Siobhan has sent us her work on this problem. Well done, Siobhan!

The problem says that

t_{8}^{2}

has two cycles, whereas

t_{8}^{3}

has one cycle and

t_{8}^{4}

has four cycles. I noticed that in each of these examples the number of cycles in

t_{8}^{k}

is the highest common factor of

8

and

k

. I decided to test this on shuffles of lengths

9

and

10

Powers of the cycle (1 2 3 4 5 6 7 8 9 10)

Here are the tables I got:

\begin{matrix} t_{9}^{0} & = & (1) \\ t_{9}^{1} & = & (123456789) \\ t_{9}^{2} & = & (135792468) \\ t_{9}^{3} & = & (147) (258) (369) \\ t_{9}^{4} & = & (159483726) \\ t_{9}^{5} & = & (162738495) \\ t_{9}^{6} & = & (174) (285) (396) \\ t_{9}^{7} & = & (186429753) \\ t_{9}^{8} & = & (198765432) \end{matrix}

\begin{matrix} t_{10}^{0} & = & (1) \\ t_{10}^{1} & = & (12345678910) \\ t_{10}^{2} & = & (13579) (246810) \\ t_{10}^{3} & = & (14710369258) \\ t_{10}^{4} & = & (15937) (261048) \\ t_{10}^{5} & = & (16) (27) (38) (49) (510) \\ t_{10}^{6} & = & (17395) (284106) \\ t_{10}^{7} & = & (18529631074) \\ t_{10}^{8} & = & (19753) (210864) \\ t_{10}^{9} & = & (11098765432) \end{matrix}

This agrees with my prediction: the number of cycles in

t_{n}^{k}

is the highest common factor of

k

and

n

. (Of course,

(1)

really means

(1) (2) (3) \dots (n)

so has

n

cycles.)

In $t_{n}^{1}$ , consecutive numbers are $1$ apart ( $1$ , $2$ , $3$ , ..., $n$ ). When we do this twice, they are two apart, because, for example, $1$ goes to $2$ which goes to $3$ so $1$ goes to $3$ . When we work out $t_{n}^{3}$ , the numbers that were $2$ apart become $3$ apart, because, for example, $1$ goes to $3$ which goes to $4$ so $1$ goes to $4$ . It's quite easy to see that this will keep happening, because each time we apply $t_{n}$ we move the numbers $1$ further apart. So in $t_{n}^{k}$ the numbers are $k$ apart. But now we can see that the number of cycles we get will be the highest common factor of $k$ and $n$ ; this is like the Stars problem mentioned in this question.