The sequence proceeds as follows:

$t_{1} = 13$
$t_{2} = 40$
$t_{3} = 20$
$t_{4} = 10$
$t_{5} = 5$
$t_{6} = 16$
$t_{7} = 8$
$t_{8} = 4$
$t_{9} = 2$
$t_{10} = 1$
$t_{11} = 4$
$t_{12} = 2$
$t_{13} = 1$

The block $2, 1, 4$ repeats ad infinitum after $t_{8}$.

When $n$ is a multiple of $3$,  $t_{n} = 2$.

Since $2007$ is a multiple of $3$,  $t_{2008} = 1$

This problem is taken from the UKMT Mathematical Challenges.