Collatz conjecture

Unsolved problem in mathematics: Does the Collatz sequence eventually reach 1 for all positive integer initial values? (more unsolved problems in mathematics)

The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem;^[1][2] the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud),^[4][5] or as wondrous numbers.^[6]

The conjecture can be summarized as follows. Take any positive integer n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO^[7]) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."^[8] He also offered $500 for its solution.^[10]

Statement of the problem

Numbers from 1 to 9999 and their corresponding total stopping time

Histogram of total stopping times for the numbers 1 to 100 million. Total stopping time is on the x axis, frequency on the y axis. Note that for all positive integers the histogram would be a completely different, exponentially-growing sequence (see #In reverse)

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.

In modular arithmetic notation, define the function f as follows:

$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$

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:

$a_i = \begin{cases}n & \text{for } i = 0 \\ f(a_{i-1}) & \text{for } i > 0 \end{cases}$

(that is: $a_i$ is the value of $f$ applied to $n$ recursively $i$ times; $a_i = f^i(n)$).

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 a_i = 1 is called the total stopping time of n.^[3] The conjecture asserts that every n has a well-defined total 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 that 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.

n = 19, for example, takes longer to reach 1: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

The sequence for n = 27, listed and graphed below, takes 111 steps (41 steps through odd numbers), climbing to 9232 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 (sequence A008884 in OEIS)

Numbers with a total stopping time longer than any smaller starting value form a sequence beginning with:

1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, … (sequence A006877 in OEIS).

The starting values whose maximum trajectory point is greater than that of any smaller starting value are as follows:

1, 2, 3, 7, 15, 27, 255, 447, 639, 703, 1819, 4255, 4591, 9663, 20895, 26623, 31911, 60975, 77671, 113383, 138367, 159487, 270271, 665215, 704511, ... (sequence A006884 in OEIS)

Number of steps for n to reach 1 are

0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, ... (sequence A006577 in OEIS)

The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,631, with 1132 steps.^[11][12]

The powers of two converge to one quickly because $2^n$ is halved $n$ times to reach one, and is never increased, but for Mersenne number M_n, they need to increase n times and usually need more steps to reach 1.

Visualizations

• 

  Directed graph showing the orbits of small numbers under the Collatz map. The Collatz conjecture is equivalent to the statement that all paths eventually lead to 1.

• 

  Directed graph showing the orbits of the first 1000 numbers.

• 

  The x axis represents starting number, the y axis represents the highest number reached during the chain to 1.

Cycles

Any counterexample to the Collatz conjecture would have to consist either of an infinite divergent trajectory or a cycle different from the trivial (4; 2; 1) cycle. Thus, if one could prove that neither of these types of counterexample could exist, then all positive integers would have a trajectory that reaches the trivial cycle. Such a strong result is not known, but certain types of cycles have been ruled out.

The type of a cycle may be defined with reference to the "shortcut" definition of the Collatz map, $f(n)=(3n+1)/2$ for odd n and $f(n)=n/2$ for even n. A cycle is a sequence $(a_0; a_1; \ldots a_q)$ where $f(a_0)=a_1$, $f(a_1)=a_2$, and so on, up to $f(a_q)=a_0$ in a closed loop. For this shortcut definition, the only known cycle is (1; 2). Although 4 is part of the single known cycle for the original Collatz map, it is not part of the cycle for the shortcut map.

A k-cycle is a cycle that can be partitioned into 2k contiguous subsequences: k increasing sequences of odd numbers alternating with k decreasing sequences of even numbers. For instance, if the cycle consists of a single increasing sequence of odd numbers followed by a decreasing sequence of even numbers, it is called a 1-cycle.^[13]

Steiner (1977) proved that there is no 1-cycle other than the trivial (1;2).^[14] Simons (2004) used Steiner's method to prove that there is no 2-cycle.^[15] Simons & de Weger (2003) extended this proof up to 68-cycles: there is no k-cycle up to k = 68.^[13] Beyond 68, this method gives upper bounds for the elements in such a cycle: for example, if there is a 75-cycle, then at least one element of the cycle is less than 2385×2⁵⁰.^[13] Therefore, as exhaustive computer searches continue, larger cycles may be ruled out. To state the argument more intuitively: we need not look for cycles that have at most 68 trajectories, where each trajectory consists of consecutive ups followed by consecutive downs. See below for an idea of how one might find an upper bound for the elements of a cycle.

