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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). 5
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; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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 )

LINKS

Table of n, a(n) for n=1..72.

FORMULA

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.

EXAMPLE

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

PROG

(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)))))

CROSSREFS

Cf. A105670, A105672, A093347, A093348.

Sequence in context: A162251 A244011 A065968 * A256963 A019657 A134791

Adjacent sequences:  A105666 A105667 A105668 * A105670 A105671 A105672

KEYWORD

nonn

AUTHOR

Benoit Cloitre, May 03 2005

STATUS

approved

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Last modified September 23 15:54 EDT 2017. Contains 292361 sequences.