Define a function $\displaystyle {f}:{\mathbb{{N}}}\rightarrow{\mathbb{{N}}}$ as follows.

$\displaystyle {f{{\left({n}\right)}}}={\left\lbrace\begin{array}{c} \frac{{n}}{{2}}\ \text{ if }\ {n}\ \text{ is even, and }\ \\{3}{n}+{1}\ \text{ if }\ {n}\ \text{ is odd.}\end{array}\right.}$ 

 We can use this function to define a recursive sequence with initial condition $\displaystyle {c}_{{1}}\in{\mathbb{{N}}}$ , and  $\displaystyle {c}_{{{k}+{1}}}={f{{\left({c}_{{k}}\right)}}}$.

For example, if $\displaystyle {c}_{{1}}={5}$, then $\displaystyle {\left({c}_{{k}}\right)}={\left({5},{16},{8},{4},{1},{1},{4},{2},{1},\ldots\right)}$.

Now you try.  Fill in the table below.

$\displaystyle {k}$	$\displaystyle {c}_{{k}}$
1	12
2
3
4
5
6
7
8
9
10

If you were to try lots of different initial conditions, you will probably guess the following.

Conjecture: For any choice of initial $\displaystyle {c}_{{1}}\in{\mathbb{{N}}}$, the sequence $\displaystyle {\left({c}_{{k}}\right)}$  will eventually output $\displaystyle {1}$.

This is called the Collatz conjecture.  Nobody knows if it is true or false, but the statement has been checked for all starting values up to 87$\displaystyle \times$2⁶⁰.

For more information, click here.