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## 'Collatz 13' printed from http://nrich.maths.org/

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.*

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