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%I #9 Sep 29 2018 18:45:57
%S 1,1,3,5,3,3,7,5,5,5,9,5,5,7,5,5,5,11,9,9,7,5,7,7,9,7,7,7,13,7,7,9,7,
%T 7,7,11,9,9,7,5,7,7,9,7,7,7,15,13,13,11,9,11,11,11,9,9,7,5,9,9,11,9,9,
%U 9,13,11,11,9,7,9,9,11,9,9,9,17,9,9,11,9,9,9,13,11,11,9,7,9,9,11,9,9,9,15,13
%N Number of runs (of equal bits) in the maximal "phinary" (A130601) 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="/A133773/b133773.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>.
%H Casey Mongoven, <a href="http://caseymongoven.com/catalogue/b522.htm">Music based on this sequence</a>.
%e A130601(3)=1101 because phi^1 + phi^0 + phi^-2 = 3; 1101 has 3 runs: 11,0,1. So a(3)=3.
%Y Cf. A133772, A130601.
%K nonn
%O 1,3
%A _Casey Mongoven_, Sep 23 2007
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