Having fun with the Collatz Conjecture

Let’s use a different approach to tackle the problem — organizing all uneven numbers in a different way.

Number are now organized according to two dimensions x and y. Every unit in the x dimension or vertical axis corresponds to an uneven number (2x - 1) and every unit in the y axis, or horizontal axis corresponds to a power of 2 (2^y). This means all uneven natural numbers can be organized by the number of time they’re divisible by two after applying Collatz (3x - 1).

For example 9 can be divided by 2 twice because y=2, (9 * 3 + 1 = 28 / 2 = 14 / 2 = 7), and so on.

c1: all uneven numbers organized by the number of times they’re divisible by two after applying 3x+1.

Now let’s go a step further and apply a step of Collatz to the above function (c1) until we get all uneven numbers again.

Notice anything? This is were the fun starts. It turns out that this second function is easy to predict, when y is uneven we get 6x - 1 and y is even we get 6x - 5.

c2: 6x — 1 and 6x — 5 repeat until infinity.

Now we can go back and forth between c1 and c2, and multiple patterns will emerge:

The first pattern is that if we are running a brute force search, we only need to calculate the first two columns of the c1 function: 4x - 1 and 8x - 7. Everything that’s after y > 2 is redundant according to c2.

Example:

• c1(x=2, y=4) = 37 -> (112, 56, 28, 14) -> c2(x=2, y=4) = 7

Instead of running the seed “37” it’s easier to shift 2 columns to the left and run “9” instead.

Example:

• c1(x=2, y=2) = 9 -> (28,14) -> c2(x=2, y=2) = 7

So applying Collatz to “37” produces the same result as applying it to “9” but with extra steps or powers of 2. The same goes for “45”and “11” or “309” and “19”.

c1: it’s only required to prove 4x — 1 and 8x — 7

There is a second pattern the green numbers in c1 are not contained in c2. They are multiples of 3 (3, 9, 15, 33, …) and represent roots or points of entry in the conjecture. This means that for example 9 will never show up as a step in another number’s sequence, the only way to get a sequence with 9 is to start with 9.

This also means all multiples of 3 will map to either 6x - 1 or 6x - 5 (non multiples of 3).

A complete example of using this method to run Collatz on seed “11”:

c1(x=3, y=1) = 11 (look for “11” in c1)

c2(x=3, y=1) = 17 (get corresponding Collatz step in c2)

c1(x=3, y=2) = 17 (look for “17” in c1)

c2(x=3, y=2) = 13 (get corresponding Collatz step in c2)

c1(x=1, y=3) = 13 (look for “13” in c1)

c1(x=1, y=1) = 3 (look for shortcut in c1, “13” has y>2)

c2(x=1, y=1) = 5 (get corresponding Collatz step in c2)

c1(x=1, y=4) = 5 (look for “5” in c1)

c1(x=1, y=2) = 1 (look for shortcut in c1, “5” has y>2)

c1(x=1, y=2) = 1 (get corresponding Collatz step in c2)

.