[seqfan] Balanced binary system

Vladimir Shevelev shevelev at bgu.ac.il
Fri Feb 5 11:28:18 CET 2016


Dear SeqFans,

Let us call the "balanced binary" system, if in the
binary representation every 2^j is replaced by 2^(j+1)-2^j. 
Then there appear an additional digit (-1). 
For example, 3=2+1 = (4-2)+(2-1) = 4-1 ={1,0,-1}_b, 
where 1,0,-1 are considered as the digits in balanced number
system.  Also 7= 4+2+1 =(8-4)+(4-2)+(2-1) = 8-1 =
(1,0,0,-1)_b.
Thus every block of 1's of length >=1 in binary 
representation of n becomes to a pair of digits (1,-1)
(and 0's).
The properties of the "balanced binary" system: 
a) the first digit is 1; 
b) the digital sum is always 0; 
c) deleting 0's, we obtain alternative sequence of 
1,-1 for every n; 
d) representation of every n>=0 is unique.
e) number of 1's (or the same number of (-1)'s)
equals the number of blocks of 1's in binary.
In binary balanced system we have the representations: 
1={1,-1}, 2={1,-1,0}, 3={1,0,-1}, 4={1,-1,0,0}, 
5={1,-1,1,-1}, 6={1,0,-1,0}, 7={1,0,0,-1}, 
8={1,-1,0,0,0},9={1,-1,0,1,-1}, 10={1,-1,1,-1,0},... 
I submitted (0,1) sequence A268411 listing the parity
of number of 1's in the balanced reprsentation of n
or the parity of the number of blocks of 1's in binary.
It is modulo 2 A069010. I proved and did as title
the following interesting property of this sequence
(similar to Thue-Morse sequence A010060-cf.):
Let A_k denote the first 2^k terms; then A_1 = {0,1}
and for k >= 1, A_{k+1} = A_k B_k, where B_k is 
obtained from A_k by interchanging the first 2^(k-1) 
0's and 1's of A_k and upheld others.
To this definition the following formula corresponds 
a(n+2^k)=1-a(n) for 0<=n<=2^(k-1)-1; 
a(n+2^k)=a(n) for 2^(k-1)<=n<=2^k-1. 
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).
These formulas are not much complicater
than the corresponding formulas for
Thue-Morse sequence.
I wonder is all of that known? 

Best regards,
Vladimir




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