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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?

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  • $\begingroup$ Regarding your first sentence: have you seen this? $\endgroup$ May 16 '16 at 19:08
  • $\begingroup$ I suppose that the cycle $1 \to 4 \to 2 \to 1$ does not count. $\endgroup$ 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. $\endgroup$ May 25 '16 at 18:59
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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.

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  • $\begingroup$ There might be a proof of the given claim, but none is known. $\endgroup$
    – Peter
    May 18 '16 at 9:34
  • $\begingroup$ Note that above, 68-cycle does not mean a cycle of length 68. Instead, it means a cycle that has 68 maximal increasing streaks, alternating with 68 maximal decreasing streaks. $\endgroup$ May 15 '20 at 9:02

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