[seqfan] Re: Balanced binary system

[Kevin Ryde]

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.