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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 '3''n'' + 1 conjecture', the 'Ulam conjecture' (after Stanislaw Ulam), the 'Syracuse problem', as the 'hailstone sequence' or 'hailstone numbers', or as 'Wondrous numbers' as per ''Gödel, Escher, Bach''. It asks whether a certain kind of number sequence always ends in the same way, regardless of the starting number.
Paul Erdős said about the Collatz conjecture, "Mathematics is not yet ready for such problems." He offered $500 for its solution. (Lagarias 1985)

Contents

Statement of the problem

Examples

Program to calculate Collatz sequences

Supporting arguments

Experimental evidence

Probabilistic evidence

Other ways of approaching the problem

In reverse

As rational numbers

As an abstract machine

As a parity sequence

Extensions to larger domains

Iterating on ''all'' integers

Iterating on rational numbers with odd denominators

Iterating on real or complex numbers

Optimizations

Syracuse Function

See also

References and external links

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:
:
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:
:
The Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.'' Or, more formally:
:
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 takes 111 steps, climbing above 9,000 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 }

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 believe intuitively that the conjecture is true. Here are two reasons for expecting this.

Experimental evidence

The conjecture has been checked by computer for all start values up to
10 × 2⁵⁸ ≈ 2.88 × 10¹⁸[1].
While impressive, such computer bounds are of very limited evidential value, with several conjectures having turned out to have exceptionally large-valued counterexamples (e.g. the Pólya conjecture, Mertens conjecture).

Probabilistic evidence

If one considers only the ''odd'' numbers in the sequence generated by the Collatz process, then one can argue that on average (specifically, the geometric mean of the ratios) the next odd number should be about ¾ of the previous one [2], which suggests that they should decrease in the long run (although this is not evidence against cycles, only against divergence).

Other ways of approaching the problem

In reverse

There is another approach to prove the conjecture, which considers the bottom-up
method of growing the Collatz graph. The Collatz graph is defined by an inverse relation,

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'', 3''n'' + 1 ≡4 (mod 6) iff ''n'' ≡1 (mod 2) and thus ''n'' ≡1, 3 or 5 (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 3''n'' + 1 of the function ''f''(''n'') is replaced by the common substitute "shortcut" relation (3''n'' + 1)/2 (see Optimizations below), the Collatz graph is defined by the inverse relation,

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:
:
(3n - d + 1)/4d & mbox{if } 3n + d + 1 ge 2d end{cases} (''n'' = numerator; ''d'' = denominator).
This works because 3''x'' + 1 = 3(''d'' + ''n'') + 1 = (2''d'') + (3''n'' + ''d'' + 1) = (4''d'') + (3''n'' - ''d'' + 1). Reducing a rational before every operation is required to get ''x'' as an odd.

As an abstract machine

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:
# Add the original with a "1" appended to the end to the original (interpreting the string as a binary integer), i.e. $3n+1\; =\; (2n+1)\; +\; n$
# Remove all trailing "0"s.

As a parity sequence

For this section, consider the Collatz function in the slightly modified form
:
This can be done because when ''n'' is odd, 3''n'' + 1 is always even.
If P(…) is the parity of a number, that is P(2''n'') = 0 and P(2''n'' + 1) = 1, then we can define the Collatz parity sequence for a number ''n'' as ''p_i'' = P(''a_i''), where ''a''₀ = ''n'', and ''a''_''i''+1 = ''f''(''a''_''i'').
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 Collatz cycles, then their corresponding parity cycles must be different.
The Collatz Conjecture can be rephrased as stating that the Collatz parity sequence for every number eventually enters the cycle 0 → 1 → 0.

Extensions to larger domains

Iterating on ''all'' integers

For 'any' integer n, 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 the size of the 'full' cycle (in parentheses): the number of members, odd or even, belonging to a cycle, counted without repetition.
a) 1 → 1 (size 3)
b) 0 → 0 (size 1)
c) -1 → -1 (size 2)
d) -5 → -7 → -5 (size 5)
e) -17 → -25 → -37 → -55 → -41 → -61 → -91 → -17 (size 18)
We may define the 'Generalized Collatz Conjecture' as the assertion that every integer, under iteration by f, eventually falls into one of these five cycles a), b), c), d), or e).

Iterating on rational numbers with odd denominators

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.
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 ''k''₀, …, ''k''_''m''-1, then the unique fraction which generates that parity cycle is
:.
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 identicle 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) →
But when reduced to lowest terms {5/7}, it is the same as that of the 4-element sub-sequence
:(1 1 0 0) →
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
:
ight)+(3z+1)sin^2left(rac pi 2 z
ight),
which simplifies to .
If the standard Collatz map defined above is optimized by replacing "3''n'' + 1" with "(3''n'' + 1)/2" (see Optimizations below), it can be viewed as the restriction to the integers of the smooth real and complex map
:
ight)+rac 1 2 (3z+1)sin^2left(rac pi 2 z
ight),
which simplifies to .
Iterating the above optimized map in the complex plane produces the Collatz fractal.

Optimizations

★ If ''n'' is a multiple of 4, it can be divided by 4.
: 'Reason': It is initially even. When it is divided by two, it is still even.
: 'Example': If ''n'' = 20, then the sequence goes
:: 20 → 10 → 5 (= 20/4).

