A174205 Write natural numbers in base 2 as a stream of digits. Moving left to right, delete odd occurrences of digit 0 and 1.

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, 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, 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

Offset: 1

Author

Paolo P. Lava and Giorgio Balzarotti, Mar 15 2010

Keywords

• easy
• nonn
• base

Comments

The natural numbers in base 2 are 0, 1, 10, 11, 100, 101, 110, 111, 1000..... and define a stream of digits if concatenated:

0110111001011101111000 Delete odd-indexed occurrences of 0 (replaced by .):

.110111.01.11101111.0. Also delete odd-indexed occurrences of 1 (replaced by .):

..10.1..01..1.01.1..0.

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.

Crossrefs

Cf. A007088, A174203 - A174210.

Programs

Sage

def A174205(N=100):
    a = [0] + flatten([n.digits(base=2)[::-1] for n in IntegerRange(1,N)])
    for bit in 0, 1:
        a = [d for i,d in enumerate(a) if not (d == bit and a[:i+1].count(bit) % 2 == 1)]
    return a # D. S. McNeil, Dec 08 2010