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

1,1

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.

Links

Table of n, a(n) for n=1..133.

Prog

(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

Crossrefs

Cf. A007088, A174203 - A174210.

Sequence in context: A287523 A288932 A155482 * A334413 A114483 A127822

Adjacent sequences: A174202 A174203 A174204 * A174206 A174207 A174208

Keyword

easy,nonn,base

Author

Paolo P. Lava and Giorgio Balzarotti, Mar 15 2010

Status

approved