★ More generally, one can write ''n'' as the product of its factors, and change the power of 2 to 2⁰.
: 'Reason': If the power of 2 of the prime factorization is greater than 0, the number is even, and the following step would produce the same factorization with one less power of 2.
: 'Example': Instead of
:: 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (17 steps),
: one can go
:: 15, 46 (2¹×23), 23, 70 (2¹×35), 35, 106 (2¹×53), 53, 160 (2⁵×5), 5, 16 (2⁴), 1 (11 steps).

★ If ''n'' is odd, one step can be skipped by finding (3''n'' + 1) / 2
: 'Reason': an odd times an odd is always an odd (neither contributes a factor of 2 and without a 2, it cannot be even), so 3''n'' is odd. Hence after two steps we have (3''n'' + 1) / 2
: 'Example': Take ''n'' = 35. Then the normal Collatz sequence is
:: 35 → 3 × 35 + 1 = 106 → 53 (= (3 × 35 + 1)/2).

★ If ''n'' ≡ 1 mod 4, then we can replace it repeatedly by (''n'' - 1)/4 to obtain an intermediate ''k'' not equivalent to 1 modulo 4, and obtain 3''k'' + 1.
: 'Reason': ''n'' = 4''m'' + 1 is always odd, hence becomes 3''n'' + 1 = 12''m'' + 4 = 4(3''m'' + 1), and by the first optimization, we may drop the 4. Note that if ''m'' is odd, 3''m'' + 1 is its Collatz successor, so that to continue the Collatz sequence of ''n'', we can instead continue that of ''m''. Hence we have effectively replaced ''n'' by ''m'' = (n - 1)/4. We may make this reduction repeatedly, until we obtain a number ''k'' which is not equivalent to 1 modulo 4. Finally, we are now looking at 3''k'' + 1, as desired. Note that ''k'' may be even, so we cannot necessarily divide by 2.
: 'Example': When ''n'' = 405 the optimized sequence is
:: 405 → 101 → 25 → 6 → 19.
: The normal Collatz sequence is:
:: 405 → 1216 → 608 → 304 → 152 → 76 → 38 → 19.
The above facts can be used to create a new version of the Collatz function :
:
g(n) & mbox{if } n equiv 0 \
3 h(n) + 1 & mbox{if } n equiv 1 \
n/2 & mbox{if } n equiv 2 \
rac{1}{2}(3n+1) & mbox{if } n equiv 3
end{cases} pmod{4}
where
:
g(rac{n}{2}) & mbox{if } n equiv 0 \
n & mbox{if } n equiv 1
end{cases} pmod 2
:
n & mbox{if } n ;
otequiv; 1 \
h(rac{n-1}{4}) & mbox{if } n equiv 1
end{cases} pmod{4}.

Syracuse Function

If ''k'' is an odd integer, then 3''k'' + 1 is even, so we can write 3''k'' + 1 = 2^a''k''′, 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''′ .
Some properties of the Syracuse function are:

★ ''f'' (4''k'' + 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'' - 1}''h'' - 1 (see here for the notation).

★ For all odd ''h'', ''f'' (2''h'' - 1) ≤ (3''h'' - 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:
We know that 1, 3, 5, 7, and 9 are 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 = 2^''p''''h'' for ''p'' ≥ 1, ''h'' odd, and ''k'' = 2^''p''''h''-1. Now we have:

★ If ''p'' = 1, then ''k'' = 2''h'' - 1. It is easy to check that ''f'' (''k'') < ''k '', so ''f'' (''k'') ∈ ''E''; hence ''k'' ∈ ''E''.

★ If ''p'' ≥ 2 and ''h'' is a multiple of 3, we can write ''h'' = 3''h′''. Let ''k′'' = 2^{''p'' + 1}''h′'' - 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 ''k'' ∈ ''E''. (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 ; 3''k''′ ∈ ''E'' we are done. (Cf.).

See also

★ Residue class-wise affine groups

★ Modular arithmetic

References and external links

★ Jeffrey C. Lagarias. ''The 3x + 1 problem: An annotated bibliography (1963--2000)''.

★ Jeffrey C. Lagarias. ''The 3x + 1 problem: An annotated bibliography, II (2001--)''.

★ Jeffrey C. Lagarias. ''The 3x + 1 problem and its generalizations'', American Mathematical Monthly Volume 92, 1985, pp. 3 - 23.

★

★ Günther J. Wirsching. The Dynamical System Generated by the $3n+1$ Function. Number 1681 in Lecture Notes in Mathematics. Springer-Verlag, 1998.

★ 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).

★

★

★ Hailstone Patterns discusses different resonators along with using important numbers in the problem (like 6 and 3^5) to discover patterns.

★ Review of progress, plus various programs

★ Ohira, Reiko & Yamashita, Michinori A generalization of the Collatz problem

★ URATA, Toshio Some Holomorphic Functions connected with the Collatz Problem

★ Matti K. Sinisalo: On the minimal cycle lengths of the Collatz sequences, Preprint, June 2003, University of Oulu

★ Paul Stadfeld: Blueprint for Failure: How to Construct a Counterexample to the Collatz Conjecture