Lagarias Collatz Bibliography Apa


Collatz Problem

A problem posed by L. Collatz in 1937, also called the mapping, problem, Hasse's algorithm, Kakutani's problem, Syracuse algorithm, Syracuse problem, Thwaites conjecture, and Ulam's problem (Lagarias 1985). Thwaites (1996) has offered a £1000 reward for resolving the conjecture. Let be an integer. Then one form of Collatz problem asks if iterating

(1)

always returns to 1 for positive. (If negative numbers are included, there are four known cycles (excluding the trivial 0 cycle): (4, 2, 1), (, ), (, , , , ), and (, , , , , , , , , , , , , , , , , ).)

The members of the sequence produced by the Collatz are sometimes known as hailstone numbers. Conway proved that the original Collatz problem has no nontrivial cycles of length . Lagarias (1985) showed that there are no nontrivial cycles with length . Conway (1972) also proved that Collatz-type problems can be formally undecidable. Kurtz and Simon (2007) proved that a natural generalization of the Collatz problem is undecidable; unfortunately, this proof cannot be applied to the original Collatz problem.

The Collatz algorithm has been tested and found to always reach 1 for all numbers (Oliveira e Silva 2008), improving the earlier results of (Vardi 1991, p. 129) and (Leavens and Vermeulen 1992). Because of the difficulty in solving this problem, Erdős commented that "mathematics is not yet ready for such problems" (Lagarias 1985).

The following table gives the sequences obtained for the first few starting values (OEIS A070165).

, , , ...
11
22, 1
33, 10, 5, 16, 8, 4, 2, 1
44, 2, 1
55, 16, 8, 4, 2, 1
66, 3, 10, 5, 16, 8, 4, 2, 1

The numbers of steps required for the algorithm to reach 1 for , 2, ... are 0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, ... (OEIS A006577; illustrated above). Of these, the numbers of tripling steps are 0, 0, 2, 0, 1, 2, 5, 0, 6, ... (OEIS A006667), and the number of halving steps are 0, 1, 5, 2, 4, 6, 11, 3, 13, ... (OEIS A006666). The smallest starting values of that yields a Collatz sequence containing , 2, ... are 1, 2, 3, 3, 3, 6, 7, 3, 9, 3, 7, 12, 7, 9, 15, 3, 7, 18, 19, ... (OEIS A070167).

The Collatz problem can be implemented as an 8-register machine (Wolfram 2002, p. 100), quasi-cellular automaton (Cloney et al. 1987, Bruschi 2005), or 6-color one-dimensional quasi-cellular automaton with local rules but which wraps first and last digits around (Zeleny). In general, the difficulty in constructing true local-rule cellular automata arises from the necessity of a carry operation when multiplying by 3 which, in the worst case, can extend the entire length of the base- representation of digits (and thus require propagating information at faster than the CA's speed of light).

The Collatz problem was modified by Terras (1976, 1979), who asked if iterating

(2)

always returns to 1 for initial integer value (e.g., Lagarias 1985, Cloney et al. 1987). This is simply the original statement above but combining the division by two into the addition step if is odd, thus compressing the number of steps. The following table gives the sequences for the first few starting values , 2, ... (OEIS A070168).

, , ...
11
22, 1
33, 5, 8, 4, 2, 1
44, 2, 1
55, 8, 4, 2, 1
66, 3, 5, 8, 4, 2, 1
77, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1

If negative numbers are included, there are 4 known cycles: (1, 2), (), (, , ), and (, , , , , , , , , , ). It is a special case of the "generalized Collatz problem" with , , , , and . Terras (1976, 1979) also proved that the set of integers has a limiting asymptotic density , such that if is the number of such that and , then the limit

(3)

exists. Furthermore, as , so almost all integers have a finite stopping time. Finally, for all ,

(4)

where

(Lagarias 1985).

A generalization of the Collatz problem lets be a positive integer and , ..., be nonzerointegers. Also let satisfy

(8)

Then

(9)

for defines a generalized Collatz mapping. An equivalent form is

(10)

for where , ..., are integers and is the floor function. The problem is connected with ergodic theory and Markov chains. Matthews obtained the following table for the mapping

(11)

where .

# cyclesmax. cycle length
0527
11034
213118
317118
419118
521165
623433

Matthews and Watts (1984) proposed the following conjectures.

