Search: id:A104325 Results 1-1 of 1 results found. %I A104325 %S A104325 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, %T A104325 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, %U A104325 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 A104325 Number of runs of equal bits in the Dual Zeckendorf (binary) representation of n. %C A104325 Sequence has some interesting fractal properties (plot it!) %H A104325 Ron Knott using Fibonacci Numbers to represent whole numbers %e A104325 The Dual Zeckendorf representation of 13 is 10110(fib) corresponding to {8, 3, 2} %e A104325 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 A104325 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:=i-2 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[i-1] then c:=c+1 fi od; c end proc: seq(countruns(dualzeckrep(n)),n=1..100); %Y A104325 Cf. A014417, A104324. %Y A104325 Sequence in context: A133334 A003603 A135227 this_sequence A133084 A118851 A112383 %Y A104325 Adjacent sequences: A104322 A104323 A104324 this_sequence A104326 A104327 A104328 %K A104325 nonn %O A104325 1,2 %A A104325 Ron Knott (enquiry(AT)ronknott.com), Mar 01 2005 Search completed in 0.001 seconds