n)) + 1 if n does not divide f(n)

converges to 1 for all polynomials.

5.1 Examples

We want to examine the behavior the polynomial representation presents. But first we need tounderstand what form the polynomials we are examining must take. The polynomials may be anydegree, the coefficents of each term must be either a 0 or 1. Degree one polynomials have only twoforms that they may take. We create a table to make sure that we have all the possible forms ofthe polynomials accounted for. Take the following table for degree 1 polynomials.(

1 11 0

)10

This table gives us the coefficient for every term in our polynomials. This table represents thepolynomials; x and x+ 1.

The matrix for degree two polynomials is:1 1 11 1 01 0 11 0 0

For any degree polynomial that we have, we can create a table like this by using 2n to know how

many rows our matrix will need. This will actually give us twice as many rows that we actuallyneed. The first column of the matrix will be such that half of the rows are 1’s and the other halfwill be zeros. We are not intersted in the part of the matrix that has 0 for the coefficient of thedegree we are looking at. So we only concern ourselves with coefficients of 1 in the first column.

5.1.1 First degree Polynomials

x:

f(x)x

=x

x= 1

(x+ 1):

(x+ 1)f(x) + 1 = (x+ 1)(x+ 1)

= x2 + 2x+ 2

= x2

= x

= 1

5.1.2 Second degree Polynomials

Now we look at second degree polynomials. The second degree polynomials are:x2, x2 + x, x2 + 1, x2 + x+ 1.

x2:

f(x)x

=x2

x

=x

x= 1

x2 + 1:

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(x+ 1)(f(x)) + 1 = (x+ 1)(x2 + 1) + 1

= x3 + x2 + x+ 1 + 1

= x3 + x2 + x

= x2 + x+ 1

(x+ 1)(f(x)) + 1 = (x+ 1)(x2 + x+ 1) + 1

= x3 + 2x2 + 2x+ 2

= x3

= x2

= x

= 1

x2 + x:

f(x)x

=x2 + x

x= (x+ 1)

We can stop here with this example because it is a first degree polynomial. By referring to theexamples of the first degree polynomials, we see that this does in fact go to 1. A similar argumentcan be made for x2 + x+ 1 since it appeared in the previous example of x2 + 1. This is in fact thepattern that we need to prove that the conjecture is true for all iterations of the theorem. We wantto show that the degree of the polynomial goes up by at least a degree of one before it returns tof(x)

x and converges to 1.

Proof. We proceed with induction.Check for first degree polynomials.Check x:x is divisible by x, so we have.

f(x)x

=x

x= 1.

Check (x+ 1):(x+ 1) is not divisible by x so,

(x+ 1)(f(x) + 1) = (x+ 1)((x+ 1) + 1)

= x2 + 2x+ 2

= x2

=x

x= 1

12

So first degree Polynomials reduce to 1.Suppose that for all k degree polynomials that the function will reduce to 1 over a number ofiterated steps.Let x be a polynomial of degree k + 1, thenxk+1

and

f(x)x

=xk+1

x

=xk · xx

= xk

...= 1

So xk+1 reduces to 1 over a number of steps.Next we look at (x+ 1)(f(x)) + 1.

We want to show that xk+1 + . . .+ xj + 1) converges to 1.Where j < k.

(x+ 1)(f(x)) + 1 = (x+ 1)(xk+1 + . . .+ xj + 1)

= xk+2 + . . .+ xj+1 + x+ xk+1 + . . .+ xj + 1 + 1

= xk+2 + kk+1 + . . .+ xj+1 + kj + x

= kk+1 + xk + . . .+ xj + xj−1 + 1

= (x+ 1)(xk+1 + . . .+ xj−1 + 1) + 1

= xk+2 + . . .+ xj + x+ xk+1 + . . .+ xj−1 + 1 + 1

= xk+2 + xk+1 + . . .+ xj + xj−1 + x

= kk+1 + xk + . . .+ xj−1 + xj−2 + 1...

= xk+1 + . . .+ x0 + 1

= xk+1 + . . .+ 1 + 1

= xk + . . .

...= 1

We can see as we proceed through iterations of the function, that the degree of the polynomialgoes up at least one degree before the next iteration brings it back down. Through the process wecan see that j continues to decrease. Eventually the degree of j will equal zero and with addittion

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modulo 2, that term will go away. After this happens our other degree terms start approachingzero as well. Continuing the process we reach our last term, xk, which will converge to 1.

