Question

Can someone help me solve the Collatz conjecture? Reward is $500,000 if you are able to solve it btw!

Answer 1

: (2^(n+1))−1 mod 2^(n+2).

Answer 2

cap

Answer 3

its 120 million Japanese yen the prize btw ~$1m

Answer 4

You don't have 500k lol

Solving Collatz is a lot harder than you think.

Answer 5

ik i donbt have 500k xD

Answer 6

dont*

Answer 7

Key points about the Collatz prize:
Offered by: Bakuage, a Japanese company.
Amount: 120 million Japanese yen.
Problem: The Collatz conjecture, a famous unsolved mathematical problem.

Answer 8

"no" yet you see the word in that reply you just made, unsolved mathematical problem

Answer 9

yes, it is the prize for SOLVING it

Answer 10

ik its not solved xD

Answer 11

Steiner's method to prove that there is no 2-cycle.[22] Simons and de Weger (2005) extended this proof up to 68-cycles; there is no k-cycle up to k = 68.[15] Hercher extended the method further and proved that there exists no k-cycle with k ≤ 91.[20] As exhaustive computer searches continue, larger k values may be ruled out. To state the argument more intuitively; we do not have to search for cycles that have less than 92 subsequences, where each subsequence consists of consecutive ups followed by consecutive downs.[clarification needed] 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 ) = { { 2 n } if n ≡ 0 , 1 , 2 , 3 , 5 { 2 n , n − 1 3 } if n ≡ 4 ( mod 6 ) . {\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1,2,3,5\\[4px]\left\{2n,{\frac {n-1}{3}}\right\}&{\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 backwards to all positive integers. For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, ⁠ n − 1 / 3 ⁠ ≡ 1 (mod 2) if and only if 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 ) = { { 2 n } if n ≡ 0 , 1 { 2 n , 2 n − 1 3 } if n ≡ 2 ( mod 3 ) . {\displaystyle R(n)={\begin{cases}\{2n\}&{\text{if }}n\equiv 0,1\\[4px]\left\{2n,{\frac {2n-1}{3}}\right\}&{\text{if }}n\equiv 2\end{cases}}{\pmod {3}}.} For any integer n, n ≡ 1 (mod 2) if and only if ⁠ 3n + 1 / 2 ⁠ ≡ 2 (mod 3). Equivalently, ⁠ 2n − 1 / 3 ⁠ ≡ 1 (mod 2) if and only if 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). Alternatively, 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, fk(n) = 1). Then in binary, the number n can be written as the concatenation of strings wk wk−1 ... w1 where each wh is a finite and contiguous extract from the representation of ⁠ 1 / 3h ⁠.[23] 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.[24] 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: Append 1 to the (right) end of the number in binary (giving 2n + 1); Add this to the original number by binary addition (giving 2n + 1 + n = 3n + 1); Remove all trailing 0s (that is, repeatedly divide by 2 until the result is odd). 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 shortcut form of the Collatz function f ( n ) = { n 2 if n ≡ 0 3 n + 1 2 if n ≡ 1 ( mod 2 ) . {\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1\end{cases}}{\pmod {2}}.} 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 (or parity vector) for a number n as pi = P(ai), where a0 = n, and ai+1 = f(ai). Which 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.[2][17] Applying the f function k times to the number n = 2ka + b will give the result 3ca + d, where d is the result of applying the f function k times to b, and c is how many increases were encountered during that sequence. For example, for 25a + 1 there are 3 increases as 1 iterates to 2, 1, 2, 1, and finally to 2 so the result is 33a + 2; for 22a + 1 there is only 1 increase as 1 rises to 2 and falls to 1 so the result is 3a + 1. When b is 2k − 1 then there will be k rises and the result will be 3ka + 3k − 1. The power of 3 multiplying a is independent of the value of a; it depends only on the behavior of b. This allows one to predict that certain forms of numbers will always lead to a smaller number after a certain number of iterations: for example, 4a + 1 becomes 3a + 1 after two applications of f and 16a + 3 becomes 9a + 2 after four applications of f. Whether those smaller numbers continue to 1, however, depends on the value of a. As a tag system For the Collatz function in the shortcut form f ( n ) = { n 2 if n ≡ 0 3 n + 1 2 if n ≡ 1. ( mod 2 ) {\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0\\[4px]{\frac {3n+1}{2}}&{\text{if }}n\equiv 1.