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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 ...
Conjecture Field Comments Eponym(s) Cites 1/3–2/3 conjecture: order theory: n/a: 70 abc conjecture: number theory: ⇔Granville–Langevin conjecture, Vojta's conjecture in dimension 1 ⇒Erdős–Woods conjecture, Fermat–Catalan conjecture Formulated by David Masser and Joseph Oesterlé. [1] Proof claimed in 2012 by Shinichi Mochizuki: n/a ...
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 ...
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;
More generally, as 1 is a square mod for all >, there can be no complete covering system of modular identities for all , because 1 will always be uncovered. [ 14 ] Despite Mordell's result limiting the form of modular identities for this problem, there is still some hope of using modular identities to prove the Erdős–Straus conjecture.
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c satisfy the equation a n + b n = c n for any integer value of n greater than 2. The cases n = 1 and n = 2 have been known since antiquity to have infinitely many solutions. [1]
The 1-factorization conjecture that if is odd or even and , respectively, then a -regular graph with vertices is 1-factorable. The perfect 1-factorization conjecture that every complete graph on an even number of vertices admits a perfect 1-factorization.
Turán also showed that a somewhat weaker assumption, the nonexistence of zeros with real part greater than 1 + N −1/2+ε for large N in the finite Dirichlet series above, would also imply the Riemann hypothesis, but Montgomery (1983) showed that for all sufficiently large N these series have zeros with real part greater than 1 + (log log N ...