# Collatz conjecture

The Collatz conjecture is an unsolved conjecture in mathematics. It is named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, as the Ulam conjecture (after Stanislaw Ulam), or as the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers.

We take any whole number n greater than 0. If n is even, we halve it (n/2), else we do "triple plus one" and get 3n+1. The conjecture is that for all numbers this process converges to 1. Hence it has been called "Half Or Triple Plus One", sometimes called HOTPO.

Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered \$500 for its solution. (Lagarias 1985)

## Statement of the problem

Consider the following operation on an arbitrary positive integer:

• If the number is even, divide it by two.
• If the number is odd, triple it and add one.

For example, if this operation is performed on 3, the result is 10; if it is performed on 28, the result is 14.

In modular arithmetic notation, define the function f as follows: $f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \pmod{2}\\ 3n+1 & \mbox{if } n\equiv 1 \pmod{2} \end{cases}$

Now, form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

In notation: $a_i = \begin{cases}n & \mbox{for } i = 0 \\ f(a_{i-1}) & \mbox{for } i > 0. \end{cases}$

or ${a_{i}} = \frac{1}{2}{a_{i-1}} - \frac{1}{4}(5a_{i-1}+2)((-1)^{a_{i-1}}-1)$

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

That smallest i such that the above holds is called the total stopping time of n. The conjecture asserts that every n has a well-defined stopping time. If, for some n, such an i doesn't exist, we say that n has infinite total stopping time and the conjecture is false.

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

## Examples

For instance, starting with n = 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1.

Starting with n = 11, the sequence takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

If the starting value n = 27 is chosen, the sequence, listed and graphed below, takes 111 steps, climbing to over 9000 before descending to 1.

{ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 }

The number less than 100 million with the longest total stopping time is 63728127, with 949 steps.

The number less than 1 billion with the longest total stopping time is 670617279, with 986 steps.

## Program to calculate Collatz sequences

A specific Collatz sequence can be easily computed, as is shown by this pseudocode example:

function collatz(n)
while n > 1
show n
if n is odd
set n to 3n + 1
else
set n to n / 2
show n


This program halts when the sequence reaches 1, in order to avoid printing an endless cycle of 4, 2, 1. If the Collatz conjecture is true, the program will always halt no matter what positive starting integer is given to it. (See Halting Problem for a discussion of the relationship between open-ended computer programs and unsolved mathematics problems.)

## Supporting arguments

Although the conjecture has not been proven, most mathematicians who have looked into the problem think the conjecture is true because experimental evidence and heuristic arguments support it.

### Experimental evidence

The conjecture has been checked by computer for all starting values up to 20 × 258 ≈ 5.764 × 1018. While impressive, such computer evidence should be interpreted cautiously. More than one important conjecture has been found false, but only with very large counterexamples. (See for example the Pólya conjecture, the Mertens conjecture and the Skewes' number.)

All initial values tested so far eventually end in the repeating cycle {4,2,1}, which has only three terms. It is also known that {4,2,1} is the only repeating cycle possible with fewer than 35400 terms.

### A probabilistic heuristic

If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average ¾ of the previous one. (More precisely, the geometric mean of the ratios of outcomes is ¾.) This yields a heuristic argument that every Collatz sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it pretends that Collatz sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has 2 division steps for every multiplication step for almost all 2-adic starting values.)

## Other formulations of the conjecture

### In reverse

There is another approach to prove the conjecture, which considers the bottom-up method of growing the so called Collatz graph. The Collatz graph is a graph defined by the inverse relation $R(n) = \begin{cases} 2n & \mbox{if } n\equiv 0,1,2,3,5 \\ 2n, (n-1)/3 & \mbox{if } n\equiv 4 \end{cases} \pmod{6}.$

