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## 'Zodiac' printed from http://nrich.maths.org/

In

Sheffuls , you will find an introduction to the cycle notation
which you'll need to use for this investigation. The

Shuffles interactivity will also help. We're going to
investigate turns within turns. Watch out for those Zodiacal
alignments!

Let's start by generating all the permutations that can be made by
repeating the cycle $(1 2 3 4 5 6 7 8)$ which we'll call $t_8$.

In the Shuffles interactivity you can do this by choosing the
length 8 in the Flip and Turn Factory. Pull down two copies of Turn
1 from the menu bar and compose them to obtain Turn 2, as
illustrated in the diagram below:

There is also a

short
tutorial that explains how to create and manipulate
shuffles.

Compose Turn 2 with another copy of Turn 1 and you will have
Turn 3. Keep going, and after tidying things up a bit you should
have all eight of them:

You can see how these shuffles relate to turns of the octagon by
pressing the bottom right play buttons.

In cycle notation you can make Turn 2 like this: $$t_8^2 = (1 2 3 4
5 6 7 8)(1 2 3 4 5 6 7 8) = (1 3 5 7)(2 4 6 8).$$ Think '1 goes to
2, and then 2 goes to 3; so 1 goes to 3'.

Then Turn 3 = $t_8^3 = (1 3 5 7)(2 4 6 8)(1 2 3 4 5 6 7 8)$ and so
on.

You should end up with a similar table like this:

$$\begin{eqnarray} t^0 &=& (1)\\ t_8^1 &=& (1 2 3 4
5 6 7 8)\\ t_8^2 &=& (1 3 5 7)(2 4 6 8)\\ t_8^3 &=&
(1 4 7 2 5 8 3 6)\\ t_8^4 &=& (1 5)(2 6)(3 7)(4 8)\\ t_8^5
&=& (1 6 3 8 5 2 7 4)\\ t_8^6 &=& (1 7 5 3)(2 8 6
4)\\ t_8^7 &=& (1 8 7 6 5 4 3 2) \end{eqnarray}$$

Notice that $t_8^2$ contains two cycles, but $t_8^3$ contains just
one, and $t_8^4$ contains 4. Why might this be?

Investigate what happens with length 9 and length 10 cycles. As a
check, you should find that $t_9^3$ contains 3 cycles, but $t_9^2$
contains just one.

Can you find a general rule? How many cycles will $t_{23}^2$
contain? How many cycles will $t_{30}^{10}$ contain?

Now look at

Stars . Notice anything?