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%I #13 Sep 29 2018 18:45:37
%S 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,
%T 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,
%U 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
%N Number of runs (of equal bits) in the maximal Lucas binary (A130311) representation of n.
%D 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.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phigits.html">Using Powers of Phi to represent Integers</a>.
%e 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.
%Y Cf. A133770, A130311.
%K nonn
%O 1,1
%A _Casey Mongoven_, Sep 23 2007; corrected Mar 23 2008
%E 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
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