A103542 Binary equivalents of A102370.

0, 11, 110, 101, 100, 1111, 1010, 1001, 1000, 1011, 1110, 1101, 11100, 10111, 10010, 10001, 10000, 10011, 10110, 10101, 10100, 11111, 11010, 11001, 11000, 11011, 11110, 111101, 101100, 100111, 100010, 100001, 100000, 100011, 100110, 100101

Offset: 0

Author

N. J. A. Sloane, Mar 23 2005

Keywords

• nonn
• easy
• base

Comments

The number of 1's in the n-th term appears to match A089400. - Benoit Cloitre, Mar 24 2005

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..2000
• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.
• D. Applegate et al., Sloping Binary Number, A New Sequence Related to the Binary Numbers, arXiv:math/0505295 [math.NT], 2005.

Programs

Mathematica

f[n_] := Block[{k = 1, s = 0, l = Max[2, Floor[Log[2, n + 1] + 2]]}, While[k < l, If[ Mod[n + k, 2^k] == 0, s = s + 2^k]; k++ ]; s]; Table[ FromDigits[ IntegerDigits[f[n] + n, 2]], {n, 0, 35}] (* Robert G. Wilson v, Mar 23 2005 *)

Python

def a(n): return '0' if n<1 else bin(sum([(n + k)&(2**k) for k in range(len(bin(n)[2:]) + 1)]))[2:] # Indranil Ghosh, May 03 2017

Extensions

More terms from Robert G. Wilson v and Benoit Cloitre, Mar 23 2005