 I acknowledge that I'm on the uncedured traditional territories of the Comox and Qualicum First Nations and I'm grateful for the use of their lands. The Colatz Conjecture. Can the 3N plus 1 problem be solved by 4N plus 1? What is the Colatz Conjecture? Well, beginning with any integer, a path is created by these rules. Rule C1, if the integer is odd, multiply by 3 and increase by 1. Rule C2, if the integer is even, divide by 2. Continue this process with each new integer created. The Colatz Conjecture is that no matter what integer that you begin with, the path will end with the 4, 2, 1 loop. When the conjecture is approached by analyzing the path that can be created a step at a time, it can be confounding and confusing. Another name for Colatz is the hailstone problem because the numbers act like hailstones in a thunderstorm, rising and falling with no predictable pattern. To avoid this difficulty, I'm going to take a different approach. I don't think these numbers should be thought of as independent hailstones, but rather as integers whose values depend on the surrounding numbers. We'll use Colatz rules to create these groups and once created, they will allow us a more manageable approach to constructing a Colatz tree. First, because I want to build my framework on positive odd integers, I will show that all positive even integers have a path that leads to a positive odd integer. Starting with an integer that has the prime factors of 2 to the np, where n is positive and p is any odd number. Colatz rules C2 of dividing even numbers by 2 will create n even values in its path, followed by the odd value p. And conversely, once the odd integers are in place, the even integers are easily generated. Now let's focus on constructing the Colatz tree with respect to the positive odd integers. For a concrete example of a group, let's start with 17. Because 17 is odd, I must use the Colatz rule C1 that odd numbers are multiplied by 3 and then have one added to them. For 17, the C1 rule results in the value 52. At this point, the next step is to apply rule C2 and divide 52 by 2 to get the result 26. But the group that I'm interested in is not as concerned with the path from 17 as it is with the numbers that precede the path from 17. So instead, I'll invert rule C2 and multiply by 2 once to get 104 and then once again to get 208. Here I notice that 208 is 3 times 69 plus 1, so I'll use the inverse of rule C1 to get the value 69. This forms the complete group based on the number 17. Some things to notice, 69 and 17 have different paths but have a consistent connection because they share the same path from 52 on. The template for this group is that odd numbers begin and end the group and there are always three even numbers between them. So for the general case, beginning with any odd number to n plus 1, we use the rule C1 to create 6n plus 4. Now we apply C2 inverted once to get 12n plus 8 and one more time to get 24n plus 16. Then inverting rule C1, we complete the group with 8n plus 5. So we've created a valid collet's path from 8n plus 5 to 24n plus 16 to 12n plus 8 to 6n plus 4 where it's joined by the path from 2n plus 1. Every odd number has a unique group constructed by this process, providing the infinite number of groups needed to construct an infinite collet's tree. What other properties does this group have? Well, if the starting number is n, the larger number is always 4n plus 1 and we will call this n to 4n plus 1 transformation rule C3 to go with the previous rules C1 and C2. A property essential to creating the collet's tree is that the group begins and ends with an odd number. This means that C3 can be executed repeatedly to create a chain of groups, each one building on the last. Using the starting point of n equals 1, we can see how the rule C3 constructs a chain in the collet's tree. Applying C3 once, we can get the group containing 5. Applying C3 again to 5, we get the group containing 21 and again to get 85. And applying C3 an infinite number of times, we create the chain of infinite powers of 2 and their associated odd limbs which form the central structure of the tree. Notice also that we've addressed our uncertainty about the distance to the next odd number by finding the position of an odd number adjacent to the path. This consistency allows us to descend the stream of even numbers to a deterministic result at the bottom of the chain. Using the standard rules C1 and C2 on the starting point of 85, we yield 85, 256, 128, 64, 32, 16, 8, 4, 2, 1 with an unbroken stretch of 8 even numbers. The rule C3 allows us to create consistent segments delimited by the odd numbers 85, 21, 5, and 1. It's important to note that only the numbers that are 5 mod 8 can use the inverted C3 rule to travel down this chain because the starting point of this chain of groups must be odd. Let's use a number, like 33, which is 1 mod 8. We can see that 4 times 8 plus 1 equals 33, so 4 plus 1 can be calculated. But 8 is an even number, so it would be divided by 2 by rule C2 and would never initiate the creation of the group with rule C1. All numbers in a Colatz chain other than the first one need to be in the form 5 mod 8. We've looked at the chain that starts with 1, but now we see the chains can begin with any odd number that is either 1 mod 8, 3 mod 8, or 7 mod 8. The chain of groups cannot begin with an odd number of the form 5 mod 8 because, as we already know, 5 mod 8 numbers are a result of applying C3 to an earlier odd number in the chain so they could not start the chain. Now that we've created these long chains, we just need to show that application of the rules C1, C2, and C3 will construct predecessor and successor values for each odd number in the tree. We've already seen that the C3 rule results in a 5 mod 8 value preceding any odd number, but if we look at the predecessor rules for C1 and C2 alone, we see that they have three outcomes for their predecessors. This is because odd numbers divisible by 3 will not remain divisible by 3 after they are reduced by 1, and so the inverse of rules C1 never applies to them. Since the predecessor rule for C3 are captured by module 8 and the predecessor rules for C1 and C2 are captured by module 6, we'll view both the predecessor and successor rules in module 24 to specify the values of both preceding and succeeding numbers for every possible odd number. Having our actions determined by module 24 combined with the additional option of C3 gives us a much more nuanced way of dealing with the odd numbers than the traditional call-outs rules C1 and C2 provide, and we are still operating within the rules of the call-outs conjecture. Starting with 1 mod 24, we note that we can calculate both the succeeding and preceding odd numbers with certainty because 1 mod 24 will only occur at the end of the chain. We also have the preceding value due to the C3 rule because 1 mod 24 is odd. 3 mod 24 is a case where we have an odd number divisible by 3, so it would have no predecessor through the traditional call-outs rules, but does have a 4M plus 1 predecessor because rule C3 does not depend on divisibility by 3 to create a group. 5 mod 24 is a case where we have an odd number within a chain that is not divisible by 3, so it does have a preceding odd number through the inverse of C1 and C2 in addition to the preceding odd because of C3. The succeeding odd number will be found through the inverse of C3 effectively moving us down the chain. As we move on through the other odd numbers in module 24, we see the variety of different behaviors that would never be captured by a simple odd or even rule that the traditional rules of the call-outs conjecture provide us. This means that given any odd number, we can always find a preceding and a succeeding odd number and we can calculate the values and positions of those numbers based on the value of our current position, mod 24. What is more, since the succeeding and preceding numbers are also odd, every odd number is tightly bound to their neighbors in the call-outs tree. You might do a thought experiment of what it would take to have an odd number not be present in the call-outs tree. Removal of any odd number from the tree violates the surrounding predecessor and successor rules because the surrounding odd numbers would continue to be linked to the value and the position of the removed value based on the three rules C1, C2 and C3. Since no odd number can be removed from the call-outs tree without violating the call-outs rules, then the call-outs conjecture is true and every integer will have a path that converges to one. Given the technique that we've used to construct these groups, you might say that we've solved a 3n plus 1 problem with the 4n plus 1 solution.