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  2. Collatz conjecture - Wikipedia

    en.wikipedia.org/wiki/Collatz_conjecture

    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 ...

  3. Talk:Collatz conjecture/Archive 2 - Wikipedia

    en.wikipedia.org/wiki/Talk:Collatz_conjecture/...

    Basically 3n+1 must, for all n, cross, intersect, or land on or something math-speak, a value that is also in 2^x where x is a natural number. How about, there is a non-empty intersection between the set of all numbers 2^x and any set of 3n+1 for any natural starting number n.

  4. Talk:Collatz conjecture/Archive 1 - Wikipedia

    en.wikipedia.org/wiki/Talk:Collatz_conjecture/...

    Suppose n is odd. Then 3n+1 is even. With probability 1/2 the next odd number is (3n+1)/2; So half of all even numbers have a single factor of 2. with probability 1/4, the next odd number is (3n+1)/4; So half the remaining even numbers have 2 factors of 2. with probability 1/8, the next odd number is (3n+1)/8;

  5. List of conjectures - Wikipedia

    en.wikipedia.org/wiki/List_of_conjectures

    As reformulated, it became the "paving conjecture" for Euclidean spaces, and then a question on random polynomials, in which latter form it was solved affirmatively. 2015: Jean Bourgain, Ciprian Demeter, and Larry Guth: Main conjecture in Vinogradov's mean-value theorem: analytic number theory: Bourgain–Demeter–Guth theorem, ⇐ decoupling ...

  6. Erdős–Straus conjecture - Wikipedia

    en.wikipedia.org/wiki/Erdős–Straus_conjecture

    However, 1 is a square mod 3 (equal to the square of both 1 and 2 mod 3), so there can be no similar identity for all values of that are congruent to 1 mod 3. More generally, as 1 is a square mod n {\displaystyle n} for all n > 1 {\displaystyle n>1} , there can be no complete covering system of modular identities for all n {\displaystyle n ...

  7. Millennium Prize Problems - Wikipedia

    en.wikipedia.org/wiki/Millennium_Prize_Problems

    The conjecture is that there is a simple way to tell whether such equations have a finite or infinite number of rational solutions. More specifically, the Millennium Prize version of the conjecture is that, if the elliptic curve E has rank r, then the L-function L(E, s) associated with it vanishes to order r at s = 1.

  8. NYT ‘Connections’ Hints and Answers Today, Saturday ... - AOL

    www.aol.com/nyt-connections-hints-answers-today...

    Get ready for all of today's NYT 'Connections’ hints and answers for #587 on Saturday, January 18, 2025. Today's NYT Connections puzzle for Saturday, January 18, 2025 The New York Times

  9. 3x + 1 semigroup - Wikipedia

    en.wikipedia.org/wiki/3x_+_1_semigroup

    The elements of a generating set of this semigroup are related to the sequence of numbers involved in the still open Collatz conjecture or the "3x + 1 problem". The 3x + 1 semigroup has been used to prove a weaker form of the Collatz conjecture. In fact, it was in such context the concept of the 3x + 1 semigroup was introduced by H. Farkas in ...