A175945 Start with the binary expansion of n; list the lengths of the runs of ones and interpret this list as lengths of runs of alternating ones and zeros in binary.

1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 4, 3, 6, 7, 15, 1, 2, 2, 4, 2, 5, 4, 8, 3, 6, 6, 12, 7, 14, 15, 31, 1, 2, 2, 4, 2, 5, 4, 8, 2, 5, 5, 11, 4, 9, 8, 16, 3, 6, 6, 12, 6, 13, 12, 24, 7, 14, 14, 28, 15, 30, 31, 63, 1, 2, 2, 4, 2, 5, 4, 8, 2, 5, 5, 11, 4, 9, 8, 16, 2, 5, 5, 11, 5, 10, 11, 23, 4, 9, 9, 19, 8, 17

Offset

1,3

Comments

Trailing binary zeros don't affect the run lengths of 1's, so a(2n)=a(n). Trailing isolated binary ones add a run of a single 1, so a(4n+1) = 2*a(n) if a(n) is odd, a(4n+1) = 2*a(n)+1 if a(n) is even. - R. J. Mathar, Dec 07 2010

Links

Paul Tek, Table of n, a(n) for n = 1..10000

Example

n=43 in binary is 101011, where the 1 has run-lengths of 1, 1 and 2. This is interpreted as a run of 1 one, run of 1 zero and run of 2 one's, 1011, which back to decimal is a(43)=11. - R. J. Mathar, Dec 07 2010

Mathematica

takelist[l_, t_] := Module[{lent, term}, Set[lent, Length[t]]; Table[l[[t[[y]]]], {y, 1, lent}]]
frombinrep[x_] := FromDigits[Flatten[Table[Table[If[OddQ[n], 1, 0], {d, 1, x[[n]]}], {n, 1, Length[x]}]], 2]
binrep[x_] := repcount[IntegerDigits[x, 2]]
onebinrep[x_]:=Module[{b}, b=binrep[x]; takelist[b, Range[1, Length[b], 2]]]
zerobinrep[x_]:=Module[{b}, b=binrep[x]; takelist[b, Range[2, Length[b], 2]]]
Table[frombinrep[onebinrep[n]], {n, START, END}]

Crossrefs

Sequence in context: A087227 A060477 A345037 * A209859 A354092 A262218

Adjacent sequences: A175942 A175943 A175944 * A175946 A175947 A175948

Keyword

base,easy,nonn

Author

Dylan Hamilton, Oct 28 2010

Extensions

Definition rewritten by N. J. A. Sloane, Dec 08 2010.

Status

approved