0
$\begingroup$

One potential counterexample of the Collatz conjecture would be if there was a number that looped back to itself. Of course, this would still not prove the conjecture because some sequences could still potentially diverge to infinity. Has it been proven that Collatz sequences cannot repeat?

$\endgroup$
  • $\begingroup$ Regarding your first sentence: have you seen this? $\endgroup$ – J. M. is a poor mathematician May 16 '16 at 19:08
  • $\begingroup$ I suppose that the cycle $1 \to 4 \to 2 \to 1$ does not count. $\endgroup$ – Rodrigo de Azevedo May 16 '16 at 19:17
  • $\begingroup$ @thecat It is widely known that proof of the two facts a) there is no non-trivial loop, and b) no sequence ascends to infinity, is sufficient to prove the Collatz conjecture. Since these are the only two ways some linear sequence might not ultimately reach 1. $\endgroup$ – samerivertwice May 25 '16 at 18:59
0
$\begingroup$

No but it has been proven that if a cycle exist it is large and has certain properties.

Simons & de Weger (2003) extended this proof up to 68-cycles: there is no k-cycle up to k = 68.[13] Beyond 68, this method gives upper bounds for the elements in such a cycle: for example, if there is a 75-cycle, then at least one element of the cycle is less than 2385×2^50... from collatz wiki

from the connection between cycle length and maximum number in cycle they can be exaustivly searched and cycle of that length disproven. but no general result.

$\endgroup$
  • $\begingroup$ There might be a proof of the given claim, but none is known. $\endgroup$ – Peter May 18 '16 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.