Please see the " Microsoft word" attachment.

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- Dec 26th 2006, 08:32 PMJenny20a squence question
Please see the " Microsoft word" attachment.

- Dec 27th 2006, 01:44 PMThePerfectHacker
There seems to be something wrong.

If $\displaystyle n=5$

Then,

$\displaystyle 2=2$

$\displaystyle 2^2=4$

$\displaystyle 2^4=16=1$

$\displaystyle 2^1=2$

This is non-constant. - Dec 27th 2006, 11:45 PMCaptainBlack
$\displaystyle 2^{2^4}=2^{16}=65536\equiv 1\ \mod\ 5$

Also

$\displaystyle

2^{4k}=(2^4)^k \equiv 1 \ \mod \ 5

$

and since the $\displaystyle \log_2$ of all the numbers in the sequence after $\displaystyle 2^{2^2}$ are divisible by $\displaystyle 4$ the result follows for $\displaystyle \mod\ 5$.

By a similar argument we can show that if any element in the sequence is ever congurent to 0 or 1 so

are all subsequent terms.

RonL