A133771 Number of runs (of equal bits) in the maximal Lucas binary (A130311) representation of n.

2, 1, 1, 2, 3, 1, 3, 2, 3, 1, 4, 5, 3, 3, 2, 3, 1, 5, 4, 5, 3, 4, 5, 3, 3, 2, 3, 1, 6, 7, 5, 5, 4, 5, 3, 5, 4, 5, 3, 4, 5, 3, 3, 2, 3, 1, 7, 6, 7, 5, 6, 7, 5, 5, 4, 5, 3, 6, 7, 5, 5, 4, 5, 3, 5, 4, 5, 3, 4, 5, 3, 3, 2, 3, 1, 8, 9, 7, 7, 6, 7, 5, 7, 6, 7, 5, 6, 7, 5, 5, 4, 5, 3, 7, 6, 7, 5, 6, 7, 5, 5, 4, 5, 3, 6

Offset: 1

Author

Casey Mongoven, Sep 23 2007; corrected Mar 23 2008

Keywords

• nonn

Examples

			A130311(19)=101110 because 11+4+3+1=19 (a sum of Lucas numbers); this representation has four runs: 1,0,111,0. So a(19)=4.
		

References

• Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Links

• Ron Knott, Using Powers of Phi to represent Integers.

Crossrefs

Cf. A133770, A130311.

Extensions

The b-file submitted by Casey Mongoven did not match the terms of the sequence, so I have deleted it. Of course it may be that the sequence is wrong and the b-file was correct. Should be rechecked. - N. J. A. Sloane, Nov 10 2010