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Recall the Collatz function given by: $$ T(n) = \begin{cases} {\dfrac{n}{2}} & n \equiv 0\pmod 2\\ & \\ 3n+1 & n \equiv 1\pmod 2 \end{cases} $$ The well-known conjecture states that $T^{(k)}(n)=1$ for all $n$, and $k$ large enough.

Is it correct that this is not known that $T^{(k)}(n)\to\infty$ does not occur for any $n$, as $k\to\infty$. What is known about this aspect of the problem?

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  • $\begingroup$ I think that it is not known is there some number starting from which the sequence will show the tendency to possibly diverge to infinity, but someone will know more than I. $\endgroup$
    – user480281
    Commented Oct 23, 2017 at 12:10
  • $\begingroup$ See en.wikipedia.org/wiki/Collatz_conjecture#Cycles $\endgroup$
    – lhf
    Commented Oct 23, 2017 at 12:15
  • $\begingroup$ Yes, this is still unproven for most numbers (those greater than about $10^{60}$). It may in fact not even be provable since even if we found some such trajectory, the fact that it would be infinitely long means we could never follow it for long enough to show it has no end. Although few skilled mathematicians think this will be the case. $\endgroup$
    – Robert Frost
    Commented Oct 25, 2017 at 18:30
  • $\begingroup$ @RobertFrost: upps, the recently marked high number is $5 \times 2^{60}$ or so but not $10^{60}$ $\endgroup$
    – Gottfried Helms
    Commented Dec 4, 2017 at 10:50
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    $\begingroup$ @Robert : well :-) so this shall be true for the upcoming historical aeras... How could I have missed that! $\endgroup$
    – Gottfried Helms
    Commented Dec 4, 2017 at 14:51

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I should have looked harder: a 2010 survey of Jeff Lagarias on p.22 conjecture (C2) exactly answers this (as of 2010, of course):

Does the 3x + 1 function have a divergent trajectory, i.e., an integer starting value whose iterates are un- bounded? This is conjectured not to be the case.

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