# A question regarding Collatz conjecture

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 $k,j$?

-
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? –  mjqxxxx Feb 21 '11 at 18:40
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. –  TCL Feb 21 '11 at 18:49
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. –  Bill Dubuque Feb 21 '11 at 19:33
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? –  Gottfried Helms Feb 22 '11 at 9:17
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. –  awllower Aug 18 '11 at 14:36

Not really an answer but there are a few things I'd like to point out:

1) $n_{j,k} = \displaystyle \sum_{i=0}^{j-1} 3^{i \cdot 2^{k}} = \frac{3^{j \cdot 2^{k}}-1}{3^{2^{k}}-1}$

2) Since $j$ is odd, $n_{j,k}$ is also odd.

3) Thus, $T(n_{j,k})=3 \cdot n_{j,k}+1$ and $T^{2}(n_{j,k})=\frac{1}{2}(3 \cdot n_{j,k}+1)$, since $3n+1$ is even for all $n$.

4) $n_{j,k}$ has an interesting form when viewed in base 3. For example,

IntegerDigits[n[7,3],3]
{1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1}


That is, for $n_{j,k}$ we get a 1 followed by $j$ zeros, etc.

5) Here is a plot of the lengths of the sequences $T^{k}(n_{j,k})$ for $1 \le j \le 16$ and $1 \le k \le 8$, although for these sequences I used a highly reduced version of the Collatz function where if $n$ is even, $T(n)=n/2^{\kappa}$, where $\kappa=\max\{k : \, 2^{k}|n \}$:

Even though these numbers have a certain form that would seem to make it easier to prove they have a downward trajectory to 1, after looking around a little bit I really didn't seem to see any patterns. $\{n_{j,k} : j \, \text{odd}, k \in \mathbb{N}^{+}\}$ is a fairly "large" set of integers, so this question might really be similar in difficultly to the full conjecture.

-