So, instead of proving that all natural numbers eventually lead to 1, we can prove that 1 leads to all natural numbers. For any integer n, n ≡ 1 (mod 2) iff 3n + 1 ≡ 4 (mod 6). Equivalently, (n - 1)/3 ≡ 1 (mod 2) iff n ≡ 4 (mod 6). Also, the inverse relation forms a tree except for the 1-2-4 loop (the inverse of the 1-4-2 loop of the unaltered function f defined in the statement of the problem above). When the relation 3n + 1 of the function f is replaced by the common substitute "shortcut" relation (3n + 1)/2, the Collatz graph is defined by the inverse relation, $R(n) = \begin{cases} 2n & \mbox{if } n\equiv 0,1 \\ 2n, (2n-1)/3 & \mbox{if } n\equiv 2 \end{cases} \pmod{3}.$

Conjecturally, this inverse relation forms a tree except for a 1-2 loop (the inverse of the 1-2 loop of the function f(n) revised as indicated above).

### As rational numbers

The natural numbers can be converted to rational numbers in a certain way. To get the rational version, find the highest power of two less than or equal to the number, use it as the denominator, and subtract it from the original number for the numerator (527 → 15/512). To get the natural version, add the numerator and denominator (255/256 → 511).

The Collatz conjecture then says that the numerator will eventually equal zero. The Collatz function changes to: $f(n, d) = \begin{cases} (3n + d + 1)/2d & \mbox{if } 3n + d + 1 < 2d \\ (3n - d + 1)/4d & \mbox{if } 3n + d + 1 \ge 2d \end{cases}$ (n = numerator; d = denominator).

This works because 3x + 1 = 3(d + n) + 1 = (2d) + (3n + d + 1) = (4d) + (3n - d + 1). Reducing a rational before every operation is required to get x as an odd.

### As an abstract machine that computes in base two

Repeated applications of the Collatz function can be represented as an abstract machine that handles strings of bits. The machine will perform the following two steps on any odd number until only one "1" remains:

1. Add the original with a "1" appended to the end to the original in binary, i.e. 3n + 1 = (2n + 1) + n
2. Remove all trailing "0"s.

This prescription is plainly equivalent to computing a Collatz sequence in base two.

Example:The starting number 7 is written in base two as 111. The resulting Collatz sequence is:

                                                    111
1111
10110
10111
100010
100011
110100
11011
101000
1011
10000


### As a parity sequence

For this section, consider the Collatz function in the slightly modified form $f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1. \end{cases} \pmod{2}$

This can be done because when n is odd, 3n + 1 is always even.

If P(…) is the parity of a number, that is P(2n) = 0 and P(2n + 1) = 1, then we can define the Collatz parity sequence for a number n as pi = P(ai), where a0 = n, and ai+1 = f(ai).

Using this form for f(n), it can be shown that the parity sequences for two numbers m and n will agree in the first k terms if and only if m and n are equivalent modulo 2k. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Collatz cycles, then their corresponding parity cycles must be different.

The proof is simple: it is easy to verify by hand that applying the f function k times to the number a 2k+b will give the result a 3c+d, where d is the result of applying the f function k times to b, and c is how many odd numbers were encountered during that sequence. So the parity of the first k numbers is determined purely by b, and the parity of the (k+1)th number will change if the least significant bit of a is changed.

The Collatz Conjecture can be rephrased as stating that the Collatz parity sequence for every number eventually enters the cycle 0 → 1 → 0.

### As a tag system

For the Collatz function in the form $f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1. \end{cases} \pmod{2}$

Collatz sequences can be computed by the extremely simple 2-tag system with production rules abc, ba, caaa. In this system, the positive integer n is represented by a string of n a's, and iteration of the tag operation halts on any word of length less than 2. (Adapted from De Mol.)

The Collatz conjecture equivalently states that this tag system, with an arbitrary finite string of a's as the initial word, eventually halts. See the linked article for a worked example.

## Extensions to larger domains

### Iterating on all integers

For any integer n, rather than just positive integers, we map it to the integer f(n), where

f(n) = 3n + 1  if n is odd;
f(n) = n/2     if n is even.


Interestingly, there are in this case a total of 5 known cycles, which all integers seem to eventually fall into under iteration of f. These cycles are listed here, starting with the well-known cycle for positive n.

