Is there a new cycle for $f(n)$ in a "$1 x+3$"-variation of the Collatz-problem?

Question

Let $f$ be defined as follows.

$$ f(n) := \begin{cases} n+3 & \text {if $n$ is odd,} \\ \frac{n}{2} & \text {if $n$ is even.} \end{cases}$$

If we start at $n=15$, we get the following sequence by successive applications of $f$:

15,18,9,12,6,3,6,3,6…

I found the two following cycles:

3,6,3,6,3,6…

and

1,4,2,1,4,2…

Is there any other cycle?

Thank you so much!

Answer 1

When you successively apply $f$ to a value, it must be divided by $2$ at least one step out of two (because when you add $3$, you get an even number). Let $n$ be an odd number strictly greater than $3$. Then $f^2(n) = \frac{n+3}{2} < n$. We continue to apply $f$ until we get an other odd number, which is then strictly less than $n$.

So every number leads to an odd number less than or equal to $3$. The two possibilities are $1$ and $3$, leading to the two only cycles.

Answer 2

$$\text{odd}(n)\land n>3\implies \text{even}(n+3)\land n>3\implies n>\frac{n+3}2$$ $$\text{even}(n)\land n>1\implies n>\frac n2$$

hence the sequence is decreasing for all $n$ but possibly $0,1,2,3$.

Then

$$0\to0\to0\to0\to0\to0\cdots$$ $$1\to4\to2\to1\to4\to2\cdots$$ $$2\to1\to4\to2\to1\to4\cdots$$ $$3\to6\to3\to6\to3\to6\cdots$$

For negative $n$, the sequence is increasing both for odd and for even.

Answer 3

For general transformation $a \gt 3, a \text{ is odd}$ we'll have one transformation (with any $k>0$): $$a=2k+3 \to b=(a+3)/2 = ((2k+3)+3)/2 = k+3 $$

• 1. Result for general transformation

$$ a=2k+3 \to b=k+3 \\ \implies \\ \text{ for all odd $a \gt 3$ each transformation $ a \to b={a+3\over2^A}$decreases} $$

• 2. Test on existence of short (1-step) cycles: $$ a = (a+3)/2^A \\ a2^A-a = 3 \\ a = {3 \over 2^A-1 } \\ \implies a \in \{1,3\} \\ \implies \text{ cycle $1$: } (1 \to 4 \to 2 \to 1 \to \cdots);\\ \text{ cycle $2$: } (3 \to 6 \to 3 \to \cdots ) $$ Since all other $a \gt 3$ decrease (by 1. ) only no more cycle is possible.