Can anyone prove or disprove this conjecture?

Question

Let f(x) be a function defined on the positive integers such that:

f(x) = x/2 if x is even
f(x) = (3*x+1)/2 if x is odd

Then the conjecture is: iterates of f(x) will eventually reach 1 for any initial value of x.

Answer 1

This is a well known conjecture (called the "3n+1 conjecture"), but nobody has been able to PROVE it, although a computer has shown it is true for insanely high numbers.

Although it is almost 100% certainly true, it has not been rigorously proven by anyone yet. There's a lot of deep math in this problem

More info: http://www-personal.ksu.edu/~kconrow/

NOTE: (Re mathematician): for varying definitions of "famous". Ask random people on the street who Andrew Wiles is. And 3n+1 is certainly smaller than Fermat.

Answer 2

The express chute!

Its a little like chutes and ladders. You go up ladders on odd squares and down chutes on even squares. Except there is an express chute that gets you straight home (back to1) every time at every power of 2 (2,4,8,16,32, and so on). It is therefore biased. If it wasn't for the express chute, it would not be true since the chances of landing on an even square up a ladder is 50/50, and the chances of landing on an even square down a chute is also 50/50. Also, every number is the destination of a chute, and every third number is the destination of a ladder.

Try adding the kicker (removing the express):
f(x) = 3*x/2 if x is a power of 2

See where that goes...

Answer 3

Infinitely lengthy ? not likely. All accurate numbers more suitable than 3 are of the kind (6n +/- a million). on your conjecture to be authentic, the first 3 want to have both 2 "+a million"s, and one "-a million"s, or any incorrect way round. even as the present sum remains accurate, say of the kind "6n + a million", the subsequent 2 primes can not both be of the kind "6n + a million". this can make the sum a distinct of three. regrettably, this can ensue infinitely many times. Likewise for the kind "6n - a million", there'll be 2 consecutive primes of the kind "6n - a million" you ought to favor to characteristic to it. The progression of the primes (and no matter if or not they're "6n+a million" or "6n-a million") is extremely abnormal. No ordinary kinds like the single you ought to favor ought to ensue for all consecutive primes from one aspect on. i imagine you'd be able to locate arbitrarily lengthy sequences of primes for which the progression holds, yet no longer infinitely.

Answer 4

try x=1 we have f(1)=2<>1

Answer 5

I'm going to try something and add to this later. I was thinking proof by induction might be the way to go.

Well, after reading the posts below mine, I guess I won't waste my time; thanks!

Answer 6

hmmm... I have the proof in my notes... but they are at school waiting for my professor to look at them before submission... and I don't want to release them early

Answer 7

This is a fairly well-known open problem. In other words, if you can answer it, you will be famous.