Week 10

Hi all,

Before I prove something I want to highlight the space race at Dropbox for U of T https://www.dropbox.com/spacerace?r=NTQ4NDE0MjI3OQ. Go to this link to get a lot of space on Dropbox. As of this post everyone gets 15GB for two years.

Anyhow back to proving. Continuing on my problem. I am going to prove that once a power of 2 appears in the sequence we have a 1 in the sequence.

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Let P(n) be if the Collatz sequence starting at n has a x_m = 2^k, for some natural numbers m and k, then it contains a 1.

Assume x_m = 2^k.

x_m = 2*2^(k-1) so it is even. Following the sequence we divide by 2.
x_(m+1) = 2^(k-1) = 2*2^(k-2), we again divide by 2.

This procedure continues k times until x_(m+k) = 2^(k-k) = 2^0 = 1

Hence P(n).

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So now I know if I can show that the sequence at some point contains a power of 2 then I am done. I know at that point the sequence contains a 1.

More next time

Wendy