Variant of the Collatz Conjecture - Why is the following happening:

Question

I am a web developer that programs in PHP which is limited to large math calculations.

I am also a math enthusiastic that likes learning math through problems that seem simple (but are anything like simple), and the best example is the Collatz Conjecture.

I am not dilusional and I am well aware that there is no chances for someone at my level to solve the Collatz Conjecture, but I like to try and solve other elements that may be related.

Let's begin with the description of the conjecture: According to the rules of the Collatz Conjecture if $𝑛$ is odd then execute $3𝑛+1$ and when $𝑛$ is even execute $𝑛/2$. Repeat until (supposedly) reaching $𝑛=1$.

So I was trying to prove the most basic variant: If $𝑛$ is odd then execute $𝑛+1$ and when $𝑛$ is even execute $𝑛/2$. Repeat until (supposedly) reaching 𝑛=1.

I was able to proof it in my own head,

So I progressed into trying to prove: If $𝑛$ is odd then execute $𝑛+𝑥$ and when $𝑛$ is even execute $𝑛/2$. Repeat until (supposedly) reaching $𝑛=1$. So obviously sometimes it reached 1 and some times it reached 𝑥.

I was able to proof it in my own head.

So I realized that I had to find a different approach for when 𝑛 is odd that is more than just adding a constant.

So I decided to add $1$ for the first odd occurrence, $3$ to the second odd occurrence, $5$ to the third odd occurrence , $7$ to the fourth odd occurrence .... (with increments of $+ 2$)

I have then ran a PHP script with a list of the first 2500 odd numbers, with a limit accepted of 10000 steps.

I immediately noticed that most numbers ended in 1, example:

$13 → 14 → 7 → 10 → 5 → 10 → 5 → 12 → 6 → 3 → 12 → 6 → 3 → 14 → 7 → 20 → 10 → e5 → o20 → 10 → 5 → 22 → 11 → 30 → 15 → 36 → 18 → 9 → 32 → 16 → 8 → 4 → 2 → 1$

but then there were numbers that had so many steps and were cut by the accepted $10000$ steps of my script.

I have then checked a specific number that seemed to have endless steps, which is the beginning number $21$.

To my surprise there were no loops but I have noticed that there seemed to be a pattern that shows growing into infinity.

Fo example the beginning number of $21$:

$21 → 22 → 11 → 14 → 7 → 12 → 6 → 3 → 10 → 5 → 14 → 7 → 18 → 9 → 22 → 11 → 26 → 13 → 30 → 15 → o34 → 17 → 38 → 19 → 42 → 21 → 46 → 23 → 50 → 25 → 54 → 27...$

I have then noticed that all the numbers that seem to have endless steps reach $6 → 3 → 10 → 5 → 14 → 7 → 18....$

Which seems to be a pattern that grows into infinity,

So my question is how can I prove that:

Starting with $6$ and then first odd occurrence ($3$) add $7$, second odd occurrence add $9$, third odd occurrence add $11$... will always grow infinitely (*Note the sequence/pattern of odd → even → odd → even...)?

Do all these numbers that seem to have endless steps, begin the endless path through $6 → 3 → 10 → 5 → 14 → 7 → 18....$ ? In other words is that the only path to infinite steps in my presented variant?

Answer 1

This is the sequence https://oeis.org/A066070, and yes it continues indefinitely. The pattern is

$$2x \to x \to 2x+4 \to x+2 \to 2x+8 \to x + 4 \to \dots$$

Notice that whenever an odd number $n$ is in the sequence, the number $n+4$ is added to it. This can be formally proven by induction.

EDIT: incomplete answer to follow-up question: this is the only loop of this "kind". If we have:

$$n \to 2n+2a \to n+a \to 2n+3a+2 \to n+\frac32a+1\to\dots$$

we have $n+2a = n + \dfrac32 a + 1$, giving $a=2$. If we have:

$$n \to 4n + 4a \to\to n+a \to 4n+4a + 2\to2n + 2a + 1$$

we already start breaking the pattern. Therefore there are no more "short, double-arithmetic-sequence loops". Whether a larger or irregular loop exists requires some more thought. For all values that fell into a loop that I have checked, it eventually falls into this loop.