To save steps, we list only the odd numbers of each cycle (except for the trivial cycle {0}). Each odd number n, when f is applied repeatedly, will next reach an odd number at (3n+1) / (the largest power of 2 that divides 3n+1); each cycle is listed with its member of least absolute value first. We follow each cycle with its full length in parentheses, full meaning that the even terms are counted as well.

a) 1 → 1   (length 3)
b) 0 → 0   (length 1)
c) -1 → -1  (length 2)
d) -5 → -7 → -5   (length 5)
e) -17 → -25 → -37 → -55 → -41 → -61 → -91 → -17   (length 18)


The Generalized Collatz Conjecture is the assertion that every integer, under iteration by f, eventually falls into one of these five cycles.

### Iterating with odd denominators or 2-adic integers

The standard Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms. The number is taken to be odd or even according to whether its numerator is odd or even. A closely related fact is that the Collatz map extends to the ring of 2-adic integers, which contains the ring of rationals with odd denominators as a subring.

The parity sequences as defined above are no longer unique for fractions. However, it can be shown that any possible parity cycle is the parity sequence for exactly one fraction: if a cycle has length n and includes odd numbers exactly m times at indices k0, …, km-1, then the unique fraction which generates that parity cycle is $\frac{3^{m-1} 2^{k_0} + ... + 3^0 2^{k_{m-1}}}{2^n - 3^m}$.

For example, the parity cycle (1 0 1 1 0 0 1) has length 7 and has 4 odd numbers at indices 0, 2, 3, and 6. The unique fraction which generates that parity cycle is $\frac{3^3 2^0 + 3^2 2^2 + 3^1 2^3 + 3^0 2^6}{2^7 - 3^4} = \frac{151}{47}$.

The complete cycle being: 151/47 → 250/47 → 125/47 → 211/47 → 340/47 → 170/47 → 85/47 → 151/47

Although the cyclic permutations of the original parity sequence are unique fractions, the cycle is not unique, each permutation's fraction being the next number in the loop cycle:

(0 1 1 0 0 1 1) → $\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{250}{47}$

(1 1 0 0 1 1 0) → $\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{{125}}{47}$

(1 0 0 1 1 0 1) → $\frac{3^3 2^0 + 3^2 2^3 + 3^1 2^4 + 3^0 2^6}{2^7 - 3^4} = \frac{211}{47}$

(0 0 1 1 0 1 1) → $\frac{3^3 2^2 + 3^2 2^3 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{340}{47}$

(0 1 1 0 1 1 0) → $\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{170}{47}$

(1 1 0 1 1 0 0) → $\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^3 + 3^0 2^4}{2^7 - 3^4} = \frac{85}{47}$

Also, for uniqueness, the parity sequence should be "prime", i.e., not partitionable into identical sub-sequences. For example, parity sequence (1 1 0 0 1 1 0 0) can be partitioned into two identical sub-sequences (1 1 0 0)(1 1 0 0). Calculating the 8-element sequence fraction gives

(1 1 0 0 1 1 0 0) → $\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^8 - 3^4} = \frac{125}{175}$

But when reduced to lowest terms {5/7}, it is the same as that of the 4-element sub-sequence

(1 1 0 0) → $\frac{3^1 2^0 + 3^0 2^1}{2^4 - 3^2} = \frac{5}{7}$

And this is because the 8-element parity sequence actually represents two circuits of the loop cycle defined by the 4-element parity sequence.

In this context, the Collatz conjecture is equivalent to saying that (0 1) is the only cycle which is generated by positive whole numbers (i.e. 1 and 2).

### Iterating on real or complex numbers

The Collatz map can be viewed as the restriction to the integers of the smooth real and complex map $f(z)=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+(3z+1)\sin^2\left(\frac \pi 2 z\right)$,

which simplifies to $\frac{1}{4}(2 + 7z - (2 + 5z)\cos(\pi z))$.

