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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 ...
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 ...
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 problem was finally solved in 1993 by Jeff Kahn and Gil Kalai, who showed that the general answer to Borsuk's question is no. [9] They claim that their construction shows that n + 1 pieces do not suffice for n = 1325 and for each n > 2014. However, as pointed out by Bernulf Weißbach, [10] the first part of this claim is in fact false.
There is no section on how 3n+1 conjecture may do anything with prime numbers; could use one. Example: proving 3n+1 conjecture means there is no infinite Nnext=(3n+1)/2 sequence of prime numbers that just won't stop appearing. Seriously, proving 3N+1 is real means proving curious sequences like this one are impossible.
Examples of this type include the Erdős–Heilbronn Conjecture (for a restricted sumset) and the Cauchy–Davenport Theorem. The methods used for tackling such questions often come from many different fields of mathematics, including combinatorics, ergodic theory, analysis, graph theory, group theory, and linear-algebraic and polynomial methods.
Bad idea, you should always include the loop cycle. True, it's easy to snip it for 3n+1 in the positive domain, but not so easy elsewhere. If you snip off the loop cycle of +(3n+1), there remains only one "trunk". But the -5 & -17 graphs in -(3n+1) have more than one "trunk", so removing the loop cycle makes the pieces of the remaining graph ...
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 ...