Smallest number of a hypothetical second Collatz Cycle

Question

Update/warning: The explaination below contains some obvious errors. These errors have been worked on and turned this research into: Generating (almost?) all odd numbers for the 3n + 1 problem

Most here are probably aware of the Collatz Conjecture. It is conjectured that every number eventually ends in a trivial cycle of 1 -> 4 -> 2 -> 1 if you follow these rules:

Take any number:

1. If it is odd, multiply by three and add one
2. If it is even, divide by two
3. Repeat.

First of all, a disclaimer: I'm not a mathematician, I'm a computer programmer that likes math (a lot) and it trying to learn more. Maybe the thing I'm doing here is very wrong or trivial, just tell me. This is what I want to find out/learn... I hope some of you will follow along and teach me new things.

Smallest number in the cycle

If a second cycle in the Collatz Conjecture exists, it must have a smallest number in the cycle. This number has some interesting properties (if it exists).

Lets call the smallest number of the cycle Y.

First of all:

Y % 2 = 1

Y can't be even because that would divide by two creating a smaller number.

This leaves Y to be either:

Y = 6*n + 1
Y = 6*n + 3
Y = 6*n + 5

Preceding numbers

Let's look at the numbers preceding Y:

Y = 6*n + 3 is impossible. The number before this step in the cycle must be Y * 2. Thus X = Y * 2 = 6 * ((n * 2) + 1), X%6=0. There is no odd number that can form such a number X%6=0 because (odd * 3) + 1 is always X%6=4. So this number can't be part of a cycle.

Y = 6*n + 5 is also impossible. The number before this step in the cycle must be Y * 2. Thus X = Y * 2 = 6 * ((n*2) + 1) + 4. This number is X%6=4, but there is an odd number W that comes before X in the cycle:

W = ((X)-1)/3
W = ((6 * ((n*2) + 1) + 4)-1)/3
W = 4*n + 3

And we know:
Y = 6*n + 5

As you can see W < Y. Thus Y isn't the lowest number in the cycle, preceding number W is smaller. Y in this form can't be the lowest point of a cycle.

This leaves Y = 6*n + 1 as only option.

Looking forward

Next we can look at the numbers after Y:

Because Y has the form 6*n + 1 we know something about the next two numbers in the sequence.

Y = 6*n + 1
Z = Y*3+1 = 18*n + 4
A = Z / 2 = 9*n + 2

Next I want to split Y into two groups, with n being even and n being odd:

Y = 12*i + 1
A = 18*i + 2

And:

Y = 6 + 12*j + 1
A = 9 + 18*j + 2

The first form 12*i + 1 can be proven to be false, except for i=0 (the trivial known cycle 1>4>2>1)

When i = 0:

    Y = 1
    A = 2
    Now A is even and there is a successor B, which is A/2:
    B = A/2 = 2/2 = 1. This is equal to Y and can make the trivial loop. But this breaks Y > 1.

When i > 0:

    Y = 12*i + 1
    A = 18*i + 2
    B = 9*i + 1

    Given i>0 the resulting B is smaller than Y, thus Y isn't the lowest number, resulting number B is smaller.

So this leaves Y = 6 + 12*j + 1 as the only possible form for a smallest number in an hypothetical second cycle?

This eliminates 91.6% of the numbers, maybe there is a trick to eliminate the remaining 8.4%? Or am I doing something very weird here, making mistakes etc?

Answer 1

Define the function $f(x)=4x+1$

If we analyse where any odd number $n$ receives branches from the odds which precede it, firstly $n\equiv0$ mod $3$ receive no branches, obviously.

$n\equiv1$ mod $3$ receive branches on their even powers of $2$ - given by $f^q(\frac{4n-1}{3}):q\in\mathbb{N}$, and

$n\equiv2$ mod $3$ receive branches on their odd powers of $2$, given by $f^q(\frac{2n-1}{3}):q\in\mathbb{N}$

The only case in which one odd rises to the next, is if it lands on the lowest odd power of $2$, i.e. if:

$n_m=(3n_{m-1}+1)/2$

And you are right, in what I think is the 2nd part of your question these numbers whose successors are greater, are in a minority. The first example is the number $3$.

However coming to the first part of your question, this is not correct because every number $n$ has an immediately preceding number which is greater. If we take the number $31$ for example, which is on the well-known long string which leads from $27$ to $1$, we see that it rises afterwards but there are many numbers immediately preceding it which fall to it, which can be obtained by the formula I described above:

$f^q(\frac{4n-1}{3}):q\in\mathbb{N}=\{41,165,661,2645,10581,\ldots\}$

Even if some immediate predecessor of $q$ was lower than it, there are other immediate predecessors to it which are above it and some loop could come via any one of those.

Answer 2

My frank suggestion is that to learn mathematics it is more profitable to play with simple number theory rather than wasting time on open problems like the Collatz conjectures.

Y = 6*n + 3 is impossible. The number before this step in the cycle must be Y * 2. Thus X = Y * 2 = 6 * ((n * 2) + 1), X%6=0. There is no odd number that can form such a number X%6=0 because (odd * 3) + 1 is always X%6=4. So this number can't be part of a cycle.

This very first step is wrong.

Here is a counter-example sequence:

$6n+3\,,12n+6\,,24n+12\,,\cdots$

You assumed incorrectly that a doubling step must be preceded by a triple-plus-one step. However, despite your wrong proof, it is in fact true that a cycle cannot have the least member being a multiple of $3$, since it must be before a larger number and hence when we can trace backwards and there must be a triple-plus-one step somewhere, but after repeated doubling a multiple of $3$ remains a multiple of $3$ and can never be the result of a triple-plus-one step.

Y = 6*n + 5 is also impossible. The number before this step in the cycle must be Y * 2. Thus X = Y * 2 = 6 * ((n*2) + 1) + 4. This number is X%6=4, but there is an odd number W that comes before X in the cycle [...]

Same error again. Counter-example: $6n+5\,,12n+10\,,24n+20\,,48n+40\,,16n+13\,,\cdots$.

I stopped reading at that point, because this approach is doomed to fail.