1. If , then all trajectories for eventually cycle.

2. If , then almost all trajectories for are divergent, except for an exceptional set of integers satisfying

(12)

3. The number of cycles is finite.

4. If the trajectory for is not eventually cyclic, then the iterates are uniformly distribution mod for each , with

(13)

for .

Matthews believes that the map

(14)

will either reach 0 (mod 3) or will enter one of the cycles or , and offers a $100 (Australian?) prize for a proof.

SEE ALSO:Hailstone Number, Juggler Sequence, Wolfram SequencesREFERENCES:

Applegate, D. and Lagarias, J. C. "Density Bounds for the Problem 1. Tree-Search Method." Math. Comput.64, 411-426, 1995.

Applegate, D. and Lagarias, J. C. "Density Bounds for the Problem 2. Krasikov Inequalities." Math. Comput.64, 427-438, 1995.

Bruschi, M. "Two Cellular Automata for the Map." http://arxiv.org/abs/nlin/0502061/. 26 Feb, 2005.

Burckel, S. "Functional Equations Associated with Congruential Functions." Theor. Comp. Sci.123, 397-406, 1994.

Cloney, T.; Goles, E.; and Vichniac, G. Y. "The Problem: A Quasi Cellular Automaton." Complex Sys.1, 349-360, 1987.

Conway, J. H. "Unpredictable Iterations." Proc. 1972 Number Th. Conf., University of Colorado, Boulder, Colorado, pp. 49-52, 1972.

Crandall, R. "On the '' Problem." Math. Comput.32, 1281-1292, 1978.

De Mol, L. "Tag Systems and Collatz-Like Functions." Theor. Comput. Sci.390, 92-101, 2008.

Everett, C. "Iteration of the Number Theoretic Function , ." Adv. Math.25, 42-45, 1977.

Guy, R. K. "Collatz's Sequence." §E16 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 215-218, 1994.

Kurtz, S. A. and Simon, J. "The Undecidability of the Generalized Collatz Problem." In Theory and Applications of Models of Computation: Proceedings of the 4th International Conference (TAMC 2007) held in Shanghai, May 22-25, 2007 (Ed. J.-Y. Cai, S. B. Cooper, and H. Zhu). Berlin: Springer, pp. 542-553, 2007.

Lagarias, J. C. "The Problem and Its Generalizations." Amer. Math. Monthly92, 3-23, 1985.

Leavens, G. T. and Vermeulen, M. " Search Programs." Comput. Math. Appl.24, 79-99, 1992.

Margenstern, M. and Matiyasevich, Y. "A Binomial Representation of the Problem." Acta Arith.91, 367-378, 1999.

Matthews, K. R. "The Generalized Mapping." http://www.numbertheory.org/pdfs/survey.pdf.

Matthews, K. R. "A Generalized Conjecture." [$100 Reward for a Proof.] http://www.numbertheory.org/gnubc/challenge.

Matthews, K. R. and Watts, A. M. "A Generalization of Hasses's Generalization of the Syracuse Algorithm." Acta Arith.43, 167-175, 1984.

Oliveira e Silva, T. "Maximum Excursion and Stopping Time Record-Holders for the Problem: Computational Results." Math. Comput.68, 371-384, 1999.

Oliveira e Silva, T. "Computational Verification of the Conjecture." Sep. 19, 2008. http://www.ieeta.pt/~tos/3x+1.html.

Schroeppel, R.; Gosper, R. W.; Henneman, W.; and Banks, R. Item 133 in Beeler, M.; Gosper, R. W.; and Schroeppel, R. HAKMEM. Cambridge, MA: MIT Artificial Intelligence Laboratory, Memo AIM-239, p. 64, Feb. 1972. http://www.inwap.com/pdp10/hbaker/hakmem/flows.html#item133.

Sloane, N. J. A. Sequences A006577/M4323, A006666/M3733, A006667/M0019, A070165, A070166, A070167, A070168, in "The On-Line Encyclopedia of Integer Sequences."

Terras, R. "A Stopping Time Problem on the Positive Integers." Acta Arith.30, 241-252, 1976.

Terras, R. "On the Existence of a Density." Acta Arith.35, 101-102, 1979.

Thwaites, B. "Two Conjectures, or How to Win £1100." Math. Gaz.80, 35-36, 1996.