If the standard Collatz map defined above is optimized by replacing the relation 3n + 1 with the common substitute "shortcut" relation (3n + 1)/2, it can be viewed as the restriction to the integers of the smooth real and complex map $f(z)=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+\frac 1 2 (3z+1)\sin^2\left(\frac \pi 2 z\right)$,

which simplifies to $\frac{1}{4}(1 + 4z - (1 + 2z)\cos(\pi z))$.

Iterating the above optimized map in the complex plane produces the Collatz fractal.

## Optimizations

The "parity" section above gives a way to speed up simulation of the sequence. To jump ahead k steps on each iteration (using the f function from that section), break up the current number into two parts, b (the k least significant bits, interpreted as an integer), and a (the rest of the bits as an integer). The result of jumping ahead k steps can be found as:

f k+c[b](a 2k+b) = a 3c[b]+d[b].

The c and d arrays are precalculated for all possible k-bit numbers b, where d [b] is the result of applying the f function k times to b, and c [b] is the number of odd numbers encountered on the way. For example, if k=5, you can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using:

c [0...31] = {0,3,2,2,2,2,2,4,1,4,1,3,2,2,3,4,1,2,3,3,1,1,3,3,2,3,2,4,3,3,4,5}
d [0...31] = {0,2,1,1,2,2,2,20,1,26,1,10,4,4,13,40,2,5,17,17,2,2,20,20,8,22,8,71,26,26,80,242}.

For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration which is due to Tomás Oliveira e Silva and is used in the record confirmation of the Collatz conjecture. If, for some given b and k, the inequality

f k+c[b](a 2k+b) = a 3c[b]+d[b] < a 2k+b

holds for all a, then the first counterexample, if it exists, cannot be b modulo 2k. For instance, the first counterexample must be odd because f(2n) = n; and it must be 3 mod 4 because f3(4n+1) = 3n+1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting value is as good as checking an entire congruence class. As k increases, the search only needs to check those residues b that are not eliminated by lower values of k. On the order of 3k/2 residues survive. For example, the only surviving residues mod 32 are 7, 15, 27, and 31; only 573162 residues survive mod 225 = 33554432.

## Syracuse function

If k is an odd integer, then 3k + 1 is even, so we can write 3k + 1 = 2ak′, with k' odd and a ≥ 1. We define a function f from the set I of odd integers into itself, called the Syracuse Function, by taking f (k) = k′ (sequence A075677 in OEIS).

Some properties of the Syracuse function are:

• f (4k + 1) = f (k) for all k in I.
• For all p ≥ 2 and h odd, f p - 1(2 p h - 1) = 2 3 p - 1h - 1 (see here for the notation).
• For all odd h, f (2h - 1) ≤ (3h - 1)/2

The Syracuse Conjecture is that for all k in I, there exists an integer n ≥ 1 such that f n(k) = 1. Equivalently, let E be the set of odd integers k for which there exists an integer n ≥ 1 such that f n(k) = 1. The problem is to show that E = I. The following is the beginning of an attempt at a proof by induction:

1, 3, 5, 7, and 9 are known to exist in E. Let k be an odd integer greater than 9. Suppose that the odd numbers up to and including k - 2 are in E and let us try to prove that k is in E. As k is odd, k + 1 is even, so we can write k + 1 = 2ph for p ≥ 1, h odd, and k = 2ph-1. Now we have:

• If p = 1, then k = 2h - 1. It is easy to check that f (k) < k , so f (k) ∈ E; hence kE.
• If p ≥ 2 and h is a multiple of 3, we can write h = 3h′. Let k′ = 2p + 1h′ - 1; we have f (k′) = k , and as k′ < k , k′ is in E; therefore k = f (k′) ∈ E.
• If p ≥ 2 and h is not a multiple of 3 but h ≡ (-1)p mod 4, we can still show that kE. (Cf.)

The problematic case is that where p ≥ 2 , h not multiple of 3 and h ≡ (-1)p+1 mod 4. Here, if we manage to show that for every odd integer k′, 1 ≤ k′ ≤ k-2 ; 3k′ ∈ E we are done. (Cf.).