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

Author

Dylan Hamilton, Oct 28 2010

Keywords

• base
• easy
• nonn

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

Examples

			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
		

Links

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

Programs

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}]

Extensions

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