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%I #11 Jun 17 2022 03:43:05
%S 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 27,28,28,22,23,23,25,54,54,53,53,51,48,48,49,49,43,44,44,46,35,35,36,
%U 36,38,41,41,40,40,88,87,87,85,82,82,83,83,77,78,78,80,69,69,70,70,72
%N 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 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 Missing numbers are the nearest integer to tau^2*n, n>=0 (cf. A004937).
%C #{k>0 : a(k) = k} = infinity.
%C 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).
%H Amiram Eldar, <a href="/A105669/b105669.txt">Table of n, a(n) for n = 1..10000</a>
%F F(2n) = F(2n+1) - F(n+1)^2 + F(n)*F(n-1) for n>0.
%F a(F(2n-1)) = F(2n)-1 for n>1.
%F 1/tau < a(n)/n < tau.
%e For 5 = F(5) < k <= F(6) = 8 we get a(6) = 8-a(6-5) = 8-a(1) = 7.
%e a(7) = 8-a(7-5) = 8-a(2) = 6.
%e a(8) = 8-a(8-5) = 8-a(3) = 6.
%t 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 *)
%o (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 Cf. A000045, A001622, A004937, A105670, A105672, A093347, A093348.
%K nonn
%O 1,2
%A _Benoit Cloitre_, May 03 2005
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