%I #18 Mar 23 2020 17:31:34
%S 1,0,1,0,1,1,0,1,1,0,1,0,1,0,1,0,1,1,0,1,1,1,0,1,1,0,0,1,0,1,0,0,1,0,
%T 1,1,0,0,1,1,0,1,0,1,1,1,0,1,0,1,0,1,1,0,1,1,0,1,1,1,1,0,1,1,1,0,0,1,
%U 0,0,1,0,0,1,0,0,1,0,1,0,0,1,1,0,1,0,0,1,1,1,0,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,1,1,0,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,1,0,1
%N Write natural numbers in base 2 as a stream of digits. Moving left to right, delete odd occurrences of digit 0 and 1.
%C The natural numbers in base 2 are 0, 1, 10, 11, 100, 101, 110, 111, 1000..... and define a stream of digits if concatenated:
%C 0110111001011101111000 Delete odd-indexed occurrences of 0 (replaced by .):
%C .110111.01.11101111.0. Also delete odd-indexed occurrences of 1 (replaced by .):
%C ..10.1..01..1.01.1..0.
%C The stream of bits that remain after these two rounds of deletion is chopped into single bits which define the entries of the current sequence.
%o (Sage)
%o def A174205(N=100):
%o a = [0] + flatten([n.digits(base=2)[::-1] for n in IntegerRange(1,N)])
%o for bit in 0, 1:
%o a = [d for i,d in enumerate(a) if not (d == bit and a[:i+1].count(bit) % 2 == 1)]
%o return a # _D. S. McNeil_, Dec 08 2010
%Y Cf. A007088, A174203 - A174210.
%K easy,nonn,base
%O 1,1
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Mar 15 2010