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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 ...
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;
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 ...
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.
The Erdős–Straus conjecture on the Diophantine equation 4/n = 1/x + 1/y + 1/z. The Erdős conjecture on arithmetic progressions in sequences with divergent sums of reciprocals. The Erdős–Szekeres conjecture on the number of points needed to ensure that a point set contains a large convex polygon.
Restriction 1: is a variable which ... Using our intuition we conjecture that the conclusion might be . Using the Rules of Inference table we can prove the conjecture ...
The perfect 1-factorization conjecture that every complete graph on an even number of vertices admits a perfect 1-factorization. Cereceda's conjecture on the diameter of the space of colorings of degenerate graphs [87] The Earth–Moon problem: what is the maximum chromatic number of biplanar graphs? [88]