%I #9 Sep 29 2018 18:45:29
%S 2,1,2,2,4,3,2,4,3,4,2,4,3,4,4,6,5,2,4,3,4,4,6,5,4,6,5,6,2,4,3,4,4,6,
%T 5,4,6,5,6,4,6,5,6,6,8,7,2,4,3,4,4,6,5,4,6,5,6,4,6,5,6,6,8,7,4,6,5,6,
%U 6,8,7,6,8,7,8,2,4,3,4,4,6,5,4,6,5,6,4,6,5,6,6,8,7,4,6,5,6,6,8,7,6,8,7,8,4
%N Number of runs (of equal bits) in the minimal Lucas binary (A130310) 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 Casey Mongoven, <a href="/A133770/b133770.txt">Table of n, a(n) for n = 1..199</a>
%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 A130310(17)=101001 because 11 + 4 + 2 = 17 (a sum of Lucas numbers); this representation has five runs: 1,0,1,00,1. So a(17)=5.
%Y Cf. A133771, A130310.
%K nonn
%O 1,1
%A _Casey Mongoven_, Sep 23 2007