

A104325


Number of runs of equal bits in the dual Zeckendorf representation of n (A104326).


3



1, 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, 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, 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
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OFFSET

0,3


COMMENTS

Sequence has some interesting fractal properties (plot it!)


LINKS



EXAMPLE

The dual Zeckendorf representation of 13 is 10110(fib) corresponding to {8, 3, 2}.
The largest set of Fibonacci numbers whose sum is n (cf. the Zeckendorf representation 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.


MAPLE

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);


MATHEMATICA

Length @ Split[IntegerDigits[#, 2]] & /@ Select[Range[0, 1000], SequenceCount[ IntegerDigits[#, 2], {0, 0}] == 0 &] (* Amiram Eldar, Jan 18 2020 *)


CROSSREFS



KEYWORD



AUTHOR



EXTENSIONS

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jan 18 2020


STATUS

approved



