- #1
firlz
- 2
- 0
Is it possible, using the Collatz hailstone sequence to ever start at a number n and end up with 2n at some point later in the sequence for values greater than 2? Can you have a sequence that goes n . . . 2n (I don't care what any of the exact values are, I want to know if using variables it is possible to ever get to a later part of the sequence that reduces to 2n from n, and what the requirements are. From there it should be easy to determine why the number 1 works, and to determine if it is possible for other values to work)?
My reason for the question comes to this. In order for the conjecture to be false there must either be a loop, or a infinitely increasing series. I think that the answer to this question should determine whether or not any loops exist.
_______________
If there is at least one loop outside of 1,2,4 (a loop being a sequence of numbers that repeat infinitely), there must be a smallest number that is part of any loop. This number must also be the smallest number in its own loop.
I'll call this number l.
Things I have reasoned must be true
l cannot be even. The reason for this is that if l is even then l/2 would be part of the looping series (which would mean that the number in question does not fit the definition of l, which requires it being the smallest number in its series).
because it is the lowest value, l cannot be reached in the loop from a smaller number (otherwise it wouldn't be l), and thus the only way to get back to l is by dividing the previous number in the sequence by 2. Thus 2l must also be a part of the sequence.
Which leads me to asking, using the hailstone sequence for values greater than 2, is it possible to get to 2n from n? Or does anyone know a method that could be used to demonstrate whether it is or isn't possible? (I suspect that it isn't possible but I would like to know if there is a way to be sure)
My reason for the question comes to this. In order for the conjecture to be false there must either be a loop, or a infinitely increasing series. I think that the answer to this question should determine whether or not any loops exist.
_______________
If there is at least one loop outside of 1,2,4 (a loop being a sequence of numbers that repeat infinitely), there must be a smallest number that is part of any loop. This number must also be the smallest number in its own loop.
I'll call this number l.
Things I have reasoned must be true
l cannot be even. The reason for this is that if l is even then l/2 would be part of the looping series (which would mean that the number in question does not fit the definition of l, which requires it being the smallest number in its series).
because it is the lowest value, l cannot be reached in the loop from a smaller number (otherwise it wouldn't be l), and thus the only way to get back to l is by dividing the previous number in the sequence by 2. Thus 2l must also be a part of the sequence.
Which leads me to asking, using the hailstone sequence for values greater than 2, is it possible to get to 2n from n? Or does anyone know a method that could be used to demonstrate whether it is or isn't possible? (I suspect that it isn't possible but I would like to know if there is a way to be sure)