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 2⁶⁰.^[16] All initial values tested so far eventually end in the repeating cycle (4; 2; 1), which has only three terms. From this lower bound on the starting value, a lower bound can also be obtained for the number of terms a repeating cycle other than (4; 2; 1) must have.^[17] When this relationship was established in 1981, the formula gave a lower bound of 35,400 terms.^[17]

This computer evidence is not a proof that the conjecture is true. As shown in the cases of the Pólya conjecture, the Mertens conjecture and the Skewes' number, sometimes a conjecture's only counterexamples are found when using very large numbers.

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 3/4 of the previous one.^[18] (More precisely, the geometric mean of the ratios of outcomes is 3/4.) This yields a heuristic argument that every Hailstone 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 assumes that Hailstone sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has two division steps for every multiplication step for almost all 2-adic starting values.)

And even if the probabilistic reasoning were rigorous, this would still imply only that the conjecture is almost surely true for any given integer, which does not necessarily imply that it is true for all integers.

Rigorous bounds

Although it is not known rigorously whether all positive numbers eventually reach one according to the Collatz iteration, it is known that many numbers do so. In particular, Krasikov and Lagarias showed that the number of integers in the interval [1,x] that eventually reach one is at least proportional to x^0.84.^[19]

Other formulations of the conjecture

In reverse

The first 21 levels of the Collatz graph generated in bottom-up fashion. The graph includes all numbers with an orbit length of 21 or less.

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

$R(n) = \begin{cases} \{2n\} & \text{if } n\equiv 0,1,2,3,5 \\ \{2n,(n-1)/3\} & \text{if } n\equiv 4 \end{cases} \pmod{6}.$

So, instead of proving that all positive integers eventually lead to 1, we can try to prove that 1 leads to all positive integers. 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). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of this article).

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,

$R(n) = \begin{cases} \{2n\} & \text{if } n\equiv 0,1 \\ \{2n,(2n-1)/3\} & \text{if } n\equiv 2 \end{cases} \pmod{3}.$

For any integer n, n ≡ 1 (mod 2) iff (3n + 1)/2 ≡ 2 (mod 3). Equivalently, (2n − 1)/3 ≡ 1 (mod 2) iff n ≡ 2 (mod 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).

Alternately, replace the 3n + 1 with n' / H(n') where n' = 3n + 1 and H(n') is the highest power of 2 that divides n' (with no remainder). The resulting function f maps from odd numbers to odd numbers. Now suppose that for some odd number n, applying this operation k times yields the number 1 (that is, $f^k(n) = 1$). Then in binary, the number n can be written as the concatenation of strings w_k w_k-1 … w₁ where each w_h is a finite and contiguous extract from the representation of 1 / 3^h.^[20] The representation of n therefore holds the repetends of 1 / 3^h, where each repetend is optionally rotated and then replicated up to a finite number of bits. It is only in binary that this occurs.^[21] Conjecturally, every binary string s that ends with a '1' can be reached by a representation of this form (where we may add or delete leading '0's to s).

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 three steps on any odd number until only one "1" remains:

1. Append 1 to the (right) end of the number in binary (giving 2n + 1);
2. Add this to the original number by binary addition (giving 2n + 1 + n = 3n + 1);
3. Remove all trailing "0"s (i.e. repeatedly divide by two until the result is odd).

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

Example

The starting number 7 is written in base two as 111. The resulting Hailstone 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

$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \\ (3n +1)/2 & \text{if } n \equiv 1. \end{cases} \pmod{2}$

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 Hailstone parity sequence (or parity vector) for a number n as p_i = P(a_i), where a₀ = n, and a_i+1 = f(a_i).

What operation is performed (3n + 1)/2 or n/2 depends on the parity. The parity sequence is the same as the sequence of operations.

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 2^k. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Hailstone cycles, then their corresponding parity cycles must be different.^[3][22]

Applying the f function k times to the number a·2^k + b will give the result a·3^c + 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.

As a tag system

For the Collatz function in the form

$f(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \\ (3n +1)/2 & \text{if } n \equiv 1. \end{cases} \pmod{2}$

Hailstone sequences can be computed by the extremely simple 2-tag system with production rules a → bc, b → a, c → aaa. In this system, the positive integer n is represented by a string of n a, 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 Tag system#Example: Computation of Collatz sequences for a worked example).

