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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 ...
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 ...
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 ...
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 ...
I will answer any questions that the commenters may have for me. ----- Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[7] Jeffrey Lagarias stated in 2010 that the Collatz conjecture "is an extraordinarily difficult problem, completely out of reach of present day mathematics".[8]
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.
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 ...
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