\end{cases}}{\pmod {2}}} Hailstone sequences can be computed by the 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 copies of 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 as the initial word, eventually halts (see Tag system for a worked example). Extensions to larger domains Iterating on all integers 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 four 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. 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 cycles above or the cycle 0 → 0. Iterating on rationals with odd denominators The 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. Then the formula for the map is exactly the same as when the domain is the integers: an 'even' such rational is divided by 2; an 'odd' such rational is multiplied by 3 and then 1 is added. 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. When using the "shortcut" definition of the Collatz map, it is known that any periodic parity sequence is generated by exactly one rational.[25] Conversely, it is conjectured that every rational with an odd denominator has an eventually cyclic parity sequence (Periodicity Conjecture[2]). If a parity cycle has length n and includes odd numbers exactly m times at indices k0 < ⋯ < km−1, then the unique rational which generates immediately and periodically this parity cycle is 3 m − 1 2 k 0 + ⋯ + 3 0 2 k m − 1 2 n − 3 m . {\displaystyle {\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 four odd terms at indices 0, 2, 3, and 6. It is repeatedly generated by the fraction 3 3 2 0 + 3 2 2 2 + 3 1 2 3 + 3 0 2 6 2 7 − 3 4 = 151 47 {\displaystyle {\frac {3^{3}2^{0}+3^{2}2^{2}+3^{1}2^{3}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {151}{47}}}as the latter leads to the rational cycle 151 47 → 250 47 → 125 47 → 211 47 → 340 47 → 170 47 → 85 47 → 151 47 . {\displaystyle {\frac {151}{47}}\rightarrow {\frac {250}{47}}\rightarrow {\frac {125}{47}}\rightarrow {\frac {211}{47}}\rightarrow {\frac {340}{47}}\rightarrow {\frac {170}{47}}\rightarrow {\frac {85}{47}}\rightarrow {\frac {151}{47}}.} Any cyclic permutation of (1 0 1 1 0 0 1) is associated to one of the above fractions. For instance, the cycle (0 1 1 0 0 1 1) is produced by the fraction 3 3 2 1 + 3 2 2 2 + 3 1 2 5 + 3 0 2 6 2 7 − 3 4 = 250 47 . {\displaystyle {\frac {3^{3}2^{1}+3^{2}2^{2}+3^{1}2^{5}+3^{0}2^{6}}{2^{7}-3^{4}}}={\frac {250}{47}}.} For a one-to-one correspondence, a parity cycle should be irreducible, that is, not partitionable into identical sub-cycles. As an illustration of this, the parity cycle (1 1 0 0 1 1 0 0) and its sub-cycle (1 1 0 0) are associated to the same fraction ⁠ 5 / 7 ⁠ when reduced to lowest terms. In this context, assuming the validity of the Collatz conjecture implies that (1 0) and (0 1) are the only parity cycles generated by positive whole numbers (1 and 2, respectively). If the odd denominator d of a rational is not a multiple of 3, then all the iterates have the same denominator and the sequence of numerators can be obtained by applying the "3n + d " generalization[26] of the Collatz function T d ( x ) = { x 2 if x ≡ 0 ( mod 2 ) , 3 x + d 2 if x ≡ 1 ( mod 2 ) . {\displaystyle T_{d}(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}},\\[4px]{\frac {3x+d}{2}}&{\text{if }}x\equiv 1{\pmod {2}}.\end{cases}}} 2-adic extension The function T ( x ) = { x 2 if x ≡ 0 ( mod 2 ) 3 x + 1 2 if x ≡ 1 ( mod 2 ) {\displaystyle T(x)={\begin{cases}{\frac {x}{2}}&{\text{if }}x\equiv 0{\pmod {2}}\\[4px]{\frac {3x+1}{2}}&{\text{if }}x\equiv 1{\pmod {2}}\end{cases}}}is well-defined on the ring Z 2 {\displaystyle \mathbb {Z} _{2}} of 2-adic integers, where it is continuous and measure-preserving with respect to the 2-adic measure. Moreover, its dynamics is known to be ergodic.