%I
%S 1,2,1,3,2,1,4,3,3,2,1,5,4,3,4,3,3,2,1,6,5,5,4,3,5,4,3,4,3,3,2,1,7,6,
%T 5,6,5,5,4,3,6,5,5,4,3,5,4,3,4,3,3,2,1,8,7,7,6,5,7,6,5,6,5,5,4,3,7,6,
%U 5,6,5,5,4,3,6,5,5,4,3,5,4,3,4,3,3,2,1,9,8,7,8,7,7,6,5,8,7,7,6,5
%N Number of runs of equal bits in the Dual Zeckendorf (binary) representation of n.
%C Sequence has some interesting fractal properties (plot it!)
%H Ron Knott <a href="http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Fibonacci/fibrep.html">using Fibonacci Numbers to represent whole numbers</a>
%H Casey Mongoven, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_41_from175to192.pdf">Sonification of multiple Fibonaccirelated sequences</a>, Annales Mathematicae et Informaticae, 41 (2013) pp. 175192.
%e The Dual Zeckendorf representation of 13 is 10110(fib) corresponding to {8, 3, 2}
%e The largest set of Fibonacci numbers whose sum is n (cf. the Zeckendorf rep is the smallest set). This is composed of runs of one 1, one 0, two 1's, one 0 i.e. 4 runs in all so a(13)=4
%p dualzeckrep:=proc(n)local i,z;z:=zeckrep(n);i:=1; while i<=nops(z)2 do if z[i]=1 and z[i+1]=0 and z[i+2]=0 then z[i]:=0; z[i+1]:=1;z[i+2]:=1; if i>3 then i:=i2 fi else i:=i+1 fi od; if z[1]=0 then z:=subsop(1=NULL,z) fi; z end proc: countruns:=proc(s)local i,c,elt;elt:=s[1];c:=1; for i from 2 to nops(s) do if s[i]<>s[i1] then c:=c+1 fi od; c end proc: seq(countruns(dualzeckrep(n)),n=1..100);
%Y Cf. A014417, A104324.
%K nonn,hear
%O 1,2
%A _Ron Knott_, Mar 01 2005
