The simplest undecided problem
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- Published on 2011. január 30. vasárnap, 09:33
If someone knows simpler, let me know!
Take the following function on the set of natural numbers: $$ f\colon\Bbb N\to\Bbb N\qquad k\mapsto f(k)=\cases{k/2 & \hbox{if $k$ is even},\\ 3k+1 & \hbox{if $k$ is odd}.} $$
Having this function and a natural number $a_0\in\Bbb N$ create the sequence $a_{j+1}=f(a_j)$ .
No matter what $a_0$ is, the members of this sequence can not be less than zero, therefore it has a smallest value among all its members..
What is the minimum value of such a sequence?.
In some cases it is very easy, to see that the mimimum is 1.
This is shown by the following chart by the starting value $a_0=$ . It can be tested by any other initial value by typing it in this box and pressing "ENTER".
By these experiences a conjecture was born
The minimum value of any such sequence is 1!.
If this was true, then it means that after a finite iteration the sequence goes into the infinite cycle $4,2,1,4,2,1,....$.
The problem above is known as Collatz-conjecture and is undecided by now.
Some have indicated that the rationality of $\pi^e$ is perhaps simpler problem, but, by my opinion, the understanding of $e$ and $\pi$ is more complicated than the above problem.