Does the function fx (If even=x/2)
(If else= 3x+1)
have an x number which when consecutiveley applied doesn't result in 1?
This is a well-known problem, known as the Collatz conjecture (http://en.wikipedia.org/wiki/Collatz_conjecture) and the answer is still unknown. It has been tested by computer up to many millions and no exception has yet been found. It refers to the natural numbers (1, 2, 3, ...etc.) only.
Example
x = 3 odd so 3x+1 = 10
x = 10 is even so x/2 = 5
x = 5 is odd so 3x+1 = 16
x = 16 is even so x/2 = 8
x = 8 is even so x/2 = 4
x = 4 is even so x/2 = 2
x = 2 is even so x/2 = 1
1 is odd so 3x+1 = 4
and clearly the cycle 4, 2, 1 now repeats forever!
.
I wrote a compter program to evaluate this ...
I haven't found a countradiction for whole numbers up through 1000, but that doesn't prove it for all cases ...
This is a well-known problem, known as the Collatz conjecture (http://en.wikipedia.org/wiki/Collatz_conjecture) and the answer is still unknown. It has been tested by computer up to many millions and no exception has yet been found. It refers to the natural numbers (1, 2, 3, ...etc.) only.
Example
x = 3 odd so 3x+1 = 10
x = 10 is even so x/2 = 5
x = 5 is odd so 3x+1 = 16
x = 16 is even so x/2 = 8
x = 8 is even so x/2 = 4
x = 4 is even so x/2 = 2
x = 2 is even so x/2 = 1
1 is odd so 3x+1 = 4
and clearly the cycle 4, 2, 1 now repeats forever!
.