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While working on the Collatz problem, I came across this answer. I understand everything except for one thing: that for $n$ odd running Collatz2($n$) is exactly like running Collatz($n+1$). I understand that the first iteration of Collatz2 takes us to $3n+3$ which is a multiple of 3. However, I don't see how this relates to the original Collatz($n+1$) sequence. For convenience, I present the definitions of both functions:

Collatz: \begin{align} f(n) &= \begin{cases} (3n + 1) / 2, & \text{for odd $n$,}\\ n / 2, & \text{for even $n$.}\\ \end{cases} \end{align}

Collatz2: \begin{align} f(n) &= \begin{cases} (3n + 3) / 2, & \text{for odd $n$,}\\ n / 2, & \text{for even $n$.}\\ \end{cases} \end{align}

DaBler
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  • Who claims that those are the same ? – Peter May 23 '23 at 12:35
  • @Peter Read the second paragraph in the answer. It works for example with Collatz2(31) and Collatz(32). – DaBler May 23 '23 at 12:43
  • @Peter: More precisely: Collatz2: 31 -> 48 -> 24 -> 12 -> 6 -> 3 Collatz: 32 -> 16 -> 8 -> 4 -> 2 -> 1 – DaBler May 23 '23 at 12:44
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    It does not say the sequences are the same, just that they are equivalent. It says that if you run a Collatz2 sequence, eventually you will hit an odd number, and apply 3n+3. From then on, the sequence will only have multiples of 3 in it. The remaining Collatz2 steps will never make it not a multiple of 3. If you look at that remaining sequence starting with 3n+3 and ending in 3, and divide all its numbers by 3, then you get a normal Collatz sequence that starts with n+1 and ends in 1. – Jaap Scherphuis May 23 '23 at 13:31
  • Note that the linked question/answers do not have the division by 2 in the odd-n branch of the function, and that is what I assumed in my previous comment. – Jaap Scherphuis May 23 '23 at 13:36
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    In short, $Collatz_2(3\cdot k)=3\cdot Collatz(k)$. – Jaap Scherphuis May 23 '23 at 14:39
  • @JaapScherphuis: I know that I get 3n+3 (multiple of three) when I do the first step of Collatz2 of odd number. I also know that the Collatz and Collatz2 sequences of multiples of three are related by Collatz2(3n) = 3 Collatz(n). That wasn't my question. I ask how Collatz2(n) can be related to Collatz(n+1), as these two sequences follow the Collatz2(3n) = 3 Collatz(n) rule. Also the division by 2 in the odd-n branch of the function does not affect the answer (it's just skipping the n/2 step for both functions). – DaBler May 23 '23 at 16:43
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    The point is that as soon as a Collatz2 sequence hits an odd number $n$ (which it will always do at some point), it produces $3n+3=3(n+1)$ in the next step and therefore this and the remaining numbers is just triple the normal Collatz sequence of $n+1$. This shows that the question of whether there is a Collatz2 sequence that does not end in 3 is equivalent to the question of whether there is a Collatz sequence that does not end in 1. – Jaap Scherphuis May 23 '23 at 18:12
  • I got it. You are welcome to submit this comment as an answer. – DaBler May 25 '23 at 13:40
  • Duplicate of the much older question : https://math.stackexchange.com/questions/211061/about-collatz-3n3 – mick Feb 11 '24 at 18:04
  • Does this answer your question? About Collatz 3n+3? – mick Feb 11 '24 at 18:05

2 Answers2

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The $3n+1$ rule and the $3n+3$ rule are special cases of the $3n+k$ rule, where $k$ can be any odd integer. Applying any of these rules is the same as applying $3n+1$ to fractions with denominator $k$, and omitting the denominators. This isn't too hard to see:

A fraction with odd denominator can be considered "odd" or "even" according to its numerator, so consider applying the usual $3n+1$ rule, starting with $\frac15$. We get:

$$\frac15 \rightarrow \frac85 \rightarrow \frac45 \rightarrow \frac25 \rightarrow \frac15 \rightarrow \cdots$$

Now, just start with $1$, and apply the $3n+5$ rule:

$$1 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 \rightarrow \cdots$$

As you can see, these are just the numerators of the previous trajectory.

This works because $1=\frac55$, so adding $1$ to the fraction $\frac{n}5$ just gives us $\frac{n+5}5$.

Anyway, if we're using the $3n+3$ rule, then we're just working with the ordinary Collatz rule, for fractions with denominator $3$. As soon as we multiply by $3$ one time, we end up with an integer, and then we're just playing the usual Collatz game. A trajectory under the $3n+3$ rule, after a single odd step, is a just a regular Collatz trajectory on integers, but we're thinking of each integer $n$ as the fraction $\frac{3n}3$, and simply writing down numerators.

The terminal loop $12 \rightarrow 6 \rightarrow 3$ is really just a shorthand for the terminal loop $\frac{12}3 \rightarrow \frac63 \rightarrow \frac33$, which is just $4 \rightarrow 2 \rightarrow 1$.

G Tony Jacobs
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$3n+3$ is not equivalent to $3n+1$. In general, the sequences for $3n+k_1$ and $3n+k_2$, where $k_1$ and $k_2$ are odd integers, will have corresponding scaled sequences in the $3n+k_1k_2$ system where $n$ is simply scaled by $k_2$ or $k_1$ respectively. The converse is not true since the $3n+k_1k_2$ system may contain sequences that correspond to neither $3n+k_1$ nor $3n + k_2$. For example, $3n+5$ has a loop that corresponds to the $3n+1$ ($1,4,2$) loop which is ($5,20,10$) but also has other loops that have no correspondence in the $3n+1$ system. None of the elements of these additional loops are a multiple of $5$. All sequences in $3n+1$ have a corresponding scaled sequence in $3n+3$ but not all sequences in $3n+3$ have a corresponding sequence in $3n+1$.

Eric Shumard
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