Vardi, I. "The Problem." Ch. 7 in Computational Recreations in Mathematica. Redwood City, CA: Addison-Wesley, pp. 129-137, 1991.

Wirsching, G. J. "Über das Problem." Elem. Math.55, 142-155, 2000.

Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, pp. 100, 122, and 904, 2002.

Zeleny, E. "Collatz Problem as a Cellular Automaton." http://demonstrations.wolfram.com/CollatzProblemAsACellularAutomaton/.

CITE THIS AS:

Weisstein, Eric W. "Collatz Problem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CollatzProblem.html

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The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integern. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

The conjecture is named after Lothar Collatz, who introduced the idea in 1937, two years after receiving his doctorate.[1] It 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;[2][4] 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),[5][6] or as wondrous numbers.[7]

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered $500 for its solution.[9]Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics."[10]

Statement of the problem[edit]

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 functionf as follows:

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:

(that is: is the value of applied to recursively times; ).

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 ai = 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[edit]

For instance, starting with n = 12, one gets the sequence 12, 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, in large font), 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 the OEIS)

Numbers with a total stopping time longer than that of 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 the 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 the 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 the OEIS)

The longest progression for any initial starting number

less than 10 is 9, which has 19 steps,
less than 100 is 97, which has 118 steps,
less than 1,000 is 871, which has 178 steps,
less than 10,000 is 6,171, which has 261 steps,
less than 100,000 is 77,031, which has 350 steps,
less than 1 million is 837,799, which has 524 steps,
less than 10 million is 8,400,511, which has 685 steps,
less than 100 million is 63,728,127, which has 949 steps,
less than 1 billion is 670,617,279, which has 986 steps,
less than 10 billion is 9,780,657,630, which has 1132 steps,[11][12] and
less than 100 billion is 75,128,138,247, which has 1228 steps.

The powers of two converge to one quickly because is halved times to reach one, and is never increased.

Visualizations[edit]

  • 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. This plot shows a restricted y axis: some x values produce intermediates as high as 2.7e7 (for x = 9663)

Cycles[edit]

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, for odd n and for even n. A cycle is a sequence where , , and so on, up to 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[edit]

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[edit]

As of 2017[update], the conjecture has been checked by computer for all starting values up to 87×260 ≈ 1020.[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 Skewes' number, sometimes a conjecture's only counterexamples are found when using very large numbers.

A probabilistic heuristic[edit]

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[edit]

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 x0.84.[19]

Other formulations of the conjecture[edit]

In reverse[edit]

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

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,

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, ). Then in binary, the number n can be written as the concatenation of stringswkwk−1w1 where each wh is a finite and contiguous extract from the representation of 1/3h.[20] The representation of n therefore holds the repetends of 1/3h, 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[edit]

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[edit]

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

111 1111 101110111 10001100011 110111011 1011011 111 1

As a parity sequence[edit]

For this section, consider the Collatz function in the slightly modified form

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 pi = P(ai), where a0 = n, and ai+1 = f(ai).

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 2k. 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·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.

As a tag system[edit]

For the Collatz function in the form

Hailstone 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 na, 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[edit]

Iterating on all integers[edit]

An extension to the Collatz conjecture is to include all integers, not just positive integers. Leaving aside the cycle 0 → 0 which cannot be entered from outside, there are a total of 4 known 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.

CycleOdd-value cycle lengthFull cycle length
1 → 4 → 2 → 113
−1 → −2 → −112
−5 → −14 → −7 → −20 → −10 → −525
−17 → −50 → −25 → −74 → −37 → −110 → −55 → −164 → −82 → −41 → −122 → −61 → −182 → −91 → −272 → −136 → −68 → −34 → −17718

The generalized Collatz conjecture is the assertion that every integer, under iteration by f, eventually falls into one of the four cycles above or the cycle 0 → 0. The 0 → 0 cycle is often regarded as "trivial" by the argument, as it is only included for the sake of completeness.

Iterating with odd denominators or 2-adic integers[edit]

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

 

 

 

 

(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

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) →
(1 1 0 0 1 1 0) →
(1 0 0 1 1 0 1) →
(0 0 1 1 0 1 1) →
(0 1 1 0 1 1 0) →
(1 1 0 1 1 0 0) →

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) →
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)
Iteration time for inputs of 2 to 10,000,000
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.
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