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

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 $$π$$ is odd then execute $$3π+1$$ and when $$π$$ is even execute $$π/2$$. Repeat until (supposedly) reaching $$π=1$$.

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

I was able to proof it in my own head,

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

I was able to proof it in my own head.

So I realized that I had to find a different approach for when π 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:

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

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 $$21$$:

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

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

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 β even β odd β even...)?

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

• +1 ; what is the importance of Collatz Conjecture ?? Feb 27, 2021 at 8:37
• @haidangel for me personally it is not important, but itβs a brain exercise. Not trying to solve the conjecture, but rather learn from it Feb 27, 2021 at 8:46

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.

• Is that the only one ? I mean starting at 6 (or 3 as the odd) Feb 27, 2021 at 8:34
• What is that sequence called ? Feb 27, 2021 at 8:35