Search: id:A105669
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%I A105669
%S A105669 1,2,2,4,7,7,6,6,12,11,11,9,20,20,19,19,17,14,14,15,15,33,32,32,30,27,
%T A105669 27,28,28,22,23,23,25,54,54,53,53,51,48,48,49,49,43,44,44,46,35,35,36,
%U A105669 36,38,41,41,40,40,88,87,87,85,82,82,83,83,77,78,78,80,69,69,70,70,72
%N A105669 A "fractal" transform of the Fibonacci numbers F(n)=A000045(n): a(1)=1, 
               then for n>1 if F(n)<k<F(n+1) we have a(k)=F(n+1)-a(k-F(n)) and when 
               k=F(n+1) we force a(F(n+1))=F(n+1)+(1+(-1)^n)*(F(n).
%C A105669 Let b denote the sequence of n such that a(n)=a(n+1), then b(n)=floor(tau^2*n) 
               where tau=(1+sqrt(5))/2
%C A105669 Missing numbers are the nearest integer to tau^2*n, n>=0 (cf. A004937)
%C A105669 #{k>0:a(k)=k}=infinity
%C A105669 This kind of "fractal" transform can be applied to any increasing monotonic 
               sequence giving true fractal properties for sequences = (m^n)_{n>
               0} with m integer >=2, specially when m is odd (cf. A093347, A093348 
               )
%F A105669 n>0 a(F(2n))=F(2n+1)-F(n+1)^2+F(n)F(n-1)
%F A105669 n>1 a(F(2n-1))=F(2n)-1
%F A105669 1/tau < a(n)/n < tau.
%e A105669 for 5=F(5)<k<=F(6)=8 we get a(6)=8-a(6-5)=8-a(1)=7; a(7)=8-a(7-5)=8-a(2)=6; 
               a(8)=8-a(8-5)=8-a(3)=6
%o A105669 (PARI) f=(1+sqrt(5))/2; a(n)=if(n<2,1,fibonacci(floor(log(sqrt(5)*n)/
               log(f))+1)-a(n-fibonacci(floor(log(sqrt(5)*n)/log(f)))))
%Y A105669 Cf. A105670, A105672, A093347, A093348.
%Y A105669 Sequence in context: A115754 A162251 A065968 this_sequence A019657 A134791 
               A095760
%Y A105669 Adjacent sequences: A105666 A105667 A105668 this_sequence A105670 A105671 
               A105672
%K A105669 nonn
%O A105669 1,2
%A A105669 Benoit Cloitre (benoit7848c(AT)orange.fr), May 03 2005

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