Extensions to larger domains

Iterating on all integers

An obvious extension is to include all integers, not just positive integers. Leaving aside the trivial cycle 0 → 0, there are a total of 4 known non-trivial cycles, which all nonzero integers seem to eventually fall into under iteration of f. These cycles are listed here, starting with the well-known cycle for positive n:

Odd values are listed in large bold. Each cycle is listed with its member of least absolute value (which is always odd) first.

Cycle	Odd-value cycle length	Full cycle length
1 → 4 → 2 → 1 …	1	3
−1 → −2 → −1 …	1	2
−5 → −14 → −7 → −20 → −10 → −5 …	2	5
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17 …	7	18

The generalized Collatz conjecture is the assertion that every integer, under iteration by f, eventually falls into one of the four non-trivial cycles above, or is the trivial cycle 0 → 0.

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, …, k_m−1, then the unique fraction which generates that parity cycle is

$\frac{3^{m-1} 2^{k_0} + \cdots + 3^0 2^{k_{m-1}}}{2^n - 3^m}.$

(1)

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).

Cycle bounds

Eq. (1) also gives a rough idea about how one can prove that cycles of certain lengths do not exist. For a hypothetical cycle of length n, the numerator is bounded above by 3ⁿ - 2ⁿ (this corresponds to a cycle of all odd numbers). A lower bound for the denominator can be obtained by letting n/m be an optimal rational approximation to log(3)/log(2). Together these give an upper bound for the unique fraction that generates a cycle of length n. If this upper bound is smaller than the largest number for which the conjecture has been verified to hold, then a cycle of length n is impossible.

Iterating on real or complex numbers

Cobweb plot of the orbit 10-5-8-4-2-1-2-1-2-1-etc. in the real extension of the Collatz map (optimized by replacing "3n + 1" with "(3n + 1)/2")

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

$f(z)=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+\frac {(3z+1)} {2} \sin^2\left(\frac \pi 2 z\right),$

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

Collatz fractal

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

The point of view of iteration on the real line was investigated by Chamberland (1996),^[23] and on the complex plane by Letherman, Schleicher, and Wood (1999).^[24]

Collatz map fractal in a neighbourhood of the real line

Optimizations

Time-space tradeoff

The section As a parity sequence 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(a 2^k + b) = a 3^c(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.^[25] 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}.

This requires 2^k precomputation and storage to speed up the resulting calculation by a factor of k, a space-time tradeoff.

Modular restrictions

For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration, used by Tomás Oliveira e Silva in his computational confirmations of the Collatz conjecture up to large values of n. If, for some given b and k, the inequality

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

holds for all a, then the first counterexample, if it exists, cannot be b modulo 2^k.^[17] For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f²(4n + 1) = 3n + 1, smaller than 4n + 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. Only an exponentially small fraction of the residues survive.^[26] For example, the only surviving residues mod 32 are 7, 15, 27, and 31.

Syracuse function

If $k$ is an odd integer, then $3k+1$ is even, so $3k+1 = 2^{a}k^\prime$ with $k^\prime$ odd and $a \geq 1$. The Syracuse function is the function $f$ from the set $I$ of odd integers into itself, for which $f(k)=k^\prime$ (sequence A075677 in OEIS).

Some properties of the Syracuse function are:

• $\forall k \in I, f(4k+1)=f(k)$. (Because $3(4k+1)+1=12k+4=4(3k+1)$
• In more generality: For all $p \geq 1$ and odd $h$, $f^{p-1}(2^ph-1)=2 \cdot 3^{p-1}h-1$. (Here $f^{p-1}$ is function iteration notation.)
• For all odd $h$,$f(2h-1) \leq \frac{3h-1}{2}$

The Collatz conjecture is equivalent to the statement that, for all $k$ in $I$, there exists an integer $n \geq 1$ such that $f^n(k) = 1$.

Undecidable generalizations

In 1972, J. H. Conway proved that a natural generalization of the Collatz problem is algorithmically undecidable.^[27]

Specifically, he considered functions of the form

$g(n) = a_i n + b_i, {n\equiv i \pmod P}$

where $a_0,b_0,\dots,a_{P-1},b_{P-1}$ are rational numbers which are so chosen that $g(n)$ is always integral.

