%I A105774
%S A105774 1,1,2,4,4,7,7,6,12,12,11,9,9,20,20,19,17,17,14,14,15,33,33,32,30,30,27,
%T A105774 27,28,22,22,23,25,25,54,54,53,51,51,48,48,49,43,43,44,46,46,35,35,36,
%U A105774 38,38,41,41,40,88,88,87,85,85,82,82,83,77,77,78,80,80,69,69,70,72,72
%N A105774 A "fractal" transform of the Fibonacci's numbers : a(1)=1 then if F(n)<k<=F(n+1) 
               a(k)=F(n+1)-a(k-F(n)) where F(n)=A000045(n).
%C A105774 Let tau=(1+sqrt(5))/2 then the missing numbers 3,5,8,10,13,16,18,21,... 
               are given by (round(tau^2*k))_{k>0} (A004937 ). Indices n such that 
               a(n)=a(n+1) are given by (floor(tau^2*k)-1)_{k>0} (A003622). n such 
               that a(n) differs from a(n+1) are given by (floor(tau*k+1/tau))_{k>
               0} (A022342). Indices n giving isolated terms (a(n) differs from 
               a(n-1) and a(n+1)) are given by (floor(tau*floor(tau^2*k))_{k>0} 
               (A003623). Remove 0's form the first differences of sorted values 
               then you get a version of the infinite Fibonacci's word (A001468). 
               I.e. sorted values are 1,1,2,4,4,6,7,7,9,9,11,12,12,..., first differences 
               are 0,1,2,0,2,1,0,2,0,2,1,0,2,0,1,..., remove 0's gives 1,2,2,1,2,
               2,1,2,1,2,2,1,2,1,2,2,1,2,...#{ k : a(k)=k}=infty
%F A105774 a(A000045(n))=A006498(n); limsup a(n)/n=tau and liminf a(n)/n=1/tau where 
               tau=(1+sqrt(5))/2
%F A105774 a(n) mod 2 = A085002(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), 
               May 10 2005
%e A105774 For 1=F(2)<k<=F(3)=2 the rule gives a(2)=2-a(1)=1 ... if 5=F(5)<k<=F(6)=8 
               the rule forces a(6)=8-a(6-5)=8-a(1)=7; a(7)=8-a(2)=7; a(8)=8-a(3)=6
%Y A105774 Cf. A105669, A105670, A105672, A093347, A093348.
%Y A105774 Sequence in context: A082515 A062855 A103622 this_sequence A130805 A023831 
               A023844
%Y A105774 Adjacent sequences: A105771 A105772 A105773 this_sequence A105775 A105776 
               A105777
%K A105774 nonn
%O A105774 1,3
%A A105774 Benoit Cloitre (benoit7848c(AT)orange.fr), May 04 2005