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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 $k,j$?

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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
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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
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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
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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

2 Answers 2

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 $\kappa=\max\{k : \, 2^{k}|n \}$:

enter image description here

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.

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It is a conjecture that every starting integer will end in the 4-2-1 loop. The integers have been investigated with computers and a counter example to the 4-2-1 loop has never been found. I collected dozens of articles on this problem for several years some time ago. It is still an open problem. "Everyone" assumes the conjecture to be true, but no one has been able to prove it. The problem has been extensively studied and resists solution, despite its seemingly simplicity.

A quick Google search on "Collatz conjecture" will find numerous references. see for example: http://en.wikipedia.org/wiki/Collatz_conjecture

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I think you missed the question at hand. So I downvoted. –  mixedmath Jan 27 '12 at 3:39
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Don't over think this. When the given equation is evaluated for some specific values of k and j it yields an integer. No matter how big the integer is, it is still an integer. The conjecture has been evaluated randomly for some really big integers by computer and it holds. The sequence of integers has also been evaluated to some large number and it holds. So making a complex integer function just makes it look more complex. –  MaxW Jan 27 '12 at 6:20
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-1: How is this related to the question? –  Did Jan 27 '12 at 8:55
    
My reading of this was that j and K were given and the function evaluated once to get a seed. You then spun that seed through function T() until you got to the 4-2-1 loop. The problem doesn't make any sense if same K is used for seed and to limit the number iterations to some exact number. If k's are same then for k=1 the function must evaluate to 2, and for K=2 must be 4. –  MaxW Jan 27 '12 at 9:33
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It specifically says "for all such k, j". The questions is not about 1 specific value of k and j. You need to prove it terminates for all numbers of this form, or find a counterexample, to answer this question. Your answer is just generic, basic info about the conjecture that the OP probably already knows. He's saying "I know the conjecture has not been proved in general, but what can we say in this special case?" If you haven't given any specific info in this case, then you haven't answered the question. –  Graphth Apr 16 '12 at 18:34

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