The standard Collatz function is given by $P=2$, $a_0 = 1/2$, $b_0 = 0$, $a_1 = 3$, $b_1 = 1$. Conway proved that the problem:

Given g and n, does the sequence of iterates $g^k(n)$ reach 1?

is undecidable, by representing the halting problem in this way. Closer to the Collatz problem is the following universally quantified problem:

Given g does the sequence of iterates $g^k(n)$ reach 1, for all n>0?

Modifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one). Kurtz and Simon^[28] proved that the above problem is, in fact, undecidable and even higher in the arithmetical hierarchy, specifically $\Pi^0_2$-complete. This hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.^[29]

The Ultimate Challenge: the 3x+1 problem

This volume,^[30] edited by Jeffrey Lagarias and published by the American Mathematical Society, is a compendium of information on the Collatz conjecture, methods of approaching it and generalizations. It includes two survey papers by the editor and five by other authors, concerning the history of the problem, generalizations, statistical approaches and results from the theory of computation. It also includes reprints of early papers on the subject (including an entry by Lothar Collatz himself).

See also

• 3x + 1 semigroup
• Residue-class-wise affine group
• Modular arithmetic

References

1. Maddux, Cleborne D.; Johnson, D. Lamont (1997). Logo: A Retrospective. New York: Haworth Press. p. 160. ISBN 0-7890-0374-0. The problem is also known by several other names, including: Ulam's conjecture, the Hailstone problem, the Syracuse problem, Kakutani's problem, Hasse's algorithm, and the Collatz problem.
2. According to Lagarias (1985),^[3] p. 4, the name "Syracuse problem" was proposed by Hasse in the 1950s, during a visit to Syracuse University.
3. Jeffrey C. Lagarias (1985). The 3x + 1 problem and its generalizations. The American Mathematical Monthly 92(1): 3-23.
4. Pickover, Clifford A. (2001). Wonders of Numbers. Oxford: Oxford University Press. pp. 116–118. ISBN 0-19-513342-0.
5. "Hailstone Number". MathWorld. Wolfram Research.
6. Hofstadter, Douglas R. (1979). Gödel, Escher, Bach. New York: Basic Books. pp. 400–402. ISBN 0-465-02685-0.
7. Friendly, Michael (1988). Advanced Logo: A Language for Learning. Hillsdale, New Jersey, USA: Lawrence Erlbaum Associates. ISBN 0-89859-933-4.
8. Guy (2004),^[9] p. 330
9. Guy, Richard K. (2004). Unsolved problems in number theory (3rd ed.). Springer-Verlag. "E17: Permutation Sequences". ISBN 0-387-20860-7. Zbl 1058.11001. Cf pp. 336–337.
10. R. K. Guy: Don't try to solve these problems, Amer. Math. Monthly, 90 (1983), 35–41. By this Erdos means that there aren't powerful tools for manipulating such objects.
11. Leavens, Gary T.; Vermeulen, Mike (December 1992). "3x+1 Search Programs". Computers & Mathematics with Applications 24 (11): 79–99. doi:10.1016/0898-1221(92)90034-F.
12. Roosendaal, Eric. "3x+1 Delay Records". Retrieved 27 November 2011. (Note: "Delay records" are total stopping time records.)
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External links

• Keith Matthews' 3x + 1 page: Review of progress, plus various programs.
• Distributed computing (BOINC) project that verifies the Collatz conjecture for larger values.
• An ongoing distributed computing project by Eric Roosendaal verifies the Collatz conjecture for larger and larger values.
• Another ongoing distributed computing project by Tomás Oliveira e Silva continues to verify the Collatz conjecture (with fewer statistics than Eric Roosendaal's page but with further progress made).
• An animated implementation that uses arbitrary-precision arithmetic.
• Weisstein, Eric W., "Collatz Problem", MathWorld.
• Collatz Problem at PlanetMath.org..
• Collatz Iterations on the Ulam Spiral grid on YouTube.
• Collatz Paths by Jesse Nochella, Wolfram Demonstrations Project.
• Collatz cycles? About cycles, very basic, contains also some unusual graphs (HTML).
• Collatz cycles? About loops, compacted text, rather basic (PDF).
• An animated example of the Collatz sequence for the number 2²⁶ + 1 in base 2.
• Collatz Conjecture A T-shirt illustrating the problem by xkcd author Randall Munroe.