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Old 2018-03-17, 09:05   #29
R. Gerbicz
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"Robert Gerbicz"
Oct 2005

1,531 Posts

Originally Posted by Auto Felix View Post
How do you know that all sequences start with 1, and end with 8 (for even) or 3 (for odd)? Is there a proof?
Thanks for your interest!
I"ve constructed all basic sequences that holds this, and then by induction we still maintain this. Note that all integers in [1,24] is free, because T(c,0) and T(c,1) doesn't use them. So we can use these small (1,3,8) integers at the endpoints.

You could ask why we haven't used a shifted and reversed representation, so the sequences starts with 1,3 and 1,8; because in that case we don't know the last term of seq0 and seq1, so we can't glue the sequences. Or why we haven't used the constant 3 at the end for every sequence, because in that case 3=seq0(n)=seq1(n+1), but (n+1-n)=1 is odd so the parity position condition wouldn't be true. There are some traps here.

Last fiddled with by R. Gerbicz on 2018-03-17 at 09:09
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