3
$\begingroup$

Prove or disprove:

Let $a_0$ be any positive integer,defining:

$$a_{n+1} = \begin{cases}\frac{a_n}{2} &, a_n \text{ even}\\ 3a_n - 1 &, a_n \text{ odd}. \end{cases}$$

Then $a_k=1,7$ or $17$(all of them forms loops) where $k$ is some integer....

$\endgroup$
1
$\begingroup$

This is the same thing as the Collatz sequences except you start at negative integers.
It is not known if those three cycles are the only cycles nor if every sequence ends up in a cycle instead of diverging to $- \infty$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The Collatz problem terminates at $1$, how is this the same thing? $\endgroup$ – Hashir Omer May 13 '14 at 9:57
  • 1
    $\begingroup$ @HashirOmer The Collatz problems starting at any negative integer (like $-1,-7$ or $-17$) don't terminate at $1$. $\endgroup$ – mercio May 13 '14 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.