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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).
6
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
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
FORMULA
F(2n) = F(2n+1) - F(n+1)^2 + F(n)*F(n-1) for n>0.
a(F(2n-1)) = F(2n)-1 for n>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.
MATHEMATICA
a[n_] := a[n] = If[n <= 1, 1, Fibonacci[(k = Floor[Log[Sqrt[5]*n]/Log[GoldenRatio]]) + 1] - a[n - Fibonacci[k]]]; Array[a, 100] (* Amiram Eldar, Jun 17 2022 *)
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)))))
KEYWORD
nonn
AUTHOR
Benoit Cloitre, May 03 2005
STATUS
approved