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I wonder what is known about variations of Collatz where $3n+1$ is replaced by $3n + 2k + 1$ where k is a fixed positive integer. In the OP ' about Collatz $3n+3$ ' it was confirmed that $3n+3$ behaves like Collatz itself. About Collatz 3n+3?

I wonder about other values of $k$.

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The title and body have conflicting versions of $k$. –  joriki Oct 11 '12 at 17:02
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The first stop for this kind of question is The 3x+1 problem and its generalizations by Lagarias. –  lhf Oct 11 '12 at 17:07
    
Did a comment get romoved ?? I got a message of another comment ... –  mick Oct 11 '12 at 20:08
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Not exactly. Consider the case $3n+3$. There $Coll_{3n+3}(2^{m}k)$ behave $2^{m-1}k,\ldots,k,3(k+1),3Coll_{3n+1}(k+1)$. For the general case $Coll_{3n+3l}(2^{m}k)$ behave $2^{m-1}k,\ldots,k,3(k+l),3Coll_{3n+l}(k+l)$. So $Coll_{3n+15}(k)$ behaves like $3Coll_{3n+5}(k+5)$ for odd $k$. –  P.. Oct 22 '12 at 20:44
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@mick If you can find $k$ such that $\operatorname{Coll}_{3n+2k+1}$ give only one cycle then you proved Collatz conjecture.Of course if there is such a $k$ then necessary $2k+1=3^a$. See my answer. –  P.. Oct 26 '12 at 19:05

1 Answer 1

up vote 4 down vote accepted
  1. (Collatz Conjecture) $\operatorname{Coll}_{3n+1}=\operatorname{Coll}$ (the original Collatz function) give only one cycle.
  2. For $\ell=3^ar, \ a\geq 1$ with $3 \nmid r , \ \operatorname{Coll}_{3n+\ell}$ behave like $\operatorname{Coll}_{3n+r}$ (meaning that iteration of $\ \operatorname{Coll}_{3n+\ell}$ beginning at $x$ terminates, for each $x \in \mathbb{N}$ iff iteration of $\operatorname{Coll}_{3n+r}$ beginning at $x$ terminates, for each $x \in \mathbb{N}$). In particular if $\ell=3^a$ then $\ \operatorname{Coll}_{3n+\ell}$ behave like $\operatorname{Coll}$.
  3. For $\ell$ such that $3 \nmid \ell , \ \operatorname{Coll}_{3n+\ell}$ give at least two cycles, the one being iteration of $\operatorname{Coll}_{3n+\ell}$ beginning at $1$, and the other iteration of $\operatorname{Coll}_{3n+\ell}$ beginning at $\ell$. Further for $k \in \mathbb{N}$ such that $\ell | k$ iteration of $\operatorname{Coll}_{3n+\ell}$ at $k$ behave like $\operatorname{Coll}$ (and eventually reach $\ell$, assuming the truth of Collatz conjecture).

For (2):

if $x=2^bk$ with $2 \nmid k$ then iteration of $\operatorname{Coll}_{3n+\ell}$ at $x$ goes: $$2^bk\rightarrow 2^{b-1}k\rightarrow \ldots\rightarrow k \rightarrow 3k+l=3k+3^ar=3(k+3^{a-1}r)\rightarrow \ldots $$ so whatever $x$ is, iteration of $\operatorname{Coll}_{3n+\ell}$ beginning at $x$ will give at some point $3y$, for some $y \in \mathbb{N}$.

If $2 \nmid y$ then $\operatorname{Coll}_{3n+\ell}(3y)=3\cdot 3y+\ell= 3( 3y+\frac{\ell}{3}) = 3\operatorname{Coll}_{3n+\frac{\ell}{3}}(y)$.

If $2|y$ then $\operatorname{Coll}_{3n+\ell}(3y)=3 \frac{y}{2}=3\operatorname{Coll}_{3n+\frac{\ell}{3}}(y)$.

And so on.

So $\ \operatorname{Coll}_{3n+\ell}$ behave like $\operatorname{Coll}_{3n+r}$.

For (3):

Let $x \in \mathbb{N}$. If $x=\ell k$ with $2 \nmid k$ then $\operatorname{Coll}_{3n+\ell}(x)=3\cdot \ell k+\ell= \ell( 3k+1) = \ell \operatorname{Coll}(y)$, while if $x=2\ell k$ then $\operatorname{Coll}_{3n+\ell}(x)=\frac{x}{2}=\ell k= \ell \operatorname{Coll}(2k)$.