[2] Define the parity vector function Q acting on Z 2 {\displaystyle \mathbb {Z} _{2}} as Q ( x ) = ∑ k = 0 ∞ ( T k ( x ) mod 2 ) 2 k . {\displaystyle Q(x)=\sum _{k=0}^{\infty }\left(T^{k}(x)\mod 2\right)2^{k}.} The function Q is a 2-adic isometry.[27] Consequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that almost all trajectories are acyclic in Z 2 {\displaystyle \mathbb {Z} _{2}}. An equivalent formulation of the Collatz conjecture is that Q ( Z + ) ⊂ 1 3 Z . {\displaystyle Q\left(\mathbb {Z} ^{+}\right)\subset {\tfrac {1}{3}}\mathbb {Z} .} Iterating on real or complex numbers Cobweb plot of the orbit 10 → 5 → 8 → 4 → 2 → 1 → ... in an extension of the Collatz map to the real line. The Collatz map can be extended to the real line by choosing any function which evaluates to x / 2 {\displaystyle x/2} when x {\displaystyle x} is an even integer, and to either 3 x + 1 {\displaystyle 3x+1} or ( 3 x + 1 ) / 2 {\displaystyle (3x+1)/2} (for the "shortcut" version) when x {\displaystyle x} is an odd integer. This is called an interpolating function. A simple way to do this is to pick two functions g 1 {\displaystyle g_{1}} and g 2 {\displaystyle g_{2}}, where: g 1 ( n ) = { 1 , n is even, 0 , n is odd, {\displaystyle g_{1}(n)={\begin{cases}1,&n{\text{ is even,}}\\0,&n{\text{ is odd,}}\end{cases}}} g 2 ( n ) = { 0 , n is even, 1 , n is odd, {\displaystyle g_{2}(n)={\begin{cases}0,&n{\text{ is even,}}\\1,&n{\text{ is odd,}}\end{cases}}} and use them as switches for our desired values: f ( x ) ≜ x 2 ⋅ g 1 ( x ) + 3 x + 1 2 ⋅ g 2 ( x ) {\displaystyle f(x)\triangleq {\frac {x}{2}}\cdot g_{1}(x)\,+\,{\frac {3x+1}{2}}\cdot g_{2}(x)}. One such choice is g 1 ( x ) ≜ cos 2 ⁡ ( π 2 x ) {\displaystyle g_{1}(x)\triangleq \cos ^{2}\left({\tfrac {\pi }{2}}x\right)} and g 2 ( x ) ≜ sin 2 ⁡ ( π 2 x ) {\displaystyle g_{2}(x)\triangleq \sin ^{2}\left({\tfrac {\pi }{2}}x\right)}. The iterations of this map lead to a dynamical system, further investigated by Marc Chamberland.[28] He showed that the conjecture does not hold for positive real numbers since there are infinitely many fixed points, as well as orbits escaping monotonically to infinity. The function f {\displaystyle f} has two attracting cycles of period 2 {\displaystyle 2}: ( 1 ; 2 ) {\displaystyle (1;\,2)} and ( 1.1925... ; 2.1386... ) {\displaystyle (1.1925...;\,2.1386...)}. Moreover, the set of unbounded orbits is conjectured to be of measure 0 {\displaystyle 0}. Letherman, Schleicher, and Wood extended the study to the complex plane.[29] They used Chamberland's function for complex sine and cosine and added the extra term 1 π ( 1 2 − cos ⁡ ( π z ) ) sin ⁡ ( π z ) + {\displaystyle {\tfrac {1}{\pi }}\left({\tfrac {1}{2}}-\cos(\pi z)\right)\sin(\pi z)\,+} h ( z ) sin 2 ⁡ ( π z ) {\displaystyle h(z)\sin ^{2}(\pi z)}, where h ( z ) {\displaystyle h(z)} is any entire function. Since this expression evaluates to zero for real integers, the extended function f ( z ) ≜ z 2 cos 2 ⁡ ( π 2 z ) + 3 z + 1 2 sin 2 ⁡ ( π 2 z ) + 1 π ( 1 2 − cos ⁡ ( π z ) ) sin ⁡ ( π z ) + h ( z ) sin 2 ⁡ ( π z ) {\displaystyle {\begin{aligned}f(z)\triangleq \;&{\frac {z}{2}}\cos ^{2}\left({\frac {\pi }{2}}z\right)+{\frac {3z+1}{2}}\sin ^{2}\left({\frac {\pi }{2}}z\right)\,+\\&{\frac {1}{\pi }}\left({\frac {1}{2}}-\cos(\pi z)\right)\sin(\pi z)+h(z)\sin ^{2}(\pi z)\end{aligned}}} is an interpolation of the Collatz map to the complex plane. The reason for adding the extra term is to make all integers critical points of f {\displaystyle f}. With this, they show that no integer is in a Baker domain, which implies that any integer is either eventually periodic or belongs to a wandering domain. They conjectured that the latter is not the case, which would make all integer orbits finite. A Collatz fractal