[seqfan] Re: Balanced binary system

[Vladimir Shevelev]

Thank you, Kevin, for this information!
You see that for me this system, of course, was
not the end in itself. By the balanced system I map
runs of 1's of n in binary to separate digits.
Therefore, I thought that A268411 should possess 
the properties closed to Thue-Morse sequence.
For me it was very interesting that such properties
indeed exist!

Regards,
Vladimir
________________________________________
From: Kevin Ryde [user42_kevin at yahoo.com.au]
Sent: 07 February 2016 10:17
To: seqfan at list.seqfan.eu
Cc: ולדימיר שבלייב
Subject: Re: Balanced binary system

shevelev at bgu.ac.il (Vladimir Shevelev) writes:

>> Let us call the "balanced binary" system, if in the
>> binary representation every 2^j is replaced by 2^(j+1)-2^j.

>Davis and Knuth in

>  "Number Representations and Dragon Curves I and II", Journal of
> Recreational Mathematics, volume 3, numbers 2 and 3, April and July
>  1970.  Reprinted in Knuth "Selected Papers on Fun and Games", 2010.

>call this a "folded" representation of n.  It arises in their
>calculations for the paperfolding Heighway/Harter dragon curve.

>For them "folded" means anything with non-zero bits alternate +1 and -1.
>That allows negative n (just by flipping all signs), and they show
>easily every non-zero n has exactly two representations, one ending +1
>and one ending -1 (when you choose the low, working upwards all the rest
>are determined).  The two differ only in tweaking the low end.

>> I submitted (0,1) sequence A268411 listing the parity
>> of number of 1's in the balanced reprsentation of n

>A geometric interpretation of that in the dragon curve could be on
>segment direction of A246960,

>   a(n) == 0 if A246960(n) == 0 or 3      segment "east or south"
>   a(n) == 1 if A246960(n) == 1 or 2   so segment "north or west"

>as long as I haven't made some horrible mistake.

>> Besides we have a recursion:
>> a(0)=0, a(2*n)=a(n); for odd n, a(2*n+1)=a(n);
>> for even n, a(2*n+1)=1-a(n).

>I suppose that might also go something by a(4*n+1), letting 1 mod 4
>notice the top of each run of 1-bits.
 
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