A puzzle on the Collatz conjecture?

Question

The Collatz conjecture is this: Start with any positive integer. If it is even, divide it by 2. If it is odd, multiply it by 3 and add 1. Repeat this process over and over, and eventually you will reach 1. For example, if you start with 3, the sequence goes 3 10 5 16 8 4 2 1, completing the path in seven steps.

This conjecture has never been proven. However, it easily seen that, assuming positive integer n satisfies the conjecture, then 2n also satisfies the conjecture, since the path for 2n must start 2n, n and then follow the same path as n.

How about the following: assuming odd positive integer n satisfies the conjecture, can it be proven that 4n+1 also satisfies the conjecture? Also, is the path longer or shorter, and by how many steps?

Answer 1

4n+1 is odd:
3(4n+1)+1 = 12n+4 is even:
(12n+4)/2 = 6n+2 is even:
(6n+2)/2 = 3n+1 (is even, but we don't have to care).

Following the path of n, which is odd, we immediately come to:
n
3n+1.

Therefore 4n+1, when n is odd, satisfies the conjecture, and the path is two steps longer.

Answer 2

4n+1 is always odd, so starting with 4n+1 the next number in the sequence would be 3(4n+1) + 1 = 12n + 4. This is even, so the next number is 6n + 2. This is even too, so the next number is 3n + 1.

Notice that if n is odd, then the very next number in the sequence is also 3n+1. So if it works for n, then it works for 4n+1. You just have to do three steps to go from 4n+1 to 3n+1.

Interesting conjecture. I never heard of this one before. On a side note, I found you can prove that the sequence always has to end in "16, 8, 4, 2, 1" unless it's one of those same numbers.

Answer 3

The conjecture has been checked by computer for all starting values up to 5 × 2^60 ≈ 5.764×10^18 To write a program which deals with this stuff one must use libraries to handle large integers repeat show n if n is odd then set n = 3n + 1 else set n = n / 2 endif until n = 1

Answer 4

It's true. If n is odd and satisfies the Collatz conjecture
so does 4n+1. The path of both numbers has exactly
the same length.
I don't remember the proof right now, but I'll
try and work it out.
But to get an idea of what's going on. Here are all the numbers with path length 1:
5 21 85 321, ... .
Note that each number in this row is 4 times the
previous one + 1.

Answer 5

3(4n+1)+1=12n+4
=4(3n+1)
=2^2(3n+1)

since odd +ve n start with 3n+1, the path is longer by 3 more step. (odd, even,even)

Answer 6

4n+1,12n+4,6n+2, 3n+1, terminates if 3n+1 terminates.

Answer 7

What are you, my freakin' doctoral advisor?

:-)