3x 1 2

It is also known as the Collatz problem or the hailstone problem. This leads to the sequence 3, 10, 5, 16, 4, 2, 1, 4, 2, 1, A sequence obtained by iterating the function from a given starting value is sometimes called "the trajectory" of that starting value. Obviously there can be no consecutive odd numbers in any trajectory, 3x 1 2, but there may 3x 1 2 be consecutive even numbers, especially when the trajectory reaches a power of 4, in which the trajectory quickly plummets to 1 after passing through all the intervening powers of 2.

The Collatz conjecture [a] is one of the most famous unsolved problems in mathematics. The conjecture asks whether repeating two simple arithmetic operations will eventually transform every positive integer into 1. It concerns sequences of integers in which each term is obtained from the previous term as follows: if the previous term is even , the next term is one half of the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence. It is named after the mathematician Lothar Collatz , who introduced the idea in , two years after receiving his doctorate.

3x 1 2

MathPapa Toggle navigation. This is a tutorial on how to use the Algebra Calculator , a step-by-step calculator for algebra. Solving Equations First go to the Algebra Calculator main page. In the Calculator's text box, you can enter a math problem that you want to calculate. Examples To see more examples of problems that Algebra Calculator understands, visit the Examples page. Feel free to try them now. To graph a point, enter an ordered pair with the x-coordinate and y-coordinate separated by a comma, e. To graph two objects, simply place a semicolon between the two commands, e. Polynomials Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x. Evaluating Expressions Algebra Calculator can evaluate expressions that contain the variable x. To evaluate an expression containing x, enter the expression you want to evaluate, followed by the sign and the value you want to plug in for x. Algebra Calculator can also evaluate expressions that contain variables x and y. To evaluate an expression containing x and y, enter the expression you want to evaluate, followed by the sign and an ordered pair containing your x-value and y-value. Here is an example evaluating the expression xy at the point 3,4 : xy 3,4.

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Use step-by-step calculators for chemistry, calculus, algebra, trigonometry, equation solving, basic math and more. Gain more understanding of your homework with steps and hints guiding you from problems to answers! Wolfram Alpha Pro step-by-step solutions not only give you the answers you're looking for, but also help you learn how to solve problems. Not just answers—step-by-step solutions buttons expand answers and explain how that answer was found. Break steps down even further with intermediate steps. These optional additional explanations for individual pieces of step-by-step solutions help guide you.

3x 1 2

Below are multiple fraction calculators capable of addition, subtraction, multiplication, division, simplification, and conversion between fractions and decimals. Fields above the solid black line represent the numerator, while fields below represent the denominator. In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole. For example, in the fraction of 3 8 , the numerator is 3, and the denominator is 8. A more illustrative example could involve a pie with 8 slices. If a person were to eat 3 slices, the remaining fraction of the pie would therefore be 5 8 as shown in the image to the right. Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined.

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Bibcode : AcAri. The resulting function f maps from odd numbers to odd numbers. 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. Mobile Apps. The numbers with a record number of steps are given by A Wonders of Numbers. There is another approach to prove the conjecture, which considers the bottom-up method of growing the so-called Collatz graph. Compare All. ISSN X. Whether those smaller numbers continue to 1, however, depends on the value of a. All School Discussions. GMAT Focus tests. For a one-to-one correspondence, a parity cycle should be irreducible , that is, not partitionable into identical sub-cycles. The starting values whose maximum trajectory point is greater than that of any smaller starting value are as follows:. Responding to this work, Quanta Magazine wrote that Tao "came away with one of the most significant results on the Collatz conjecture in decades".

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Numbers with a total stopping time longer than that of any smaller starting value form a sequence beginning with:. Theoretical Computer Science. We noticed you are actually not timing your practice. 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. The proof is based on the distribution of parity vectors and uses the central limit theorem. Thus the term "hailstone" is sometimes used to refer to this problem. Quantitative Questions. Practice Pays we will pick new questions that match your level based on your Timer History. Note that since the odd-indexed powers of 2 are congruent to 2 modulo 3, they are only reachable from halving a power of 4. The function Q is a 2-adic isometry. Journal of Integer Sequences. 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. As k increases, the search only needs to check those residues b that are not eliminated by lower values of k. 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.