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).

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

Amiram Eldar, Table of n, a(n) for n = 1..10000

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)))))

Crossrefs

Cf. A000045, A001622, A004937, A105670, A105672, A093347, A093348.

Sequence in context: A162251 A244011 A065968 * A338917 A256963 A355306

Adjacent sequences: A105666 A105667 A105668 * A105670 A105671 A105672

Keyword

nonn

Author

Benoit Cloitre, May 03 2005

Status

approved