centered at the origin, with real parts from -5 to 5. Most of the points have orbits that diverge to infinity. Coloring these points based on how fast they diverge produces the image on the left, for h ( z ) ≜ 0 {\displaystyle h(z)\triangleq 0}. The inner black regions and the outer region are the Fatou components, and the boundary between them is the Julia set of f {\displaystyle f}, which forms a fractal pattern, sometimes called a "Collatz fractal". Julia set of the exponential interpolation. There are many other ways to define a complex interpolating function, such as using the complex exponential instead of sine and cosine: f ( z ) ≜ z 2 + 1 4 ( 2 z + 1 ) ( 1 − e i π z ) {\displaystyle f(z)\triangleq {\frac {z}{2}}+{\frac {1}{4}}(2z+1)\left(1-e^{i\pi z}\right)}, which exhibit different dynamics. In this case, for instance, if Im ⁡ ( z ) ≫ 1 {\displaystyle \operatorname {Im} (z)\gg 1}, then f ( z ) ≈ z + 1 4 {\displaystyle f(z)\approx z+{\tfrac {1}{4}}}. The corresponding Julia set, shown on the right, consists of uncountably many curves, called hairs, or rays. 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 is given by fk(2ka + b) = 3c(b, k)a + d(b, k). The values of c (or better 3c) and d can be precalculated for all possible k-bit numbers b, where d(b, k) is the result of applying the f function k times to b, and c(b, k) is the number of odd numbers encountered on the way.[30] For example, if k = 5, one can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using c(0...31, 5) = { 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, 5) = { 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 2k 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 fk(2ka + b) = 3c(b)a + d(b) < 2ka + b holds for all a, then the first counterexample, if it exists, cannot be b modulo 2k.[13] 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.[31] For example, the only surviving residues mod 32 are 7, 15, 27, and 31. Integers divisible by 3 cannot form a cycle, so these integers do not need to be checked as counter examples.[32] Syracuse function If k is an odd integer, then 3k + 1 is even, so 3k + 1 = 2ak′ with k′ odd and a ≥ 1. The Syracuse function is the function f from the set I of positive odd integers into itself, for which f(k) = k′ (sequence A075677 in the OEIS). Some properties of the Syracuse function are: For all k ∈ I, f(4k + 1) = f(k). (Because 3(4k + 1) + 1 = 12k + 4 = 4(3k + 1).) In more generality: For all p ≥ 1 and odd h, fp − 1(2ph − 1) = 2 × 3p − 1h − 1. (Here fp − 1 is function iteration notation.) For all odd h, f(2h − 1) ≤ ⁠ 3h − 1 / 2 ⁠ The Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer n ≥ 1 such that fn(k) = 1. Undecidable generalizations In 1972, John Horton Conway proved that a natural generalization of the Collatz problem is algorithmically undecidable.[33] Specifically, he considered functions of the form g ( n ) = a i n + b i when n ≡ i ( mod P ) , {\displaystyle {g(n)=a_{i}n+b_{i}}{\text{ when }}{n\equiv i{\pmod {P}}},}where a0, b0, ..., aP − 1, bP − 1 are rational numbers which are so chosen that g(n) is always an integer. The standard Collatz function is given by P = 2, a0 = ⁠ 1 / 2 ⁠, b0 = 0, a1 = 3, b1 = 1. Conway proved that the problem Given g and n, does the sequence of iterates gk(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 gk(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[34] proved that the universally quantified problem is, in fact, undecidable and even higher in the arithmetical hierarchy; specifically, it is Π0 2-complete. This hardness result holds even if one restricts the class of functions g by fixing the modulus P to 6480.[35] Iterations of g in a simplified version of this form, with all b i {\displaystyle b_{i}} equal to zero, are formalized in an esoteric programming language called FRACTRAN.

Answer 12

Gotta love copy/paste.

Answer 13

nah i typed all that in google docs, used parahraser, then copy and pasted my work (parahrased)