A103185 a(n) = Sum_{ k >= 0 such that n + k == 0 mod 2^k } 2^(k-1).

0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 2, 1, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 2, 17, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 2, 1, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 34, 17, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 2, 1, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1, 0, 1, 2, 17, 8, 5, 2, 1, 0, 1, 2, 1, 0, 5, 2, 1

Offset

0,3

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 [pdf, ps].

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.

Formula

a(n) = (A102370(n) - n)/2.

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 - 1)]; k++ ]; s]; Table[ f[n], {n, 0, 103}] (* Robert G. Wilson v, Mar 18 2005 *)

Prog

(PARI) A103185(n)=(sum(k=0, ceil(log(n+1)/log(2)), if((n+k)%2^k, 0, 2^k))-1)/2 \\ Benoit Cloitre, Mar 20 2005
(Python)
def a102370(n): return 0 if n<1 else sum([(n + k)&(2**k) for k in range(len(bin(n)[2:]) + 1)])
def a(n): return (a102370(n) - n)/2 # Indranil Ghosh, May 03 2017

Crossrefs

Cf. A102370(n).

Sequence in context: A243821 A143445 A133727 * A130513 A114596 A083417

Adjacent sequences: A103182 A103183 A103184 * A103186 A103187 A103188

Keyword

nonn,easy,base

Author

N. J. A. Sloane, Mar 18 2005

Extensions

More terms from Robert G. Wilson v, Mar 18 2005

Status

approved