15
$\begingroup$

The Collatz function $T$ is defined on the set $\mathbb{Z}^+$ of positive integers as: $T(n)=n/2$ if $n$ is even, and $T(n)=3n+1$ if $n$ is odd. Let $T^k$ be the $k$th iteration of $T$. We say $n$ terminates if $T^k(n)=1$ for some $k$.

Let $n$ be an integer of the form $$3^{2^k(j-1)}+3^{2^k(j-2)}+\cdots +3^{2^k}+1$$

where $k,j\in \mathbb{Z}^+$ and $j$ is odd.

Question: Will $n$ terminate for all such $k,j$?

  • 3
    What is the motivation for considering these $n$ in particular? Do you have a reason to think that this restriction makes the Collatz conjecture easier/different in nature?2011-02-21
  • 0
    I came up with these integers in my research, for reasons too tedious to describe here. I am hoping someone may know a reference or proof to this question.2011-02-21
  • 4
    There are various 3-adic analyses by Wirshing, Applegate, Lagarias et al. that may yield your answer. A web search should turn up much of interest.2011-02-21
  • 1
    Hmm, possibly I misread something, but isn't this be expressible much simpler? I read it like: let $a=3^{2^k}$ then let $n=\frac{a^j-1}{a-1}$ ? If I heve this right, then for increasing *j* the sequences of increasing *k* are subsequences of each other and we need only show the problem for *k=0*. Numerically I have the sequence of *n* having *k=0* and *j* increasing as *[1,4,13,40,121,...]* where we have $n_{j+1}=3*n_j+1$. Did I get this right so far?2011-02-22
  • 0
    Note that $k\ge 1$ in my assumption.2011-02-22
  • 0
    @TCL: so what is the final set of values *n* which have to be considered? Should be some ordered subset of *[4,40,364,...]* with $n_{j+1} = 9*n_j + 4 $ ? (I don't mind to proceed thinking about this unless this form of the assumed problem's simplification is confirmed to be correct so far.)2011-02-22
  • 0
    No. All the numbers in my list are odd. Perhaps you meant $n_{j+1}=9n_j+1$? But then my $n$ for $k=2,j=3$ (and others) is not in the list.2011-02-23
  • 0
    Hmm, I wrote a longer comment as an answer, but had an error, so deleted that answer. I try to get things working, sorry...2011-02-23
  • 1
    Do your researches turn up somewhat helpful in the direction? I am quite interested in this question, and will leave no effort to keep an eye on it. Is the reason for you to consider such a sequence also helpful in proving this special case? Thanks for sharing your result here.2011-08-18
  • 0
    In my deleted answer I attempted to begin with a simplification of the list of the numbers to be tested, but without making progress in really solving the problem. That simplification can correctly be stated as $ \displaystyle n_k={9^k-1 \over 9^{2^{ \lt k,2 \gt } } - 1} $ where the notation ** means the exponent to which the primefactor *p* occurs in *m* . It seems to me, that this description of the set of numbers involved is somehow better designed for induction or other types of proving (we have now only one parameter *k*, for instance)...2011-12-09

1 Answers 1