So for $x \in \ell \mathbb{N}$ the iteration of $\operatorname{Coll}_{3n+\ell}$ at $x$ behave like $\operatorname{Coll}$.

This give at least one cycle and shows that for $x \in \ell \mathbb{N}$, iteration of $\operatorname{Coll}_{3n+\ell}$ at $x$ will eventually reach $\ell$.

Now working $\pmod{\ell}$ we see that for $x \notin \ell \mathbb{N}$ if iteration of $\operatorname{Coll}_{3n+\ell}$ at $x$ give a cycle this cycle is disjoint from the cycle obtained by $\operatorname{Coll}_{3n+\ell}(\ell)$(Or by $\operatorname{Coll}_{3n+\ell}(y)$ for any $y \in \ell \mathbb{N}$, if we assume the truth of Collatz conjecture). This is because for $x \not\equiv 0 \pmod \ell$ we have $3x+\ell \equiv 3x \not\equiv 0 \pmod \ell$ and (for even $x$) $\frac{x}{2} \equiv x \frac{l+1}{2} \not\equiv 0 \pmod \ell$.

We see that case (2) is similar to the case (3),with the exception that the three in $3n+\ell$ makes the iteration of $\operatorname{Coll}_{3n+\ell}$ at $x$ to produce a multiple of $3$ whatever $x$ is.


In the following table I list:

In the first column values of $\ell$. In the second how many cycles $\operatorname{Coll}_{3n+\ell}$ I believe give and in third pairs $(y,z)$(see the example). The integers $y$ are those minimal integers s.t. iteration of the various $\operatorname{Coll}_{3n+\ell}(y)$ give all cycles of $\operatorname{Coll}_{3n+\ell}$. The integers $z$ are the smallest positive integers such that iteration of $\operatorname{Coll}_{3n+\ell}$ beginning at $z$ reach $y$.

$$ \begin{array}{|c|c|c|} \hline \ell & \text{Number of cycles} & (y,z) \\ \hline 1 & 1 & (1,1) \\ \hline 5 & 6 & (1,1),(19,3),(5,5),(23,23), (187,123),(347,171)\\ \hline 7 & 2 & (5,1),(7,7)\\ \hline 11 & 3 & (1,1),(13,3),(11,11) \\ \hline 13 & 10 & (1,1),(13,13),(131,19),(211,99),\\ & & (259,123),(227,147),(287,159),\\ & & (251,163),(283,283),(319,319) \\ \hline 17 & 3 & (1,1),(23,9),(17,17)\\ \hline 19 & 2 & (5,1),(19,19)\\ \hline 23 & 4 & (41,1),(5,5),(7,7),(23,23)\\ \hline 25 & 8 & (7,1),(17,3),(5,5),\\ & & (95,15),(25,25),(115,115),\\ & & (935,615),(1735,855)\\ \hline 29 & 5 & (1,1),(11,3),(29,29),\\ & & (3811,2531),(7055,5859)\\ \hline 31 & 2 & (13,1),(31,31)\\ \hline \end{array}$$

For example for $\ell=17$ we have three cycles. They are:

  • $1\rightarrow20\rightarrow10\rightarrow5\rightarrow32\rightarrow\ldots \rightarrow 1$ of length 9,
  • $23\rightarrow146\rightarrow73\rightarrow \ldots \rightarrow46\rightarrow23$ of length 49 and
  • $17\rightarrow68\rightarrow34\rightarrow17$

For the second one $19$ is the smallest integer for witch $\operatorname{Coll}_{3n+17}$ will give this cycle.

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Nice answer. But I believe 3n+5 gives an unbounded amount of cycles. –  mick Nov 1 '12 at 20:18
    
That's interesting. And why do you believe that? Do you know any cycle different from the six I gave? –  P.. Nov 1 '12 at 21:31
    
I would need to check my data again ... later ... –  mick Nov 1 '12 at 22:26
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For this particular case I made an exhaustive checking for all values smaller than $10000000$ and I didn't find other cycles. –  P.. Nov 1 '12 at 22:33
    
Then I am mistaken sorry. –  mick Nov 1 '12 at 23:20

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