We ask now, does the result still hold if we replace x+ 1 with other polynomials? We begin bylooking at a few examples, if we use x2 + 1 as our polynomial, will the other polynomials reduce to1.

Theorem 5.2. For all polynomials in field 2, the function

g(x) =

{f(x)

x if x divides f(x)(x2 + 1)(f(x)) + 1 if x does not divide f(x)

converges to 1.

Check for degree one polynomials:It is trivial that x goes to 1.Check for x+ 1.

x+ 1→ (x2 + 1)(f(x) + 1) = ((x2 + 1)((x+ 1) + 1)

= (x2 + x)

Now this goes back to: f(x)x

x2 + x→ f(x)x

=(x2 + x

x= x+ 1

After iterating the function twice we arrived back with our starting polynomial, creating a loop.So x2 + x does not converge to 1. There are three possibilities that can happen here, the polyno-mials can reduce to 1, they can loop, or they can tend to infinity. The previous example creates aloop. Here is an example of one that goes to infinity:

Theorem 5.3. g(x) =

{f(x)

x if x divides f(x)(x2)(f(x)) + 1 if x does not divide f(x)

It is trivial that x2 converges to 1.Next we use, x+ 1:

x+ 1→ x2(x+ 1) + 1 = x3 + x2 + 1

x3 + x2 + 1→ x2(x3 + x2 + 1) + 1 = x5 + x4 + x2 + 1

x5 + x4 + x2 + 1→ x2(x5 + x4 + x2 + 1) + 1 = x7 + x6 + x4 + x2 + 1

Each iteration of the function leads to larger exponents. By adding 1 at the end of everyiteration, the function can never go to a polynomial that is divisible by x. Each iteration of thefunction raises the degree of the polynomial by 2, so the polynomial tends to infinity.

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Theoretically, there exist other polynomials that can be used for the leading coefficient that willin fact reduce to 1. None were found in the research of this paper though, it seems that to find anypolynomials that do go to one, requires trial and error to discover polynomials that will reduce to1. No discernible pattern emerged that a conjecture could be stated that polynomials of a certainform would reduce to 1.

6 Conclusion

We have seen that there are many different ways to view this problem. We try many differentmethods of approach on the collatz conjecture to see if we can find a pattern that may help provethe theorem. The difficulty with the 3n+1 problem is that it seems to simulate ”random” behavior.There seems to be no discernible pattern in the conjecture that leads to us being able to say thatevery number ends with 1.

This is why we look at different models for the same problem, to see if there is any pattern thatcan be discovered. We looked at the real valued equation to determine if there was any discerniblepattern there. We saw that the function was bounded between the linear functions 3n+ 1 and n

2 .Transforming the conjecture into a function that existed on R allowed us to once again convert theconjecture into another equation that existed in C.

From R we went to the complex plane. We viewed the fractal and in doing so, found aninteresting similarity in the Collatz conjecture and that of the Mandelbrot set. Having two unrelatedfunctions exhibit the same behavior makes one wonder if there is indeed an underlying relationshipthat has yet to be discovered.

So why do we study this problem in the first place? Paul Erdos said that “mathematics is notready for such problems”. We study generalizations of the 3n + 1 problem in hopes that we mayfind new phenomena, these new phenomena may indicate limits of validity of known results for theproblem.[3] By viewing the Collatz conjecture in different forms we discover intersting results thatmay be applicable to other areas of Mathematics.

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References

[1] J.J. Connor, E.F. Robertson, Collatz Biography, http://www-groups.dcs.st-and.ac.uk / his-tory/Biographies/Collatz.html

[2] Jean Paul Van Bendegem, The Collatz Conjecture: A Case Study in Mathematical ProblemSolving, Logic and Logical Philosophy, Volume 14 (2005), 7-23.

[3] Jeffrey C. Lagarias, The 3x+1 problem and its Generalizations,

[4] Christopher D.J. Jones, The Collatz Problem: An Iterative Algorithm, International Bac-calaureate Extended Essay, Mathematics, March 3, 1998.

[5] E.B. Saff, A.D. Snider, Fundamentals of Complex Analysis: with Applications to Engineeringand Science, 3rd ed.Pearson Education Inc, 2003.

[6] E. Di Giacomo, Computing Radial Drawings on the Minimum Number of Circles, Journal ofGraph Algorithms and Applications, http://jgaa.info/ vol.9, no.3 pp. 465-389, 2005.

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