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Search: id:A105669
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| A105669 |
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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). |
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6
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| 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, 27, 28, 28, 22, 23, 23, 25, 54, 54, 53, 53, 51, 48, 48, 49, 49, 43, 44, 44, 46, 35, 35, 36, 36, 38, 41, 41, 40, 40, 88, 87, 87, 85, 82, 82, 83, 83, 77, 78, 78, 80, 69, 69, 70, 70, 72
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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
Missing numbers are the nearest integer to tau^2*n, n>=0 (cf. A004937)
#{k>0:a(k)=k}=infinity
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 )
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FORMULA
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n>0 a(F(2n))=F(2n+1)-F(n+1)^2+F(n)F(n-1)
n>1 a(F(2n-1))=F(2n)-1
1/tau < a(n)/n < tau.
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EXAMPLE
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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
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PROGRAM
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(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)))))
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CROSSREFS
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Cf. A105670, A105672, A093347, A093348.
Adjacent sequences: A105666 A105667 A105668 this_sequence A105670 A105671 A105672
Sequence in context: A115754 A162251 A065968 this_sequence A019657 A134791 A095760
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), May 03 2005
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